A beginner question

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Date: 09 Jun 2007 12:56:18
From: actuary@mchsi.com
Subject: A beginner question

To All:

I reently purchased a Canon PowerShot SD1000 for strictly home use. I
was out the other night trying to locate Vesta and I did but then I
wondered if I could capture Vesta (magnitude 5.8) on a 2 second
exposure (maximum exposure time) using the Canon 1000SD camera.

I don't have a tripod so I didn't try. What is the applicable
mathematics to answer the question?

Thanks.

Larry

Date: 11 Jun 2007 09:29:00
From: actuary@mchsi.com
Subject: Re: A beginner question

On Jun 11, 11:00 am, Chris L Peterson <c...@alumni.caltech.edu > wrote:
> On Mon, 11 Jun 2007 21:06:25 +1000, "Peter Webb"
>
> <webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
> >Essentially then digital ISOs are analogous to push-processing on film. To
> >somewhat extend the analogy, its like a "digital zoom" - no extra
> >information is actually added.
>
> I don't think the analogy is precise, but I'd agree that digital ISO is
> more like push processing than native film ISO. However, as I noted
> earlier, if the gain is applied in the sensor, before the readout stage,
> it's possible to see a slight improvement in S/N because the readout
> noise contribution isn't amplified (or isn't fully amplified).
>
> >Sure. My issue was that I had two different "scales" in use in my head, and
> >now way to calibrate between them. I know its much the same principles for
> >astro and terrestrial photography (of course, no depth of field issues,
> >point sources only, some others), but I had no way "in my head" of equating
> >the two scales. When I looked at the question concerning a magnitude 5.8
> >object on a terrestrial camera, I had no scale to equate this to daytime
> >photography. Unlike, for example, photographing the moon which is directly
> >illuminated by the Sun. I don't even know how to work out the exposure etc
> >for photographing the Sun directly (and nor would I try!); if I did, I could
> >pull out an inverse square law and work out what would be needed for the Sun
> >as a distant point source to be "resolved".
>
> Yeah, I don't think there's any way to really equate the calculation of
> exposure for terrestrial imaging with the calculation for astronomical
> imaging.
>
> >Presumably the camera optics are not much better than are needed to resolve
> >to physical pixel size on the CCD array (what would be the point)? So
> >presumably also the point source can easily spill over into adjacent pixels.
> >Most digital cameras will let you reduce resolution - I wonder whether this
> >improves sensitivity (as it does for film grain size). This is presumably
> >whether they sum the baseline data from adjacent pixels to produce the lower
> >resolution image. If they do, halving the total image size should provide a
> >massive 1.5 dB gain.
>
> >Of course, this can as easily be done in post-processing, I guess?
>
> There is only one noise source that is pixel dependent, and that's
> readout noise. For short exposures (typically less than a few seconds
> with ordinary digital cameras, and up to many minutes with cooled
> cameras under dark skies) the readout noise is dominant, so there is a
> value to having less pixels involved. However, this advantage is only
> realized when the camera is capable of binning the pixels within the
> sensor, so a single readout operation is involved. If you reduce the
> resolution after the sensor is read, either in the camera or during post
> processing, there is no reduced noise.
>
> For exposures long enough that readout noise is fairly insignificant, it
> makes no difference how many pixels are involved.
>
> >Is the intent to capture in each JPEG frame the "dark background" in
> >sufficient resolution to add multiple images and decrease noise by summing?
> >So is the JPEG really better considered as a map of the background noise
> >rather than the foreground point sources, and the digital artefacts don't
> >matter as much?
>
> >And if the JPEG contains the data to exactly reconstruct the original
> >bitmap, then its hard to see how it could be much smaller than the bitmap.
> >You have got to be chucking information away somewhere.
>
> Well, a lossless JPEG compression (which isn't an option with any camera
> I've seen, although the standard supports it) could still be highly
> compressed with respect to a bitmap for many astronomical images, given
> that much of the background may be essentially zero. Naturally, if you
> fully encode the noise, the JPEG will have to be the same size as the
> bitmap.
>
> >I assume that RAW is just that, the pixel values. Every camera I have owned
> >in the last five years has it, wouldn't that be heaps better? (Although I
> >have never used it, and until recently was prohibitively expensive in
> >storage).
>
> That's what you'd like RAW to be, although with most cameras there's
> still a certain degree of processing that has occurred before the raw
> image is saved. But it's close enough to raw pixel values in most cases
> that typical image processing operations such as stacking can be
> performed. Besides the lack of compression artifacts, what's very
> important with RAW is that the original dynamic range of the sensor is
> largely preserved. The data will be 10-12 bits deep (60dB-72dB),
> compared with at best 8 bits (48dB) with JPEG encoding.
>
> _________________________________________________
>
> Chris L Peterson
> Cloudbait Observatoryhttp://www.cloudbait.com

To: All

I'm still struggling to simply take a 15 second exposure on my SD
1000. I set the time delay to 10 seconds. I set the number of
exposures to 2 and I set the long shutter speed to 15 seconds. I
depress the shutter button and the lens retracts and the camera seems
to shut off.

