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Equipment Discussions >> Refractors

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GlennLeDrew
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Re: Reflector/Refractor equivalence formula new [Re: jhayes_tucson]
      #5788902 - 04/09/13 06:38 PM

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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Jon Isaacs
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Re: Reflector/Refractor equivalence formula new [Re: GlennLeDrew]
      #5789042 - 04/09/13 08:10 PM

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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CounterWeight
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Re: Reflector/Refractor equivalence formula new [Re: Jon Isaacs]
      #5789562 - 04/10/13 03:12 AM

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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Jon Isaacs
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Re: Reflector/Refractor equivalence formula new [Re: CounterWeight]
      #5789840 - 04/10/13 09:29 AM

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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CounterWeight
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Re: Reflector/Refractor equivalence formula new [Re: Jon Isaacs]
      #5790400 - 04/10/13 01:56 PM

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/arcsec˛\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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GOLGO13
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Re: Reflector/Refractor equivalence formula new [Re: CounterWeight]
      #5790488 - 04/10/13 02:42 PM

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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jhayes_tucson
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Re: Reflector/Refractor equivalence formula new [Re: GOLGO13]
      #5790627 - 04/10/13 04:08 PM

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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JKoelman
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Re: Reflector/Refractor equivalence formula new [Re: jhayes_tucson]
      #5791729 - 04/11/13 02:26 AM

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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Jon Isaacs
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Re: Reflector/Refractor equivalence formula new [Re: jhayes_tucson]
      #5791993 - 04/11/13 09:29 AM

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 new [Re: Jon Isaacs]
      #5793724 - 04/12/13 02:05 AM

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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Jon Isaacs
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Re: Reflector/Refractor equivalence formula new [Re: CounterWeight]
      #5793944 - 04/12/13 08:28 AM

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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jrbarnett
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Re: Reflector/Refractor equivalence formula new [Re: CollinofAlabama]
      #5794426 - 04/12/13 01:14 PM

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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DJCalma
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Re: Reflector/Refractor equivalence formula new [Re: jrbarnett]
      #5794533 - 04/12/13 01:57 PM

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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Asbytec
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Re: Reflector/Refractor equivalence formula new [Re: Jon Isaacs]
      #5795486 - 04/12/13 10:11 PM

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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RodgerHouTex
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Re: Reflector/Refractor equivalence formula new [Re: Asbytec]
      #5797272 - 04/13/13 06:13 PM

Refractor Equivelant (inches) = Reflector Primary Size (inches) - Obstruction Diameter (inches)

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CounterWeight
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Re: Reflector/Refractor equivalence formula new [Re: RodgerHouTex]
      #5799722 - 04/15/13 12:30 AM

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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Jon Isaacs
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Re: Reflector/Refractor equivalence formula new [Re: Asbytec]
      #5800180 - 04/15/13 10:31 AM

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

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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GlennLeDrew
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Re: Reflector/Refractor equivalence formula [Re: Asbytec]
      #5801021 - 04/15/13 05:05 PM

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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Asbytec
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Re: Reflector/Refractor equivalence formula new [Re: GlennLeDrew]
      #5803647 - 04/16/13 10:28 PM

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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