Another Focal Length Question - Fast/Slow

[mikeyjames (Mick)]

Hi all,
I'm just asking this out of curiosity.

I have been trying to understand why a 'fast' scope is fast. I've scoured the net and found a thousand pages that tell me the following:

F Number = Focal Length/Aperture
F5 = Fast
F10 = Slow
Fast = shorter exposure time.

I know all of this, but I'm finding that an explanation about 'why' this is so is a lot more difficult to find. The two explanations I have seen that seem to make sense are (as I understand them):

1) That at same aperture, a longer focal length = higher magnification. This in turn spreads the same amount of photons over a larger area on the sensor, which means less information captured at each pixel and therefore longer exposures are needed to fill the gaps. I suppose just like resizing a small pic to much larger and it looks like crap (unless you're on CSI Miami where they can resolve a suspect's fingerprint from a low quality video feed at 100m).

OR

2) Someone gave an example of standing in a 10m diameter pipe. When standing 40m back = F4, 60m back = F6, 100m back - F10, and so on. The person asked us to think what the target would look like at the F4 stop, and so on at each stop. As we go further back the field of view gets smaller and the image loses brightness, until maybe several hundred metres where you can barely make our anything.

Both of these make a bit of sense to me. I like the second one because I can actually visualise what is happening, but the fact that I like it doesn't mean it's right.

Are either/both/neither of these on the right track? If not, can someone explain why a 'fast' scope is actually fast.


Thanks
Mick

[Atmos (Colin)]

It can actually be quite complicated but here is a quick example. You have a 10" F/5 and a 10" F/10. On both telescopes you use a KAF8300 which has 5.4 micron pixels.

On the 10" F/5 you are imaging at 0.877"/pixel while on the 10" F/10 you are at 0.44"/pixel.

This means that every pixel is capturing four times the area (four times as many photons) on the F/5 than the F/10. Now the confusing part is as follows. Let's say the 10" F/4 used the ASI1600 with 3.8 micron pixels and you had one of the older KAF 1024x1024 15 micron sensors (Amdor still work with them I believe).

In this case the 10" F/4 is at 0.617"/pixel while the 10" F/10 is at 1.218"/pixel. In this instance the slower telescope is in essence near four times faster as each pixel captures near four times the surface area.

[Cimitar (Evan)]

Hi Mick, I've been wondering the exact same thing lately (I have an F/10 SCT and often wonder whether imaging would be any faster with a Newt).

As a follow-up question to Mick's (however I don't wish to hijack the thread) - is there a standard rule (or ratio) as to how much faster an F/6 is compared to say an F/4? (e.g. is it twice as fast?), and say for F/10 to F/5 will it be 5 times as fast? Or is it a bit more involved than that? I'm guessing it is - I liked Colin's example where you outlined how a camera chip needs to be considered

I'll be following this thread with interest

Cheers, Evan

[Wavytone]

Its a square law.

F/4 is (6/4)^2 = 2.25 times as fast as F/6.

The notion of fast/slow originates from photography and is referring to the shutter speed required for a correctly exposed image.

But... here's the catch:

The exposure depends on the amount of light reaching the sensor, which is proportional to the AREA of the telescope objective. Area is proportional to diameter squared.

But F/ratio is a ratio of two lengths = focal length/diameter.

So for two telescopes having focal ratios F1 and F2, the relative speed of one vs the other is (F1/F2) squared.

[mikeyjames (Mick)]

Originally Posted by Atmos

It can actually be quite complicated but here is a quick example. You have a 10" F/5 and a 10" F/10. On both telescopes you use a KAF8300 which has 5.4 micron pixels.

On the 10" F/5 you are imaging at 0.877"/pixel while on the 10" F/10 you are at 0.44"/pixel.

This means that every pixel is capturing four times the area (four times as many photons) on the F/5 than the F/10. Now the confusing part is as follows. Let's say the 10" F/4 used the ASI1600 with 3.8 micron pixels and you had one of the older KAF 1024x1024 15 micron sensors (Amdor still work with them I believe).

In this case the 10" F/4 is at 0.617"/pixel while the 10" F/10 is at 1.218"/pixel. In this instance the slower telescope is in essence near four times faster as each pixel captures near four times the surface area.

I think I understand what you mean. thank you. Out of interest, from my understanding, what you have described is from the 1st example in my earlier post. Is that correct?

Also, is there any truth in the analogy of the pipe in my example #2?

[Camelopardalis (Dunk)]

Putting elements of all this together, if you take two telescopes with the same aperture, one f/5 and one f/10, then the image will reach the same "brightness" (accumulation of photons resulting in electrical signal on the sensor) 4x faster with the f/5 scope than the f/10 scope.

Of course, if you're using the same sensor in this example, your resolution (per pixel) will also be 1/2 in each dimension, i.e. each pixel is seeing 4x the amount of sky as when attached to the f/10 scope...

Depending on your target, there are pros and cons to each

[ChrisV (Chris)]

I always have trouble with this. How does it fit with this view of the "f-ratio myth" ...
http://www.stark-labs.com/help/blog/...ioAperture.php
and the embedded link to
http://www.stanmooreastro.com/f_ratio_myth.htm

Totally confused, Chris

[Camelopardalis (Dunk)]

Dunno about myth...I took some test shots with my C8 at f/10 and f/7 and the shots at f/7 were approximately 2x the intensity of the f/10 shots...which is what I'd expect for one f-stop "faster"

[Wavytone]

There's a whole photographic industry that swears by that "myth".

[traveller (Bo)]

The way I used to explain to photo students goes something like this
Imagine you wan to fill a bucket with sand through a funnel. If the bucket contain a set amount of sand (exposure value) and you have a funnel with a wider opening (larger aperture, lower f ratio), the bucket will fill up faster than a funnel with a smaller opening.
This example can also explain depth of field in that a wider funnel will result in a larger pile of "flatter" sand than narrow funnel with a taller pile of sand.
Bo

[Tropo-Bob (Bob)]

The longer the focal length is of a lens/mirror, the more magnified the image. (So with a F10 lens, the image is magnified compared to a F5 lens by a factor of 2)

However, the light being received by the lens/mirror is the same, therefore the longer focus lens will spread/magnify the SAME light over a larger area.

So the F10 lens is slow because a longer exposure time is needed to brighten the fainter (spread-out) image to the same level as an F5 lens.

The F10 lens needs 4 times the exposure time as a F5 lens to brighten to the image to the same extent (The required exposure times relate to the magnification difference (2)squared)

It worked that way for photography and perhaps to a lessor extent for imaging, but this is not applicable to visual observations.

[Cimitar (Evan)]

Originally Posted by Wavytone

Its a square law.

F/4 is (6/4)^2 = 2.25 times as fast as F/6.

The notion of fast/slow originates from photography and is referring to the shutter speed required for a correctly exposed image.

But... here's the catch:

The exposure depends on the amount of light reaching the sensor, which is proportional to the AREA of the telescope objective. Area is proportional to diameter squared.

But F/ratio is a ratio of two lengths = focal length/diameter.

So for two telescopes having focal ratios F1 and F2, the relative speed of one vs the other is (F1/F2) squared.

Thanks mate , I understand now. Really appreciate your input and from others on this topic

Cheers, Evan