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CollinofAlabama
Carpal Tunnel

Reged: 11/24/03

Loc: Lubbock, Texas, USA
Reflector/Refractor equivalence formula
#5764738 - 03/29/13 01:41 PM
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Folks,

I decided to post this in the Refractor forum, too, to get people's input here. Jarad Schiffer, in the thread "Long focus Newtonian Vs refractor" in the Reflector forum wrote the oft quoted formula that to determine a Newtonian's performance to an equivalent refractor, do this ...

Reflector Primary - Secondary = Refractor Primary Equivalent

I noted Jon Isaacs many times writing that the real difference is more like

Reflector Primary - 1" = Refractor Primary Equivalent

I propose, to quantify Jon's assertion, perhaps the formula should look like this ...

Reflector Primary - (0.5 * Secondary) = Refractor Primary Equivalent

As one got above a 40% obstruction, this formula might not apply, but I'm not as concerned about astrographs functioning as visual instruments

I take the example of the Celestron Omni XLT 150, which has a 150mm objective with a 46.5mm CO. Now according to Jarad's formula, that should make the Omni XLT equivalent to a 103.5mm refractor for planetary performance. But in my proposed reworked formula, the "Refractor Equivalence" of the Omni XLT should be a 126.75mm scope. Now, people may say this is wishful thinking, and I cannot say with any certainty that it isn't, but it would be nice to see how an Omni XLT 150 compares with 4" ED, 110mm ED, and 120mm ED (as well as the equivalent achromatic) scopes. I own neither the 110 nor 120mm ED scopes, nor any achromat, though there's a STRONG temptation for me to buy the XLT and compare it to my 102mm ED scope. According to Jarad's formula, it should still surpass it, but really, just barely, and the planetary performance should really be about the same. Stay tuned, but anyone else wishing to evaluate my recalculation of that old Reflector ~ Refractor Equivalence formula is more than welcome to educate me and the rest of the CN Brotherhood.

For the record, the Omni XLT delivers the same number of photons to your retina as a 142.6mm refractor -- (Pi * r squared of the Primary) minus (Pi * r squared of the Central Obstruction). Obviously, those photons are maligned a bit for human vision by the CO, but how much is the question.

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Eddgie
Postmaster

Reged: 02/01/06

Re: Reflector/Refractor equivalence formula [Re: CollinofAlabama]
#5764854 - 03/29/13 02:18 PM Attachment (115 downloads)
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The old "Primary minus Secondary obstruction) offers at very best, a very crude approximation of how the instruments will perform visually, and only visually.

And if you want to make the comparison and not get an approximation, try using Abberator 3.0 to generate your own MTF plots. It is easy to use.

The MTF plot tells the truth about how much contrast is lost.

Here is an MTF for an 8" Newtonian with 23% obstruction (Dashed red) vs a perfect 8" aperture (solid red) and a perfect 6" unobstructed aperture. I have included the spider vanes (which means the scope started with a 21% secondary mirror and I added 2% to that for the spider vanes).

The Green line is for a perfect 6" Aperture with the assumption that the color correction is perfect (which is rarely the case by the way, but I am assuming a best case.

As you can see, the 8" scope offers contrast that is on par with the perfect 6" aperture for larger details, but for the smallest details, the reflector pulls away, still showing small detail after the perfect 6" aperture has run out of steam.

So, no need to apply coarse formulas when you can plot it pretty closely and know a more accurate picture.

Abberator will only allow you to plot one scope.

I plot the larger instrument, then hand draw the curve for the smaller perfect instrument.

In this case, the 6" has .75 of the maximum spatial response of the 8" scope, so it is plotted to the .75 index on the frequency line at the bottom of the chart.

Also, MTF ignores the higher illumination you get out of the larger aperture for a given exit pupil. A brighter image helps the eye better see the lowest contrast detail on the target.

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hfjacinto
I think he's got it!

Reged: 01/12/09

Loc: Land of clouds and LP
Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5764882 - 03/29/13 02:33 PM
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I think it should be this:

Premium APO > Everything else

Ha!

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Eddgie
Postmaster

Reged: 02/01/06

Re: Reflector/Refractor equivalence formula [Re: CollinofAlabama]
#5764895 - 03/29/13 02:37 PM Attachment (111 downloads)
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Here you go..

I did it for you.

This is a plot showing a perfect 150mm aperture (red line), and 150mm aperture with a 33% obstruction (the secondary obstruction is only 31%, but the spider vanes add about 2% and dude, you have to count that too, though it is often not mentioned in these dialogs).

The green line shows the equivalent aperture for a scope that is a perfect 90mm aperture.

Now remember, just like most people don't count the spider vane, refractor people think refractors are all perfect, but fast ED scopes do indeed loose some contrast because of chromatic aberration, but at 90mm, it is usually not enough to matter (though at larger apertures it can indeed be meaningful).

As you can see, for the larger details (the left side of the graph, your scope will only equal about a 90mm unobstructed scope.

As the detail gets smaller though, your scope draws even, but at the important "Visual frequencies" that are generally represented by the left half of the graph, the scope is not really performing any better than a 90mm unobstructed scope.

For the right part of the graph (past about .5 on the frequency scale), the 6" easily pulls ahead.

The problem is that this represents detail that is very small, and better caught by a CCD camera. Visuallly, the exit pupil gets so small by the time the image gets large ehough that the brightness falloff causes the image to suffer.

A camera would show the difference in performance, but visually, the contras would be no better.

Ah, but once again, the larger scope has a huge brightness advantage. The human eye finds it easier to see low contrast detail when the image is better illuminated.

This means that when you magnified the image to 150x in both scopes, it would be much brighter in the 6" making the lowest contrast detail easier to see. Both scopes will show the detail with the same contrast, but all of the detail will be brighter in the larger scope.

I personally find that illumination is an important factor when planetary observing.

The last issue is quality. Don't expect your scope to have optics of the same quality as is common in even inexpensive ED scopes made today. I gave your instrument the benefit of the doubt and made it perfect.

Bottom line? It is very complicated, but I think you are being optimistic.

Expect planetary performance below what the best 4" APOs can deliver.

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junomike
Pooh-Bah

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Loc: Ontario
Re: Reflector/Refractor equivalence formula [Re: CollinofAlabama]
#5764921 - 03/29/13 02:49 PM
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My main issue with all of this (formula's and graph's) is It's only in "theory"! In the field the main contributor to how my C11 is gonna compare to my Apo (in this case an AT111EDT ~ 4.37") is.........SEEING! IME the Central Obstruction in a Reflector is much more harmful in regards to Contrast in poor seeing.

Last year I had both in the field and on a few nights the much smaller Apo brought in more detail than the larger scopes (my C11 and an excellent 12" Dob).

When seeing did permit the use of higher magnification, the smaller Apo was easily bested by the larger scopes. This occurred less than 50% of the time however!

This is the main reason I believe there is so much of the
"this scope beat that much larger scope" and Refractor vs. Reflector controversy.

It's all about the four rights: The right scope of the right size used on the right target in the right Seeing Conditions.

This is also why one scope is not usually sufficient for everything.

Mike

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Sean Puett
Carpal Tunnel

Reged: 09/06/10

Loc: always cloudy, washington
Re: Reflector/Refractor equivalence formula [Re: hfjacinto]
#5764924 - 03/29/13 02:50 PM
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That formula is supposed to be for planetary performance. Jon correct me if I am wrong, I think the -1" is for everything else.
My personal opinion is that the CO affect is a bit overstated if it is under 25% dso or 20% planetary. I was considering lengthening my ota and using a low profile (moonlite) focuser to allow me to get my secondary under 20%. Jon Isaacs told me to make secondary masks larger by the same percentage to see if it would be worth all the time, money, and effort. What I noticed on Jupiter was that the 5% make very little difference and so I used an even larger one to make the difference more obvious. I decided that it wasn't worth it.
Later I found out that the center of the refractor lens does very little if anything anyway. Now that I typed this, I think I should have answered in the reflector forum... Go easy on me.

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Eddgie
Postmaster

Reged: 02/01/06

Re: Reflector/Refractor equivalence formula [Re: hfjacinto]
#5764930 - 03/29/13 02:54 PM
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You know, I consider myself to be a pretty balanced kind of guy.

A 6" APO is indeed a very difficult scope to beat.

Even the Ultimate Newtonian 8" (which is what I plotted the first time) is only on par with a perfect 6" APO from a raw MTF perspective.

To get a clearly superior image, you would need to go to perhaps a 10" reflector.

That actually speaks volumes about the excellence of a high quality 6" APO.

I own one and can attest to their astonishingly good performance.

But they can be beat. It just takes more than an inch of aperture in most cases.

And often it takes more than even a few inches to equal the contrast transfer. I would put my 6" APO up against a good quality C11 and expect it to hold its own.

And the difference in the quality of the view between my C14 and my 6" APO is not "Breath-taking." Yes, better in the C14. I routinely observe detail that is out of reach of the 6" APO, but only with great patience and good seeing.

Many people think I am a refractor hater and nothing could be further from the truth. I admire them greatly, Though I have no use for small ones.

They can be beat though, but below about 8", and they really do have the upper hand unless you go to something like the MN 78 which is highly specialized and of superb optical quality.

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Sean Puett
Carpal Tunnel

Reged: 09/06/10

Loc: always cloudy, washington
Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5765161 - 03/29/13 04:25 PM
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The above described situation from my last post came as a result of getting my first decent refractor and wanting my 12" do more "refractor like". I started reading about secondary sizes and effects on contrast and since I have the two scopes I want at this time, I thought seriously about trying to improve the reflector. The project may still happen at some point just because I may rebuild my OTA and dob base. It is hard to have near perfection in a refractor and not want it for your big scope.
After seeing that 10" apo video (with seeing issues) and comparing it to a memory of a d14 at a low angle, conditions limited both too much for any real comparison. I haven't ever used an apo big enough to compare to my 12" dob on a good night. I have had better views than that video, but what does that mean? So I guess in a very long way, I have said that I cannot confirm or dispute the formula.

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GlennLeDrew
Postmaster

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Re: Reflector/Refractor equivalence formula [Re: Sean Puett]
#5765542 - 03/29/13 07:02 PM
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Such 'formulae' are exit pupil dependent. At larger exit pupils, resolving power and contrast transfer are not impacted by a central obstruction.

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Eddgie
Postmaster

Reged: 02/01/06

Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5765589 - 03/29/13 07:32 PM
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This is a point that I strongly concur with.

While much of the attention is focused on the contrast transfer the instruments, (and it is so much more than "Theory") people often omit the role of illumination.

In a thread going on right now, someone said that they looked through a 400mm f/15 achromat and had one of the best views that they have ever had (though they did not do a side by side comparison with a similarly good 400mm reflector at the 5000 foot elevation of the observatory).
With 400mm of aperture, the brightness from Jupiter is probably so high at 300x that the eye is moving well into mesoptic mode, where the eye's contrast sensitivity threshold is much better than in scotopic mode that is common in much smaller instruments.

I almost always say that the bigger aperture will always have the advantage of having a brighter image at a given magnification, and that this added illumination can make low contrast details easier to see.

My own experience has been that the amount of detail I cold see on planets was almost dead linear to the clear aperture of the scopes I have owned. In almost every case, the more aperture I used, the more detail I was able to see.

Until I owned the C14, I never really was able to resolve any detail on the Jovian moons for example. Ganymede would show variations in shading but not really show specific details with specific shapes.

Only when I moved to the C14 did I start to resolve this kind of detail.

The smallest refractor that I have heard a similar report of resolving Osiris and Galilee Regio where from an 8" APO.

Clear aperture rules.

Anyone wanting better planetary performance should not be looking at "Planetary eyepieces." They should be looking for more good quality clear aperture.

And the illumination that comes with aperture is in itself an important advantage.

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timps
super member

Reged: 02/24/13

Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5765660 - 03/29/13 08:14 PM
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So should I buy the 152mm Explore Scientific Apo or a 14" Meade/Celestron?

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Scott Beith
SRF

Reged: 11/26/03

Loc: Frederick, MD
Re: Reflector/Refractor equivalence formula [Re: timps]
#5765693 - 03/29/13 08:32 PM
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Quote:

So should I buy the 152mm Explore Scientific Apo or a 14" Meade/Celestron?

Both!

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Jon Isaacs
Postmaster

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Re: Reflector/Refractor equivalence formula [Re: CollinofAlabama]
#5765820 - 03/29/13 09:52 PM
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Quote:

I noted Jon Isaacs many times writing that the real difference is more like

Reflector Primary - 1" = Refractor Primary Equivalent

Honestly, I don't remember writing this. What I remember are calculations for light through put based on reflectivity and surface area measurements and mentions of rule of thumb estimations of planetary contrast based on "clear aperture", the aperture minus the secondary diameter.

Jon

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CollinofAlabama
Carpal Tunnel

Reged: 11/24/03

Loc: Lubbock, Texas, USA
Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5765878 - 03/29/13 10:41 PM
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First, Jon, I'm pretty sure you recently wrote this, but I'm on an iPhone right now, not the tool to find your posts on reflectors' performance (you have quite a few in the regard, mind you!)

Ed, first, thanks for doing the graph. This kind of information is enjoyable to look at. Something tangible, and yet ...

Sean made the point about "in the field". And here's where your graphs run afoul of reality. Joe Bergeron wrote this review in 2007, comparing his 92mm apo, 150 XLT and AP 155. In his estimation they lined up as one would expect, with the AP 155 best, the 92mm the worst, and the 150 XLT performing in between. Of course, this sheds little light on where and to what degree the 150 XLT would fall -- above a 102mm? above a 110mm? above a 120mm? -- since all those certainly fall between a 155mm and a 92mm scope. But Bergeron specifically notes it performing better than the 92mm, thus implying the graph you supply, Ed, may not apply in the field. "In overall performance, the \$400 Omni was midway between these two fabled refractors, and closer to the big one than the little one." You may say this proves nothing, and you may be right. But you could be wrong, too.

I'd just like to hear from someone who has used the Omni XLT and one (or more) of the 4", 110mm or 120mm refractors.

I have always liked refractors, but can't forget the way I saw an 8" LightBridge put an Orion 120 ED to shame on Saturn one evening.

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timps
super member

Reged: 02/24/13

Re: Reflector/Refractor equivalence formula [Re: Scott Beith]
#5765959 - 03/29/13 11:57 PM
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That would be ideal but if you could only have one, which would it be?

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JKoelman
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Reged: 05/16/11

Loc: Bangalore, India
Re: Reflector/Refractor equivalence formula [Re: timps]
#5766025 - 03/30/13 01:20 AM
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A single formula will not capture all performance aspects. Assuming diffraction-limited optics and perfect seeing, for planetary detail the aperture-based formula:

NewtApert - ObstructDiam = EquivApertAPO

seems reasonable. For faint fuzzies, I'd go with a straightforward light gathering-based equation:

NewtApert^2 - ObstructDiam^2 = EquivApertAPO^2

which is more favorable towards Newtonians.

Purely from a theoretical perspective, a formula like

NewtApert - ObstructDiam/2 = EquivApertAPO

can't be correct for the full range of secondary obstructions, as for secondaries approaching the size of the primary the equivalent APO aperture should drop to zero. For small percentage obstructions, however, the equation with the squares is mathematically equivalent to

EquivApertAPO = NewtApert - ObstructDiam^2/2NewtApert

Therefore, as a rough rule:

NewtApert - ObstructDiam < EquivApertAPO < NewtApert - ObstructDiam^2 / 2NewtApert

where the leftmost expression is more relevant for planetary, and the rightmost expression more for DSO.

Edited by JKoelman (03/30/13 03:25 AM)

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azure1961p
Postmaster

Reged: 01/17/09

Loc: USA
Re: Reflector/Refractor equivalence formula [Re: JKoelman]
#5766318 - 03/30/13 08:59 AM
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I agree the rule of thumb about subtracting CO size to arrive at a clear aperture visual observation equivalent is a very very rough approximation. This had always been a rather nebulous notion among observers and where there were voids in proof it was - and still is - a place where ego filled in. Now we have MTF. I think I was one of the last people to actually refer to it or use it but alas its about as real as you're ever going to get. It might seem like an abstraction of logic but once I got in the hang of it then it all made sense and with logic clarity.

I know what you are looking for - a simple equation somewhat like the CO subtraction - and in lieu of MTF - its a fair if rough call, but the graphic mapping out of it all sais it so handedly (and ego free) its probably the uncontested as the most revealing way to compare - though the author could tweak things a but to be quite frank. Too there are other variables Id like to see addressed in the graph. Environmental effects like fan versus no fan, cool down versus insufficient cool down. Being able to model these into the MTF graphic would be engaging stuff. Adding Pickerings scale of seeing effect as well could produce some great comparisons - along with the ability to overlay multiple scope MTF curves.

Anyway its worth downloading Aberrator even if at first it might seem a little wonky. Its all to a good end. And strangely like some other top notch programs - its free.

Pete

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Cotts
Just Wondering

Reged: 10/10/05

Loc: Toronto, Ontario
Re: Reflector/Refractor equivalence formula [Re: azure1961p]
#5766405 - 03/30/13 10:06 AM
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The trouble with a rule of thumb is that people's thumbs vary greatly in length.....

Dave

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Eddgie
Postmaster

Reged: 02/01/06

Re: Reflector/Refractor equivalence formula [Re: CollinofAlabama]
#5766419 - 03/30/13 10:13 AM
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Quote:

Sean made the point about "in the field". And here's where your graphs run afoul of reality.

Clearly you missed the part where I said that Illumination is an important characteristic of the telescopes.

The MTF is concrete. It is defined by diffraction and makes the outcome absollute in terms of performance at the focal plane.

But in the same post, I very clearly said that illumination is usually ignored in these dialogs (I never ignore it) and said that while the obstructed apeture might not have any better contrast transfer, that the better illumination would make whatever detail was present easier to see.

So while the instruments are peformeing the same at the focal plane, the observer's eye has a stronger signal to work with, which again, is an advantage to apeture that is often not included in these converstaions.

In other words, I agree with you in one sense, that the field result will be a bit different, but the MTF is absolute. Nothing violates it.

And that is why I have said that clear aperture has for me been the most important attribute of how two telescopes would perform on planets and extended targets. Even if the contrast is the same, the observer always benefits from the brighter image in the larger scope.

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Eddgie
Postmaster

Reged: 02/01/06

Re: Reflector/Refractor equivalence formula [Re: azure1961p]
#5766437 - 03/30/13 10:23 AM
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Quote:

It might seem like an abstraction of logic but once I got in the hang of it then it all made sense and with logic clarity.

Yes, once one really understands MTF, the clarity it brings to how an instrument will perform is very satifying.

And it is no more of an empty threory than the theories on the wave nature of light.

It is the wave nature of light that causes contrast loss, and it is absolute and easy to model, and can predict with very high accuracy how the instrument will behave.

That is why almost all professionals will use MTF to define instrument performance

It is all inclusive from design, quality, contrast transfer, and angular resolution.

One graph tells you the truth, the whole truth, and nothing but the truth of how the image will be formed at the focal plane.

But as in my previous post, it is important to say that the human eye's contrast sensitivity has to be factored in when using the instrument visuallly, and illumination of a larger aperture aids in detecting and resolveing the smallest detail on the target.

Even when seeing limits my to 250x, the image is so much brighter in my C14 that all of the details are much easie to see than in my 6" APO.

Illumiantion is a huge key to our ability to resolve detail.

Take a black and white picture from a newspaper out at night and see what happens. The contrast on the target doesn't change but the illumination changes.

And the more illumination you put on that picture, the more details will emerge.

Simple experiment and conclusive. Increase the illumination 50% and you will see more on the target.

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Eddgie
Postmaster

Reged: 02/01/06

Re: Reflector/Refractor equivalence formula [Re: timps]
#5766509 - 03/30/13 11:00 AM
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Quote:

So should I buy the 152mm Explore Scientific Apo or a 14" Meade/Celestron?

Two very different instruments.

I own both a 6" APO and a C14.

I use the 6" APO mostly for wide field observing during the summer and winter Milky Way, and for doubles during this time of the year.

It works well for planets and if it is already set up, I will use it for planets, but if there is no scope set up and I want to look at planets, I bring out the C14.

But I get the appeal of the big refractor. For wide field work, the view is the best I have ever had. At all powers, stars are sharp right to the edge of the field stop.

And I get the appeal there. I love the pinpoint stars offered by the 6" APO.

But for must general use, if it fits into the field of the C14, I use that.

If it doesn't fit into the field of the C14, but fits into the field of the EdgeHD 8", I use that.

Only if it doesn't fit into the field of the EdgeHD 8" am I inclined to pull out the big refractor.

So, to me, they are different scopes for different uses.

What do you want to do most?

If it is deep sky, the C14 is clearly a better choice even with the off axis curvature and coma. For most deep sky targets, you don't notice the off axis abberations of the C14.

So, I cannot advise you really on what you should get. Think about what you want to observer and how important off axis performacne is to you.

Good luck.

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GlennLeDrew
Postmaster

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Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5767092 - 03/30/13 04:01 PM
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Eddie,
One must be careful of treating the greater aperture as delivering a brighter image, this being the main reason for seeing more detail. Aperture, image brightness and resolving are inextricably intertwined. The larger instrument delivers better resolution because of the diameter of the entrance pupil, extra light is part of the equation.

To assess performance differences most objectively, and to minimize the variables, comparisons must be conducted at identical exit pupil. The exit pupil is the arbiter of the visual appearance of diffraction effects. A 1" and 100" aperture, both working at the same exit pupil, will provide identical appearances for a star (the stars chosen being of suitable brightness for the aperture, of course.) And for extended objects, like planets, at the same exit pupil the object surface brightness is the same, and the degree of resolution *as perceived on the retina* will be the same. It's just that the larger scope delivers a commensurately larger image and a concomitantly more detailed view. But on the retina, the *apparent* angular resolving power is the same.

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Eddgie
Postmaster

Reged: 02/01/06

Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5767178 - 03/30/13 05:06 PM
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This does not differ at all from what I have said.

I have said that a larger aperture will enjoy a better illuminated image for a given magnification.

And of course that means that at the same exit pupil, the image will be larger, but just as bright.

And the consequence of that is that you can magnify the image more in the larger aperture which makes it easier to resolve the detail.

And I have repeatedly stated here that these dialogs usually only talk about what happens at the focal plane and ignore the working of the human eye.

It would appear that we are not in dis-agreement here.

But I disagree that at the retina there will be no difference.

If the larger aperture is allowed by seeing to transfer more contrast, that contrast will show in more detail at all powers in the scope with the larger clear aperture. Angular resolution is quite a bit different than contrast ttransfer. You can resolve 192 line pair per millimeter in even a very poor f/10 telescope.

One observer might see a particular detail rendered with 6% contrast (right at the edge of perception for the dark adapted eye) while the other might see it with 15% contrast. The one that sees the higher contrast will say that the view is "Sharper". After all, that is what contrast transfer is. It is image sharpness. But the loss of contrast for mid-frequency detail does indeed make it less crisp in the eyepiece for that same instrument,

The MTF of the instrument allows the instrument to simply transfer more of the low contrast detail to the focal plane.

If there is more on the focal plane to see, then the observer will see it.

