How does "F" ratio affect imaging

[Striker]

Ok guys I dont think this has been aswered before as I am very interested to see what come's out of this.

For example if you have a SCT F10 and fit a Focal reducer how can this possibly decrease imaging time....wouldn't aperture be the defining point to how many photos hit the ccd.

I have spoken to a few people about this with varied replies.

If the focal reducer can decrease exposure time how does this provide more light to the chip.????..the light path distance is still the same as F10 but now has to futher travel through another lense such as a Focal reducer.

I dont know the answer thats why I am asking.

Thanks in advance

[Starkler]

Its all about photons per unit area.

Consider a 10" f5 and a 5" f5 scope. The exposure time for an extended object will be the same but the image will be smaller on the ccd chip for 5" scope due to its shorter focal length.

If you take an sct of f10, then put a focal reducer in to bring it to f6.3, you are aiming the collected photons for a given object to a smaller number of pixels on the camera.

So, if you keep f-ratio as a constant and vary the aperture of a scope, the photon density per pixel wont change. As aperture increases, the field of view will be decreased due to longer focal length and the image of an object will appear larger.

I notice a similar effect with visual observing. For a given exit pupil size, brightness of extended objects appears the same regardless of aperture. Having a great big truss dob means that you can get a closer look at higher mags whilst still maintaining that brightness.

I think thats all correct :whistle:

[Lester]

Hi Tony,

A focal reducer that halves the f ratio of the telescope is actually doing the opposite of a 2x barlow. Where a barlow doubles the magnification the FR in this instance will halve the magnification. When you reduce the magnification the image becomes brighter, and because of the brighter image you can reduce the imaging time.

[janoskiss]

Yep. What Geoff said.

[Striker]

Thanks guys,

This was the paragraph that made the most sence.


Thanks Geoff

[jase]

Starkler summed in up well. Ultimately the focal ratio would have to be one of the most important aspects when imaging. It should match the detector you are using to optimise the pixel per arcsecond which should also include your seeing conditions. There is not point trying to image at high resolution, say .25 arcseconds per pixel if you don't have the steady skies to do it. Generally, aim for the 1.5 to 3 arcseconds per pixel with any given combination.

It is true though that a 40" (1m) scope operating at F/7 and a 10" SCT operating at F/7 will capture the sky at the same speed, i.e. identical exposures would capture the same amount of nebulosity. The advantage the 40" scope has is image scale. The image will be much larger, this is purely based on focal length. 40" (1000mm) at F/7 give a 7000mm focal length, while the 10" SCT at the same focal ratio gives a 1750mm focal length - so the SCT image scale is smaller.

A telescope with a fast focal ratio will provide a wider field of view
A telescope with a slow focal ratio will provide a higher level of magnification
Light gathering power is not all that important with the sensitivity of CCD chips todayThere are some other interesting points, but can't remember them now.:thumbsup:

[Striker]

Thanks Jase.

Just shows me how important I need to reduce the focal length of the SCT.

[Lee]

I was under the impression that for two scopes of same f ratio, that the bigger aperture scope would give you a brighter image, and thus cut down imaging time - very interesting thread so far.

[iceman]

Get a newt! Buy back your 8" dob! ;)

[Dennis]

Hi Tony

Here is a nice, simple 3 liner:

F-ratio determines speed at which a nice non-grainy image can be formed.
Focal length determines field of view (for a given chip size).
Aperture determines Rayleigh resolution, although in practice resolution is usually limited by seeing.

This is a direct copy/paste from the informative article at:
http://www.princeton.edu/~rvdb/images/ApertureFacts.html
By Robert J. Vanderbei at Princeton University.

Cheers

Dennis

[Lee]

Is this range what we can expect given average atmospheric conditions, or is this a range of "good conditions" to image under...????

[sheeny]

Geoff, that about the best layman's description I've seen.:thumbsup:

The simplified explanation I have in my head from an astrophotography book I've read was along the lines of "aperture affects the exposure time for point sources like stars, and F ratio affects the exposure time for extended sources like nebulae, the moon, terrestrial photos etc".

Of course, the two are related.

Al.

[jase]

This depends on what you want to do and is open to interpretation. If you are imaging the planets, you'd want to try and get under .75 arcsec per pixel for optimal resolution. Atmospheric conditions are your problem here. Some sites located near mountains or the sea will never provide this level of stability.

Deep sky objects don't require such high resolution work and in most cases are extremely dim compared to planets. A higher arcsec per pixel is then desirable. I'm not saying you can't go high res on deep sky objects, you can, but you'll have better luck around the 1.5 to 3.0 range. Always try and match the camera with your telescope (or vice-versa - depending on budget). If you're really keen, get yourself an adaptive optics unit or similar.