Help -- What am I doing wrong.

Larry

Date: 10 Jun 2007 11:33:56
From: actuary@mchsi.com
Subject: Re: A beginner question

On Jun 10, 10:13 am, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au > wrote:
> "Chris L Peterson" <c...@alumni.caltech.edu> wrote in messagenews:2u3m63t0fjdp69llarrpoml092s9jcgr5e@4ax.com...
>
>
>
>
>
> > On Sat, 09 Jun 2007 12:56:18 -0700, "actu...@mchsi.com"
> > <actu...@mchsi.com> wrote:
>
> >>I reently purchased a Canon PowerShot SD1000 for strictly home use. I
> >>was out the other night trying to locate Vesta and I did but then I
> >>wondered if I could capture Vesta (magnitude 5.8) on a 2 second
> >>exposure (maximum exposure time) using the Canon 1000SD camera.
>
> >> I don't have a tripod so I didn't try. What is the applicable
> >>mathematics to answer the question?
>
> > I assume you mean using just the lens in the camera? Then yes, it might
> > just barely be possible. You need to consider the number of photons
> > received. In the case of a mag 5.8 object, and your ~3mm aperture,
> > that's about 1000 photons per second. Between atmospheric losses, and
> > issues of camera sensitivity, you might actually record about 25% of
> > these, or 250 photons (500 in 2 seconds). That would give you an ideal
> > S/N of about 22, which is more than good enough to show a visible dot.
> > However, there are other noise sources not so easy to calculate, such as
> > dark current noise, readout noise, and sky noise. Considering all these,
> > there's a good chance that your 500 photons of signal might be dropping
> > down into the mud.
>
> > BTW, doesn't the SD1000 actually have a 15 second maximum exposure time?
> > That should be long enough to easily get a mag 5.8 object, even with the
> > increase in dark current noise.
>
> > _________________________________________________
>
> > Chris L Peterson
> > Cloudbait Observatory
> >http://www.cloudbait.com
>
> For a "Beginner question" its a remarkably good one.
>
> Whilst I have not taken astronomical photographs, I have a pretty good
> understanding of optics. So I was surprised to find that I could find no way
> of calculating exposure times of astronomical objects (other than the moon)
> by comparison to terrestrial ISOs, f-stops etc. I note above that this is
> not the route you took to your solution. Can it be done that way?
>
> Anyway, that's a useful computation, and I now know that a 3mm aperture
> aperture will see the same objects as a human observer with a 2 second
> exposure. Did you assume an equivalent ISO rating for the CCD, or are you
> just assuming it is cranked up to the maximum in the calculation?
>
> Anyway, a good question and a good answer - thanks.- Hide quoted text -
>
> - Show quoted text -

Thanks for the input. I couldn't get my Canon SD1000 to take a 15
second exposure. I'm doing something wrong. I went back to my other
interest, visually observing of variable stars. While I'm beginner
with the Canon SD1000, I've been observing variable stars visually
many years ago but haven't done much until a few weeks ago. It's as
if I went to sleep in the days of film and woke up in the era CCDs.
I'm playing catch-up now. I used to use a tripod mounted equitorail
drive with a SLR. I took many photos of Halley's comet back in 1986
(?).

What I would like to buy is a digital SLR that allows me to take
exposures of 30 sec to a minute of wide field portions of the sky.

