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Equipment Discussions >> Refractors

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GOLGO13
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Re: Reflector/Refractor equivalence formula new [Re: Jon Isaacs]
      #5780110 - 04/05/13 04:36 PM

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

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

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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GOLGO13
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Re: Reflector/Refractor equivalence formula new [Re: Jon Isaacs]
      #5781556 - 04/06/13 11:00 AM

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

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

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

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

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

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

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

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

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

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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azure1961p
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Re: Reflector/Refractor equivalence formula new [Re: Asbytec]
      #5784611 - 04/07/13 07:35 PM

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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GlennLeDrew
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Re: Reflector/Refractor equivalence formula new [Re: Asbytec]
      #5784666 - 04/07/13 08:03 PM

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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Asbytec
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Re: Reflector/Refractor equivalence formula new [Re: GlennLeDrew]
      #5784872 - 04/07/13 09:55 PM

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

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

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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GlennLeDrew
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Re: Reflector/Refractor equivalence formula [Re: Jon Isaacs]
      #5785078 - 04/07/13 11:52 PM

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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Asbytec
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Re: Reflector/Refractor equivalence formula [Re: GlennLeDrew]
      #5785259 - 04/08/13 03:45 AM

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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