But once again, in my very first post on this topic, I said it was very complex. We are far more in agreement than we differ I think,

Edited by Eddgie (03/30/13 05:13 PM)

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t.r.
Post Laureate

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Loc: Upstate NY
Re: Reflector/Refractor equivalence formula [Re: CollinofAlabama]
#5767360 - 03/30/13 06:28 PM
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Quote:

First, Jon, I'm pretty sure you recently wrote this, but I'm on an iPhone right now, not the tool to find your posts on reflectors' performance (you have quite a few in the regard, mind you!)

Ed, first, thanks for doing the graph. This kind of information is enjoyable to look at. Something tangible, and yet ...

Sean made the point about "in the field". And here's where your graphs run afoul of reality. Joe Bergeron wrote this review in 2007, comparing his 92mm apo, 150 XLT and AP 155. In his estimation they lined up as one would expect, with the AP 155 best, the 92mm the worst, and the 150 XLT performing in between. Of course, this sheds little light on where and to what degree the 150 XLT would fall -- above a 102mm? above a 110mm? above a 120mm? -- since all those certainly fall between a 155mm and a 92mm scope. But Bergeron specifically notes it performing better than the 92mm, thus implying the graph you supply, Ed, may not apply in the field. "In overall performance, the \$400 Omni was midway between these two fabled refractors, and closer to the big one than the little one." You may say this proves nothing, and you may be right. But you could be wrong, too.

I'd just like to hear from someone who has used the Omni XLT and one (or more) of the 4", 110mm or 120mm refractors.

I have always liked refractors, but can't forget the way I saw an 8" LightBridge put an Orion 120 ED to shame on Saturn one evening.

I can't speak about a 150XLT...but I can assure you that my 1/6-1/7th wave C-6XLT handily beat out my old Tak Sky 90, Megrez 90 and TMB 92L! Take it to the bank. So I have concluded that 6" of sct aperture trumps 3.5" of unobstructed aperture consistently. In addition, for a season I compared a 4" apo to a C5, and the C5 fell short. Bracketing would put a 3.5" apo in line with a C5.

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timps
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Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5767419 - 03/30/13 07:18 PM
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I would like to use the telescope for most things. (Nebula, star clusters, galaxy, planetary and lunar).Visual and imaging. I believe that a 14" SCT, be it Meade or Celestron, is the best telescope for this. I think they are the best "all rounder". They are good value for money too when you consider the price of even a "cheap" 6" Apo.
However, I would still like a 6" Apo. Maybe I should start off with the SCT & get the Apo further down the track.

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GlennLeDrew
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Re: Reflector/Refractor equivalence formula [Re: timps]
#5767492 - 03/30/13 08:06 PM
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This originally was the first portion of the complete post following below. How and why it got sent while I was composing it (with a short distraction taking me away for a bit) is a mystery...

Edited by GlennLeDrew (03/30/13 09:06 PM)

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Fomalhaut
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Re: Reflector/Refractor equivalence formula [Re: junomike]
#5767540 - 03/30/13 08:21 PM
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Quote:

My main issue with all of this (formula's and graph's) is It's only in "theory"! In the field the main contributor to how my C11 is gonna compare to my Apo (in this case an AT111EDT ~ 4.37") is.........SEEING! IME the Central Obstruction in a Reflector is much more harmful in regards to Contrast in poor seeing.

Last year I had both in the field and on a few nights the much smaller Apo brought in more detail than the larger scopes (my C11 and an excellent 12" Dob).

When seeing did permit the use of higher magnification, the smaller Apo was easily bested by the larger scopes. This occurred less than 50% of the time however!

This is the main reason I believe there is so much of the
"this scope beat that much larger scope" and Refractor vs. Reflector controversy.

It's all about the four rights: The right scope of the right size used on the right target in the right Seeing Conditions.

This is also why one scope is not usually sufficient for everything.

Mike

Mike,
According to Anton Kutter, the inventor of the original TCT, the diameter of the first diffraction ring being ~1.8 times the one of the airy disk, together with the percentage of energy in the first diffraction ring increasing with increasing size of obstruction, causes the light in the airy disk to merge with the one in the 1st diffraction ring. This effect increases with decreasing quality of seeing (or optics!!!) and causes an ~1.8times "enlarged virtual airy disk" in moments of poor seeing.

According to these thoughts (and independant of what we can derive and quantify more exactly by means of comparing contrast transfer graphs), in poor seeing an unobstructed telescope shows an ~1.8 times smaller "airy disk" compared to a substantially obstructed telescope in bad seeing:

=> Refractor ~ as good as an 1.8 times bigger, substantially obstructed reflector during moments of insufficient seeing. (The susceptibility to this effect increases with size of obstruction.)

Kutter said these considerations showed him the way to the TCT... (Anton Kutter, "Mein Weg zum Schiefspiegler", lecture given at the Swiss Astronomical Society in 1962.)

Chris

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GlennLeDrew
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Re: Reflector/Refractor equivalence formula [Re: timps]
#5767611 - 03/30/13 08:59 PM
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Eddie,
Indeed, we agree upon much more than we differ.

Perhaps I was not so clear. What I was stressing is this. Other things being equal (e.g., quality, transmission efficiency, etc.), any two telescopes working at the same exit pupil deliver images having identical characteristics as regards the scale of diffraction on the retina.

A larger aperture does not deliver higher contrast; it merely delivers a larger image. In essence, the contrast transfer scales as the exit pupil diameter. The following should illustrate...

Consider two otherwise equal and equally good telescopes, one being simply scaled up so as to be 10X larger. We have the Jupiter we all know and love in the field of the smaller scope, and a hypothetical *identical* Jupiter 2.0 which happens to be located 10X farther away (and identically illuminated, so that it has identical surface brightness) and in view with the 10X bigger scope.

With the same eyepiece installed, the exit pupil would be the same, and so the views through each scope, aimed at its respective planet, would be identical (assuming perfect seeing, naturally.) One would have no idea whatsoever which scope one is peering through. Image size, surface brightness, degree of resolution and contrast transfer would be precisely the same. The only difference is that the 10X larger scope reveals 10X smaller details. But to the eye, at given exit pupil diffraction effects and hence perceived image quality are the same.

The MTF chart for each of these same f/ratio and same optical quality scopes is identical when considered in terms of linear scaling at the focus (e.g., line pairs per millimeter), which is of direct import for imaging. And when used afocally (with an eyepiece), linear resolving power and contrast transfer are directly related to the exit pupil diameter.

I stress these points so as to disabuse anyone of the false notion that a larger aperture has better contrast transfer.

If one considers performance differences on any one particular target whose angular dimensions are fixed, then certainly will the larger aperture afford better performance. But in terms of the *perceived* effects introduced by diffraction, aperture by itself has no impact whatsoever. Exit pupil diameter is the prime arbiter here.

For those who tend to skim longish posts like this, I can already hear the protests generated by the previous paragraph (and much else preceding). Just to be clear, this concerns not *absolute* angular resolving power, but the perception of image quality and resolution *on the retina*, which is entirely exit pupil dependant (again, assuming perfect seeing and otherwise similar optical configuration and quality.)

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CollinofAlabama
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Re: Reflector/Refractor equivalence formula [Re: Fomalhaut]
#5768980 - 03/31/13 02:54 PM
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Well, Chris and other theorists, you now have the written 'in the field' testimony of two CNers that a 6" F/5 reflector and a 6" SCT consistently performed better than 90 (and 92mm) clear aperture scopes. Tim writes you can take it "to the bank". Pretty definitive, I'd say. Now, perhaps you think Tim and Joe Bergeron don't know what they're doing and are not to be trusted compared to charts and "definitive publications from optical experts". I don't think this way and, bumpkin that I am, trust them and their analysis as somehow more valid.

All this leads me to believe that, in the field, "aperture illumination" Ed writes of is more valuable than charts and essays when it comes to what you'll actually see at the eyepiece comparing scopes. Of course, these antecedotes still don't tell me where the XLT 150 would fall compared to a 102mm/110mm/120mm refractor, but I do believe it would outperform any 90mm (or 92mm) apo. I already figured that was true (Joe wrote his article in 2007, btw), and wasn't very swayed by theories, any way. But my fundamental comparison scheme is still a mystery. Anyone else with more experience in these matters is more than welcome to join in.

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t.r.
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Re: Reflector/Refractor equivalence formula [Re: CollinofAlabama]
#5769077 - 03/31/13 03:54 PM
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There are NO absolutes Collin...I still prefer the stellar images in the 90 apos!

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sg6
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Re: Reflector/Refractor equivalence formula [Re: t.r.]
#5769389 - 03/31/13 06:03 PM
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People don't seem to buy scopes on the basis of what is equivalent to one another, they buy because they like one or another. You have posted this in the refractor section, so people here will generally prefer refractors immaterial of the diameter. If they wanted large apertures then they wouldn't be here.

I didn't buy the Megrez 90 because it was equivalent to anything, I bought it because I liked it.

I have read several of these "equivalences" and you seem to be able to choose 1:1 or as mentioned here 1:1.8 and so any ratio in between. That really is if you care. Will say I for one don't care.

If I had a 10" reflector and a 80mm refractor which would I use, easy, both, and on the same things, DSO's, planets, doubles. Would I throw away one because I had the other? Not a chance in hell.

Is a 102mm refractor equivalent to a 150mm reflector? I don't know, and in honesty the whole question is of no interest. They are different instruments and the one thing I would expect is they give different results and by that they cannot be equivalent.

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Fomalhaut
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Re: Reflector/Refractor equivalence formula [Re: sg6]
#5769449 - 03/31/13 06:31 PM
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Living in a region close to the northern Alps in Europe with often quite unsteady seeing, 20 years ago I "upgraded" (sorry, but that's how I felt and still feel) from an 8"-SCT to a perfect (!) 4"-apo performing most of times better or then at least more beautifully. I've never regretted this, but supplemented the apo with a 7"-DK later on in order to have "the best of both worlds".

I know (from own experience) that a good 7 to 8 inch SCT or DK CAN sometimes do a little bit better on the planets, but most of times doesn't in my climate. As compared to a 33 to 34% obstructed 6"-scope, the same apo virtually always produces better (more crisp and contrasty) planetary images with at least as much detail.

By the way, I have the utmost respect for good Newtonians with reduced obstructions (not more than 20%): Such a 6-incher definitely can do what an SCT or DK with its bigger obstr. cannot, i.e. outperforming a 4-inch apo even on planets (good seeing provided).
And this all isn't theory but practice first of all (having been more or less in this pastime for over 50 years now).

Chris

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Sean Puett
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Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5772064 - 04/02/13 01:23 AM
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Quote:

Quote:

Sean made the point about "in the field". And here's where the graph runs afoul of reality.
Quote:

Sean did not make this point. Mike made this point. I happen to agree with Eddie. He is far more knowledgeable about this hobby and the gear used than I am. I have never experienced the "smaller apo showed more than the larger mirrored telescope" phenomena. Some people just prefer refractors, and that is fine but, I don't want to be converted (in the religious use).

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csrlice12
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Re: Reflector/Refractor equivalence formula [Re: Sean Puett]
#5772889 - 04/02/13 01:02 PM
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In my endeavour to find the "one"scope; I've discovered it is has green lettering and rectangular shaped, and comes in 1, 5, 10, 20, 50,&100 sizes. There's also the rare sought after 2. If you go to the scope store they'll trade you all kinds of nice stuff for it! Anybody know where I can get more of these?

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Ziggy943
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Re: Reflector/Refractor equivalence formula [Re: csrlice12]
#5773114 - 04/02/13 03:06 PM
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There is no reflector equivalent to a 9" Clark refractor period.

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Eddgie
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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5773233 - 04/02/13 04:05 PM
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Quote:

during moments of insufficient seeing.

This is why I always say that the most important factor in doing good planetary observing is patience.

I never really use the 6" APO for planets anymore. I just found that under my own local conditions, I just about always see more in the C14.

And I like seeing more. Resolving features on Ganymede, or seeing spokes in the rings of Saturn, or seeing lots more detail on Mars are all things that make me prefer the C14.

And now, I will add "Binoviewers" to that.

While it is true that larger apertures suffer more from seeing, I would say that there is when using both scopes side by side, there was rarely a night that I did not see more detail in my C14 than in my 6" APO.

This has always been the case with the most serious planetary observers I think. They had the patience to endure maybe an hour of viewing to get maybe 10 minutes of seeing that is sufficient for getting the most detail that conditions will permit.

And the reason I am so in favor of binoviewers is because going from even the best "Planetary" eyepieces to binoviewers made a consistently larger difference in my ability to be patient and catch those moments of good seeing. I don't know if I can resolve more detail using binoviewers or not, but I know that when I use them, at the end of the session, I always feel like I have seen more than before using them under typical conditions.

So for me, it is simple. I would rater be patient and see more detail than I can see with a 6" APO rather than look through he 6" APO and be limited to what that scope will show.

Even on the best nights, I have not seen detail using the 6" APO that I routinely see in the C14.

And when seeing is so poor that I can't see more than in the 6" APO, then it just is not worth doing, because even the 6" APO will suffer on such nights.

So the choice is to be content with never seeing more than 6" can show, or seeing more, but investing more time to do it.

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Re: Reflector/Refractor equivalence formula [Re: Ziggy943]
#5773241 - 04/02/13 04:10 PM
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respectfully ziggy, your post needs a pic of such a glorious instrument!

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Eddgie
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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5773346 - 04/02/13 04:59 PM
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Quote:

A larger aperture does not deliver higher contrast; it merely delivers a larger image. In essence, the contrast transfer scales as the exit pupil diameter. The following should illustrate...

I hope that you take the time to read this complete post.

I think that your statement is not really correct.

Here is the way it works. All of this is detailed in two sources.. The book "Telecope Optics" and the Book "Star testing.."

Linear resolving power in line pair per millimeter at the focal plane is a function of focal ratio and only focal ratio.

Any two f/10 telescopes will have a linear resolution at the focal plane of 192 line pair per millimeter in green light, regardless of aperture or obstruction.

This is a function of the focal ratio and the focal ratio alone.

The maximum spatial frequency (width of the smallest line pair that can be resolved) is related to the aperture though, and not the focal ratio.

Here is where many people get confused. They believe that since any two f/10 telescopes can resolve 192 line pair per millimeter at the focal plane that they both have the same contrast transfer.

That is not the case. Not at all.

Suppose you have a 6" f/10 scope and a 12" f/10 scope.

Also, suppose that you have a special eyepiece with a 1mm field stop.

Suppose you start with a chart with and MTF chart and you start moving it away from each scope.

At a point that is some distance from the focal plane, the 6" scope will start to show the lines as blurring together. This happens when the chart is far enough away that 192mm line pairs now occupy the space between the edges of your 1mm field stop.

Ah, but if you now place the chart a the same exact distance from the 12" scope, you are seeing at only half of that scopes maximum spatial frequency.

So, this means that in this scope, rather than seeing 192 line pair, you are only seeing 96 line pair in your one millimeter field stop when the chart is at the same distance.

Same chart, same lines, same distance, but now, your scope is seeing the same exact detail at only .5 of its maximum spatial frequency.

At this point, the lines in the 5" scope have lost all contrast. The chart shows as a gray chart with no lines resolved at all.

You have reached the 6" scopes maximum spatical frequency and run out of contrast.

But the 12 inch scope easily shows 96 lines.

So, the image on the focal plane is rendered at twice the size because the aperture is twice as large, so every detail is rendered in the larger scope at .5 of the maximum spatial frequency of the larger scope when it is rendered at 100% of the maximum frequency of the smaller scope.

All telescopes loose contrast, and that contrast loss is a function of diffraction. The smaller the aperture, or any obstruction lowers the contrast transfer, but all things being equal, the bigger instrument will always transfer more contrast because there is less diffraction caused by the aperture.

Here is another way to look at it.

The formula for spatial frequency is s' max = 1/f x wavelength [cycles/length]. I am sorry, I don't know how to do formulas on a computer.

Anyway, suppose you have a 152mm aperture that is f/8.12
If you do the formula (and I am taking this from Suiter's book), the linear resolving power would be 220 cycles per line pair at the focal plane.

But.. And this is important... The S' Max (max spatial frequency) would be 1.32 cycles per arc second. This represents a line pair that where the lines are alternating from black to white with 1.32 crests for each arc second of true field.

At this point, the 6" scope is out of gas. It cannot present linear detail smaller than this size.

The 12" f/8.12 scope though has a S' max of twice this (2.64 cycles per arc second)

Another way to look at this is that the line width for a single line would be 1/2 of the S' max.

In the 6" aperture, all contrast would be lost when each line was .66 arc seconds in width.

In the 12" aperture, the line would be .33 arc seconds before all contrast was lost.
.

A line on the target that was .66 arc seconds wide would still have 42% contrast in the 12" aperture when all contrast was lost in the 6" aperture.

I needed to add that for MTF, sinusoidal lines are used.

For true black and white lines, the S' max will be somewhat off the chart and represented by the fact that often the Sparrow Criterion is referenced at a bit higher than S' max.

But for Sinusoidal lines on a distant target, the 6" scope will loose contrast when the lines 1.32 arc seconds peak to peak, but the 12" will not loose contrast until the lines are .66 arc seconds peak to peak.

Edited by Eddgie (04/02/13 06:03 PM)

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Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5773492 - 04/02/13 05:36 PM Attachment (14 downloads)
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Here is an example from above.

In this chart, we see that a 12" f/8.12 scope will resolve 2.64 lines per arc second and this is represented by the 1 on the X axis.

The 6" f/8.12 scope will resolve 1.32 line pairs per arc second and this is represented but the green line ending a .5 on the X axis.

At 1.32 lines per arc second on the target, the 6" aperture is loosing all contrast.

In the larger scope though, these lines are half the S' max of the 12" scope. If 2.64 lines per arc second is the S' max, then 1.32 lines per arc second is .5 of the scopes S' max.

This means that the lines that have lost contrast in the 6" scope are still being shown with 42% contrast in the 12" scope.

so, any detail that is smaller than about 1.32 arc seconds in the small scope is almost out of contrast.

But for the larger scope (assuming a perfect aperture), the detail is shown with 42% contrast. You can see this by following the line up from .5 on the X axis to where it intersects the contrast loss line on the Y axis.

A perfect 12" aperture can resolve 2.64 line pair per arc second and will loose 42% contrast on 1.32 line pair per arc second.

The 6" aperture will loose 100% contrast at 1.32 line pair per arc second.

Now of course a central obstruction can (and does) lower that contrast transfer, but even with an obstruction, the 12" scope is still transferring contrast while the 6" aperture is completely done and cannot render detail smaller than this.

Contrast transfer is about the light that is taken from a geometric point (star or point on an extended target) and transferred away from the geometric center of this point into the area around it.

It should be common sense that if the aperture has a much larger Airy disk to start with, then light is deposited much further from the center of that geometric point in a smaller scope than in a larger scope.

And that is what reduces contrast as the aperture gets smaller. The light from the center of that geometric point is spread further away from it.

If that geometric point is a point on the edge of a white line against a black background, then that light will spread further away from the edge of the line in a 6" scope than in a 12" scope because the Airy Disk in a 12" scope is smaller (half the size in a telescope twice as large).

And this is why all things being equal, a larger aperture preserves more contrast than a smaller aperture.

And by the way, the C14 has a specified resolving power (according to Celestron's literature) of .39 arc seconds.

Guess what you get when you divide 1 arc second by 2.64 which is the S' max for a 14" aperture? You get .37 arc seconds. But that is for a Dawes split where there is no separation. A Raleigh Spit is listed at .39 by Celstron for the C14, so the MTF plot is right between these two at .38 arc seconds

For a C6, Celestron lists the resolving power of the C6 as .77 arc seconds.

Guess what you get when you divide 1 arc second by 1.32. About .76 arc seconds.

The difference of course is that these are obstructed apetures which very slightly changes the size of the Airy Disk, which in turn changes the contrast transfer very slightly.

MTF shows how much contrast loss will occur for a detail of a given linear size. The bigger the detail in terms of S' max, the less contrast that will be lost.

But all things to do with design and quality being equal, a bigger aperture starts with a smaller limit on detail size, and for any given size detail up to 1 line pair per millimeter at the focal plane, the larger instrument will always show that detail with more contrast.

Edited by Eddgie (04/02/13 07:44 PM)

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Fomalhaut
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Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5773655 - 04/02/13 06:38 PM
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Quote:

While it is true that larger apertures suffer more from seeing, I would say that there is when using both scopes side by side, there was rarely a night that I did not see more detail in my C14 than in my 6" APO.

Even on the best nights, I have not seen detail using the 6" APO that I routinely see in the C14.

And when seeing is so poor that I can't see more than in the 6" APO, then it just is not worth doing, because even the 6" APO will suffer on such nights.

Yours really must be a fantastic C14!
And I envy you your obviously paramount seeing conditions.

Chris

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Eddgie
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Re: Reflector/Refractor equivalence formula [Re: Fomalhaut]
#5773696 - 04/02/13 07:00 PM
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well, it has pretty good optics I would say. Star test shows very smooth optics. There is a zone about half way out which used to be very typical in Celestron SCTs bigger than 8".

The star test shows very little spherical aberration.

So yes, I would say that as large SCTs go, it is one of the best I have owned, thought the optical quality is not as good as my EdgeHD 8" which is stunningly good.

But as my post above indicates, the contrast transfer of the C14 is quite a bit better than a 6" APO, so with patience, it is easy to see more detail.

My seeing is rarely perfect, but unless seeing is less than about 2 arc seconds, I prefer to look at other things besides planets anyway.

When seeing is below about 2 arc seconds, usually is all it takes is patience for moments of sub arc second seeing, and when that happens, subtle details come into view.

About a year ago, I posted on resolving the Ray system from the feature Osiris on Ganymede and also the curve and flat rim of Reggio Galilee on Ganymede on the same night using the C14.

That didn't just happen. Seeing was maybe 2 arc seconds, so not that good.

I was using about 340x in mono-vision when I made the sighting (11mm plossl).

At the time, Ganymede was I think only 1.7 arc seconds, so this represents detail that was on the order of one arc second and less in size.

Anyway, I did not just put in the eyepiece and suddenly see these features. I probably spent 30 minutes looking at Ganymede before I got maybe 20 Seconds where the seeing steadied out.

Over maybe another 45 minutes, I saw both features two more times.

So, out of perhaps an hour and a half of viewing, I noted these features for a grand total of maybe 90 seconds.

But I saw them. I went in to validate my sighting using my astronomy program and sure enough, they were in the position and presented the shapes I saw.

So, there you have it. Maybe 90 minutes of viewing time.

I have seen shading on Ganymede using the 6" APO, but I have never really resolved specific shapes in that scope. It just doesn't have the contrast of the much bigger C14.

Planetary observing with larger apertures will show you more but only if you have excellent seeing or a lot of patience.

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GlennLeDrew
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Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5773706 - 04/02/13 07:04 PM
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Eddie,
I in no way dispute all you've stated regarding resolving power, both linear and angular.

My argument, which could have more pointedly stated, is that resolving power and contrast transfer are not equivalent. To merely say that the larger aperture delivers higher contrast in untrue. All optical instruments of *perfect* quality deliver the same contrast as that of the eye alone. At the same exit pupil, all deliver the same quality of image.

