10" f10 vs 12" f8

[allan gould]

I can't work out how much extra exposure I would have to do with a 10" f10 compared to a 12" f8 scope to give an equivalent ccd exposure.
Any geniuses out there that can assist?
Allan

[clive milne]

Approximately 40% all else being equal.

[allan gould]

Thanks for that Clive. Much appreciated.
Allan

[cometcatcher]

That doesn't sound right. f10 is 2/3 of an f stop dimmer than f8. A full f stop would require double the exposure, so 2/3 f stop make 66% more exposure needed?

[RickS]

Approximately 1.57 times assuming a 50% obstruction by diameter.

[allan gould]

Kevin
I think you didn't see that I was comparing a 10" and 12" at different f ratios. But thanks any way.



Thanks Rick, I thought that it would be in the ball park you and Clive have given.
Allan

[cometcatcher]

I did, but if you're shooting extended objects, the exposure is determined by f ratio not aperture. For instance a 3" f6 or a 20" f6 have the same exposure time since they're both f6.

But I've got a funny feeling I'm missing something. ;) Someone explain it to me.

[RickS]

You're correct. The difference in exposure time (ignoring central obstruction) only depends on the f/ratios: 10*10/8*8 = 1.5625

However, if you're an astro imager you'll also want to think about image scale as well as exposure time. This will be very different when comparing a 3" f/6 and a 20" f/6.

[cometcatcher]

Yes. Image scale is very different but the "average" exposure will be the same. It will of course vary as the 20" could pick off bright and dimmer areas within the field of the 3". Allan's image scale is roughly the same, 100" for the 10" vrs 96" FL for the 12".

Aperture makes greatest impact on point sources like stars, although I've read f ratio affects this also.

[Regulus]

F ratio: f8 is f8 is f8 regardles of the lens size.

[RickS]

I still don't buy the argument that f/ratio is terribly relevant for imaging. The two things that do matter are aperture (which determines how many photons you can capture) and image scale (which determines how those photons are distributed into pixels). Those two parameters determine exposure time and resolution - both fundamental concerns of the imager.

Cheers,
Rick.

[cometcatcher]

It's just part of the equation. I got a little confused comparing a decimal to percentage but other than that, we're all correct lol.

[Geoff45]

I agree with Rick here. This topic keeps on recurring. People should read this article for some very useful info http://www.stanmooreastro.com/f_ratio_myth.htm.; There is a nice pic of the same field taken at 10 min with an 8.2" scope at f/12.4 and f/3.9. If anything, the longer f-ratio gives a better image. Here is another useful link to the same topic http://www.stark-labs.com/help/blog/files/FratioAperture.php.
Geoff

[RickS]

Focal length and pixel size determine image scale. f/ratio has no role to play here ;)

[Geoff45]

Don't forget the camera here. A f8 scope and a camera with 8micron pixels will gather exactly the same amount of light per pixel as the same aperture f4 scope and a camera with 4micron pixels. Both systems will have the same "speed".

Don't forget binning. Binning 2 x 2 effectively halves your focal ratio.
Geoff

[RickS]

Poor Allan's original question (sorry, mate!) was about imaging. We're talking about image scale in arcsec per pixel as it is conventionally used by imagers.



Exactly, Geoff! Pixel size is hugely important.

[allan gould]

Cr@p, Im glad I asked such a simple question!!!!
I assumed but did not state I would be using the same camera.
Essentially the image scale is the same for each scope, with the 12" F8 having a slightly larger obstruction than the 10" f10; which still makes the 12" gather more photons for the same length of exposure as its the bigger bucket. Very roughly a 10 min sub with the 12" would be equivalent to a 15 min sub with the 10" scope.
Think Ive got a handle on it now but very interesting all the same and thanks to all who replied.
Allan

[RickS]

Not trying to be an ass here, but the original question was not about the abstract brightness relationship between two scopes. It was specifically about CCD exposure times:



QED... I'll shut up now. :D

[Peter.M]

My terrible napkin math says for a given camera you would have to expose the 10 inch 1.5 times the length to get the same exposure, assuming that there is not obstruction.

[clive milne]

Teach me to write a 10 second reply... As I walked out the door I realised the error in my post.

So for clarity;
For (under-sampled) star images the exposure times will be approximately proportional to the aperture ratio squared;
ie) the 12" will collect 44% more light so the 10" will take 56% longer for the same depth of exposure. The unspoken assumption here is that a combination of background sky brightness and detector noise will determine the absolute magnitude limit. There is quite a difference between sensors (as much as 2 magnitudes) The Sony cmos chips being probably best in class within the amateur budget.