Larry

Date: 09 Jun 2007 15:57:30
From: actuary@mchsi.com
Subject: Re: A beginner question

On Jun 9, 5:37 pm, Editor <c...@yahoo.edu > wrote:
> sheesh! try it and tell us! assume you already did and arent saying
> so you posted here to see if you could after already trying so you
> want to try agin and get some tips? Got any recipes for broiled buffalo?
>
>
>
> "actu...@mchsi.com" wrote:
> > To All:
>
> > I reently purchased a Canon PowerShot SD1000 for strictly home use. I
> > was out the other night trying to locate Vesta and I did but then I
> > wondered if I could capture Vesta (magnitude 5.8) on a 2 second
> > exposure (maximum exposure time) using the Canon 1000SD camera.
>
> > I don't have a tripod so I didn't try. What is the applicable
> > mathematics to answer the question?
>
> > Thanks.
>
> > Larry- Hide quoted text -
>
> - Show quoted text -

No, I haven't tried it. I just got the camera yesterday but I will
try it tonight.

Larry

Date: 11 Jun 2007 02:42:17
From: LaFouine
Subject: Re: A beginner question

On Sat, 09 Jun 2007 15:57:30 -0700, "actuary@mchsi.com"
<actuary@mchsi.com > wrote:

>On Jun 9, 5:37 pm, Editor <c...@yahoo.edu> wrote:
>> sheesh! try it and tell us! assume you already did and arent saying
>> so you posted here to see if you could after already trying so you
>> want to try agin and get some tips? Got any recipes for broiled buffalo?
>>
>>
>>
>> "actu...@mchsi.com" wrote:
>> > To All:
>>
>> > I reently purchased a Canon PowerShot SD1000 for strictly home use. I
>> > was out the other night trying to locate Vesta and I did but then I
>> > wondered if I could capture Vesta (magnitude 5.8) on a 2 second
>> > exposure (maximum exposure time) using the Canon 1000SD camera.
>>
>> > I don't have a tripod so I didn't try. What is the applicable
>> > mathematics to answer the question?
>>
>> > Thanks.
>>
>> > Larry- Hide quoted text -
>>
>> - Show quoted text -
>
>No, I haven't tried it. I just got the camera yesterday but I will
>try it tonight.
>
>Larry

Hi,

It should work
Here http://img46.imageshack.us/my.php?image=m45iris2006112426pf3.gif
An animation of 2 images of Iris (Mag 7.0, in the right corner)
i've done it in Nov 2006.

Canon Powershot G5 on tripod
15 x 5s shots in Raw
add/registered with Iris
http://www.astrosurf.com/buil/us/iris/iris.htm

The most difficult was to point at it and find
the right time exposure to compensate the Earth rotation with
empiric method ;o)

I try to shoot Vesta too, but weather and
Light pollution in low horizon where vista is
are not so good conditions here.
ex 23-05-2007 :
http://img181.imageshack.us/my.php?image=shot0001wt5.jpg

@+

Date: 09 Jun 2007 15:56:29
From: actuary@mchsi.com
Subject: Re: A beginner question

On Jun 9, 4:19 pm, Chris L Peterson <c...@alumni.caltech.edu > wrote:
> On Sat, 09 Jun 2007 12:56:18 -0700, "actu...@mchsi.com"
>
> <actu...@mchsi.com> wrote:
> >I reently purchased a Canon PowerShot SD1000 for strictly home use. I
> >was out the other night trying to locate Vesta and I did but then I
> >wondered if I could capture Vesta (magnitude 5.8) on a 2 second
> >exposure (maximum exposure time) using the Canon 1000SD camera.
>
> > I don't have a tripod so I didn't try. What is the applicable
> >mathematics to answer the question?
>
> I assume you mean using just the lens in the camera? Then yes, it might
> just barely be possible. You need to consider the number of photons
> received. In the case of a mag 5.8 object, and your ~3mm aperture,
> that's about 1000 photons per second. Between atmospheric losses, and
> issues of camera sensitivity, you might actually record about 25% of
> these, or 250 photons (500 in 2 seconds). That would give you an ideal
> S/N of about 22, which is more than good enough to show a visible dot.
> However, there are other noise sources not so easy to calculate, such as
> dark current noise, readout noise, and sky noise. Considering all these,
> there's a good chance that your 500 photons of signal might be dropping
> down into the mud.
>
> BTW, doesn't the SD1000 actually have a 15 second maximum exposure time?
> That should be long enough to easily get a mag 5.8 object, even with the
> increase in dark current noise.
>
> _________________________________________________
>
> Chris L Peterson
> Cloudbait Observatoryhttp://www.cloudbait.com

Thanks. I just discovered how to set the exposure time to 15
seconds. I borrowed a tripod and the sky is clear in Central
Illinois. I'll give it a shot tonight.