For example, if when you place a series of aperture stops in front of your eye you find that the diffraction pattern on a point source just becomes apparent at 1.5mm, then with any excellent scope you will see the same degree of diffraction at the same 1.5mm exit pupil.

Another example, an expansion upon one supplied earlier. One makes up a series of test charts, all identical, but placed at a *distance appropriate to the magnification* of each instrument. One is for the unaided eye, another for a small scope, and a third for a big scope. All are to be viewed at the same 1mm pupil. In all cases, with the target subtending the same angular size on the retina, the visual appearance as regards 'softening' due to diffraction effects is identical.

This is why I say that the *appearance* of contrast scales with the exit pupil. Other things being equal, diffraction is purely exit pupil dependent. After all, the exit pupil is merely the shrunk-down entrance pupil. And to the eye, it's the aperture of its own pupil, or the exit pupil, whichever is the smaller, which ultimately controls diffraction.

So again, to the observer's retina, aperture alone has no role to play in contrast. That's controlled by the exit pupil.

It's only when restricting to the limited situation of directing attention on just one target of fixed angular size do we get the impression that a larger aperture improves contrast. But that's a 'red herring', really, for then the argument is merely stating what is fundamentally obvious. Namely, that a larger aperture reveals finer detail. But it does not improve contrast.

And this is perhaps where our wires are crossing. You're working in terms of *absolute resolving power*, while I'm reducing to the relative resolving power *as detected by the retina*.

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GlennLeDrew
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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5773767 - 04/02/13 07:22 PM
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To say that a C14 has better contrast transfer than a 6" APO implies a poorer-than-expected quality in the refractor. The Cat's central obstruction introduces additional diffraction which impacts resolving power at certain scales, thus reducing contrast transfer. If the refractor were to be optically perfect, or good enough in that respect which matters, it would offer better contrast transfer, in spite of its much smaller aperture.

If contrast transfer is merely tied to resolving power in the larger sense, as when changing the aperture of the entrance pupil, then the term 'contrast transfer' is largely superfluous.

As I understand it, 'contrast transfer' describes the actual contrast as compared to that of a perfect, unobstructed system of the same aperture, as is usually illustrated in the MTF chart. If so, we must be careful to not conflate the better resolving of a larger aperture with better contrast transfer.

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5773782 - 04/02/13 07:31 PM
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Ziggy, if I did not believe that you are completely and absolutely sold on Celestron telescopes I would find your posts to be more objective. As it is, you, me and most others on these forums have a bias toward one type scope or another and find ways to show how the telescope we chose, the one we spent big money on, is the best. The old saying "there are lies and there are statistics" is true.

If mathematical equations were all that was needed we could just read a book and never observe to see for ourselves how our scopes actually perform.. For example, I suspect all your equations "assume" perfect collimation, scope at thermal equilibrium, etc.

I own a 20" F/3.3 Mike Lockwood Quartz mirror Starmaster, a 10" F/9 TMB LZOS triplet apochromatic refractor and an AP 10" F/14.6 Maksutov. Aperture is not the do all end end all to observing. I have to collimate my 20" Every time I use it. I have never had to collimate either 10" scopes. A friend of mine, Bob, contacted me about how I should get an auto collimator to tweek the collimation of the 20". Go to the Catseye website and read all about using an auto collimator. When your headache goes away finish reading my post.
I love all my scopes. They are all very special and I would not give up Any of them. My advice is Quit worrying about Aperture and which scope is Better than the one you have. It's not!

JimP

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5773860 - 04/02/13 07:58 PM Attachment (15 downloads)
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Not so. You need to read my post.

A 6" aperture transfers far less contrast to the focal plane than a 14" aperture.

Please read my post above, or ask anyone that has read Suiter's book.

There are examples that show this clearly.

A larger aperture (all other things being equal) preserves more contrast from the target than a smaller aperture.

The S' max for a 6" scope is only 1.32 lines per arc second (apparent size of the detail on the target).

The S' max for a 14" aperture is almost 2.5 times that figure.

This means that a C14 preserves far more contrast than a 6" aperture.

Here is the MTF plot for each scope.

Notice that the S' Max for the 6" is only .43 of the S' max of the 14" (about 1.32 lines per arc second at the target for the 6" vs almost 2.8 lines per arc second in the C14).

Even with the big central obstruction and 1/8 wave of spherical aberration, the C14 is still transferring much more contrast than the 6" APO.

You should read Suiter's Book.

The book "Telescope Optics" will also confirm this data.

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Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5774024 - 04/02/13 08:59 PM
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And that is why I will buy a 14" SCT.

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Re: Reflector/Refractor equivalence formula [Re: timps]
#5774091 - 04/02/13 09:14 PM
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It is good to read peoples opinions. Sometimes they assist and sometimes they confuse but really, it all comes down to what one sees through the eyepiece. If Eddgie has both scopes and obviously has the oportunity to compare them through the eyepiece then I will tend to agree with what he is saying.

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5774296 - 04/02/13 10:14 PM
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Glen, I really encourage you to get Suiter's Second Edition of Star Testing.

Suiter addresses this topic in detail.

In Chapter 3, Suiter goes to great lengths to explain how a larger aperture (design and quality being equal) will have better contrast for all size details.

In Appendix B, he presents alternate was of calculating contrast transfer.

In all models, the larger aperture (again, quality and design being equal) always has better contrast for every size detail.

The book "Telescope Optics" also has a lengthy explanation of contrast transfer, though in this book, they don't directly present diagrams showing the contrast transfer for different apertures. Instead, they only show the formulas for calculating S' Max.

I found that it was necessary to read both a few times to get the concept, but Suiter's book gave the key on page 59.
People that think that refractors always have better contrast than other designs are mistaken. This is only true when the quality is the same or better, and the clear aperture is more than in the case of the obstructed instrument.

In the same way that larger refractors have better contrast transfer than smaller refractors, a bigger reflector, even if not optically perfect and with a large central obstruction, can still have better contrast transfer than a smaller refractor.

And that is what the MTF chart exists for. I allows you to express in lines per arc second (at the target) how much contrast will be preserved from the starting contrast.

I really encourage you to read Suiter's book.

Anyone that has the book can confirm from the diagram at the bottom of page 59 that the graphs I presented above tell the store.

A 12" scope will have far better contrast transfer at the focal plane than a 6" scope, and a 32% obstructed 14" scope with 1/8th wave of spherical aberration will have better contrast transfer at the focal plane than the most perfect 6" APO ever made.

Anyone that looks at page 59 in Suiter's book and spends the time reading the chapter will concur with this.

Edited by Eddgie (04/02/13 11:10 PM)

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Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5774445 - 04/02/13 11:35 PM
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Eddgie,
I was recently given Suiter's 1994 edition by a friend. In it, the MTF chart is always normalized to the *fraction* of the maximum spatial frequency. This is appropriate when considering a telescope in the afocal configuration, where diffraction effects scale as the exit pupil.

My emphasis in this discussion is on the appearance of the image at the eyepiece and at given exit pupil. The better quality instrument, irrespective of aperture, will deliver the better quality view.

The eye sees an image, and knows not what is the aperture delivering it. All it 'knows' is whether the image is good or not so good as regards such things as diffraction and aberrations. If at some particular exit pupil one rates the view less afflicted by diffraction and aberrations for a smaller aperture, then that smaller aperture, in some respect at least, delivers by definition better contrast transfer.

Any two telescopes, no matter how much they may differ in aperture but which have identical MTF charts, will deliver identical contrast at the same exit pupil.

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Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5774447 - 04/02/13 11:36 PM
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I assume that the references to Suiter, and page 59, are to his book on star testing?
Anyway, i just grabbed my copy.
On page 163:
9.1 Central Obstruction
The most obvious and potentially most damaging kind of transmission change is caused by the centrally placed diagonal or secondary mirror. ... However, the negative consequences of central obstruction can be readily and precisely calculated. We will see that they worsen considerably beyond a linear obstruction of 20 to 26 percent of the aperture. As long as the obstruction is kept inside that fraction, the image closely approximates that of an unobstructed telescope."

The section continues, exploring various aspects of the central obstruction, equivalence formula, etc.
It appears that (from that section, and his formula on page 163), that all other things being equal (including the crucial element of seeing...) that the unobstructed equivalent of a 14" with a 32% would be 9.52", according to Suiter. On that score, the 34% CO equivalent of a 6" unobstructed aperture would be a tad more than 9 inches?

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Re: Reflector/Refractor equivalence formula [Re: Cotts]
#5774496 - 04/03/13 12:09 AM
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Quote:

The trouble with a rule of thumb is that people's thumbs vary greatly in length.....

Dave

Dave nailed it with his analogy. A vast majority of the arguments summarized by the thread title are pure bunk.

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Re: Reflector/Refractor equivalence formula [Re: moynihan]
#5774519 - 04/03/13 12:22 AM
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When they refer to the percentage of the central obstruction, Is that a percentage by area or diameter?
Celestron 14" for example: secondary mirror obstruction by area is 10.3% but by diameter is 32.1%.

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Re: Reflector/Refractor equivalence formula [Re: timps]
#5774546 - 04/03/13 12:38 AM
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Quote:

When they refer to the percentage of the central obstruction, Is that a percentage by area or diameter?

Could be either (percentage by area is figured by taking the square root of the percentage by diameter; nobody actually measures the area). You can tell which has been stated by the figure. It will only make sense as one or the other (10%-15% range would be area while 30%-45% would be diameter). Which is chosen generally depends on the context. Light loss varies as the area while contrast varies as the diameter.

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5774551 - 04/03/13 12:43 AM
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Quote:

Eddgie,
I was recently given Suiter's 1994 edition by a friend. In it, the MTF chart is always normalized to the *fraction* of the maximum spatial frequency. This is appropriate when considering a telescope in the afocal configuration, where diffraction effects scale as the exit pupil.

My emphasis in this discussion is on the appearance of the image at the eyepiece and at given exit pupil. The better quality instrument, irrespective of aperture, will deliver the better quality view.

The eye sees an image, and knows not what is the aperture delivering it. All it 'knows' is whether the image is good or not so good as regards such things as diffraction and aberrations. If at some particular exit pupil one rates the view less afflicted by diffraction and aberrations for a smaller aperture, then that smaller aperture, in some respect at least, delivers by definition better contrast transfer.

Any two telescopes, no matter how much they may differ in aperture but which have identical MTF charts, will deliver identical contrast at the same exit pupil.

Glenn:

I am with Eddgie on this one... One needs to step back from what is "always done" and consider what one wants to know when attempting to analyze the difference between two scope that differ in aperture. This is the question as I see it:

For a given spacial frequency, which scope will provide the superior contrast. All we want to know is for a given detail on the surface of Jupiter, which scope will show it with more contrast. If you normalize by aperture, then the scale of the object must be normalized and that is not what we want, we are looking for absolute numbers for comparison.

In the case of the C-14 and the 6 inch, the same exit pupil occurs at 2.3 times the magnification, the details with the same contrast are much finer in the larger scope.

Jon

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5774684 - 04/03/13 03:42 AM
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Interesting stuff. here is my 2c worth....

A crude rule of thumb for the equivalent obstructed and unobstructed equivalent aperture is take away the obstruction from the aperture for an obstructed scope. This seems to hold for up to 10-11 inches. And for obstructions from 20-35 %.

However obstructed scopes need to be well cooled, well baffled, perfectly collimated and have a very good figure if they are to produce similar contrast as an unobstructed "apochromat" equal to their aperture minus the obstruction. With this reasoning a 10 inch planetary Newtonian with a 2.6 inch secondary should be able to hold it's own against a 7 inch apochromat.

The apochromat could well seem to be more appealing visually and have a higher contrast to brightness than the Newtonian.

I have an 8 inch f-6 Dobsonian which has a 2 inch secondary. It does indeed show similar views to a good five inch apochromat when it is well cooled. The missing inch may be because of the less than perfect control of scattered light.

I think once achromatic refractors start to obey the f rule of 3 times the aperture in inches they also behave close to an obstructed aperture minus the obstruction.

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Re: Reflector/Refractor equivalence formula [Re: Kevin Barker]
#5774720 - 04/03/13 05:22 AM
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Jon,
It's axiomatic that at given spatial frequency a larger aperture delivers better contrast. This is the essence of the aperture race. And as I stated, I am in complete agreement.

My emphasis is on what the observer sees *at the scale of the retina's resolving power*, so to speak.

We have two otherwise identical telescopes. A 2" is examining a 1st magnitude star and a 20" is examining a 6th magnitude star. Each is working at, say, the same 1mm exit pupil. The view through each eyepiece will be indistinguishable.

We have two optically excellent telescopes, one a 2" refractor and the other a 20" Cat. As before, each looks to a 1st and 6th magnitude star, respectively. At the same exit pupil, the Cat's image suffers additional diffraction. An observer conducting a blind test would judge it inferior.

This is the crux of my argument. Discounting the obvious increase in resolving power (and light grasp) which aperture affords, what is the *quality of the image on the retina* at any given exit pupil diameter?

We all know that resolving power scales directly as the linear aperture. This is so fundamental that once learned can be relegated to the back of the mind. The MTF chart is not concerned whatsoever with the actual, absolute resolving power, being normalized as it is to the theoretical maximum. The MTF chart is all about representing the departure from perfection of a same-size circular aperture. Two widely disparate apertures delivering the same optical quality will have identical MTF charts.

Concentrating on the *absolute* differences in resolving power resulting naturally from differences in aperture is all well and good. But that this scales linearly with aperture is so easily appreciated it hardly bears more than the briefest thought.

What we *really* desire to know is this; how far does my telescope depart from perfection? The MTF chart tells us clearly, and this is of direct relevance to the quality of the view over the range of exit pupils useable.

For an afocal instrument (with eyepiece, used visually), the quality of the view *as perceived on the retina* is what ultimately matters. In this context, then, contrast transfer must be rated with respect to the perfect wavefront emerging through a given exit pupil.

The importance of the telescope and eye *as a system* is too easily overlooked. The exit pupil is the coupling interface which locates the entrance pupil (objective) at the eye's pupil. The eye cares not a whit what other optics lie in front of it. All that matters is the pupil diameter and the wavefront passing through it.

If that wavefront is not aberrated, the dimension of the Fresnel pattern on the retina scales precisely as the f/ratio of the light cone defined by the exit pupil (or iris, if the smaller.) This is irrespective of the telescope aperture. And if any particular aberration or aberrations is present (say, 1/2 wave if spherical aberration), no matter the aperture, for given exit pupil it will have identical apparent extent on the retina.

This is why the exit pupil is so important. It is the normalizer for image surface brightness, diffraction and extent of aberrations.

If we assume for the moment that both systems are otherwise essentially perfect, how can we can say that the C14 delivers better contrast than a 6" APO when we know from its MTF chart that the additional obstruction impacts contrast??? At given exit pupil (smaller ones, of course) we will see in the C14 the degradation of contrast compared to the slightly cleaner image in the 6".

Yes, I know. The much larger image in the C14 WELL MORE THAN COMPENSATES for its slightly poorer contrast transfer. But just because that greatly larger aperture, with its commensurately better resolving piwer, so handily bests a superior but smaller instrument in no way means it delivers better contrast transfer.

If it did, then contrast transfer is to a *very* great degree merely interchangeable with resolving power, scaling almost purely with aperture. That's superfluous, don't you think?

To me, contrast transfer, at least in the context of an afocal system, is kind of like the Strehl ratio; it's related only to what a perfect, same-size, (in this case unobstructed) aperture could deliver.

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5774747 - 04/03/13 06:31 AM
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About the ganymede observation with 6+14" scopes---
***
Isn't the resolving power of a 6" around 1 arc-second?
I would think if so, an investigation why the features weren't seen is mandatory.
(Daves, and Marks remarks about rule of thumb- right on the mark!)
M.

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5774808 - 04/03/13 07:52 AM
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Quote:

Concentrating on the *absolute* differences in resolving power resulting naturally from differences in aperture is all well and good. But that this scales linearly with aperture is so easily appreciated it hardly bears more than the briefest thought.

Glenn:

It seems that all too often, the fact that "it's axiomatic that at given spatial frequency a larger aperture delivers better contrast" is simply forgotten. This is particularly true in this forum, I see it stated time and time again that refractors, regardless of aperture, have better contrast than other scopes...

This is the issue I believe Eddgie is addressing, this is the issue I constantly address: That contrast for a given spacial frequency is a function of aperture, that in terms of planetary viewing, a 10 inch Newtonian with a 20%CO will have better contrast than a 4 inch anything. To try to explain this, I use phrases like "all scopes are obstructed", "by far, the most important diffraction effect that affects contrast is the result of the most important obstruction, the Outer Obstruction, more commonly known as the Aperture."

When evaluating a telescope, I am an observer, I am looking to see what I can see, the telescope itself, it ought to be a black box. I don't really care if it's a 4 inch scope that is as perfect as a 4 inch scope can be or a 10 inch scope that has average optics, what I want to know is which one will provide me with the more detailed views of Jupiter, which one will provide me with best view of Stephan's Quintet... which one will provide me with the best view of the North American Nebula..

When one normalizes by aperture, it's pretty apparent that an apochromatic refractor, because of the simple, unobstructed, will provide better contrast. What is not so clear though is that a somewhat larger scope with a central obstruction can overcome that advantage and provide equal or superior contrast to the refractor. An "Equivalance Formula" should be able to provide a basis for this relationship.

The value of MTFs, contrast transfer, is that if they are not normalized, then they can provide a basis for analysis to determine some sort of equivalency relationship. It's pretty apparent that a 12.5 inch F/6 Newtonian is going to have a big advantage in spacial contrast and resolution over the most perfect 6 inch refractor ever built, but how about an 8 inch Newtonian, how does it fit?

And then when one includes the practical aspects, thermal management in all it's many aspects, optical quality, sensitivity to seeing.. it gets really complicated.

For me, to understand that the contrast scales with exit pupil is interesting and worth understanding but it is also academic. What counts happens at the eyepiece, when comparing two scopes, I am not using the same exit pupil in both scopes, I am using the optimal exit pupil in each scope... What do I see looking at Jupiter with my 16 inch Newtonian versus my 4 inch apo?

I think we both know the answer.

Jon

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Re: Reflector/Refractor equivalence formula [Re: Mark Harry]
#5775059 - 04/03/13 09:41 AM
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Because while the resloving power is about 1 arc second, the detials will have lost most of their contrast at this size.

That is the whole point of MTF.

If you start out with a detail that has 100% contrast (Black on white) that is only one arc second across, it will have lost almost all of that contrast in the 6" scope!

That is what MTF is about.

All telescopes loose contrast.

By the time you get to detail this small, it will have so little contrast left that it is impossible to see unless it started with 100% contrast to begin with.

Lets take the example of the curve of Reggio Galilee. This is a Mare like feature on Ganymede. It describes a curve that is ony about 1 arc second in width and one side is flattened.

But the contrast of this area is very poor.. Perhaps ony 20% against the surrounding landscape (dark gray against light gray).

Becaucause this curve starts with low contrast, in the scope that can ony resolve a one arc second feture, 98% of teh contrast would be lost at the focal plane. A feature that started with 20% of the contrast at the target now ony has about .5% contrast at the focal plane. It has become invisible to the human eye, which requires between 2.5% contrast (brightly illluminted target) and perhaps 5% contrast.

In the C14 that has twice the spatical response, this feature that started with 20% contrast still at about .4 of the scopes maxiumum spatial frequency.

This means that a 1 arc second detail in this scope will still possess over 50% of the contrast that it started with.

A one arc second feature on Ganymede in this scope (regardless of the magnification used) will still show with 10% contrast at the focal plane.

As you can see, the feature has lost too much contrast to be visible in the 6" scope at all. If it is less than one arc second in size, it completely merges with the background.

But in the C14, it is still visible with 10% contrast. Difficult, but well within the eye's ability to resolve.

Contrast transfer has nothing to do with the observer or the eyepiece or the magnification.

This is the image at the focal plane being formed by the optics.

Resolition figures published with telescopes give you the resolving power for Double Stars.

A double star is a case where the contrast starts with 100% and reprsents the best possible case for the telescope because you are looking at a black seperation between two bright Airy Disks. That black "line" that appears between the two Airy disks represents the size of a feature where the contrast is almost completly gone. We only see it becuase the the contrast starts with 100%.

So, all of this other talk about magnification and image scale for different size telscope when viewed visually miss the point.

The point is that at the focal plane, different instruments transfer contrast (get detail from the target that starts with a given contrast and form the image on the focal plane with a lower contrast than on the target) at different rates.

By the time the 6" APO shows the image, unless it starts with 100% contrast (like the line between a double star) a one arc second detail will have lost so much contrast as to be inpossible to see.

The 14" aperutre will only have lost about 50% of the contrast detail for a target this size. If the detail started with 20% contrast, 10% will still remain.

That is how MTF charts work.

They tell you how much contrast a line pair of a given width will lose when it is rendered at the focal plane.

This can be used to infer how much contrast for a similar sized detail will be preserved.

Io in the 6" APO is almost featureless even on the very best of nights.

In the C14, it takes great patiance and excellent seeing, but there are occasions where I can resolve detail.

And I should be able to. THe C14 preserves far more contrast for these small sized details than the 6" APO does.

In fact, the C14 preserves more contrast for any size detail than the 6" APO does.

And that is what the MTF chart show you. They show that both instruments loose contrast on detail, but the sloping line shows you how much contrast is lost for a given size detail

The C14 has contrast transfer that is sperior across the entire range of detail size.

All planetary detail is shown with better contrast in the C14 than in the 6" APO.

Even on nights when seeing is less than great, large scale detail still stands out better in the C14 than the 6" APO. While small scale detail may be blurred out, the observer still enjoys the better contrast of the larger instrument.

Edited by Eddgie (04/03/13 09:41 AM)

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Re: Reflector/Refractor equivalence formula [Re: moynihan]
#5775217 - 04/03/13 10:48 AM Attachment (9 downloads)
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Quote:

The section continues, exploring various aspects of the central obstruction, equivalence formula, etc.
It appears that (from that section, and his formula on page 163), that all other things being equal (including the crucial element of seeing...) that the unobstructed equivalent of a 14" with a 32% would be 9.52", according to Suiter. On that score, the 34% CO equivalent of a 6" unobstructed aperture would be a tad more than 9 inches?

To harken back to the OPs question, the formula is really only general, and works better for visual observing than for imaging. For imaging, it is not at all accurate because the obstruction doesn't matter for the finest detail the scope can resolve. Even a large obstruction doesn't loose much additional contrast past about .7 on the MTF plot meaning that for the finest detail, the size of the obstuction doesn't matter.

Here is a plot that shows what is going on.

This is a plot for a perfect 9.25" 36% obstructed instrument

The red line would show how much contrast would be lost for even a perfect 9.25" apeture.

The dashed red line shows how much contrast would be lost for the obstructed instrument.

The blue line shows how much contrast would be lost for the perfect 6" instrument.

Now the forumla for S' Max would indicate that the 9.25" scope will have lost almost all contrast for a line frequencey of 2 line pair per arc second on the target.

The S' Max for the 6" apeture (if you have been following along) is 1.32 arc seconds per line pair on the target.

