10" f10 vs 12" f8

[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 (Troy)]

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

[cometcatcher (Kevin)]

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 (Rick)]

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 (Marcus)]

Originally Posted by RickS

Approximately 1.57 times assuming a 50% obstruction by diameter.

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 (Geoff)]

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 (Rick)]

Originally Posted by marc4darkskies

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

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.

Originally Posted by ghsmith45

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

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 (Marcus)]

Originally Posted by 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.

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 (Rick)]

Originally Posted by 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?

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 (Rick)]

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 (Marcus)]

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]

Originally Posted by 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.

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]

Originally Posted by 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.

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 (Rick)]

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]

Originally Posted by 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.

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 (Rick)]

Originally Posted by 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

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.