For extended objects the exposure times are proportional to the ratios of the f-numbers squared.
In this instance it will be the same as the numbers above (derived from relative aperture) but this isn't always the case so it is probably worth noting the distinction.

c

[clive milne]

At the risk of labouring the point... exposure times on bright object will be a function of QE and aperture... You have to factor in sky background brightness and detector noise to determine how faint you can ultimately go. For point sources you have to include the effect of sampling (linear size) as well.

I hope that's a bit clearer?

[troypiggo]

Just make sure you capture your images with Sequence Generator Pro and it doesn't matter what scope you use.

[cometcatcher]

Andy, from what the guys are saying, (and this page http://www.stanmooreastro.com/f_ratio_myth.htm) the rules are different for film cameras which is where I get the figures from. I'm also pretty sure my Pentax DSLR will act more like a film camera than an astro CCD camera. My images get lost in noise at higher f ratios for a given exposure.

[RickS]

Feel free to believe what you like Andy, but if you think that f/ratio is relevant to CCD imaging speed and that pixel size isn't then I reckon you're a bit confused.

Cheers,
Rick.

[marc4darkskies]

Correct (I got 1.56). And if the 12" has say a 10% larger obstruction (i.e. 55%) then the ratio is 1.48 (ideally).

Cheers, Marcus

PS: my idealized calculation is based on aperture, central obstruction, focal length and pixel size.

[Geoff45]

Maybe we should clarify a few things. When the OP asked about getting an equivalent exposure, what did he mean by "equivalent"?
1. Same number of photons from the DSO
2. Same average ADU for the DSO
3. Same signal to noise ratio in both images.
They are all different and that's partly why we are going around in circles here.
Geoff

[RickS]

I have a similar model, Marcus. It wouldn't make a difference in this case but I also include QE and optical efficiency (loss from reflecting and refracting elements) in the calculations.



Good point, Geoff. For imaging, I reckon equivalent means the same number of photons per pixel (hence same pixel SNR) which I assume is your #3 above? Need to be careful since there are different varieties of SNR, e.g. the object SNR than Stan Moore talks about.

Cheers,
Rick.

[Peter.M]

I dont think we need to be careful, we are only changing one variable the rest should be assumed to be controlled.

Camera, location (and therefore sky noise), and target I would have thought to be held constant for the different scopes for comparison purposes. Likewise scope design should be kept constant because it gets too complex when you factor in reflectivity of all surfaces and transmission of filters/correctors.

Maby the sky noise is such that you can only expose for 5 minutes with the 10 inch. That isnt the question asked, in that case the 12 inch would get the same exposure in just over 3.

[marc4darkskies]

Yes, I account for peak QE too but assumed it to be constant. I don't bother with optical losses - assuming "good" optics it will have only a very minor impact and most won't know what the real values of their scopes are anyway. One can consider a wide range of other variables but why over complicate what only needs to be a first order approximation?

[RickS]

Yes, I suppose optical efficiency is not a big deal but it's not hard to estimate based on number of lenses and mirrors. I see differences of 5% or 6% at most.

[Peter.M]

These two statements contradict each other.

A telecompressor shortens the effective focal length which is what you are saying determines image scale.

Binning is the electronic equivalent of using a telecompressor.

Binning is a change in the size of each pixel.

Therefore pixel size changes image scale. The units commonly used for image scale are arcseconds.pixel^-1 which in its essence implies that pixel size is involved.

[RickS]

Andy,

We're obviously talking at cross purposes here. How do you define "image scale" and what physical units does it have? I'm talking about the angular size of a pixel, typically measured in arcsec per pixel.

I maintain my earlier statements too :) Clear aperture, focal length and pixel size are relevant to imaging exposure time (by determining the rate at which photons are collected in each pixel and hence how quickly pixel SNR increases). Focal ratio is not relevant.

What Peter says makes sense also. Binning increases the size of a pixel (call it a pseudo-pixel if you wish) so photons are collected more quickly and pixel SNR rises more rapidly. You trade off resolution for this improvement in SNR performance, of course. Nobody is saying this is magic. Similarly, if you look at the performance of a camera with smaller pixels you will get increased resolution (potentially - maybe seeing limited) at a cost of lower SNR for the same exposure time.

Cheers,
Rick.