First, no one should be able to argue that the angular resolution is better for a 9" scope than a 6" scope. There are numerous formulas out there that show that, and contrast transfer is after all, a function of angular resolution. It describes how far away from the geometric center of the Airy Disk light energy will fall. Everyone knows that the Airy Disk is smaller as apeture grows (all other things being equal).

Back to the chart. The 1 on the X axis indicates that the 9.25" instrument can resolve 2 line pair per arc second on the target.

The 6" instrument cannot reasolve lines this small at all. It can only reslove line paris that are 1.32 arc seconds indicated by the fact that the S' max is only 65% of the S' Max of the 9.25" scope.

Now, follow the graph over to where you see the little green line between the obstructed and perfect aperture lines.

This represents line pairs that are about .33 of the S' Max of the 9.25" Scope, or lines that are about .66 lines per arc second (in other words, line pairs about 1.5 arc seconds wide at the target).

What the MTF chart shows for line pairs (or detail about 1.5 arc second in size at the traget) is this.

If these lines started as a 1.5 arc second wide line pair with 100% contrast on the target as black and white alternating sine waves, in the perfect 9.25" scope at .33 S' Max, they will have lost about 39% of their contrast.

They will appear as very dark gray lines alternating with very light gray lines, but the contrast is high ehough that they will still appear to the eye as almost black and white.

Continue down to where the green line is between the two scopes in question. Note that the perfect 6" aperture will show those same lines with only about 46% contrast. Now, rather than looking more black and white, they are starting to look more medium dark gray and medium light gray. They are starting to blend together because the light from the Airy Disks coming from the points along the white line are spreading into the area of the black lines, causing the contrast to lower.

Now, follow the green line down to the obstructed apeture.

In this case, the contrast as fallen to 43% maybe.

Now this is not a lot of contrast difference at all. Even the very very best observers will struggle to see a difference this small, though on a brightly illuminated target, people can often judge contrast differences as small as about 2.5%.

This graph shows though that a 9.25" 36% obstruted instrument is not quite as good as a perfect 6" apeture for details that are bigger than about 1.4 or 1.5 arc seconds in size on the target.

And the 6" aperture retains that contrast advantage for all lareger size detail though as the detail gets larger, the advantage dwindles. It peaks for line pairs about 1.3 to 1.5 arc seconds wide though.

But past about .5 of the S' Max of the bigger scope notice that it quickly gets back to even footing with the perfect apeture.

This represents contrast for the smallest possible detail the scope can resolve.

And this should make perfect sense if you use a double star as an example.

The space between the double stars can represent a feature on the surface of a planet.

Suppose you had a feature that was the same length as the spacing between a pair of double stars that was 1.5 arc seconds from geometric center to geometric center.

In a small aperture, the big extension of the Airy Disks would cause them to meet very near the center of this line.

If you used a bigger scope, the Airy Disk would be much smaller, and as a consequence, you would see more of the dimension of that line that was not covered by the Airy Disk of a star at either end.

And that is what you get when you viewin a planet. Each point of every feature emits ligth that speads out the distancee of the Airy Pattern based on the size, quality and obstruction of the telscope that forms the image.

If the telescope that forms the image conentrasts that energy in a smaller circle, or keeps more of the energy in the Airy Disk itself rather than in rings far away, it may preserve more enegy from the point that formed it.

MTF describes how this energy is distrubuted around the geometric point that formed it.

So, a 9.25" 36% obstructed aperture, when used visually, will have conrast transfer that is fairly close to a 6" perfect apeture.

This is a crucial qualification though. Again, this is only if the obstucted scope is perfect.

Finding 6" APOs with perfect optics is not difficult. No leanding manufacturer is going to sell you an expensive APO that is less than near perfect.

Near perfect C9s though are not the norm. And when you add the typeical inperfections, it can cause the MTF line to further sag.

And it doesn't take much for the line to sag enough so that the contrast falls to a 10% difference.

And when it does, the difference starts to show at the eyepeice.

If you know the amount of error though, you can model it.

Edited by Eddgie (04/03/13 10:54 AM)

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Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5775244 - 04/03/13 10:59 AM Attachment (14 downloads)
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Now it gets interesting.

Here, I have modeled the same scope as appears above.. A 9.25" 36% obstructed apeture.

This one though, is less than perfect. I have modeled in a bit of spherical abberation and a bit of astigmatism, both of which are not at all uncommon in mass produced scopes.

By compraison, just about any 5" APO from a specialty provider you can buy these days will have optics that consistently border on perfect.

Notice now that the sag has incresed a bit. While it does not seem like much, suddenly, at the important visual frequenceis, the 9.25" 36% obstructed aperure is not transferring contrast any better than perhaps a perfect 5" instrument.

And this is perhaps why accounts when comparisons are made differ.

The "Rule of tumb" formula is only really good for visual observing. For imaging, obstruction is not usualy an issue.

All defects or design quailites (secondary obstruction, and optical defects) add to the sag of the MTF line.

When the scope is unobstructd, even a little quality error does not affect the contrast enought to be a concern.

But when the obstrucion is large, the quaity becomes much more critical.

Any meaningful amount of spehrical abberation or astigmatism can quickly lower the MTF performance so that it is reduce to contrast transfer no better than an apeture half its size.

So now, you have one person that says thier C9.25 is better than a 5" APO and one that says theirs is not as good.

Sample to sample quality varations could easlily account for that.

The purple line shows that with a little sperhcial abberation and a little astigmatism, the 9.25" apeture is now performing with less contrast transfer at the lower (visually important) frequencies than a perfect 5" instrument!!!!

And that is the beauty of MTF.

Edited by Eddgie (04/03/13 11:24 AM)

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Re: Reflector/Refractor equivalence formula [Re: Fomalhaut]
#5775316 - 04/03/13 11:20 AM
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Quote:

Quote:

While it is true that larger apertures suffer more from seeing, I would say that there is when using both scopes side by side, there was rarely a night that I did not see more detail in my C14 than in my 6" APO.

Even on the best nights, I have not seen detail using the 6" APO that I routinely see in the C14.

And when seeing is so poor that I can't see more than in the 6" APO, then it just is not worth doing, because even the 6" APO will suffer on such nights.

Yours really must be a fantastic C14!
And I envy you your obviously paramount seeing conditions.

Chris

I live in an area with bad seeing as the norm. I have set up both my 4" refractor and 12" reflector together numerous times. When seeing is bad, both scopes are always affected. Occasionally, low mag images in the refractor are slightly better. Most of the time this is not the case. Seeing affects magnification limits, from my experience, and has little to do with aperture or telescope design. The only reason refractors sometimes seem to perform better in bad seeing is that they can operate at a much lower minimum magnification.

These contrast debates seem to ignore the fact that you need great contrast with a small scope to see what is easily seen in a larger scope. My 4" refractor shows the belts on Jupiter as a light pinkish to brownish tone that is low contrast. My 12" reflector shows them as dark reddish brown and it shows that there are more than 2 of them. Detail can be seen in the bands as well and again, it is easily seen. The downside is that it is very bright and you don't want to be fully dark adapted when looking at Jupiter.
I do get the contrast side of the debate in that on the moon, in my refractor, I can see more detail in the bright area than with my reflector. Maybe because I am completely blinded by looking at the moon with a 12" scope, maybe not. But, it is a more pleasing view in my refractor.

I really enjoy refractors. That being said, if my wife threatened to divorce me unless I cut down to one telescope, I would choose my 12"newt. It shows more detail on everything. There are things that do not fit in the fov though. I really hope she never says that to me...

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GlennLeDrew
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Re: Reflector/Refractor equivalence formula [Re: Sean Puett]
#5775537 - 04/03/13 01:14 PM
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In the context of finding the equivalence between an unobstructed and obstructed telescope, one is naturally concentrating on the image representation in the range out to about 5 times the resolution limit (0.2 on the MTF's ordinate). It is in this region, after all, where the most significant differences lie. And so the comparative representation of image contrast is of paramount importance.

What inspired my more generalized approach was the greatly unfair juxtaposition of a C14 and a 6" refractor. Unless that big Cat has horrid optics, it naturally will present better contrast over the full range of resolution provided by the much smaller refractor.

I digressed from the specific requirements of the thread and presented a more 'holistic' way to appreciate the images as presented by more widely disparate apertures. That is, to consider contrast transfer of a system as normalized to the exit pupil.

In a fashion, this approach still has validity here. Each telescope has the same exit pupil limit as regards the realization of its theoretical resolving power. That is, if for observer X some telescope realizes its resolution limit at an exit pupil of 1.1mm, so will all others. If aberrations are bad this limit might be reached at a somewhat larger exit pupil. But at given exit pupil, the image more afflicted by diffraction and aberrations belongs to the poorer telescope as regards optical quality. But then, this is so obvious, ain't it?

Getting back on track... A poorer but larger telescope often bests a smaller, "perfect" scope, by virtue of its greater resolving power. But then, this is so obvious, ain't it?

The \$64,000 question: Can a general equivalence formula be derived? If we restrict to the limit of several times the resolution limit, and consider only a limited variance on quality, I shouldn't see why not. That is, if we discard the poorer specimens from consideration, and consider only the small scale regime of the image, we should be able to predict with some confidence.

As long as it's always borne in mind that the problem is restricted to contrast transfer as regards the rendering of small details. In terms of image brightness, the larger obstructed aperture (nominally) is the better...

...which may have some small contribution to make, partially mitigating against its reduction in contrast.

Outside the small scale regime of several times the maximal resolving power, the larger obstructed scope is the better, for the contrast reduction becomes of little importance, with the brighter image handily making up for this.

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KaStern
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Re: Reflector/Refractor equivalence formula [Re: CollinofAlabama]
#5775901 - 04/03/13 04:11 PM
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Hi,

to answer your question right one has to know
what type of refractor and what type of reflector
you are talking about?

I can tell you that my 8"f/6 newt was able to give equally good views
of Jupiter and M13 when I compared it to an 7"f/6 TMB
during the german ITT astrofest. Secondary sice is 39mm.

An f/8 achromat falls short due to it`s colour aberration
that lowers contrast transfer.

But there are some unobstructed reflectors out there an the best of these
can rival an equally apertured apochromat.

And there are some reflector types wich are very much compromised.

In addition one can find very differing optical qualities in real world telescopes.

And not at least there are very many miscollimated reflectors
amd some miscollimated refractors too.

So in the end you can only do the following:
Get a scope of sufficiently good quality, collimate it,
let it cool down and observe the objects the scope is suited for.

Cheers, Karsten

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jhayes_tucson
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Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5776927 - 04/04/13 01:47 AM
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Now it gets interesting.

Here, I have modeled the same scope as appears above.. A 9.25" 36% obstructed apeture.

This one though, is less than perfect. I have modeled in a bit of spherical abberation and a bit of astigmatism, both of which are not at all uncommon in mass produced scopes.

By compraison, just about any 5" APO from a specialty provider you can buy these days will have optics that consistently border on perfect.

Notice now that the sag has incresed a bit. While it does not seem like much, suddenly, at the important visual frequenceis, the 9.25" 36% obstructed aperure is not transferring contrast any better than perhaps a perfect 5" instrument.

And this is perhaps why accounts when comparisons are made differ.

The "Rule of tumb" formula is only really good for visual observing. For imaging, obstruction is not usualy an issue.

All defects or design quailites (secondary obstruction, and optical defects) add to the sag of the MTF line.

When the scope is unobstructd, even a little quality error does not affect the contrast enought to be a concern.

But when the obstrucion is large, the quaity becomes much more critical.

Any meaningful amount of spehrical abberation or astigmatism can quickly lower the MTF performance so that it is reduce to contrast transfer no better than an apeture half its size.

So now, you have one person that says thier C9.25 is better than a 5" APO and one that says theirs is not as good.

Sample to sample quality varations could easlily account for that.

The purple line shows that with a little sperhcial abberation and a little astigmatism, the 9.25" apeture is now performing with less contrast transfer at the lower (visually important) frequencies than a perfect 5" instrument!!!!

And that is the beauty of MTF.

One of the things that impresses me about this group is how much many of you guys know about optics; however, some of this discussion has veered completely into the weeds so I want to clarify a few points:

1) ALL optical systems have wavefront errors. Over many years, I've tested hundreds of components and systems with state-of-the-art PSI and dynamic interferometers and it is extremely rare to see any component with PV errors less than about 1/25 wave (and even that is a difficult level to measure on an absolute accuracy scale--but that is another subject.) Any contention that refractors have fewer errors than reflectors is a generalization that just ain't so. There are high quality reflecting, refracting, and catadioptric systems. It all depends on the design, manufacturing tolerances, and alignment of the system. Keep in mind that on-axis performance isn't all that counts either. Few (if any) of these systems are shift invariant and the size of the aplanatic patch is generally small, which means that spatial frequency response will vary significantly with field angle.

2) A theoretically perfect reflector with a 32.5% obscuration meets the Rayleigh (as well as the Marechal) criteria for diffraction limited performance in both Strehl and MTF. In the real world, it is possible to balance the degradation in performance due to wavefront errors against the obscuration ratio such that the performance remains diffraction limited but that will require a smaller obscuration.

3) You guys are getting way too hung up on individual frequency response components and forgetting that actual image detail is composed of the Fourier sum of the transmitted frequency components (assuming a zero PTF.) A larger aperture always allows a wider range of frequency components to exist in the details of an extended image. That produces sharper edge response making small details easier to see. Furthermore, an obscured aperture actually has slightly better high frequency response (meaning higher contrast at higher spatial frequencies) than an unobscured aperture. Small losses in contrast in the middle frequencies generally have little effect on the perceived "sharpness" of an extended object. It is the high frequency response that drives the "clarity" of small details in an extended object. Obviously, if the mid-frequency transmission falls "too-much", image quality can begin to suffer, which is why we use MTF as a tool to evaluate system performance in the first place.

Johnson's law correlates the cycles per dimension required for observing features on a target (with incoherent illumination.)

Detect = 1 cyc/dimension
Orient = 1.4 cyc/dimension
Recognize = 4.0 cyc/dimension
Identify = 6-8 cyc/dimension
Recognize (50% accuracy) = 7.5 cyc/dimension
Recognize (90% accuracy) = 12 cyc/dimension

The more cycles/dimension that you can produce is what allows better imaging and that always requires bigger aperture.

BTW, if you want to get hung up on the shape of the MTF curve, you might as well start building telescopes with square apertures. A square aperture has a (non-diagonal) linear decrease in MTF. Get over it, keep the MTF curve within the diffraction limit, and you’ll be fine.

4) I not sure that I understand what it means to "normalize the MTF to an exit pupil." MTF is commonly normalized to spatial frequency with the maximum at 1/lambda*F...where F= F/#. That means that all F/5 telescopes will have contrast fall to zero at 0.4 cyc/micron at a wavelength of 0.5 microns--regardless of aperture. That .4 cyc/micron projects onto the sky from the focal plane (not the exit pupil) at different angles which depends on the focal length of the system. For a 6", F/5 telescope that translates to 1.48 cyc/arc-sec and for a 12", F/5 telescope, 2.96 cyc/arc-sec.

5) Finally, while it is completely valid to analyze the MTF performance of the telescope objective, you guys haven’t included the eyepiece in the discussion. Remember, MTF can only be cascaded for incoherently connected components. An eyepiece is coherently coupled to the objective and must be considered a part of the system to correctly analyze full system performance. That means that you can completely destroy the performance of even a perfect objective with a poor eyepiece.

John

Edited by jhayes_tucson (04/04/13 01:50 AM)

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GlennLeDrew
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Re: Reflector/Refractor equivalence formula [Re: jhayes_tucson]
#5776976 - 04/04/13 03:08 AM
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John,
That was an excellent contribution to the discussion.

Regarding your point #4. The phrase "normalize the MTF to the exit pupil" is a clumsy way of mine to try to get across the following point. Any equally good telescope (as a system, with eyepiece of course) working at a given exit pupil will present to the eye an identical degree/scale of visible diffraction on an image point. If the Fresnel pattern subtends some angular scale in a small scope at some particular exit pupil, it will be the same for a larger scope at the same exit pupil.

In other words, the resolving power as a fraction of the maximum is a function of the exit pupil.

Again, a clumsy way to place a basic principle of the afocal system into some perspective, so as to provide something of a more complete picture.

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Sasa
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Re: Reflector/Refractor equivalence formula [Re: CollinofAlabama]
#5777013 - 04/04/13 05:14 AM
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Quote:

I take the example of the Celestron Omni XLT 150, which has a 150mm objective with a 46.5mm CO. Now according to Jarad's formula, that should make the Omni XLT equivalent to a 103.5mm refractor for planetary performance. But in my proposed reworked formula, the "Refractor Equivalence" of the Omni XLT should be a 126.75mm scope. Now, people may say this is wishful thinking, and I cannot say with any certainty that it isn't, but it would be nice to see how an Omni XLT 150 compares with 4" ED, 110mm ED, and 120mm ED (as well as the equivalent achromatic) scopes. I own neither the 110 nor 120mm ED scopes, nor any achromat, though there's a STRONG temptation for me to buy the XLT and compare it to my 102mm ED scope. According to Jarad's formula, it should still surpass it, but really, just barely, and the planetary performance should really be about the same. Stay tuned, but anyone else wishing to evaluate my recalculation of that old Reflector ~ Refractor Equivalence formula is more than welcome to educate me and the rest of the CN Brotherhood.

I had Orion Optics N150/750 with about 33% central obstruction with lambda/8 optics and Hilux coating. Quite often it was outperformed on planets by my 80mm apochromat (Lomo 80/480 triplet). There is no comparison with my ED100. This 4" refractor showed me much more on Mars and Jupiter than 150mm Newton ever did (but in this case it was not side-by-side comparison as was in case with 80mm lens).

I think the obvious difference has nothing to do with theoretical expectations. In my case, I'm storing telescopes at home at room temperature. And I was often travelling for dark skies with former 150mm Newton. I think that thermalization and constant need of colimation were the main reasons why even the small 80mm refractor was outperforming much bigger brother on planets. This is the reason why I sold 150mm Newton at the end and bough ED100. I still consider it, after few years of using ED100, as one of the best moves that I made. On the other hands, I have no doubts that a wisely used 150mm Newton with properly designed OTA would outperform 100mm refractor on planets (not necessary at f/5, it would probably require something longer, like f/8).

Edited by Sasa (04/04/13 05:40 AM)

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Jon Isaacs
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Re: Reflector/Refractor equivalence formula [Re: jhayes_tucson]
#5777123 - 04/04/13 07:32 AM
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Quote:

5) Finally, while it is completely valid to analyze the MTF performance of the telescope objective, you guys haven’t included the eyepiece in the discussion. Remember, MTF can only be cascaded for incoherently connected components. An eyepiece is coherently coupled to the objective and must be considered a part of the system to correctly analyze full system performance. That means that you can completely destroy the performance of even a perfect objective with a poor eyepiece.

Indeed.. but these days, there are very few "poor eyepieces" and very good eyepieces are not expensive... The differences between eyepieces is subtle, the difference between apertures is not.

Jon

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GlennLeDrew
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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5777244 - 04/04/13 09:47 AM
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The difference between observers' eyes is greater than found across the bulk of available modern (and not so modern) eyepieces. On axis, at any rate.

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olivdeso
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Re: Reflector/Refractor equivalence formula [Re: Eddgie]
#5777513 - 04/04/13 11:59 AM
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Quote:

So, a 9.25" 36% obstructed aperture, when used visually, will have conrast transfer that is fairly close to a 6" perfect apeture.

This is a crucial qualification though. Again, this is only if the obstucted scope is perfect.

Finding 6" APOs with perfect optics is not difficult. No leanding manufacturer is going to sell you an expensive APO that is less than near perfect.

Near perfect C9s though are not the norm. And when you add the typeical inperfections, it can cause the MTF line to further sag.

And it doesn't take much for the line to sag enough so that the contrast falls to a 10% difference.

And when it does, the difference starts to show at the eyepeice.

If you know the amount of error though, you can model it.

having compared the TEC160ED to a C9 and a C11, I can confirm this.

I would add, that the C9 should be perfectly colimated and at thermal equilibriuum. I mean collimated on the Airy pattern, which already requires a excelent seeing for these diameter.
So on the field, the TEC160ED will often perform at its maximum while the C9 will require much more care and good seeing condition to deliver similar visual performances.

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timmbottoni
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Re: Reflector/Refractor equivalence formula [Re: olivdeso]
#5777756 - 04/04/13 01:37 PM
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Fascinating topic!

I keep seeing the comparison of mirrored optics that are perfect used in the mtf and formula comparisons. I would love to see an objective way to compare not perfect mirrored optics whether newtonian or sct which is much more realistic, to a high quality apo refractor.

And I like the approach I see by some who are taking into consideration all of the factors including the human eye, seeing, cool down, etc. Somewhere I remember reading that our eyes can resolve the most detail with a 2mm exit pupil which should also be a factor

Timm

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timmbottoni
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Re: Reflector/Refractor equivalence formula [Re: timmbottoni]
#5777765 - 04/04/13 01:43 PM
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Maybe I missed it but is anyone considering the difference in the amount of light lost due to the two reflective surfaces required for reflectors?

Wouldn't this significantly reduce light gathering whereas a refractor loses much less light throughput?

I have no idea how much or how this is calculated by the way

Timm

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Jon Isaacs
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Re: Reflector/Refractor equivalence formula [Re: timmbottoni]
#5777804 - 04/04/13 01:58 PM
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Quote:

Fascinating topic!

I keep seeing the comparison of mirrored optics that are perfect used in the mtf and formula comparisons. I would love to see an objective way to compare not perfect mirrored optics whether newtonian or sct which is much more realistic, to a high quality apo refractor.

And I like the approach I see by some who are taking into consideration all of the factors including the human eye, seeing, cool down, etc. Somewhere I remember reading that our eyes can resolve the most detail with a 2mm exit pupil which should also be a factor

Timm

I think the same assumption is being made about the Newtonian optics as are being made for the refractor optics. High quality apo optics are about 1/8th-1/10th wave. This is possible with a Newtonian...

As far as maximum resolution of the eye occurring at a 2mm exit pupil, I think that is a number mentioned in terms of seeing faint DSOs. Consider resolving a double star. The 2.3 arc-second pairs of the double-double are easily resolved in an 80mm but not at a 2mm exit pupil. A 2mm exit pupil corresponds to 40x, in my experience, the cleanest splits of these pairs are over 100x in an 80mm and if I am pushing the limits of an 80mm, a 0.5mm exit pupil or smaller is desirable.

The eye works better with more light...

Seeing, that's important but variable... some folks live where seeing is rarely excellent, then a small, fast cooling scope might be good. Some folks live where the seeing is good and often excellent. Then a bigger scope that can take advantage of the seeing is useful.

Last night, I got a nice clean split of 52 Orionis, "wide enough to drive a car through"... Skytools 3 tells me it's 1.0 arc-seconds... that didn't happen with a 4 inch or a 6 inch...

Jon

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olivdeso
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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5777834 - 04/04/13 02:14 PM
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There is another factor which may count: the vitreous floaters.

They are much more annoying at x2D on a refractor than at x1D on a reflector

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jhayes_tucson
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Re: Reflector/Refractor equivalence formula [Re: timmbottoni]
#5777983 - 04/04/13 03:48 PM
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Quote:

Maybe I missed it but is anyone considering the difference in the amount of light lost due to the two reflective surfaces required for reflectors?