Some auto guiding software/camera combinations will actually tell you what your seeing will provide. As the guiding is taking place it can measure not only the mount corrections, but variations in atmospheric turbulance (provided on a arcsec measurement).

[Dennis]

Hey, Tony,

Here’s a nice experiment:
Fit a 200mm telephoto lens to your very old manual 35mm SLR camera.
Go outside and put it on a tripod and view a scene.
Open up the iris to max (F4 on mine).

Result: the viewfinder is very bright.


Stop down the iris to min (F32 on mine).

Result: the viewfinder is now very dim.


What we observe:

The focal length has not changed and thus the field of view has not changed – we see the same image, only brighter or dimmer.
Therefore, the focal ratio is what affects the “brightness” of an image.


Cheers

Dennis

[bird]

Focal Ratio = image brightness.
Focal Length = image size.

So, two scopes with the same focal ratio will produce images that are the same brightness with a given EP or camera.

Two scopes with the same focal length will produce images that are the same size with a given EP or camera.

The two properties have to be considered seperately. Anytime you want to consider brightness, use the F/ratio. When you consider image size, use the F/length.

regards, Bird

[jase]

Playing devils advocate here... (not sure if I thoroughly understand that there is no correlation between the FR and FL)
So are you telling me that:
An 8" SCT @ F/10 = 2000mm FL
An 8" SCT @ F/6.3 = 1260mm FL

Focal length (in mm) by the aperture (in mm). For example, a telescope with a 2032mm focal length and an aperture of 8" (203.2mm) has a focal ratio of 10 (2032/203.2 = F/10).

As you can see the FR changes the FL. So a faster ratio provides a wider view. Have I missed something?

[leon]

gee, i'm staying out of this one, i think i'll just read... :D

Leon

[Dennis]

Hi Jase

If you take an 8”, 2000mm, F10 SCT and add an F6.3 Reducer/Corrector, you have effectively “zoomed” out to a shorter focal length of 1260mm at F6.3. Therefore, you now have a larger field of view as well as a “faster” ‘scope, and you have simultaneously changed 2 parameters, the FL and the F Ratio.

If you keep the focal length at 2000mm, but change the F Ratio to F6.3, you would now have a 12.5” SCT. That is, 2000/6.3 = 317mm = 12.5 in

Now in this case, the 12.5” F6.3 FL being the same, i.e. 2000mm, the FOV should be the same as the 8” SCT F10, but it will appear brighter at F6.3 than at F10.

The 12.5” SCT will also be able to resolve more detail i.e. split closer double stars due to its larger aperture.

At least that’s how I interpret this.

Cheers

Dennis

[jase]

Hi Dennis,
Yes. Fully concur with your examples of both 8" and 12.5" FL/FR.
So in order to get the same image scale (2000mm) as an 12.5" @ F/6.3, you need an 8" at F/10. Similarly a 40" requires FR of F/2 and a 5" requires a FR of F/15.75 to reach the image scale (2000mm) as both 8" and 12.5" scopes. Obviously the larger aperture instruments have a higher angular resolution (central obstructions aside).

Cheers:thumbsup:

[Dennis]

Yeah, it’s a little counter-intuitive isn’t it? Reading more and more, I now understand the following to be true - I think!

When imaging a galaxy through a 12” F10 scope versus an 8” F10 scope, you could be forgiven for believing that the 12” image would automatically record the galaxy to be brighter than in the 8”.

But, they would record as the same brightness in either scope. The main difference is that the 12” (FL of 3000mm) would have a narrower field of view versus the 8” (FL of 2000mm), and thus the galaxy would be captured at a larger scale (more magnified) in the 12”, but of the same brightness.

However, visually, at the same magnification, the bigger light bucket will make the galaxy appear brighter if your eye can accommodate the full exit pupil of the eyepiece.

Cheers

Dennis

[Starkler]

To expand on this, if you have the same exit pupil size (aperture/mag) any size scope will show the galaxy at the same brightness (assuming transmission losses are equal), but at a bigger image scale (smaller fov) in the larger scope.

The limiting factor of course is the maximum size that your pupil can dilate to when dark adapted.

[astroboy]

Most this sounds fine for extended objects but back in the film days it was understood for stars aperture is the main thing , this is why I had to expose a 20 mm F5.6 shot for 90min on film to get enough stars for a good milky way shot.
I dont think CCD makes much differance but if you take the focal length too far the stars may behave like extended objects.

Zane