[allan gould]

Boy, this certainly got a bit of attention and I feel I should explain why I asked my question.
I use a QSI583WSG camera for imaging and currently have a 10" GSO RC f8 and a Meade 10" f10 SCT. I'm thinking of upgrading the 10" GSO RC to the truss version of a 12" CSO RC f8 to gain a little more light and hopefully increase the signal to noise ratio of my imaging so that I can get a bit more of the faint detail in my galaxies.
From my calculations the 10" SCT and the 12" GSO RC have almost the same image size at 0.45 and 0.46 arc sec/pixel, respectively. Thus it would come down to the extra light gathering capability of the 12" over that of the 10" to put photons into each pixel.
My question was, under these circumstances, how much longer would I have to expose with the 10" scope to get an equivalent sub as that from the 12" scope?
Of course I'm assuming that a 15 minute exposure through the 12" scope will lift the signal from the background noise more than the 10" scope would and thus make processing a little easier to pick up faint detail.
Allan

[Peter.M]

At less than a half an arc second per pixel I would have thought you would be better off using the scope you have with a camera with larger pixels. Putting a 11000 chip on your 10 inch RC gives around 1 arc second per pixel, someones going to say you can just bin the 8300 but with its shallow wells I don't think that's a good option, and of course you can bin the 11000 for 2 arc second per pixel.

At 2 arc seconds per pixel you gather light 16 times as fast ( ideally) as the same scope with your 8300 on it, and I doubt you will lose much resolution because of your seeing.

[marc4darkskies]

So your new scope would be the 12" F8. Okay, then don't you really want to know what the equivalent exposure duration would be going from the 10" to the 12" (not the other way around), right? In that case a 15 minute exposure with the 10" F10 would become a 10 min exposure on your 12" F8 (0.64 ratio). If you were to bin 2x2 with the new 12" its exposure equivalent would be 2 minutes at 0.91 arcsec/pixel.

[allan gould]

You are right Marcus, but it doesnt really matter which way round it was stated as the ratio is the same, but point taken.
Thanks for the input and binning the camera with the 12" appears to be valid except for the well depth of the 8300 chip as Peter has pointed out, however a lot of the galaxies I want to image are not that bright so it may not matter in a practical sense.

[Peter.M]

The main reason I mention the small wells , particularly for galaxy imaging is not for the galaxy to look nice because it will probably not saturate the chip but for the stars in the frame. Greg Bradley is using a small well chip on a 17 inch so it might not be anything to worry about, but personally I find that I enjoy the look of images from the 11000 more because most stars don't overexpose.

[RickS]

The well depth of the KAF8300 itself isn't an issue (when you bin 2x2 you effectively get 4 times the well depth) but Kodak in their wisdom made the horizontal shift register too shallow to cope with the full binned well depth when reading out the data. This causes horizontal blooming on bright stars. The blooming can be reduced or eliminated by reducing gain (I think the QSI cameras do this automatically) but that comes at a cost in dynamic range.

Cheers,
Rick.

[allan gould]

Rick
Ive not seen blooming with the QSI583 chip at 2x2 binning but I have seen it on very bright stars at 4x4 binning (but even then its minimal).
Im going to set up tonight (cloud permitting) to see what I can do on NGC1398 and will have a play with some parameters.
Allan

[RickS]

I think that's because the QSI camera automatically drops the gain when binned, Allan. On my SX camera you have to physically adjust a trim pot inside the camera which is somewhat less convenient! Let us know how the experiments go.

Cheers,
Rick.

[cometcatcher]

This is how I understand "binning". http://www.spotimaging.com/iq/binning.html

[Geoff45]

Well, this i how I understand binning
http://www.starrywonders.com/binning.html

[RickS]

Andy,

The article by Craig Stark that Geoff linked explains how binning works more eloquently than I could. Binning does effectively give you larger pixels and the ability to trade off reduced resolution for increased SNR. The article also reinforces what I've been trying to say - pixel size is very important in imaging and is directly relevant to exposure time.

OK, so we are talking about basically the same thing. Your arcmin/mm is calculated using exactly the same formula as my arcsec/pixel. The angular size corresponding to a distance on the focal plane is derived through basic trigonometry:

angular size = 2 * arctan(d/2F) radians [multiply by 3438 for arcmin or 206265 for arcsec]

The angular size depends on two things: d, which is 1mm for you and the pixel size for me, and F which is the focal length. You'll notice that it does not depend on the focal ratio! So I say again (for the last time in this thread, I hope) image scale does not depend on focal ratio.

Cheers,
Rick.