Wouldn't this significantly reduce light gathering whereas a refractor loses much less light throughput?

I have no idea how much or how this is calculated by the way

Timm

Throughput due to transmission issues is generally not a big issue--it depends on the quality of A/R coatings for the refractor and the reflective coatings for the reflector. Modern, multi-layer reflective coatings can easily achieve 96-99% reflectivity values. To find the total throughput simply multiply the reflectivity of each mirror together. So, for a two-mirror system, the throughput will be in the range of 92-98%. The biggest light loss in the reflector is due to the shadow of the secondary. Just remember that throughput is not what determines peak brightness in the Airy pattern. That is determined by the Strehl ratio. For circular apertures with zero wavefront errors it is easy to show (using a little Fourier optics) that the Strehl relative to an unobscured aperture is given by the square of the throughput (due to the obscuration.) So for an obscuration ratio of 32.5%, the throughput will be just shy of 90% and the relative Strehl will be 0.80 (all relative to a refractor.)

For a refractor system, the quality of the A/R coatings and the absorptivity of the glass are what matters. For small systems (<10"), the absorptivity of the glass is generally negligible so we assume the transmission to be 1.0. Broadband A/R coatings range in efficiency from about 2% to 0.25% (very difficult even at one wavelength!). If we assume that the broadband transmission at each surface is about 99% (probably a bit high), then we can compute the total transmission. For an air-spaced, double element achromat or apochromat, there are four surfaces and the total transmisision will be about 95%. Three element apochromatic objectives may be oiled together or cemented so they will likely have a little higher transmission value; but, that’s harder to compute on the back of an envelope. In either case, if the coatings are good, the light loss due to the coatings is roughly equivalent so it’s not worth worrying about.

John

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GlennLeDrew
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Re: Reflector/Refractor equivalence formula [Re: jhayes_tucson]
#5778203 - 04/04/13 05:43 PM
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The referring to the ~2mm exit pupil as about the 'optimum' for the eye goes to the point I've been making earlier. For example. In all telescopes (of reasonable quality), the observer will find an exit pupil where for his eye the effects of diffraction *just* becomes apparent. The exit pupil just marginally larger than this might be considered the 'best' compromise between image resolution and a clean view not visibly marred by diffraction.

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Jon Isaacs
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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5778538 - 04/04/13 08:53 PM
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Quote:

The referring to the ~2mm exit pupil as about the 'optimum' for the eye goes to the point I've been making earlier. For example. In all telescopes (of reasonable quality), the observer will find an exit pupil where for his eye the effects of diffraction *just* becomes apparent. The exit pupil just marginally larger than this might be considered the 'best' compromise between image resolution and a clean view not visibly marred by diffraction.

I am not quite following... Are you suggesting giving up resolution to have a prettier image, operating at a lower magnification than will allow the observer to resolve the maximum detail or most closely separate binary?

Jon

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timmbottoni
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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5778575 - 04/04/13 09:10 PM
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The 2mm exit pupil comes from the book, "The Backyard Astronomer" but it basically means that your eyepiece that you are likely to find looks the best to your eye, and you are likely to use the most is one that produces a 2mm exit pupil. Just double your focal ratio to pick the eyepiece in that sweet spot. For example, an F/10 SCT a 20mm eyepiece is likely to look the best. I found a complex link that explains the biology behind it here - http://www.telescope-optics.net/eye.htm so for planets it makes sense sort of. I'll let you decipher this at your own leisure.

Timm

Edited by timmbottoni (04/04/13 09:11 PM)

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jhayes_tucson
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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5778577 - 04/04/13 09:11 PM
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Quote:

Regarding your point #4. The phrase "normalize the MTF to the exit pupil" is a clumsy way of mine to try to get across the following point. Any equally good telescope (as a system, with eyepiece of course) working at a given exit pupil will present to the eye an identical degree/scale of visible diffraction on an image point. If the Fresnel pattern subtends some angular scale in a small scope at some particular exit pupil, it will be the same for a larger scope at the same exit pupil.

In other words, the resolving power as a fraction of the maximum is a function of the exit pupil.

Again, a clumsy way to place a basic principle of the afocal system into some perspective, so as to provide something of a more complete picture.

Glen,
I had to sit down and work it out, but you are correct. The angular size of the Airy disk in the exit pupil will always be the same for equal size exit pupils regardless of the diameter of the primary aperture. I find it a bit easier to think of it as follows: The Airy disk will appear to subtend the same angle in two telescopes when the ratio of the magnifications equals the ratio of the diameters.
John

Edited by jhayes_tucson (04/04/13 09:12 PM)

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Jon Isaacs
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Re: Reflector/Refractor equivalence formula [Re: timmbottoni]
#5778595 - 04/04/13 09:21 PM
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Quote:

The 2mm exit pupil comes from the book, "The Backyard Astronomer" but it basically means that your eyepiece that you are likely to find looks the best to your eye, and you are likely to use the most is one that produces a 2mm exit pupil. Just double your focal ratio to pick the eyepiece in that sweet spot. For example, an F/10 SCT a 20mm eyepiece is likely to look the best. I found a complex link that explains the biology behind it here - http://www.telescope-optics.net/eye.htm so for planets. I'll let you decipher this at your own leisure.

Timm

Timm:

Certainly a 2mm exit pupil does not provide the most detail for bright extended objects nor resolve the closest binaries. At 2mm it is unlikely one can actually see the Airy disk.

Jon Isaacs

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timmbottoni
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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5778698 - 04/04/13 10:15 PM
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Hi Jon,
It's been a long time since I got my degree in Biology so I'm not going to try to explain it but the link below has a more readable explanation in the section titled "Optimum Magnification"

http://starizona.com/acb/basics/observing_theory.aspx

It's complex but this attempts to explain what you perceive and why. It's just not as simple as the way you stated it, but no worries, let your eyes tell you what is best for viewing based on your scopes aperture and focal ratio when viewing different types of objects.

I only brought it up in the discussion because it has to be considered if you are trying to compare reflectors of different sizes to refractors of different sizes because the human eye has to be taken into consideration, as some of the posts have pointed out here.

Theory is great, but what you actually see, and how you see it, is often a much more personal and practical experience. That is why some people will make comments about how a smaller high quality APO will appear better than a larger reflector or SCT. Their eye is the biggest factor that the theories aren't considering.

Timm

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mgwhittle
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Re: Reflector/Refractor equivalence formula [Re: timmbottoni]
#5778706 - 04/04/13 10:19 PM
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Quote:

The 2mm exit pupil comes from the book, "The Backyard Astronomer" but it basically means that your eyepiece that you are likely to find looks the best to your eye, and you are likely to use the most is one that produces a 2mm exit pupil. Just double your focal ratio to pick the eyepiece in that sweet spot. For example, an F/10 SCT a 20mm eyepiece is likely to look the best. I found a complex link that explains the biology behind it here - http://www.telescope-optics.net/eye.htm so for planets it makes sense sort of. I'll let you decipher this at your own leisure.

Timm

I believe you are confusing exit pupil of eyepieces with actual pupil size of the eye and the effects on resolution.

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GlennLeDrew
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Re: Reflector/Refractor equivalence formula [Re: mgwhittle]
#5778721 - 04/04/13 10:36 PM
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Quote:

I believe you are confusing exit pupil of eyepieces with actual pupil size of the eye and the effects on resolution.

The effect of the pupil diameter and the exit pupil diameter are one and the same. The eye cares not a whit what it is which is stopping it down; if the pupil feeding light to the retina is 2mm in diameter, whether it be its own natural one or that produced by a telescope eyepiece, the perceived diffraction (and image brightness) is identical.

This is why I have belaboured, to the annoyance of some, I suspect , this whole business of 'normalizing to the exit pupil.'

Even one as obviously knowledgeable as John "had to work it out." From a whole lot of discussions and conversations I've witnessed over the years, I often get a feeling that a focus has been brought to bear on the finer details before having fully absorbed the fundamentals.

My mission, insofar as my meagre knowledge permits, is to ensure the basics are understood.

Jon,
Indeed, the notion of the so-called 'sweet' ~2mm exit pupil is predicated upon a willingness to forgo ultimate resolution in the quest for a view which has not *very* much less detail, but is more aesthetically pleasing.

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Re: Reflector/Refractor equivalence formula [Re: mgwhittle]
#5778723 - 04/04/13 10:36 PM
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Mark,
If your eye is dilated to 5mm and the exit pupil of the eyepiece that your eye is pressed up against letting in light is at a 2mm exit pupil, do you think your retina sees 5mm of light or 2mm of light?

Timm

Edited: Glenn beat me to the reply with a much more detailed explanation but the point is the same - the exit pupil of the eyepiece is the limiting factor unless it is larger than your pupil dilation.

Edited by timmbottoni (04/04/13 10:39 PM)

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mgwhittle
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Re: Reflector/Refractor equivalence formula [Re: timmbottoni]
#5778738 - 04/04/13 10:46 PM
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Timm,

You mentioned planets. Please correct me if I am wrong, but a 1mm exit pupil has always been touted as giving the most resolution on planetary detail without being affected by the aberrations in the eye. My apologies, I should have been more specific in what I was questioning.

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Re: Reflector/Refractor equivalence formula [Re: mgwhittle]
#5778780 - 04/04/13 11:17 PM
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Quote:

Timm,

You mentioned planets. Please correct me if I am wrong, but a 1mm exit pupil has always been touted as giving the most resolution on planetary detail without being affected by the aberrations in the eye. My apologies, I should have been more specific in what I was questioning.

Could be, if you find 1mm to be better for looking at planets, then that is the right exit pupil for you. It could even be smaller! If you read the article below, where it talks about Maximum Magnification and it refers to planets, you might find it useful. It really depends on the scope, the seeing conditions, and your eyes...

http://starizona.com/acb/basics/observing_theory.aspx

I don't honestly worry about what exit pupil I'm at, I just posed it as a consideration in the discussion. I just use trial and error on any given night and change eyepieces until I find the magnification that seems best for the scope and viewing conditions. Basically, who cares about theory, what looks best to you is what matters, right? There are reasons that things look better at different magnifications and exit pupils, which is why in my C8 I like to use the WO Zoom eyepiece. It allows you to control it easily without having to fumble with switching out eyepieces.

Timm

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Re: Reflector/Refractor equivalence formula [Re: timmbottoni]
#5778809 - 04/04/13 11:37 PM
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Timm,

While my knowledge on the subject comes from the writings of more intelligent people like you and Glenn, I am still getting confused regarding the general assumption about a 1mm exit pupil, planetary detail and where the 2mm exit pupil falls into all this. I have read the link you provided in the past and I will peruse it again. I actually do worry about exit pupils and try to match eyepieces to individual scopes in order to have at least one at 2mm and one at 1mm exit pupils when observing. I do this because of the points you and Glenn have been talking about regarding a 2mm exit pupil, which I have found in practice to match theory. But I have also found the practice of a 1mm exit pupil and planetary detail to be valid.......now how do I make that correspond to the theory regarding 2mm? Insert exasperated sigh at this point......no need to respond directly, just clarifying my thoughts....and do some more reading over my head.

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Re: Reflector/Refractor equivalence formula [Re: mgwhittle]
#5778861 - 04/05/13 12:43 AM
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So... ae we after a rational function here? Straying from the idealism for a moment, as apeture increases - each also has potential for errors from figuring/finishing, how do you porpose to quantify them(I think they are all rational to begin with), I don't see them as scalar values that somehow stand still. I can understand using the Airy function as a component, or the Fresnel-Huygens, but what exactly is the function you are using for non point source? I think most if not all the formula/functions are based on point source? If looking for something as a 'predictor' it should have some reasonable error function tandem to the specific design, IOW I'd expect a +/- from some average and not ideal perfect which I think useless. It's all that needs to be put in a harness that I think difficult, you'd need for each type and sub type.

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Re: Reflector/Refractor equivalence formula [Re: CounterWeight]
#5778936 - 04/05/13 02:38 AM
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Mark,
I wouldn't overly fret over quantizing to exact and precise values for the exit pupil. Between the useful range (0.5-0.7mm up to about 7mm), it's a continuum. For example, the approximately 2mm moderately high power 'sweet spot' will not be quite the same from one person to another. And even for an individual, going to a different scope, and certainly seeing conditions, will slightly modify this.

Getting a better handle on what works best for you will come with experience.

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5779028 - 04/05/13 06:53 AM
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Quote:

For example, the approximately 2mm moderately high power 'sweet spot' will not be quite the same from one person to another.

Absolutely!
As a physician: As exit pupil decreases, a smaller and smaller axial sector of the physiologic pupil is being utilized. A such, individual irregularities and defects within that sector of imperfect living tissue come to occupy a greater percentage of the entrance pupil of the eye, now being defined by the exit pupil of the EP. This reintroduces and exacerbates the whole issue of imperfect and obstructed aperture (and it and it impact on the MTF), but now at the level of the eye as a an optical instrument. Floaters are but one physiologic example but there are others as well. While generalization across a population regarding optimal small exit pupil can and should be made, these must be distinguished from statements about such optimization for a given individual.
Joe

Edited by jpcannavo (04/05/13 07:06 AM)

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Re: Reflector/Refractor equivalence formula [Re: jpcannavo]
#5779051 - 04/05/13 07:30 AM Attachment (10 downloads)
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Here it is:

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t.r.
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Re: Reflector/Refractor equivalence formula [Re: orion69]
#5779071 - 04/05/13 07:55 AM
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Re: Reflector/Refractor equivalence formula [Re: orion69]
#5779094 - 04/05/13 08:16 AM
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Quote:

Here it is:

You left something out!
Completing the formula!

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t.r.
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Re: Reflector/Refractor equivalence formula [Re: t.r.]
#5779101 - 04/05/13 08:21 AM
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Quote:

Indeed, the notion of the so-called 'sweet' ~2mm exit pupil is predicated upon a willingness to forgo ultimate resolution in the quest for a view which has not *very* much less detail, but is more aesthetically pleasing.

Indeed! All one has to do is look through the myriad of observer reports out there and note where the claimed "best" image is attained with any given scope. THe vast majority conclude without realizing it, that it happens at a magnification that corresponds to a 2mm EP! For planetary on occasion, personally I profitably use .4 to 1mm to see obscure details in small apertures more easily. Yes, I know that no additional detail is resolved, but if the bigger image even if stretched out abit and dimmer, helps me see a small feature more easily , I do it! Many, many times I ran a TV Genesis (with CA to boot!) up to 260x with the Televue marketed, TV 2.5x barlow and 4.8 nagler combination, which was advertised as this scopes highest power capability. I had many memoriable views of planets and doubles at that overkill .39mm exit pupil, rules of thumb be damned!

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Re: Reflector/Refractor equivalence formula [Re: timmbottoni]
#5779118 - 04/05/13 08:34 AM
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Quote:

Hi Jon,
It's been a long time since I got my degree in Biology so I'm not going to try to explain it but the link below has a more readable explanation in the section titled "Optimum Magnification"

Tim:

I am reasonably well versed in this subject, this something I have spent some time studying, both in literature, here on the web and at the eyepiece. The article makes my point very nicely:

" Empirical evidence suggests that an exit pupil of 2mm produces very detailed views of most deep-sky objects."

2mm is an empirical number for deep sky objects, not for double stars or planetary viewing where more magnification will clearly reveal more detail. I believe this was what I pointed in in my first post. For example, in an 100mm scope, a 2mm exit pupil is only 50x, hardly an optimal magnification for planetary observation and the article supports that. In my experience, in a 4 inch, something more than 100x is best for showing planetary detail.

There are some things in the article I disagree with, they suggest that for small low surface brightness objects, one should use between 150x and 250x in a 6 inch, 80x-150x in a 12 inch. Since one can see smaller, low surface brightness objects with a 12 inch than a 6 inch, working at the limit, the same exit pupil is appropriate in both scopes.

But, as you say, in the final analysis, your eye should tell you, the rest of this stuff is just guidelines. If you are getting the best view of Jupiter at 50x in a 4 inch apo, go for it... doesn't work for me though.

Jon

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5779136 - 04/05/13 08:52 AM
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Quote:

Jon,
Indeed, the notion of the so-called 'sweet' ~2mm exit pupil is predicated upon a willingness to forgo ultimate resolution in the quest for a view which has not *very* much less detail, but is more aesthetically pleasing.

Glenn:

I find two or more closely separated central disks surrounded by the interacting diffraction ring structure to be about the most aesthetically pleasing view possible. At a 2mm exit pupil, a Dawes limit double star is certainly not resolved and may not even be detectable.

Planets are a different story, at least in a larger scopes because one is rarely working at the limit of resolution, it's more a combination of optimizing the magnification in terms of brightness and seeing. Since it is generally seeing limited, Jupiter most often at it's best somewhere between 160x and 240x pretty much regardless of aperture. The eye will definitely see more contrast and detail in the larger scope at the same magnification. For viewing the planets exit pupil is not so useful because the limit is the seeing.

How does this all apply to an equivalence formula: Optical quality is most important at higher magnifications, smaller exit pupils where one can actually resolve the differences. Wandering around the Milky Way with a 6mm exit pupil is mostly about off-axis aberrations in the scope and the eyepiece. Even at a 2mm exit pupil, the eye (at least my eye) will not resolve the Airy disk structure of a star, to see the difference in on-axis performance requires pushing the magnification.

Jon

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5779391 - 04/05/13 11:01 AM
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Quote:

Quote:

Hi Jon,
It's been a long time since I got my degree in Biology so I'm not going to try to explain it but the link below has a more readable explanation in the section titled "Optimum Magnification"

Tim:

I am reasonably well versed in this subject, this something I have spent some time studying, both in literature, here on the web and at the eyepiece. The article makes my point very nicely:

" Empirical evidence suggests that an exit pupil of 2mm produces very detailed views of most deep-sky objects."

2mm is an empirical number for deep sky objects, not for double stars or planetary viewing where more magnification will clearly reveal more detail. I believe this was what I pointed in in my first post. For example, in an 100mm scope, a 2mm exit pupil is only 50x, hardly an optimal magnification for planetary observation and the article supports that. In my experience, in a 4 inch, something more than 100x is best for showing planetary detail.

There are some things in the article I disagree with, they suggest that for small low surface brightness objects, one should use between 150x and 250x in a 6 inch, 80x-150x in a 12 inch. Since one can see smaller, low surface brightness objects with a 12 inch than a 6 inch, working at the limit, the same exit pupil is appropriate in both scopes.

But, as you say, in the final analysis, your eye should tell you, the rest of this stuff is just guidelines. If you are getting the best view of Jupiter at 50x in a 4 inch apo, go for it... doesn't work for me though.

Jon

Yes Jon,

I would agree with you. My initial use of the planet was probably not well thought out because if you don't have enough magnification to see the details then 2mm is not the best exit pupil.

Timm

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Re: Reflector/Refractor equivalence formula [Re: orion69]
#5779661 - 04/05/13 01:48 PM
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Quote:

Here it is:

thanks for clearing that up. haha

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Re: Reflector/Refractor equivalence formula [Re: buddyjesus]
#5779674 - 04/05/13 01:53 PM
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Going back to comparing the two and thinking out loud, to be useful formula (though i think you need to derive one for each optic and it becomes more a function or matrix or ratio(see below))a requirement IMO is the usefulness would be bound somewhat by initial conditions as marketed. Though i see no way to avoid so called 'magic numbers, you'd need to plug in some values based on optics theory. Holding apeture and or focal ratio constant for the two scopes I see as an unnecessary burden, but should be useful for that. Should give a realistic expectation for where the person will observe. Though expressed algebraic, should be expressed arithmatic with real values as plug-ins, with capability to graph variance if desired suited to obsevers location, look for tipping points and diminishing returns.

qualities tables-

refractor, doublet refractor>| apeture | corrected to what end or not, poly strehl 'efficiency | focal ratio | quality of figure, polish, coatings | overall build quality | any added correction

refractor, triplet or quad refractor>| apeture| | poly strehl 'efficiency' | focal ratio | quality of figure and polish and coatings | overall build quality | any added correction

reflector Newtonian>| apeture | size of central obstruction (as percent of apeture area) | poly strehl 'efficiency' |quality of figure, polish,coatings | overall build quality | any added correction

Reflector 'other' | anything more to add for an SCT or Maksutov or DK/CDK/??...

Values/units of measure-

Poly strehl efficiency would be 98% of ideal "for the design as measured" corrected for materials/methods used then 'quality' could be a number .2 to .98 and used as a decimal multiplier. Higher=better.

Apeture a positive real number, just keep both measurements same units.

Focal length a positive real number, units same.

Focal ratio a positive real number.

Quality of figure and polish and coatings - a decimal multiplier 0.2 to 0.98, positive real number. Higher = better.

Overall build quality, non optical, lens cell or mirror support, focuser, baffle/shroud, fans? - a decimal multiplier 0.2 to 0.98 positive real number. Higher=better.

Many folks use a corrector on a faster Newtonian, a four lens fast refractor has it built in. Some sort of bounding box necessary here.

Magnitude/size of reference object. Here use Luna, Jupiter, a point source reference star, and something very low magnitude 'distributed' (ex nebula or galaxy disk) for real world upper lower bounding. This unfortunately would be multi valued?, but I feel important to traverse utility for object types - planets, doubles and multi star, globulars, emission and reflection nebula, galaxies.

Sky conditions of where used (LP, atmosphere stability and clarity), again I'd use a decimal multiplier that would decrease the final value for the optic - but I'd keep it the same for both scopes. Again something like 0.2 and 0.98. Higher number=better.

No small amount of fudge factor implied by the efficiency and quality measures and probably best to go on past performance as measured or 'rule of the club'? Example being if Zambuto/Lockwood/... or A-P/TEC/... vs others.

Expressed as a matrix, rational, or as a ratio I'd expect to represent known observed and be valid without ever using the term 'ideally'...

poor quality vs. poor quality different type.
poor quality vs. best quality of same type.
poor quality vs. best quality different type.
best quality vs. best quality different type.

Areas of difficulty.

-Says nothing about observers visual acuity, eyepiece, personal preference or aesthetics.
-Allows for overly optimistic values where subjective to location and quality / efficieny.
-Using 0.2 to 0.98 possibly too granular, and instead keep open the option to use integer divisors instead?
-need to include ground effect?

.

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Re: Reflector/Refractor equivalence formula [Re: CounterWeight]
#5779994 - 04/05/13 04:04 PM
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Quote:

Expressed as a matrix, rational, or as a ratio I'd expect to represent known observed and be valid without ever using the term 'ideally'...

I think the reality of the matter is that such a matrix would be a lot of work and not much use in comparing reflectors to refractors because there is very little overlap. The difficulties with reflectors disappear with increasing aperture, the difficulties with refractors increase with increasing aperture.

Such a matrix might be more useful in comparing refractors to refractors and reflectors to reflectors but in the long run, by the time one knows enough about ones own preferences and observing habits, one has looked at and through enough telescopes to already know...

Jon

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5780110 - 04/05/13 04:36 PM
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Quote:

The difficulties with reflectors disappear with increasing aperture, the difficulties with refractors increase with increasing aperture.

