Math/Optical Help?

[PRejto (Peter)]

Inspired by some formulas given in a video about Dragonfly
(http://www.iceinspace.com.au/forum/s...d.php?t=135928) I am trying to work out the effective aperture and focal length of my 2 telescope system. In Dragonfly the lenses all are the same diameter so the formulas given have the diameter as a constant:

Aperture = square root of Number of lenses x Diameter

Focal ratio: f = f/ square root of Number of lenses

So, for my 2 f7 refractors:

f = 7/1.4142 = 4.95

However, my two lenses have different diameters so the formula doesn't hold. I've worked it out by taking an average but I'm quite unsure if this is sound reasoning.

Aperture = 1.4142 x 140mm = 197mm (TEC140)
1.4142 x 180mm = 254 mm (TEC180)

Average: 451/2 = 225 mm (at f 4.95)

Thanks for any input about this!

Peter

[julianh72 (Julian)]

By my understanding, the effective aperture is the diameter of the circle with the same area as your compound lens:

Total Area = pi/4*140^2 + pi/4*180^2 = 15394 + 25447 = 40841 mm^2

A circle with a diameter of 228 mm would have the same area, so your compound lens would have an effective aperture of 228 mm by this method.

However, I am confused by how you would combine the two images into a single image. It must be hard enough to get multiple identical lenses aligned with each other so that the images can be combined, but I am not sure how you would do it with two or more dissimilar lenses without introducing "noise".

You will presumably have a different image scale on each OTA, meaning you will need to enlarge or reduce one of the images to be able to combine into a merged image (which is analogous to increasing or decreasing the focal length of that OTA). I imagine this would introduce all sorts of "dithering" artefacts when you try to combine the scaled and un-scaled images.

Or am I missing something?

[PRejto (Peter)]

Thanks Julian.

I understand what you have done mathematically. I'm surprised my ad hoc method was only 3mm different than the correct calculation.

The telescopes are precisely aimed at the same point in space using an adjustable plate under the TEC140 (Optec Libra). I chose two different cameras with different pix sizes such that the images are nearly the same size and similar resolution. The TEC180/KAF8300 CCD = .88 arc-sec, The TEC140/Trius H694 = .95 arc-sec. I have no difficulty whatsoever combining these in CCDStack. The lower resolution TEC140 image is matched to the TEC180 image as a reference.

I take your point re focal ratio since the TEC140 is effectively not f7 when the resolution is increased. I need to calculate a revised fl on the TEC140 to give a resolution of .88***. However, now I'm not sure how I would calculate the effective fl of the combined system. Do you know how to do this?

Thanks!

Peter

*** f7.57 = .88 arc-sec when matched to the TEC180 image

[Atmos (Colin)]

Correct me if I am wrong but I think you're thinking too deeply into it Mind you, I've been working 16 hour days this week so my mind is a bit kaput right now.

If I have a Nikon D810a with a nifty fifty using at F/4... This is what it is. Let's say now that I have 100 of these setup all pointed at the same thing. It is now effectively working as a 50mm F/1.26.

All it does is explain the light gathering power of an image, a way of expressing it in a way of single exposure. This means, 100 cameras all taking a single 120s exposure at F/4 would have the same SNR as a 120s exposure at F/1.26.

[PRejto (Peter)]

Hi Colin,

Well, you are the astronomer! But, so is the guy giving the Dragonfly lecture in my first post. So, those are his formulas about effective aperture and fratio. If I understand your post you seem to be saying that only the fratio changes. But using multiple lenses changes the total area of light gathering, doesn't it?

Yes, I'm over thinking this but I'm writing an article about my 2 scope system and trying to come up with some supporting math to justify my determination that imaging with two cameras at one time has some real benefit. It's obvious that I'm imaging 2x as fast. Is it not reasonable to equate this to a hypothetical single camera system that would give the same performance?

Peter

[Atmos (Colin)]

There are three parts that make up an imaging system (optically). The aperture, focal length and f/ratio. If any two of these change the third does as well.

Using my previous example, a 50mm @ F/4 has a clear aperture of 12.5mm. 100 of these have an effective clear aperture of 39.5mm (((12.5)^2)*10)^0.5. 39.3*1.26= 49.8mm (ive rounded off a few times).

Yours is slightly more complicated to calculate as you are dealing with different telescopes and cameras but the principle is the same. What it means is that over a two hour imaging period you would get about an extra 40% SNR increase over someone that only has one setup going for two hours.

[Steffen]

Aperture is a purely geometrical property of an optical system. The term means either absolute aperture – the diameter of the objective lens, or, more generally, the diameter of the virtual image of the entry pupil (the white circle you see when you hold the objective lens up to the sky and look through the front).

The relative aperture (also called f-stop or focal ratio) is focal length of the objective divided by absolute aperture.

The multiplicative "square root of number of lenses" term makes no sense to me. It would mean the more lenses you have the bigger the aperture. That's clearly not the case. If anything, a larger number of lens elements reduces transmission.

Sometimes, esp. in photography, a corrective factor ("filter factor") is applied to the relative aperture, in order to account for filters or transmission losses (on a multitude of air-glass surfaces or in the glass itself).

However, no matter how much transmission is reduced, this only affects photographic exposure calculations and "overall brightness" of the visual image, but not aperture-related properties like Airy disc size, resolving power, etc.

[Atmos (Colin)]

The reason for the square root of the lens' is purely to do with SNR increase. To use a real world example, let's say I collect 5 hours of Ha data with my 130mm F/5 refractor. I would do that in 30x600s Ha subs. As I combine the subs together the SNR of the image increases by the square root of the number of subs, this equates to about 550% (5.5x) increase in signal over a single image.

Now to get the same sort of SNR from a single 600s exposure I would need a 12" F/2.1 telescope. So the benefit of having 30 telescopes collecting data is that it would be the equivalent of having a 12" F/2.1 telescope collecting that data. What would take me to collect five hours of data, it would do every 10 minutes.

Another added benefit of having 30 telescopes is that you can increase its resolving power FAR beyond what a 5.1" or even a 12" telescope can achieve. If I placed each telescope 3 metres from the next, all in a line, it would have the resolving power of a 90m telescope. Think that's for fetched? It's what SKA (Square Kilometre Array) does in WA but with radio telescopes. It's not complete yet but eventually they're going to work in conjunction with South Africa so that we can use the Earth as a base line to have the resolving power of an approximate 6000m telescope

Don't get me wrong, it would be pointless having 30x5.1" telescopes for resolving power as the atmosphere makes anything over the 8-10" range redundant.

Want an incredibly deep all sky survey telescope? An AP3600 with a hundred QSI690 cameras, Nikon F Mounted with Samyang 135mm F/2 lens. It would effectively be a 213mm F/0.64 lens with a 36.8mp camera and resolution of 2.82"/pixel.

Of course if you want to study planetary nebula, nothing beats a good ol' 10m mirror