M33/ modifed webcam, M33 Atik 314L

[Rocket Pooch]

It was a Televue Genesis, bit more expensive than a Celestron.  It was also Olly's kit EQ6 etc, and AA guiding, basically the same setup except the scope.

[MarkS]

Hold on - I think we're in danger of confusing two concepts here and drawing the wrong conclusions.

For astrophotography, just like terrestrial photography, the ONLY thing affecting image brightness at the film/CCD is the F-ratio (i.e. focal length / aperture). Any two lenses with the same F-ratio will give the same brightness - this is precisely the reason why F-ratio is so important for photographers.  When you put your camera on the back of a telescope you are simply treating the 'scope as a giant lens.  So an 18mm wide angle lens at F10 will give exactly the same image brightness as a Celestron C11 at F10.

So what changes as we jump from 18mm to the the 2800mm of the Celestron?  Magnification!  A 1 minute exposure of the Orion Nebula taken with the 18mm (at F10) will have just the same brightness as the 1 minute exposure of the Orion Nebula taken with the C11 (at F10).  But the C11 will have have a much smaller field of view i.e. vastly increased magnification.

As the F-ratio is reduced, brightness increases and the the signal-to-noise ratio of the CCD improves with it.  Hyperstar is the logical extreme of this - taking the F-ratio down to approx F2.  F2 will give an image 25x brighter than F10 (the square law applies) so vastly reduced exposure times can be used. 

So, going back to a previous question, given two telescopes with the same F-ratio, does the aperture make any difference?  The answer is that the aperture will affect the magnification in the final image but both scopes will deliver the same brightness to the CCD.

Mark

[MarkS]

Rick,

I agree with Chris on this one.

Once we have eliminated the optics (which apart from collimation and focusing, you have no control over)  the main issue for image quality is reducing signal-to-noise ratio (SNR). 

The main reasons for noise in the image are:
1) Light pollution (street lamps or moonlight)
2) Dark current of the CCD (cooled chips deal effectively with this)
3) Read noise - this is an inherent characteristic of the combination of CCD and amplifier.

The best way to reduce SNR is to reduce the F-ratio (by buying a different scope or attaching things to your present scope e.g. reducers or hyperstar )
The other way is to use much longer exposure times - but then we have to start using guided exposures and this introduces a whole new set of additional technical problems.
To a certain extent, light pollution can be dealt with by using narrowband filters but the only real solution is to go somewhere dark.

Once you acquired good, noise-free raw data then processing this into the best possible final image is another art which takes a long time to learn.  And processing is not free from it's own controversial issues.  It depends very much on your goal.  If your intention is to perform photometry or some other science on the final image then your processing sequence will be different to the one required to create a beautiful poster for you living room.

Mark

[Mac]

true but the two scopes in the question, were not the same F ratios.

do you know how much light gathering power the 14" at F1.9 would have over the 120ED at F7.5 im using right now

So if you work out the light gathering power difference, as this is only the difference in size of glass, it works out at 12 times the light gathering power.

So if the two scopes were the same F number, then the 120ED would take 12 times longer to record the same amount of photons, due to the size of the optics.

and the difference in F stop, as you said relects the brightness, but the two differences is a factor of 4, ~F2 against ~F8

if you compare like for like, then an F8 scope will take 4 times as long to collect the same amount of photons as a F2.

so you do have to take in to consideration the size of the object lens as well as the F ratio.

As for the image size, dont forget the focal length of a scope is its objective size * F ratio. The F number is normally fixed. F10, 6.3 ect

as the F has now been changed from F10 to F1.9 its focal length also changes.

The 14" would now be a 674mm focal length telescope. (355 * 1.9) where as the ED would still be (100 * F7) 700mm focal length scope.


ps
this is the maths for the keck. http://ceres.hsc.edu/homepages/classes/astronomy/fall97/Mathematics/sec15.html

[Daniel]

Hi Guy's thanks for the explainations, sounds like I should be able to get my exposure times right down then, I was worried light polution would be a big factor at this F ratio, but since the mirror will collect more photons in less time, I guess this will all balance Itself out, I'll still be getting the light gathering power of a long exposure on my refractor even if Im only imaging for low exposure times.

Thanks

Daniel
:O)

[MarkS]

Mac,

I think I'm beginning to see where our arguments differ.  I think  there may be 2 cases that need to be considered.

1)  For an extended object (e.g. nebula, planet, moon crater) I still argue that the F-ratio is the only meaningful comparison between telescopes for astrophotography purposes.  The inverse of the square of the F-ratio determines the number of photons hitting each pixel of the CCD.  For instance an F8 scope will take 16x  as long as an F2 scope to collect the same number of photons/pixel.

2)  On the other hand, for an isolated point source (i.e. a star) the limiting magnitude seen through the eyepiece is definitely related to the size of the objective i.e. the light gathering power.

I need to think about this further  ...

Mark

[RobertM]

Another way of thinking of it Mark...

Pixel scale is inversely proportional to focal length so for a set focal length the light falling on one pixel is proportional to the area of collection (squares of the diameters or radii) of the objectives.