Jon, Can you explain what you meant here? Seems like difficulties with both designs increase with aperture. Or maybe you are talking more chromatic abberation here. There are certainly aspects of difficulty which increase with reflectors as well. Cool down, weight, eyepiece position (depending on focal ratio), size, etc.

There are people who like big refractors as well as big reflectors. I think many people would think both camps are somewhat nutty. And it's a somewhat individual opinion on what size scope is perfect and what size is too big. Each person would have their own bell curve.

I saw an interesting post on this site comparing many scopes of different types. I'm not sure how scientific it was, but I thought it made some sense. here is a link: http://www.cloudynights.com/item.php?item_id=1972

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Re: Reflector/Refractor equivalence formula [Re: CounterWeight]
#5781412 - 04/06/13 10:14 AM
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Quote:

I think most if not all the formula/functions are based on point source?

This is why Modulation Transfer Function (MTF) is difrerent.

While MTF has it's roots in point function, point function is after all, mostly a function of diffraction related to the size of the aperture.

The role of MTF is to extrapolate that data and apply it to how contrast is transferred on extended targets with different size and contrast features.

Among optical engineers and testers, MTF is considered the universal standard for comparisions of how an apeture will perform against a perfect aperture or apertures of differnt sizes.

People don't like it, but I usually let an MTF plot speak for me.

Here is a great example of MTF in action. This is a bench test done by Rohr at Astro-foren.

The MTF is near the bottom of the web page so you have to scroll down to see it.

I can look at this test and tell that the telesocpe is so poor (and the MTF includes the central obstruction) that this telescope would have contrast no better than a good ED refractor that was only 1/4 of the aperutre (perfect line that would end at about .25 on the x axis if it were plotted).

With this MTF plot for this particular telescope under test, a 152mm MCT, I can easily tell that for extended targets, I would do almost just as well (from a contrast persective) with a 38mm refractor

That is the power of MTF. You can easily tell how one telescope would compare to another.

152mm MCT that is no better than a 38mm refractor!!!!!

I just can't for the life of me understand why people on Cloudy Nights would be so resistent to learning how to use MFT?????????????????? It is such a powerful way to visualize how two instruments would compare!

Edited by Eddgie (04/06/13 10:23 AM)

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Re: Reflector/Refractor equivalence formula [Re: GOLGO13]
#5781505 - 04/06/13 10:44 AM
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Chris

If you try to design a 3 inch f/5 Newtonian that does a reasonable job of illuminating a 2 inch widefield eyepiece, or even a 1.25 inch eyepiece, you will find it's essentially impossible with a real focuser because the secondary needs to be so large. As you increase the aperture, this becomes less and less of a problem. Coma is a function of focal ratio and independent of aperture. The other disadvantage is the reflectivity losses with the mirrors. With a small scope this is important because one is using smaller exit pupils to push the magnification. But with a larger scope, larger exit pupils and higher absolute magnifications are possible so again this is less of a problem.

Refractors are the other way around, the primary aberration is chromatic and it becomes more and more difficult to correct as aperture increases. Lens fabrication costs increases with something closer to the cube of the aperture, transmission losses increase and thermal issues increase because lens thickness increases and possibly more elements make cooling more difficult.

So.. in a small scope, a faster apo refractor is a do it all, low power and high power scope and a Newtonian is a serious compromise, in a larger scope, the tables are turned. A 12.5 inch F/4 Newtonian with top notch optics can provide both big, wide low power views and be an excellent planetary and doubles scope.

Where the transition is can be seen in the market, refractors stop quite abruptly at 6 inches. There are larger refractors but they are few and generally single purpose. And at 6 inches, apos are very expensive and achromats are either long focal lengths and unwieldy or shorter but less than optimal for planetary and double star viewing.

Jon

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5781556 - 04/06/13 11:00 AM
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Ok...I don't disagree with any of that. Still, I don't want people to think that refractors from 3-6 inches are not useable because they are smaller than the same (or much less) costing SCTs and reflectors. They are just different animals in some ways.

Then again I see a lot of threads claiming that they are somehow magical items which outperform scopes twice or three times the aperture. I do think those observations are probably more based on something being wrong with the larger scope. But, it's one of those "you have to be there" to really know the situation.

Me I haven't met many telescopes I have not liked.

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5781874 - 04/06/13 01:30 PM
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Quote:

Refractors are the other way around, the primary aberration is chromatic and it becomes more and more difficult to correct as aperture increases. Lens fabrication costs increases with something closer to the cube of the aperture, transmission losses increase and thermal issues increase because lens thickness increases and possibly more elements make cooling more difficult.
Jon

It is not necessarily harder to control chromatic issues as aperture increases. Among the issues you've listed, I want to add another problem with larger refracting elements is material homogeneity. Depending on the glass type, homogeneity can become a concern for apertures as small as 8-10". One of the reasons that the Alvin Clark and sons were so successful at making large refracting objectives was that they developed techniques to apply surface corrections to handle large slope errors in the wavefront due to index variations. Changes in shape of a lens due to thermal variations tends to be less of a concern than for mirrors because the wavefront changes as delta z*(n-1) for the refractor vs 2*delta z for the mirror. However, you also have to worry about how index varies with temperature (delta n/delta t). Cost and weight are the two biggest issues with refracting objectives as size increases and focal ratio decreases. It also gets harder to mount large refracting components and you are stuck with edge mounting, which drives the weight up (because the elements have to be thick to minimize gravity sag.) So, it gets really expensive and heavy to build large aperture, fast refractors. Those are just some of the reasons why refractors have never grown beyond 40" aperture.
John

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Re: Reflector/Refractor equivalence formula [Re: jhayes_tucson]
#5782642 - 04/06/13 09:22 PM
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Quote:

It is not necessarily harder to control chromatic issues as aperture increases.

Certainly more effort is required, more design compromises required to control the chromatic aberration. Better glass, more elements, slower focal ratios. A 40 mm f/8 achromat will have very little. The 40 inch F/17 you mention has focuses for the different colors that are a quarter of an inch apart.

Jon

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Re: Reflector/Refractor equivalence formula [Re: GOLGO13]
#5783571 - 04/07/13 11:13 AM
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Quote:

I don't want people to think that refractors from 3-6 inches are not useable because they are smaller than the same (or much less) costing SCTs and reflectors. They are just different animals in some ways.

And I agree 100% with you.

Different telescopes clearly have the strengths.

For the smaller refractor, it the excellence of performacnce for a small apeture per inch of aperture. Below about 5", and the refractor is king of the hill doing virtually everything better than any reflecting design made of similar small apetures. While I personally have no use for a refractor smaller than 6", that does not mean at all that they cannot be highly useful instruments by others.

From 5" to 7", the refractor enjoys contrast that is equal to a slightly larger reflector or a considerably larger SCT, and provides wider, come free and well illuninated fields (all excellent characteristics). The tradoff is size and cost.

Beyond 7", and the refractor starts to struggle with CA unless it is made with exotic glasses, or length and weight if it is not, and this comes with a sacrifice in true field, which becomes a point at where a slightly larger reflector becomes a more realistic alternative.

I don't think anyone is saying that refractors from 3" to 6" are not usable. In fact, my favorite wide field telescope is a 6" refractor. Nothing I own does better for wide field scanning or for framing large, rich star clusters.

But for deep sky or planets, I have larger scopes that seem to serve me better.

But that in no way diminishes the usefulness of my 6" refactor because it does things none of my other telescopes can do.

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Re: Reflector/Refractor equivalence formula [Re: CollinofAlabama]
#5784150 - 04/07/13 03:52 PM
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Quote:

This kind of information is enjoyable to look at. Something tangible, and yet ...

Sean made the point about "in the field". And here's where your graphs run afoul of reality.

You can include field conditions in the MTF. Most do not include those aberrations, so we fight over them. "In the field...blah blah blah." Whose field? Yours or mine? You have to apply the same conditions to both scopes, otherwise any comparison is absolutely meaningless. Pointless. And a source of frustration. If the graphs deviate from reality it's simply because those real world variables are not included in the graph. They can be.

And you can graph seeing conditions. Pretty smart fellers have done that, taking data and developing some cursed theory. The effects of seeing are complex an complicated and often skewed by subjective preferences. It is possible for a larger, obstructed aperture to perform poorly rather quickly in deteriorating seeing, but the smaller refractor is similarly affected at some point.

Smaller apertures are certainly resistant to seeing, but they are not not immune. It all has to do with the affect on the Airy pattern which the MTF can plot with some degree of accuracy. That's the theory, the science.

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5784210 - 04/07/13 04:16 PM
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Quote:

Such 'formulae' are exit pupil dependent. At larger exit pupils, resolving power and contrast transfer are not impacted by a central obstruction.

Ya, true in principle. The impacts exist at smaller scales and not readily discernible at large exit pupils. But they are still there. Just crank up the magnification and see. But, you point it well noted, a RoT would be affected in terms of scale and, as Eddgie points out, in terms of illumination when you get beyond the focal plane to the retina.

By the way, following you and Eddgie, I nearly bit all my finger nails off. You guys are on the same sheet of music, best I can tell, you're just using different language and talking past one another from different vantage points. You're seeing the same thing from different angles. Contrast is on the vertical axis...that's the same. Resolution is along the horizontal axis, that improves with aperture.

At the same frequency, contrast does improve with aperture. At the same contrast, resolution improves with aperture. They are intertwined, as you say, and not the same thing. I think...

Edited by Asbytec (04/07/13 05:39 PM)

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Re: Reflector/Refractor equivalence formula [Re: Cotts]
#5784223 - 04/07/13 04:22 PM
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Quote:

The trouble with a rule of thumb is that people's thumbs vary greatly in length.....

Dave

Yup. Thanks for the chuckle.

I was helping Wilfried (WRAK) with his RoT for double stars (obstructed vs unobstructed.) Rules of Thumb can become pretty complicated as his work, among others, can attest to. If they become too complicated and include all pertinent variables, they cease becoming a RoT to become a cursed theory.

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5784306 - 04/07/13 05:00 PM
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Quote:

I stress these points so as to disabuse anyone of the false notion that a larger aperture has better contrast transfer.

I wish I understood these things in the same way you do. At same exit pupil, same illumination but different scale. Correct? The larger aperture should not have better contrast transfer, I believe you are correct. Both apertures fall to zero at their maximum spacial frequency.

However, a larger aperture is capable of resolving a higher spacial frequency and exit pupil dependent. That's higher resolution at the same level of contrast (from the zero to 100 percent scale all scopes are graphed against.)

And an obstructed scope is capable of even higher resolution, which is why the MTF curve deviates beyond perfect unobstructed aperture. But, those are on very small scales (alluding to your previous post on large exit pupils.)

The image cast onto the retina is a result of the image projected from the exit pupil. At the same exit pupil, a larger aperture will put a larger image on the retina. At the same magnification, a larger aperture will put a brighter image on the retina. Ya?

So, aperture does rule, even obstructed...over much of the contrast range and image scales. At large scales, aperture rules because of light gathering power and diffraction effects limited to smaller scales. At smaller scales, obstructed and larger apertures rule in terms of raw resolution.

Somewhere in the mid range, unobstructed (and same) clear apertures rules all else being equal (including seeing, thermal equilibrium, Strehl, and every other variable that - is assigned as a detriment inherent in some designs and - can definitely impact a RoT.)

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Re: Reflector/Refractor equivalence formula [Re: CollinofAlabama]
#5784323 - 04/07/13 05:08 PM
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Quote:

Of course, these antecedotes still don't tell me where the XLT 150 would fall compared to a 102mm/110mm/120mm refractor, but I do believe it would outperform any 90mm (or 92mm) apo.

Anyone else with more experience in these matters is more than welcome to join in.

I can produce sketches of Jupiter in a 6" CAT that compare well with CCD IMAGES taken with a 4 to 4.5" APO. If I could produce sketches and images (just have to dig them up from CN archives), would that be definitive? It's pretty close to what theory states, given the Strehl and CO involved, and both images and sketches are taken "in the field." How about a sketch of Ganymede albedo features?

Point being, I think the Aperture - CO = APO equivalent is a reasonable RoT. Not perfect, just a RoT. I get the feeling you like "in the field" experiences, and rightfully so. But you cannot toss theory out the window, you have to use it correctly and within its limits. I'm sure MTF limits are beyond most observer's abilities to notice, so it remains an effective tool and can model "in the field" to a high degree of accuracy. If you use it that way.

Edited by Asbytec (04/07/13 05:20 PM)

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Re: Reflector/Refractor equivalence formula [Re: Mark Harry]
#5784427 - 04/07/13 06:00 PM
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Quote:

About the ganymede observation with 6+14" scopes---
***
Isn't the resolving power of a 6" around 1 arc-second?
I would think if so, an investigation why the features weren't seen is mandatory.
(Daves, and Marks remarks about rule of thumb- right on the mark!)
M.

Ganymede features can be resolved in a 6" glass, you just gotta try it with all induced variables minimized in good seeing. I say resolved because one can see light and dark features just like resolving light and dark line pairs...at low contrast and very small exit pupils. Eddgie resolves them better in his C14 than I do in my 6" MCT, but both scopes can do the job.

For a tougher challenge, try elongating Io. It can be done in a 6" glass, too. Ask Pickering. Ask me. Try it. A few CNers have done it.

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Re: Reflector/Refractor equivalence formula [Re: Asbytec]
#5784611 - 04/07/13 07:35 PM
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To date the smallest aperture I know of that pulled this off was a 6" apo. The, Norme with his Mak. Somewhere inbetween those two examples I did it as well with my 8". This past apparition, I never had seeing good enough to pull it off. For it,needs 8/10 or better and 400x.

Pete

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Re: Reflector/Refractor equivalence formula [Re: Asbytec]
#5784666 - 04/07/13 08:03 PM
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Norme,
Regarding my earlier statement you quoted, which says that at given exit pupil the larger aperture does not by itself deliver higher contrast. This is from the viewpoint of the observer, and the 'quality' of the view at the scale of resolving power of the retina.

If we ignore the false flag of concentration on *a particular, fixed target*, we are then considering the extent to which the diffraction present impacts contrast transfer as a function the resolving power of the *detector*, which is the all-important eye for the visual observer.

The dichotomy here results from only from the change of perspective. One viewpoint examines the absolute resolving power at the focus, and the other considers the resolution as a function of the retina's resolving power.

At large exit pupils, the image is significantly under sampled by the retina, and so the full capability in instrument resolving power is not realized. At some point, depending significantly on image brightness and to some extent on intrinsic contrast, an exit pupil is reached where diffraction effects just become perceptible. At ever smaller exit pupils one gets farther into the realm of 'empty magnification'.

Where 'empty magnification' begins depends sensitively on image brightness. For sunlit objects, once the Airy disk subtends the resolved limit of a couple of arcminutes, the image is effectively fully sampled. For a dim nebula, the Airy disk could subtend a couple of *degrees* and the image would not be over sampled. Of course, this would require utterly miniscule exit pupils of few hundredths of a millimeter, which would result in complete invisibility.

In the more realistic realm, where the brightest nebulae reside, one can employ quite small exit pupils--0.5mm and even smaller--to good effect. The image has surface brightness low enough that the eye's poor resolving power permits almost ridiculous magnification, but at the same time high enough to retain visibility.

The point of this digression is to show that a refractor/reflector equivalence depends sensitively on image surface brightness. For extended DSOs (not star clusters), there is no functional difference between instruments, for the subject has insufficient surface brightness to allow the detection of contrast loss at those small scales due to diffraction.

And this is yet more fodder for the consideration of the 'quality' of the view as experienced by the eye, its own limits having a profound impact.

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5784872 - 04/07/13 09:55 PM
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Glenn, I always learn a lot form your perspective. (I have a kid screaming in my ear and some Skype chat ongoing, so I hope I read you correctly. Will come back to it momentarily.)

Yes, I'd agree the CO has little effect on larger scale DSOs and quality is important. More later, just wanted to thank you for replying.

EDIT: "At large exit pupils, the image is significantly under sampled by the retina, and so the full capability in instrument resolving power is not realized. At some point, depending significantly on image brightness and to some extent on intrinsic contrast, an exit pupil is reached where diffraction effects just become perceptible...For sunlit objects, once the Airy disk subtends the resolved limit of a couple of arcminutes, the image is effectively fully sampled."

Yes, makes sense.

Edited by Asbytec (04/08/13 02:19 AM)

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5784941 - 04/07/13 10:25 PM
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Quote:

The point of this digression is to show that a refractor/reflector equivalence depends sensitively on image surface brightness. For extended DSOs (not star clusters), there is no functional difference between instruments, for the subject has insufficient surface brightness to allow the detection of contrast loss at those small scales due to diffraction.

Glenn:

There is an easily done and quite interesting demonstration of the importance of the surface brightness of the image to the resolving power of the human eye.

A couple of months ago I happened upon some members of a local club doing public outreach with their solar scopes. I took my turns at the eyepieces. At one point I heard one of the members say that you could not see the moon through a solar filter. Having made the calculations and having done so in the past, I mentioned that I was quite sure you could see the moon through a solar filter. This resulted in a lively discussion, with skeptical looks... Of course, they didn't know me from the next Joe.

It's true, a solar filter has an attenuation of 100,000:1, about 12.6 magnitudes. The surface brightness of the full moon is 3.4 magnitudes/square-arc-second, a solar filter changes that to 16 magnitudes/square arc-second. The filter transforms the -12.6 magnitude full Moon to a magnitude 0 object with a diameter of 1/2 degree. For comparison's sake, M57, a relatively bright DSO, has a surface brightness of about 18.7 magnitudes/square arc-second, almost 3 magnitudes dimmer.

So, what does this look like at the eyepiece? Only having a solar filter for my 4 inch and under scopes, I set out that evening with the solar filter and some eyepieces to view the moon. Using my non-observing eye to find and roughly focus on the moon, I then slid the solar filter in place, the exit pupil was about 5mm.

With my eye dark adapted, I could clearly see the moon as a large half crescent (it was not yet full) but with essentially no detail. There was no detail to be seen and it was very difficult to focus, stars don't shine through the solar filter. Increasing the magnification showed nothing more and it was very difficult to focus because there was nothing to focus on. (If I focused using my observing eye, then I could watch the initial stages of my eye dark adapting, a worthy lesson in itself. The moon started out very faint but quickly gained brightness.)

The lesson this shows is the importance in surface brightness, the moons details were at the focal plane and a long exposure with a camera could have shown the detail but I could not see any because it was not bright enough. The "contrast transfer" of the telescope had no bearing on what I saw because my eye could not resolve details at such low light levels.

A larger scoper could have provided a larger image and my eye might have eeked out some major details but the same effect could have been achieved by using a filter that was less aggressive and allowed for that same surface brightness at the same magnification.

Extended exposures using a small scope will resolve details in faint DSOs that are not visible using large scopes visually. This is because the photographic process allows the full use (or at least closer to it) the resolution and contrast possible. It is not limited by the eye... those details on the moon are there in the exit pupil, the eye cannot see them.

For these dim images, it's not the contrast of the scope, any scope has plenty of contrast at the scales that the eye can resolve, it's the eye's abilities that are important.

Try it... solar filter + moon...

Jon

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Re: Reflector/Refractor equivalence formula [Re: GOLGO13]
#5784970 - 04/07/13 10:37 PM
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Quote:

Ok...I don't disagree with any of that. Still, I don't want people to think that refractors from 3-6 inches are not useable because they are smaller than the same (or much less) costing SCTs and reflectors. They are just different animals in some ways.

My point was and is simply that refractors make good small scopes, reflectors make good big scopes... that's basic common sense, I just added some reasons...

Jon

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5785078 - 04/07/13 11:52 PM
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Jon,
That solar filter used on the moon test is something that every amateur should experience. To actually *see* how truly poor is the eye's resolving power at brightness levels still well above that for the brighter DSO fuzzies really drives home the fact. Then it would be amply evident that for nebulae and galaxies, optical quality can be surprisingly bad to no ill effect; even the execrable 1/2 to 1 wave light bucket is more than good enough. Of course, stars would look bad.

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5785259 - 04/08/13 03:45 AM
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Jon, Glenn, great points. Never thought of the moon as a DSO. I guess it can be with a solar filter. So, the contrast and resolution seem to be on the focal plane as evidenced by the ability of an imaging detector to record it over time. We've all see images of M51 that best visual appearances, so this seems to hold true.

So, how does one reconcile that with the MTF? The image on the focal plane should remain consistent with MTF. It's easier to resolve the Whirlpool's spiral arms than individual stars, the lower spacial frequency arms are set off - contrast wise - by darker space between them and the core. So much so, we can see them visually.

The MTF does not involve image brightness, only scale and contrast. So, even a timed exposure of M51 will obey the resolution and contrast according to "theory." The brightness thing, as you say, is on the retina: a human factor. The MTF contrast and resolution relationships still apply to the moon through a solar filter just as it does to M51 or Jupiter.

Gonna chew on that example for a bit...fascinating brain food. What does that mean for a catadioptric/dioptric equivalence formula? Probably nothing, really, as the MTF can still be used to express that aperture to obstruction relationship as Eddgie points out. (aperture - CO, approximately.)

Edited by Asbytec (04/08/13 04:45 AM)

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Re: Reflector/Refractor equivalence formula [Re: Asbytec]
#5785454 - 04/08/13 08:54 AM
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Quote:

So, how does one reconcile that with the MTF? The image on the focal plane should remain consistent with MTF. It's easier to resolve the Whirlpool's spiral arms than individual stars, the lower spacial frequency arms are set off - contrast wise - by darker space between them and the core. So much so, we can see them visually.

Glenn is pointing out that the MTF of the telescope has little to do with what we see at the eyepiece when looking at faint objects because it's our eye that is the limitation rather than the telescope, any telescope provides sufficient contrast, sufficient resolution. Viewing the moon through a solar filter demonstrates this very nicely. I know the detail is there and clearly the telescope can resolve the detail with little loss of contrast.

But I can't see that detail, that contrast because my eye is a very poor sensor at low light levels. All that is necessary to see that detail is to brighten up the image maybe 10,000 times or even more. Unfortunately, a telescope cannot do this, the brightness of M51 is limited by the 7mm (or so) entrance pupil of the eye. When I look at M51 in my 25 inch with a 6mm exit pupil, it is no brighter than it is when I look at with a 60mm refractor with a 6mm exit pupil, it is just larger so I see more detail.

Jon

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5785739 - 04/08/13 11:31 AM
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Jon, again that's interesting. I think the MTF is still at play, but I believe you're correct about brightness. The detail is there, the eye's response is too low. And the MTF does not model brightness. I ma still trying to understand the brightness and contrast relationship, I think it's in the eye. But, does that mean the MTF does nto accurately describe contrast and resolution. Not sure, the relationship between contrast and resolution still exist as evidenced by time exposures.

Still, both a 6" refractor and a C14 will have the same trouble with brightness, so any RoT would hold at a given exit pupil. <CLICK> It just clicked with me the relation between exit pupil and MTF. what we're really taking about is the translation of full contrast of the actual object to the focal plane with the resulting loss of contrast through any telescope. It's just a brightness issue.

But, thinking out loud, its curious a larger scope will brighten the filtered moon and allow some better detail to be resolved with contrast closer to the eye's response. Still trying to figure out what that means.