Larry

Date: 09 Jun 2007 17:46:05
From: lal_truckee
Subject: Re: A beginner question

actuary@mchsi.com wrote:
>
> Thanks. I just discovered how to set the exposure time to 15
> seconds. I borrowed a tripod and the sky is clear in Central
> Illinois. I'll give it a shot tonight.

The earth moves (AKA the sky rotates) about 15deg/hour, so 1/4 deg per
minute, so 1/16 deg per 15 sec exposure. If you get an image it will be
a little blurred from rotation of the heavens, but it'll be fun anyway,
so have at it.

If you're getting images, the next step is a barn door tracker. Simple,
cheap, and plans are ubiquitous on the web. The cheapest version amounts
to two planks, a couple of hinges, a t-nut, a threaded rod, and some
means of turning the rod (by hand works fine - put a knob on the rod.)

Date: 16 Jun 2007 01:01:06
From: Eric
Subject: Re: A beginner question

lal_truckee wrote:

> actuary@mchsi.com wrote:
>>
>> Thanks. I just discovered how to set the exposure time to 15
>> seconds. I borrowed a tripod and the sky is clear in Central
>> Illinois. I'll give it a shot tonight.
>
> The earth moves (AKA the sky rotates) about 15deg/hour, so 1/4 deg per
> minute, so 1/16 deg per 15 sec exposure. If you get an image it will be
> a little blurred from rotation of the heavens, but it'll be fun anyway,
> so have at it.
>
> If you're getting images, the next step is a barn door tracker. Simple,
> cheap, and plans are ubiquitous on the web. The cheapest version amounts
> to two planks, a couple of hinges, a t-nut, a threaded rod, and some
> means of turning the rod (by hand works fine - put a knob on the rod.)

I wonder if this "barn door tracker" could be adapted to swing a large
dob...
Eric

Date: 09 Jun 2007 15:48:55
From: RMOLLISE
Subject: Re: A beginner question

On Jun 9, 5:37 pm, Editor <c...@yahoo.edu > wrote:
> sheesh! try it and tell us! assume you already did and arent saying
> so you posted here to see if you could after already trying so you
> want to try agin and get some tips? Got any recipes for broiled buffalo?

>
> - Show quoted text -

One of the purposes of s.a.a. has _always_ been to answer beginners'
questions. Astrophotography is hard enough as it is, and most newbies
will need all the advice they can get. If I can help even one person
over the bumps on the road to taking images I dern sure am gonna do
it. Getting a roll (or SD card) full of blank images doesn't do much
for the learning curve or the self-confidence.

Why do you have a problem with them asking for it?! What did you
think you'd see in an article titled "Begginer's Question"? "Sheesh"
is right. ;-)

Unk Rod

Date: 09 Jun 2007 17:37:52
From: Editor
Subject: Re: A beginner question

sheesh! try it and tell us! assume you already did and arent saying
so you posted here to see if you could after already trying so you
want to try agin and get some tips? Got any recipes for broiled buffalo?

"actuary@mchsi.com" wrote:

> To All:
>
> I reently purchased a Canon PowerShot SD1000 for strictly home use. I
> was out the other night trying to locate Vesta and I did but then I
> wondered if I could capture Vesta (magnitude 5.8) on a 2 second
> exposure (maximum exposure time) using the Canon 1000SD camera.
>
> I don't have a tripod so I didn't try. What is the applicable
> mathematics to answer the question?
>
> Thanks.
>
> Larry

Date: 09 Jun 2007 21:19:52
From: Chris L Peterson
Subject: Re: A beginner question

On Sat, 09 Jun 2007 12:56:18 -0700, "actuary@mchsi.com"
<actuary@mchsi.com > wrote:

>I reently purchased a Canon PowerShot SD1000 for strictly home use. I
>was out the other night trying to locate Vesta and I did but then I
>wondered if I could capture Vesta (magnitude 5.8) on a 2 second
>exposure (maximum exposure time) using the Canon 1000SD camera.
>
> I don't have a tripod so I didn't try. What is the applicable
>mathematics to answer the question?

I assume you mean using just the lens in the camera? Then yes, it might
just barely be possible. You need to consider the number of photons
received. In the case of a mag 5.8 object, and your ~3mm aperture,
that's about 1000 photons per second. Between atmospheric losses, and
issues of camera sensitivity, you might actually record about 25% of
these, or 250 photons (500 in 2 seconds). That would give you an ideal
S/N of about 22, which is more than good enough to show a visible dot.
However, there are other noise sources not so easy to calculate, such as
dark current noise, readout noise, and sky noise. Considering all these,
there's a good chance that your 500 photons of signal might be dropping
down into the mud.