Assuming the 120ED and C14@f/1.9 have the same f/l then the ratio of exposure lengths =  350^2/120^2 = 8.5

Assuming 10um pixels for simplicity:

ED120@f/7 has an pixel scale of:  206265 * 0.010 / (120 * 7) = 2.45 arcsec/pixel
C14@f/1.9 has an pixel scale of:  206265 * 0.010 / (350 * 1.9) = 3.10 arcsec/pixel

206265 = number of arcsec in a radian.

The C14@f/1.9 will collect 3.1^2/2.45^2 = 1.6 times the background signal of the ED120@f/7

So all in all the C14@f/1.9 will be 13.6 (8.5 * 1.6) time faster or will need 1/13th the exposure time.

For point sources like stars, correct focus has almost as much bearing on signal strength (and therefor s/n) as anything else.  At focal lengths of 1.9 the critical focus window will be tiny and I would think a robotic focus or bahnitov mask will be essential.

Think that makes sense.

[Daniel]

I bought some acetate for printing out a Bahnitov mask just yesterday, right now Im using both a hartman mask as well as Live view to zoom in 10x for focusing. Actually, with the greater light gathering power I wonder how much more will show up in my live view now, it would be nice not to have to focus on really bright stars for a change.

[MarkS]

Another way of thinking of it Mark...

[argument snipped by me]

Think that makes sense.

No, it doesn't make sense.  If you apply that argument to two different camera lenses (instead of two differennt telescopes) then the result violates what every photographer takes for granted i.e. that equivalent F-ratios give equivalent illumination.

I'll try to come up with a simple explanation this evening ...

Mark

[MarkS]

Robert,

I owe you an apology:  you did arrive at the correct result i.e. C14@F1.9 requires 1/13th the exposure time of ED120 at F7

I was slightly confused by your workings and misinterpreted the result. 
However, you can get the result much quicker by working directly with the F-ratios:

Ratio of exposure times = square(Ratio of F-ratios) = (7 / 1.9)^2 = 3.684^2 = 13.573

Mark

[RobertM]

Mark,
Yes, sorry it was a but obtuse but I wanted to work it all out in my mind using a worked example.  I was thinking about your last response and came to the conclusion that we were all talking about the same thing in different ways.  Thanks for proving my calculations and for the very useful shortcut.

Daniel,
Good luck, you'll have one heck of a rig when you get that all going perfectly.  Been tempted for a while with the Starizona system but I'll see how yours goes with the new model before making any decision.

[Mac]

If you apply that argument to two different camera lenses (instead of two different telescopes)

I aggree perfectly, but dont forget, with a lens, the focal length of the lens is completly fixed.
as is the F number an f2 lens is always an f2 lens., you can make it an f10 or f22, by stopping it down..

but with a camera lens, you are not changing the focal length by changing the F stop.
a 50mm lens @ f8 is the same focal length as the 50mm lens @ f2, but the amount of light has been changed by the diaphram at the back.

a 300mm F2 lens will give you exactly the same shutter speed as a 50mm F2 lens, but the size of the front optics, is greatly enlarged, to allow you to do this.

The F stop has been changed by stopping down the amount of light by making the exit hole smaller, thus increasing the depth of field
and increasing the exposure time, you havent actually changed the f of the camera lens. The 50mm @f 8 is still f2, but its just been stopped down.


With a telescope, the focal length although fixed normally is the size of the optics * F stop.
so the celeston 14" F10 has a focal length of 355 * 10 = 3550mm focal length. normally.

as a lens it would be decsribed as a 3550mm f10

but fitting the hyper star changes the F from F10 to F1.9
so its focal length changes from 355* 10 to 355 * 1.9 = 674mm focal length

its still a 14" telescope, but its focal length is 674mm as against 3550mm.

as a lens it would be describled as a 674mm f1.9

for the ED120 as a lens it would be described as 700mm f7.5

so effectivly, your 14" would be a ~700mm f1.9 lens as against the 700mm f7.5

The problem is (which i've just noticed sorry), by fitting the hyperstar, you have converted your 14" 3550mm into a super fast ~700mm scope
so the difference in speed is just the difference in ratios of the f. which is about 8, as you have already taken in to consideration
the change in focal lengths and object lens size ect.

i think somewhere back in the thread, i've combined the light and the f together.

So
if you bought an F1.9 700mm scope, you would get the 14" + hyper star.
and if you went for the 700mm F7.9 you would get the ED120.
and if you wanted an F1.9 3550mm scope would need an objective lens about 1.8m

[MarkS]

if you wanted an F1.9 3550mm scope would need an objective lens about 1.8m

Sounds good - where can I get one?

[Mac]

I bought some acetate for printing out a Bahnitov mask just yesterday,

I assume that you are going to print it and then place it on the front of the optics.

Not too sure if the affect of the focus would be that critical,
but dont forget you are introducing an extra light path through the acetate,
so the focus would be slightly different, then without it, (it might not even be noticable)
although if you removed the mask and replaced it with a clear acetate,
it would still keep the same extra light path.

is it me, or have we managed to combine three different subjects in this one thread?

[JohnP]

http://en.wikipedia.org/wiki/F-number