Thanks, Glenn for bringing that up in the first place and both of you for discussing it.

Edited by Asbytec (04/08/13 01:56 PM)

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Re: Reflector/Refractor equivalence formula [Re: Asbytec]
#5785793 - 04/08/13 11:54 AM
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Quote:

I ma still trying to understand the brightness and contrast relationship, I think it's in the eye. But, does that mean the MTF does nto accurately describe contrast and resolution. Not sure, the relationship between contrast and resolution still exist as evidenced by time exposures.

Indeed, it is in the eye, the ability of the eye to detect the contrast differences, to resolve the details is what governs the response of the telescope/eye system. The contrast is still there, the resolution is still there at the focal plane, in the exit pupil.

But it doesn't matter because the eye cannot detect it. The MTF describes the response of the telescope/eyepiece system, it does not describe the response of the Telescope/eyepiece/eye system. That's the critical factor, a poor mirror, a poor objective, a good lens, a good mirror, the MTF describes what is present at the focal plane but it doesn't matter because you just can't see it, it's too dim for the eye to see. All those details on the moon, they are still there, they are clean and sharp, you just cannot see them.

Jon

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5785970 - 04/08/13 01:08 PM
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Sure, still a MTF BASED RoT, as plotted by Eddgie above, can be derived because brightness applies to all scopes and apertures pretty much equally and relatively. I would think, anyway.

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5785972 - 04/08/13 01:08 PM
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Jon,

A fine testament to the contrast present at the focal plane but undetected is visual versus CCD, with a 10" for example. A decent 10" f/4 system can turn out a pic that you'd swear was from a huge apo of at least that aperture. Seen or not the CCD detects it and the post processing enhances it - even utterly invisible things simply because they were again, at the focal plane.'

Pete

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Re: Reflector/Refractor equivalence formula [Re: azure1961p]
#5786094 - 04/08/13 01:52 PM
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For virtually all MTF plots, only the optics ahead of the focus are considered; in other words, the eyepiece is not usually modelled.

The MTF chart is completely independent of target brightness; whether the photon flux be sparse or intense, the optical effect is the same.

The MTF is only fully meaningful when the detector can fully sample down to the lowest resolution of the optic. A CCD with really big pixels (more so when pixels are binned into groups), or visually when the exit pupil is large and/or the target is dim, one is not sampling the regime of the smallest scales (in the lower right of the chart, as usually constructed.)

Nevertheless, whatever extra resolution the optic delivers is there, in spite if the inadequacies of the detector to sample it.

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Re: Reflector/Refractor equivalence formula [Re: Asbytec]
#5786108 - 04/08/13 02:01 PM
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Quote:

Sure, still a MTF BASED RoT, as plotted by Eddgie above, can be derived because brightness applies to all scopes and apertures pretty much equally and relatively. I would think, anyway.

I am not sure of the need to continue to discuss the MTF when viewing objects with faint surface brightnesses. I believe Glenn's point is that at these low light levels, remembering that a telescope does not increase the surface brightness of an extended object, the eye simply does not have the ability to take see the contrast, to resolve the differences that the MTF might show.

If the eye could detect 5 arc-minute details of low surface brightness objects, Andromeda would be spectacularly detailed naked eye...

The MTF just doesn't apply to the response of the human eye and it is the that determines what you see. The other night we were looking at the Leo Triplet in my 25 inch and after battling the wind 20mph+ wind we gave up and pulled out the 16 inch. The 25 inch definitely showed it better but neither showed the detail that an 80mm would show were the eye able to detect it.

Leo Triplet in an 80mm

Jon

Edited by Jon Isaacs (04/08/13 02:08 PM)

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5786194 - 04/08/13 03:16 PM
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Yes, the MTF applies to the focal plane, our eyes detect what's there somewhat differently.

Image brightness is not modeled. Right, it's important because image brightness appears to affect contrast. But that's not true, it's the detector that is suffering. And we all suffer in nearly similar ways with regards to image brightness. So, the MTF should still be useful as a model. It puts the theoretical contrast and resolution on the focal plane and our nearly similar visual acuity takes it from there.

I think the MTF is still meaningful despite the acuity of the detector,provided the detectors are at least similar: two pairs of eyeballs, or two CCD chips, for example. It might get confusing because dim objects "appear" to the eye to have less contrast, but they don't.

Quote:

The 25 inch definitely showed it better but neither showed the detail that an 80mm would show were the eye able to detect it.

Yes, correct me if I am wrong, but the 25" and 16" still put better contrast and resolution on the focal plan than the 80mm giving our eyes the "opportunity" to get at that detail...despite the wind.

I never thought of it that way - the filtered moon concept, so that paradigm is interesting. Still, the (aperture - CO) modeling we see plotted on the MTF is still meaningful even for faint fuzzes, ya? Its not perfect because our eyes differ slightly, but close enough for a RoT because all scopes and eyes treat brightness relatively. Lemme sleep on it...interesting.

Edited by Asbytec (04/08/13 03:23 PM)

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5786215 - 04/08/13 03:28 PM
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Quote:

remembering that a telescope does not increase the surface brightness of an extended object, the eye simply does not have the ability to take see the contrast

We discuss this a lot...but I wonder if this is somewhat misleading. Or maybe seeing color in the object requires it to be bigger...but that would also explain why DSOs look brighter in the bigger scope (more of the object hitting your eye).

I've seen color in the Ring Nebula with a 30 inch dob. Is that possible with a smaller scope with a matched exit pupil? A better example for me is the Blue Snowball...I've seen color in many scopes with this one...but it's much more obvious in larger scopes. I guess I'd have to match exit pupils to do an exact test. In my friend's 16 inch it's blue at high magnifications as well as a 30 inch. In fact, in the 30 inch it looked like it does in photographs.

Found this source which seems to have a lot of good info: Surface brightness

Edited by GOLGO13 (04/08/13 03:38 PM)

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Re: Reflector/Refractor equivalence formula [Re: Asbytec]
#5786444 - 04/08/13 05:15 PM
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Quote:

Yes, correct me if I am wrong, but the 25" and 16" still put better contrast and resolution on the focal plan than the 80mm giving our eyes the "opportunity" to get at that detail...despite the wind.

It was not the contrast or the resolution at the focal plane, the 80mm has plenty of that as witnessed by the photo. Rather, the larger aperture allowed the identical brightness at a greater magnification, the details covered more of the retina.

If I had used a 40% transmission filter on the larger scope the brightnesses would be the same at equal magnifications. The two scopes could operating at the same magnification and image brightness. Two would show essentially identical images.... If the 80mm, with it's contrast and resolution, could be made equally bright, it be right there too.

Try the moon test... when you see how little you can see of the moon when it's surface brightness is reduced by a 12.6 magnitudes, it's instructive.

Jon

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5786662 - 04/08/13 07:15 PM
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I have observed the dark side of the moon being earth lit, that may be the same effect as using a strong filter. Yes, no craters were resolved, not that I remember anyway. They were there, but not illuminated.

Still, my 150mm and your 25" deliver more contrast and resolution to the focal plane. I'm trying to imagine it like this. Take an image of Jupiter in a small refractor and a larger obstructed scope. Its bright enough we might agree it will closely follow the MTF in terms of scale and contrast. Now, dim it down to barely detectable on the retina. The information is still on the focal plane following the MTF.

At the same low power magnification, a large exit pupil though different, the MTF effects are not noticeable. This is because, I'm sure you agree if I am saying it correctly, the diffraction effects are most prominent on small scales out to about 2x the Airy disc, give or take. At low power, those effects are not easily seen and the dominant larger scale effects are almost identical. I believe Glenn was alluding to this effect earlier.

At the same exit pupil, the views would be similar in image scale with the larger aperture providing better resolution. Here it gets confusing, but I believe both scopes will show the same contrast at larger scales, then the obstructed instrument will show a bit less at medium scales...and more at higher frequencies.

Ah, my time is gone, gotta run. Thinking about getting this concept right, so I'll leave the above as is even if it's not right. Thanks.

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5786695 - 04/08/13 07:31 PM
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Quote:

Quote:

Yes, correct me if I am wrong, but the 25" and 16" still put better contrast and resolution on the focal plan than the 80mm giving our eyes the "opportunity" to get at that detail...despite the wind.

It was not the contrast or the resolution at the focal plane, the 80mm has plenty of that as witnessed by the photo. Rather, the larger aperture allowed the identical brightness at a greater magnification, the details covered more of the retina.

If I had used a 40% transmission filter on the larger scope the brightnesses would be the same at equal magnifications. The two scopes could operating at the same magnification and image brightness. Two would show essentially identical images.... If the 80mm, with it's contrast and resolution, could be made equally bright, it be right there too.

Try the moon test... when you see how little you can see of the moon when it's surface brightness is reduced by a 12.6 magnitudes, it's instructive.

Jon

I agree that if the image gets too dim, you won’t see much; but, you need to separate what the telescope is doing from what you can see. First, the resolution and MTF response of the telescope is unchanged with the solar filter. In fact, the image has exactly the same contrast with and without the filter. You have simply reduced the brightness of the image so that much of image intensity is below the visual threshold--effectively reducing the contrast of the “signal” from the eyeball—not the telescope. In the limiting case, you can make all telescopes equal by putting a card in front of the telescope (ND=infinity) and you won’t see anything—no matter what size objective it has! The issue with faint extended objects is that much of the detail is at a brightness below what you can visually detect and that’s why you don’t see it. That’s the whole reason to use a camera—to integrate photons, which you can’t do with your eye. For a given telescope aperture, the peak intensity of the image of a point object at the focal plane increases with the inverse of the focal ratio. At a given focal ratio, the peak intensity increases with the square of the diameter. So, for stars, big and fast is better. For an extended object with an eyepiece, the surface intensity will vary as the inverse square of the magnification. You will get the maximum surface brightness when the diameter of the telescope exit pupil is matched to the diameter of the iris in your eye, which is the low magnification limit for any telescope. Go lower and your eye becomes the stop, which raises other issues. (As has been pointed out, the sharpest visual acuity doesn’t occur when the iris is completely filled, but that probably doesn’t matter so much at low magnification.) When the object is sufficiently bright, your iris closes and the minimum magnification increases.

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GlennLeDrew
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Re: Reflector/Refractor equivalence formula [Re: jhayes_tucson]
#5786704 - 04/08/13 07:38 PM
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When a telesco

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Re: Reflector/Refractor equivalence formula [Re: jhayes_tucson]
#5786748 - 04/08/13 08:06 PM
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When a telescope is used afocally, with an eyepiece, the objective f/ratio is of no fundamental importance. (In the real world, it's important only insofar as the availability of eyepiece focal lengths, and designs, is concerned.) The exit pupil is key, not objective f/ratio.

The 'rule' which posits an equivalence of aperture - CO is applicable only in the brighter regime. Its purpose is to find an equivalent aperture where the *small scale* diffraction effects are made similar. For dim objects, the eye fails to resolve in this regime, and so one is no longer concerned with *small scale* diffraction. Now the only concern is system transmission.

The central obstruction does remove some light, but it's not of great concern. If one wishes to be rigorous, that light removed by the CO, and and that not transmitted due to the (usually) one or two less-efficient mirror coatings, can be taken into account in order to find the aperture equivalence as regards image brightness. Coincidentally, the aperture ratio as derived from consideration of diffraction could be somewhat similar to that derived from system transmission; in such case, the one rule is applicable in a broader sense.

The important thing in the end is this; for visual application, below some surface brightness threshold, diffraction effects become quite irrelevant. To the point that in this specific regime of the dim, extended object the MTF chart (for otherwise reasonably decent telescopes) becomes largely meaningless in practice. Naturally, for stars, and any other sufficiently bright sources, the MTF is quite relevant.

The interesting thing about the visual system is that it's effectively 'multi-modal.' In the *same* field of view one can have a wide range of object type and brightness. For all of them, the brain simultaneously applies the required processing in order to eak out the best detail. The mon-faint stars will be seen and resolved to the eye's photopic limit (1-2 arcminutes on the retina), whereas a galaxy might have its core resolved to, say 5 arcminutes (again, on the retina), its larger bulge region to, say 20 arcminutes, and the disk to 1-2 degrees. All at the same time, and over a smoother continuum than suggested in this 'quantized' example.

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Re: Reflector/Refractor equivalence formula [Re: jhayes_tucson]
#5786751 - 04/08/13 08:07 PM
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Quote:

I agree that if the image gets too dim, you won’t see much; but, you need to separate what the telescope is doing from what you can see. First, the resolution and MTF response of the telescope is unchanged with the solar filter. In fact, the image has exactly the same contrast with and without the filter. You have simply reduced the brightness of the image so that much of image intensity is below the visual threshold--effectively reducing the contrast of the “signal” from the eyeball—not the telescope.

Exactly, I believe I wrote almost exactly that in my first post. The image exists at the focal plane, in the exit pupil with the same contrast, the same resolution. But the eye cannot see it.

However the importance of this is not that I have reduced the brightness of the moon to the point where I can no longer detect the contrast and resolution that exist at the focal plane. Rather, the importance is that most extended deep space objects are even dimmer than the moon viewed through a solar filter, that most of the details that exist at the focal plane, in the exit pupil, are below the "visual threshold."

The importance of viewing the moon through a solar filter is that it provides a first hand example of how poorly the eye does resolve detail at low light levels. It is just not an academic discussion with calculations and discussions of surface brightnesses and visual thresholds and the need for a detail to be degrees in dimension at the retina.

Rather it is a tool to see first hand, the workings of the eye... It's no longer academic, it's a real experience that one can learn from. I definitely recommend it for those who have not tried it.

Jon

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Re: Reflector/Refractor equivalence formula [Re: JimP]
#5786902 - 04/08/13 09:23 PM
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Quote:

Ziggy, if I did not believe that you are completely and absolutely sold on Celestron telescopes I would find your posts to be more objective. As it is, you, me and most others on these forums have a bias toward one type scope or another and find ways to show how the telescope we chose, the one we spent big money on, is the best. The old saying "there are lies and there are statistics" is true.

JimP

Jim,

I'm not sure what I said that made you think that I am "completely and absolutely sold on Celestron telescopes." I'm not. In my experience I have seen some very fine samples and IMHO have been consistently better than Meades but I don't own any SCT and I am not sold on them. They do fill a great niche for amateur astronomy and are one of the great innovations of the last century.

I admit to having a bias toward certain types of telescopes. You have most of them

Edited by Ziggy943 (04/08/13 09:32 PM)

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5787233 - 04/09/13 01:25 AM Attachment (23 downloads)
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Quote:

When a telescope is used afocally, with an eyepiece, the objective f/ratio is of no fundamental importance. (In the real world, it's important only insofar as the availability of eyepiece focal lengths, and designs, is concerned.) The exit pupil is key, not objective f/ratio.

The 'rule' which posits an equivalence of aperture - CO is applicable only in the brighter regime. Its purpose is to find an equivalent aperture where the *small scale* diffraction effects are made similar. For dim objects, the eye fails to resolve in this regime, and so one is no longer concerned with *small scale* diffraction. Now the only concern is system transmission.

The central obstruction does remove some light, but it's not of great concern. If one wishes to be rigorous, that light removed by the CO, and and that not transmitted due to the (usually) one or two less-efficient mirror coatings, can be taken into account in order to find the aperture equivalence as regards image brightness. Coincidentally, the aperture ratio as derived from consideration of diffraction could be somewhat similar to that derived from system transmission; in such case, the one rule is applicable in a broader sense.

The important thing in the end is this; for visual application, below some surface brightness threshold, diffraction effects become quite irrelevant. To the point that in this specific regime of the dim, extended object the MTF chart (for otherwise reasonably decent telescopes) becomes largely meaningless in practice. Naturally, for stars, and any other sufficiently bright sources, the MTF is quite relevant.

The interesting thing about the visual system is that it's effectively 'multi-modal.' In the *same* field of view one can have a wide range of object type and brightness. For all of them, the brain simultaneously applies the required processing in order to eak out the best detail. The mon-faint stars will be seen and resolved to the eye's photopic limit (1-2 arcminutes on the retina), whereas a galaxy might have its core resolved to, say 5 arcminutes (again, on the retina), its larger bulge region to, say 20 arcminutes, and the disk to 1-2 degrees. All at the same time, and over a smoother continuum than suggested in this 'quantized' example.

The F/# of an optical system appears in too many performance parameters to list so we'll have to disagree on the importance of F/# as it relates to telescope performance—with or without an eyepiece. Be careful, MTF is relevant ONLY for extended, spatially incoherent sources—not stars (which are spatially coherent)--but that’s an explanation best left for another thread. The discussion here appears to have evolved into a discussion about visual acuity and perception, which I believe has little relevance for comparing the performance of reflectors to refractors. Visual acuity varies greatly between observers and perception issues lead to biased opinions, which are not comparable in any measurable way. In my view, this comparison is possible but it has to be based on physical optics so without including the effects of diffraction, my participation isn't going to be very valuable so I'll wind down active involvement in this discussion.

One thing that I want to mention is that I see a number of confusing references to “the exit pupil" that go beyond simple semantics. So, I want to clarify some first order optics so that everyone is on the same page. The exit pupil is defined as the image of the stop in the "image space." For a simple Newtonian or refractor without an eyepiece, the stop, entrance pupil and exit pupil all lie in the same physical location on top of the mirror or objective lens. With an eyepiece (with positive power,) the exit pupil lies behind the eyepiece at a location where the eyepiece forms an image of the objective. It has a diameter (given by the diameter of the stop divided by the magnification) and a specific axial location. When the eyepiece is focused at infinity, the focus of the eyepiece sits at the focal plane of the objective and the beam of light coming out of the telescope will be collimated (i.e. the marginal rays are parallel) and there will be no "image plane" for objects at infinity (like stars.) Your eye refocuses the collimated beam onto the retina to form the image you see. Placing your eye at the location of the exit pupil allows collimated light from the objective at different field angles to enter the iris all at once so that you see a wide field of view. That’s where eye-relief comes in: It’s the distance of the exit pupil behind the last component in the eyepiece. It needs to be at a comfortable distance or your eye can bang into the lens and that gets uncomfortable. I’ve attached a schematic diagram.

I apologize to all you guys who already know this stuff; but, since we aren't using diagrams, I just want to make sure that the words line up with the meaning or it gets confusing. It screws me up when I see references to images “in the exit pupil” because it sounds like a reference to an image of the primary mirror (or lens) and that makes no sense in this context.

John

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Re: Reflector/Refractor equivalence formula [Re: jhayes_tucson]
#5787386 - 04/09/13 05:09 AM
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John,
That was a fine summary of the exit pupil!

I take it as given (perhaps too readily?) that all involved here well enough understand at least the basic geometry of the exit pupil. I suspect/hope that its merely idiomatic when we see such as, "in the exit pupil," when it's more correct to say, "through the exit pupil."

I look at the exit pupil this way. It's the coupler between eye and objective. It functionally allows the scope/eye system to behave as though the eye's lens/iris/cornea has been replaced by a lens having the diameter of the objective (as long as the iris is larger than the exit pupil), and a focal length equal to

Obj. diam. * eye f.l. / exit pupil, or

100 * 20 / 2 = 1,000mm f.l., yielding f/10

This makes sense, since the exit pupil is making the eye's own lens work at f/10 (c.f. 20 / 2 = 10.)

For comparison, a 7mm iris makes the eye (at 20mm f.l.) work at f/2.86. In such case, our 100mm scope working at a 7mm exit pupil results in a 'replacement' lens for the eye of 100mm diameter and 286mm f.l.

Please firgive the digression. I tend to become a little apostolic in my zeal to provide perspective. At any rate, it seems the business of diffraction effects and their representation in the MTF chart has about run its course. I'm allowing myself the luxury of exploring other factors which have a bearing on the visual experience.

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Re: Reflector/Refractor equivalence formula [Re: jhayes_tucson]
#5787425 - 04/09/13 05:57 AM
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Quote:

The F/# of an optical system appears in too many performance parameters to list so we'll have to disagree on the importance of F/# as it relates to telescope performance—with or without an eyepiece.

A telescope system, that being an objective plus an eyepiece, is afocal and so to a first order, it's performance can be described without regard to focal ratio, magnification and exit pupil are sufficient to describe the performance parameters.

It is not necessary to know the focal ratio to understand the brightness of the stars, the surface brightness of an extended object, the resolution possible, this all can be determined from the exit pupil and/or and magnification alone.

Higher order issues, aberrations of all sorts can depend on the focal ratio, design issues, eyepiece choices, there are many places where the focal ratio is important but it is not necessary to know the focal ratio of the telescope to discuss what one sees in the most basic terms because an objective with an eyepiece is afocal, a black box, and its function can be described by the exit pupil and the magnification alone.

Jon

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5788288 - 04/09/13 02:26 PM
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Quote:

Higher order issues, aberrations of all sorts can depend on the focal ratio, design issues, eyepiece choices, there are many places where the focal ratio is important ...

Ok, on this point, it sounds like we are in violent agreement.

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Re: Reflector/Refractor equivalence formula [Re: jhayes_tucson]
#5788902 - 04/09/13 06:38 PM
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Quote:

Ok, on this point, it sounds like we are in violent agreement.

If I might interject on Jon's behalf, and certainly my own.. There was no disagreement to begin with. Any amateur worth half his weight in salt is well aware of the significant role the objective f/ratio plays in the performance of the system.

But suppose we had on hand an array of very well executed systems of the same aperture, ranging from, say, f/4 to f/20. All perform equally well, and have identical MTF charts. Furthermore, we match eyepieces so as to deliver the same exit pupil in all cases. And these now complete afocal systems all deliver identical quality.

To the observer, the most intense scrutiny through the eyepiece will in no way enable to differentiate between the fast and slow objective. The views are identical at given exit pupil.

This was the point being made when 'downplaying' the role of objective f/ratio. *By itself*, it has no bearing on the image, as regards diffraction at least, and image surface brightness when an appropriate eyepiece is selected. The observer has no way to ascertain the objective f/ratio based only on the view through the eyepiece; it requires additional information not possible to derive from the image. Hence Jon's referring to the telescope system as a 'black box'.

I am an inveterate foe of the pursuit of knowledge in detail at the expense of the fuller, holistic picture. Hence my diversions into territory seemingly peripheral.

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5789042 - 04/09/13 08:10 PM
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Quote:

If I might interject on Jon's behalf, and certainly my own..

On my behalf, I think the simplest explanation that is not too simple is the best. The relationship between exit pupil and surface brightness of an extended object is fundamental and yet in it's subtle ramifications, contains the essence of why we see what we see.

And there are certainly many surprises. It takes a while to get one's mind around the fact that the Leo Triplet is equally bright in an 80mm refractor with a 6mm exit pupil and a 25 inch with a 6mm exit pupil..

But surprises, as a rule, are instructive. The wandering of the planets was, after all, the key to understanding our place in the solar system and eventually our place in the universe.

Jon

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5789562 - 04/10/13 03:12 AM
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Quote:

If I might interject on Jon's behalf, and certainly my own.. There was no disagreement to begin with. Any amateur worth half his weight in salt is well aware of the significant role the objective f/ratio plays in the performance of the system.

Glenn, I completely disagree with that notion. There are a significant number of folks that post in here with questions relating to exactly this. But I'm not worth half my salt either.

Quote:

A telescope system, that being an objective plus an eyepiece, is afocal and so to a first order, it's performance can be described without regard to focal ratio, magnification and exit pupil are sufficient to describe the performance parameters.

It is not necessary to know the focal ratio to understand the brightness of the stars, the surface brightness of an extended object, the resolution possible, this all can be determined from the exit pupil and/or and magnification alone.

Higher order issues, aberrations of all sorts can depend on the focal ratio, design issues, eyepiece choices, there are many places where the focal ratio is important but it is not necessary to know the focal ratio of the telescope to discuss what one sees in the most basic terms because an objective with an eyepiece is afocal, a black box, and its function can be described by the exit pupil and the magnification alone.

Jon - This in ways seems self contradictory. Sure in it's most basic form it's an amplifier of sorts (here I refer to the ICO 'input-conversion-output' black box concept), but then there are different types of amplifiers. I think the real issue if discussing equivelence in a realistic sense to couple the 'mapping function' of the ep/exit pupil to everything up to and including the atmosphere.

My TEC 160 KATN's my small 80/85mm scopes on planets and everything else, and it's not all due to exit pupil, which in my 160 I'm usually at or below half that goofy '2' hypothesis.

Exit pupil is certainly an important 'human factors' part of the consideration, but it's is neither the end all or be all of things - it is a result, an atmosphere filtered result, and that result will depend on what is being mapped, and the granularity of the 'mesh' to the eye and how.

I don't think anyone is confusing absolute or intrinsic brightness of the the object up there by the telescope they choose, but there is a curve as to what is up there that is reasonable to to try and resolve (to a certain magnitude) and I do think that apeture plays an important part in it all (so it's a function with limits or asymptotic). At least my experience shows this. Just how and on what it matters I feel varies by object. I also feel there is a clear difference in the way an 12" f/10 system will 'map it' vs. say an f/4 system.

I usually try to use something as a standard candle, a globular like M13, a planet like Jupiter, a nebula like M42, or a galaxy (depends). Incresing apeture certainly brings something to the party as does the focal length (or the mapping granularity) and certainly the atmosphere.

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Re: Reflector/Refractor equivalence formula [Re: CounterWeight]
#5789840 - 04/10/13 09:29 AM
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Quote:

My TEC 160 KATN's my small 80/85mm scopes on planets and everything else, and it's not all due to exit pupil, which in my 160 I'm usually at or below half that goofy '2' hypothesis.

It's due to exit pupil and magnification... of course one can back calculate the aperture from those two, a 1mm exit pupil at 160x can only be provided by a 160mm aperture.

Remembering the context of this discussion, I don't know why there is any contradiction. We were discussing the importance of focal ratio as it relates to surface brightness and object brightness at the retina.

Quote:

I usually try to use something as a standard candle, a globular like M13, a planet like Jupiter, a nebula like M42, or a galaxy (depends). Incresing apeture certainly brings something to the party as does the focal length (or the mapping granularity) and certainly the atmosphere.

Increasing aperture does bring something to the table but when viewing extended objects, it does not bring brighter, more intense images... M65 is no brighter in my 25 inch at a 6mm exit pupil than it is my 80mm with a 6mm exit pupil. It sure seems more intense but it's only larger, about 8 times by diameter, 64x by area. Sometimes mapping needs to be complicated and all those little Airy Disks with their diffraction rings need to be summed up, sometimes though, somewhat sophisticated concepts like the exit pupil make it all very simple...

Jon

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5790400 - 04/10/13 01:56 PM
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Jon it's this part
Quote:

Increasing aperture does bring something to the table but when viewing extended objects, it does not bring brighter, more intense images... M65 is no brighter in my 25 inch at a 6mm exit pupil than it is my 80mm with a 6mm exit pupil. It sure seems more intense but it's only larger, about 8 times by diameter, 64x by area.

where I think significant. But here I think good to be careful too.

M65: in my SkyTools3 lists the two magnitudes as-
Mean Surface Br. 21.3 Mag/arcsec²\and\Magnitude: 10.10 B

Cycling through various scope /ep combinations from here today it gets about 60* above horizon and so i think well placed in ways for viewing in that it's well above the 2x airmass and i don't need to try and view at zenith. The various descriptors it generates are are from very challenging to difficult to detectable. though I do not own one yet, if I select the 14" F/4.9 dob it changes to 'easy'. But what here is easy? I cannot see down to mag 21.3 so it's then about the parts that fit the harness and the area they are displayed.

Going back to the equivance formula at same apeture and focal length the descriptors don't change in a meaningful way.

The point of that outline I gave was to derate each scope according to it's strengths and weakness and arrive at a higher number equal better system, but then not necessarily as a one size fits all from a practical standpoint. here I just mean that in larger scope Jupiter can be quite bright and large to the point of irritation and in a smaller scope maybe a more plesant experience though with far less 'possible' resolution. M65 may be an invers in ways where in the smaller scope a big challenge and the larger scope very satisfying.

I'm not disagreeing about exit pupil, it's just that if comparing same apeture and focal length and optic quality and collimation, acclimation, and seeing it becomes more a common denominator. Vary the apetures under scrutiny and then it could be held constant within reason but in important ways what is contained 'in the exit pupil harness' is very different and the eye/brain response accordingly different.

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Re: Reflector/Refractor equivalence formula [Re: CounterWeight]
#5790488 - 04/10/13 02:42 PM
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Instead of all the technical formulas and talk of exit pupils...we could keep it simple and say a larger scope will provide a bigger image. And at the same magnification it will be a heck of a lot brighter. And for some objects, a small refractor will only show a extremely small object at the same exit pupil. What good is that going to do for you other than make the argument that it's technically the same "brightness" as the bigger scope.

This all is a bit off topic, but still interesting. I think there probably is good math for a reflector/refractor equivalance...but most of the time (based on money/mounting/etc) they don't really need to be equated...but used as compliments.

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Re: Reflector/Refractor equivalence formula [Re: GOLGO13]
#5790627 - 04/10/13 04:08 PM
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I thought I was done here, but I guess not. Let me see if I can explain the brightness/exit pupil issue being discussed here a little more completely for the benefit of those who might not understand it as well as a few of you.

Let’s consider extended objects first. If you ignore transmission losses, a telescope (in air) preserves a radiometric quantity called radiance. Radiance is the amount of power emitted per area per solid angle (which is the same as intensity per solid angle) by an extended object. The daytime sky provides a good example of a relatively uniform extended object. You might think that since a larger aperture gathers more light, the sky should appear brighter through a telescope; but, if you make the exit pupil match the size of the iris of the eye, the increase in brightness due to the size of the telescope is exactly offset by spreading the light out over a larger area due to the magnification of the system. This means that the radiance of the illuminated exit pupil (remember that’s an image of the primary objective) is the same as the sky. No matter what the diameter of the telescope is, as long as the size of the exit pupil matches the size of the iris, the eye can’t tell that whether it is looking directly at the sky or through the telescope—the brightness will be the same. This fact holds for ANY extended object—the sky, the moon, or a nebula.

When you increase the magnification, you decrease the size of the exit pupil and spread the same amount of light over a larger area on the retina, which makes the image appear dimmer than directly viewing the sky. This is a consequence of something called the optical constant—but we’ll ignore that for now. The basic fact is that when you make the diameter of the beam smaller at the exit pupil you are increasing the size of the resulting image on the retina, which decreases the image intensity and makes it appear less bright. (It is equally valid to view this as decreasing the angle subtended by the exit pupil as viewed from the retina, which decreases the power received per unit area.) This means that any two telescopes with the same size exit pupils (and transmission values) will produce the same brightness for a given extended object. An obscured reflector with light loss due to the secondary shadow is at a slight disadvantage compared to a refractor when it comes to image brightness on extended objects simply because less light gets transmitted.

None of this applies to point sources like stars where the image fills only a few retinal receptors regardless of the magnification. In the case of a star image, the brightness is determined solely by the size of the aperture. Up until you begin to significantly increase the size of the Airy disk on the retina, magnification has no effect on the brightness of star images seen through the eyepiece. This is why you can view stars during the daytime at high magnification. The background sky brightness is decreased while stars remain at the same apparent brightness. For stars, aperture rules: the bigger the better.

So, you may be asking yourself, if all extended objects like nebula appear just as bright in any telescope with equal size exit pupils, why bother with the telescope at all—especially a big one? If you ignore photography, the answer lies in visual response curves and optical resolution. The angular resolution of a telescope is inversely proportional to its diameter so that a larger telescope can resolve smaller features. So, the first reason is that a larger telescope allows smaller features to be seen. Edges are sharper and that contributes to the visibility of low light level features. The second reason relates to the filter experiment that Jon talked about. At low light levels, the eye can only discern contrast levels down to about 2% for an object that subtends “the right angle.” It turns out that if a low brightness object appears too small or too large, the eye has trouble seeing it. It is below the contrast threshold characteristics of the eye. So, the telescope is there to present the image at about the “right size” so that you can see it. That’s why there is a “sweet spot” for magnification that makes diffuse extended objects look the best. All of this is true for both reflectors and refractors and shouldn’t be considered as much of a figure of merit when making any performance comparisons, which is why I didn’t want to wade into this particular subject in the first place.

Hopefully, this will help get everyone on the same page--or at least a little closer.

John

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Re: Reflector/Refractor equivalence formula [Re: jhayes_tucson]
#5791729 - 04/11/13 02:26 AM
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The above touches upon a very fundamental principle for optical systems, akin to the 2nd law of thermodynamics. In thermodynamics there is a quantity called entropy that can not decrease. Similarly, in optical systems we have a quantity etendue that can not decrease.

In essence, etendue is the size of the illuminated area (pupil area) times the solid angle that measures the directional spread of the light entering that area.

Image deteriorations such as scattering and diffusion of light cause etendue to increase. Such an increase is perceived as a decrease in image brightness. So, at best (in an ideal optical system without irreversible aberrations) the etendue stays constant: etendue at exit pupil = etendue at entrance pupil (at aperture).

Bottom line: you can't defy the laws of physics - an optical system capable of increasing the brightness of an object is as much a fantasy as a perpetual motion machine.

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Re: Reflector/Refractor equivalence formula [Re: jhayes_tucson]
#5791993 - 04/11/13 09:29 AM
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Quote:

All of this is true for both reflectors and refractors and shouldn’t be considered as much of a figure of merit when making any performance comparisons, which is why I didn’t want to wade into this particular subject in the first place.

Nice summary.

I think the reason to wade into the subject is because it is fundamental and must be clearly understood before trying to formulate a reasonable reflector/refractor equivalence formula. If such a formula is to be meaningful, the response of the human eye must be considered because in many/most situations, it's the human eye that is the governing factor.

Jon

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CounterWeight
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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5793724 - 04/12/13 02:05 AM
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Jon, I agree but it must be kept in context, and again I think if keeping apeture and focal length equal it (exit pupil) is a common denominator, then the quality of the rest of parameters along with maybe central obstruction and other aspects of the mirrored systems that bears scrutiny. I feel it important to discern between planets and point sources (especially WRT globulars, double/multi and loose clusters) and diffuse gradient reflected or emission where a certain numerical difference might be meaningful as I feel here it also not a one size fits all.

As good as my TEC 160ED is, my C11 body bagged it on M13, at a fraction of the cost. I think any realistic equivalence formula should embody that along with practical useable magnification and all the rest. I think there is quite enough theory out there, what is lacking is a practical easy to understand rating or derating system where the important differences become obvious to the typical person on the street. As in how does an 11" Celestron at \$1,800(USD) compare to a \$9,000(USD) TEC scope on different objects under varying conditions over a spread of object types. Or the same TEC compare to a custom Newt at same apeture and focal length with exceptional (or at least same care in figure and finish/coatings as the TEC) optics.

What I've noticed is that there seems to be here in the refractor forum a tendancy to overly derate mirrored optics in a broad brush fashion to where a very expensive 5" triplet can somehow out perform a far less expensive mirrored system to the point of what I feel somewhat irrational broad brush of bettering mirrored system of a larger apeture on all objects, this in any seeing conditions.

I am not trying to slam theory or refractors or anything else, and agree as with you previous example it's critical to tie things to reality like M65 or Jupiter or M13 or Alberio - and then under what conditions they are observed. In a way it's more about tying down just what to expect could be seen and how. An ideal 130'apo' vs. say a C14(run of the mill quality) on M13, it should be a day and night difference even given a wide latitude of exit pupil.

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Re: Reflector/Refractor equivalence formula [Re: CounterWeight]
#5793944 - 04/12/13 08:28 AM
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Quote:

I think any realistic equivalence formula should embody that along with practical useable magnification and all the rest. I think there is quite enough theory out there, what is lacking is a practical easy to understand rating or derating system where the important differences become obvious to the typical person on the street.

I think there is enough theory. But I also think that how it all applies to the telescope/eyepiece/eye/night sky is poorly understood by the much of the amateur community and that there are so many variables that are as important or more important than the telescope, that it is just better to have an understanding, any equivalence formula will be an over simplification and to actually apply it, you will have to understand not only all the variables involved but the workings of the formula itself.

What further complicates the issue is that execution of a particular design is critical so two telescopes with the exact same optical specifications can perform quite differently, that the performance of a telescope is not static.. 15 minutes into a session, a C-11 is probably not out performing a 160mm refractor. 3 hours in, maybe...

So I think it better to understand that for sufficiently dim objects, the contrast difference between a scope with a central obstruction and one without does not matter. If one understands that source of contrast differences, then it's easy to see why that is.

Jon

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Re: Reflector/Refractor equivalence formula [Re: CollinofAlabama]
#5794426 - 04/12/13 01:14 PM
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Know what? I keep getting a "divide by zero" error any time I try to construct a formula under which any reflector achieves refractor equivalency.

Regards,

Jim

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Re: Reflector/Refractor equivalence formula [Re: jrbarnett]
#5794533 - 04/12/13 01:57 PM
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My 5" modified Mak has a 160% reflective primary and only a 3% central obstruction. It has also been refigured to 1/245 wave PV and a 1.02 Strehl ratio. Never seen a 6" apo that could even come close. I guess you could say my telescope is supercharged. Either that or my embelishment is supercharged.

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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5795486 - 04/12/13 10:11 PM
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Quote:

So I think it better to understand that for sufficiently dim objects, the contrast difference between a scope with a central obstruction and one without does not matter. If one understands that source of contrast differences, then it's easy to see why that is.

Jon

Jon, Glenn, John in Tucson and et al, thank each of you for an interesting discussion. It turns out we began talking about MTF and ended with the basic funamental of image brightness with aperture...and exit pupil...exploring some equivelence factor between unobstructed and obstructed optics. In the end, we simply explored the simple concept of image brightness (telescopes not making images brighter per unit area than they actually are and at optimum exit pupil.)

In that sense, it does not seem an equivelence factor is possible and would have to deal with brightness and scale (on the retina)...turning a larger telescope into a smaller one mathematically - what equivelence would give a similar image. How small would a 12" Newt have to be to provide image scale and brightness equivelent to a 100mm refractor? I think the answer is 100mm (using the same exit pupil.)

The exit pupil way of looking at it is a correct and simple, if not a different, way to connect exit pupil to contrast (at small exit pupils to MTF.) The latter discusson on contrast was interesting, spurring some thought, in that the contrast should be on the focal plane for dim objects, but most of the range is below the visual threshold and therefore not readily seen with the average human eye.

I am still curious about the strongly filtered moon, however. Such filtering does seem to reduce the signal to noise ratio, as John Hayes touched on. However, he mentioned the reduced SN was on the eye. I might argue SN was reduced on the focal plane, as well, since the moon is filtered prior to the image forming objective. I believe SN ratio is a form of contrast. Bright features can be made dimmer, but already black features cannot be made more black.

Reducing very bright lunar features toward pure black (or natural sky background) does seem to be a form of reduced contrast and explain why craters are not visible, as well. Of course the eye cannot discern them in low light, but image contrast of an otherwise brightly lit subject seems to have been effected, too.

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Re: Reflector/Refractor equivalence formula [Re: Asbytec]
#5797272 - 04/13/13 06:13 PM
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Refractor Equivelant (inches) = Reflector Primary Size (inches) - Obstruction Diameter (inches)

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Re: Reflector/Refractor equivalence formula [Re: RodgerHouTex]
#5799722 - 04/15/13 12:30 AM
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Jon,

I agree, that is why external factors deserve to be 'on the table', though that requires no small amount of expect knowledge and possible 'fudge factors'. I'd include the fan assist, thin and composite and conical mirrors on many modern designs and those that Ed at Deep Space Products and others like Orion provide. Finally the issue of acclimation is being adressed on open and closed tubes. I'm not disagreeing with OTF, PSF, MTF, (and a boatload of named others) - just that they are more theoretical. Great for designers, practicality for uses IMO limited. Theory is the only well behaved part of it.

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Re: Reflector/Refractor equivalence formula [Re: Asbytec]
#5800180 - 04/15/13 10:31 AM
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Quote:

I am still curious about the strongly filtered moon, however. Such filtering does seem to reduce the signal to noise ratio, as John Hayes touched on. However, he mentioned the reduced SN was on the eye. I might argue SN was reduced on the focal plane, as well, since the moon is filtered prior to the image forming objective. I believe SN ratio is a form of contrast. Bright features can be made dimmer, but already black features cannot be made more black.

I just say that out under the night sky, certainly in the context of light coming from the surface of the moon and nearby the moon, there is nothing that is "black", in this context, black is a abstract concept notion, not something that we experience. Furthermore, the filtering can be done at any point, after the focal plane but before the eyepiece or after the eyepiece but before it enters the eye, the result is the same.

In terms of contrast, the filtered moon against the night sky must certainly be much greater than a typical DSO against even the darkest skies. The filter itself is worth 12.6 magnitudes, a magnitude 2 sky is about 15.9 MASAS, through the filter, it would be 28.5 far darker than any skies here on earth.

It's an interesting experiment, give it a try... and keep in your mind that you are looking at the moon and that the detail is there in that image...

Jon

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Asbytec
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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
#5800553 - 04/15/13 01:37 PM
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Jon, thank you. Yes, I agree the idea the detail is there is intriquing. The concept that black is abstract might be key to figuring this out.

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Re: Reflector/Refractor equivalence formula [Re: Asbytec]
#5801021 - 04/15/13 05:05 PM
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Even from space the sky would not be black. If you were on the Moon's far side during lunar night, the sky would be some 4 times, or 1.5 magnitudes darker than the darkest earthly sky. The primary source of light would be sunlight scattered by dust (zodiacal light), which would of course result in a considerable variation in brightness, depending on both ecliptic longitude and latitude. In any event, even in the darkest regions, toward the ecliptic poles and somewhat in the anti-solar direction, you would readily see your hand in silhouette.

The oft-quipped descriptors like "coal-black" and "inky" for a terrestrial night are wildly wide of the mark. There is quite a lot of light emanating from the celestial hemisphere (starlight being minor). At even the darkest sites, no flashlight is needed to walk about and avoid bumping into your gear.

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5803647 - 04/16/13 10:28 PM
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Thanks, Glenn, still pondering this solar filtered moon concept. I kind of understand it, but still coming to terms with contrast, resolution, and illumination on the eye as compared to the focal plane and what that might mean in terms of applicability to MTF. I always appreciate your comments.

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Re: Reflector/Refractor equivalence formula [Re: Asbytec]
#5803965 - 04/17/13 05:43 AM
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Norme,
In a nutshell, when the subject is bright enough for the eye to work in photopic mode, the eye can easily realize the full potential of the instrument's resolving power.

As subject brightness decreases, a threshold is crossed where the eye's resolving power is just matched with that of the scope (where the Airy disk cannot quite be resolved.)

Below this threshold, further dimming results in ever worsening visual system resolving power, the same level of detail is there as when the subject is bright--and the MTF chart has not changed one iota--but the observer's eye has very much become the weak link.

A solar filter, if of very good optical quality, does not really alter the MTF (meaningfully.) while observing the Sun, all manner of detail is seen, as expected due to the still sufficiently bright solar disk.

But the Moon is dimmed by the solar filter to such an extent that it's like a DSO in terms of surface brightness. All the detail present without the filter in place is there, as a time exposure with a camera would reveal. And the MTF chart for the scope is unchanged. It's the eye's awful resolving power at this low light level which makes the perception of *any* detail--even along the ultra high contrast terminator--problematic.

As Jon says, this is a most worthy test to conduct. More than any amount of verbiage possibly could, it will drive home just how limited is the eye at low light levels. Which is why a little 60mm f/6 can reveal faint DSO detail in images which cannot be visually detected with scopes of 10X the aperture.

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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
#5804563 - 04/17/13 01:34 PM
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Quote:

Norme,
In a nutshell, when the subject is bright enough for the eye to work in photopic mode, the eye can easily realize the full potential of the instrument's resolving power.

As subject brightness decreases, a threshold is crossed where the eye's resolving power is just matched with that of the scope (where the Airy disk cannot quite be resolved.)

Below this threshold, further dimming results in ever worsening visual system resolving power, the same level of detail is there as when the subject is bright--and the MTF chart has not changed one iota--but the observer's eye has very much become the weak link.

A solar filter, if of very good optical quality, does not really alter the MTF (meaningfully.) while observing the Sun, all manner of detail is seen, as expected due to the still sufficiently bright solar disk.

But the Moon is dimmed by the solar filter to such an extent that it's like a DSO in terms of surface brightness. All the detail present without the filter in place is there, as a time exposure with a camera would reveal. And the MTF chart for the scope is unchanged. It's the eye's awful resolving power at this low light level which makes the perception of *any* detail--even along the ultra high contrast terminator--problematic.

As Jon says, this is a most worthy test to conduct. More than any amount of verbiage possibly could, it will drive home just how limited is the eye at low light levels. Which is why a little 60mm f/6 can reveal faint DSO detail in images which cannot be visually detected with scopes of 10X the aperture.

Great post. It helps to understand the selective pressures under which our nighttime visual system evolved. Discerning fine detail was not important, as evidenced by the distribution of rods in the retina. Where our night vision excels is picking up movement, particularly with peripheral vision, a critical survival task in an environment filled with large nocturnal predators.

It's not particularly good and differentiating small differences in brightness or contrast, no selective advantage in being able to do so. For example, for a light source to be perceived as being 100% brighter it has to be 900% brighter (response compression).

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Re: Reflector/Refractor equivalence formula [Re: Paul G]
#5805777 - 04/17/13 10:51 PM
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Glenn, again, thanks for the explanation. Yes, I understand pretty much everything you said.

It seems, however, some folks use the eye's low light lack of resolution as an argument against using MTF as a standard tool for telescope comparisons. To which, I agree. The eye does loose something at low light levels and the MTF does not accurately show what is observed visually. But, the point made is the information (contrast and resolution) is on the focal plane, so MTF can be used even if the eye (which varies widely, anyway) can still be a useful tool even at low light levels.

When checking my scope without the secondary baffle, I did observe the moon's dark side looking for whatever I could see to include stray light and glare. No craters. It was interesting and I suspect the same effect as a solar filter. There were patches of light albedo here and there, but no fine detail

Edited by Asbytec (04/17/13 10:53 PM)

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