BTW, doesn't the SD1000 actually have a 15 second maximum exposure time?
That should be long enough to easily get a mag 5.8 object, even with the
increase in dark current noise.

_________________________________________________

Chris L Peterson
Cloudbait Observatory
http://www.cloudbait.com

Date: 11 Jun 2007 01:13:18
From: Peter Webb
Subject: Re: A beginner question

"Chris L Peterson" <clp@alumni.caltech.edu > wrote in message
news:2u3m63t0fjdp69llarrpoml092s9jcgr5e@4ax.com...
> On Sat, 09 Jun 2007 12:56:18 -0700, "actuary@mchsi.com"
> <actuary@mchsi.com> wrote:
>
>>I reently purchased a Canon PowerShot SD1000 for strictly home use. I
>>was out the other night trying to locate Vesta and I did but then I
>>wondered if I could capture Vesta (magnitude 5.8) on a 2 second
>>exposure (maximum exposure time) using the Canon 1000SD camera.
>>
>> I don't have a tripod so I didn't try. What is the applicable
>>mathematics to answer the question?
>
> I assume you mean using just the lens in the camera? Then yes, it might
> just barely be possible. You need to consider the number of photons
> received. In the case of a mag 5.8 object, and your ~3mm aperture,
> that's about 1000 photons per second. Between atmospheric losses, and
> issues of camera sensitivity, you might actually record about 25% of
> these, or 250 photons (500 in 2 seconds). That would give you an ideal
> S/N of about 22, which is more than good enough to show a visible dot.
> However, there are other noise sources not so easy to calculate, such as
> dark current noise, readout noise, and sky noise. Considering all these,
> there's a good chance that your 500 photons of signal might be dropping
> down into the mud.
>
> BTW, doesn't the SD1000 actually have a 15 second maximum exposure time?
> That should be long enough to easily get a mag 5.8 object, even with the
> increase in dark current noise.
>
> _________________________________________________
>
> Chris L Peterson
> Cloudbait Observatory
> http://www.cloudbait.com

For a "Beginner question" its a remarkably good one.

Whilst I have not taken astronomical photographs, I have a pretty good
understanding of optics. So I was surprised to find that I could find no way
of calculating exposure times of astronomical objects (other than the moon)
by comparison to terrestrial ISOs, f-stops etc. I note above that this is
not the route you took to your solution. Can it be done that way?

Anyway, that's a useful computation, and I now know that a 3mm aperture
aperture will see the same objects as a human observer with a 2 second
exposure. Did you assume an equivalent ISO rating for the CCD, or are you
just assuming it is cranked up to the maximum in the calculation?

Anyway, a good question and a good answer - thanks.

Date: 10 Jun 2007 23:19:43
From: Chris L Peterson
Subject: Re: A beginner question

On Mon, 11 Jun 2007 01:13:18 +1000, "Peter Webb"
<webbfamily@DIESPAMDIEoptusnet.com.au > wrote:

>Whilst I have not taken astronomical photographs, I have a pretty good
>understanding of optics. So I was surprised to find that I could find no way
>of calculating exposure times of astronomical objects (other than the moon)
>by comparison to terrestrial ISOs, f-stops etc. I note above that this is
>not the route you took to your solution. Can it be done that way?

For the most part, no. The ISO setting on a digital camera is at best
loosely related to the concept of film ISO values. In fact, using high
ISO values with dim astronomical targets is rarely advantageous. That's
because the important components of noise are amplified at the same rate
as the signal. There can be some advantage to shooting at high ISO if
you save to JPEG, but if you save to RAW (which you should always do
with astronomical objects if your camera supports it) there is almost no
difference between high and low ISO settings. Or to put it a little
differently, you can shoot at low ISO and multiply all the pixel values
by 2, 4, 8 etc in your image processing program, and get the same result
as if you had shot at high ISO. That's _not_ the case with film.

You can consider the effect of f-stops to be quite similar to ordinary
terrestrial photography or imaging (assuming you are using the camera's
lens). Each stop doubles the time required to get the same amount of
signal, and more than doubles the time required to get the same S/N.

>Anyway, that's a useful computation, and I now know that a 3mm aperture
>aperture will see the same objects as a human observer with a 2 second
>exposure. Did you assume an equivalent ISO rating for the CCD, or are you
>just assuming it is cranked up to the maximum in the calculation?

As noted, the ISO is pretty meaningless. The basic S/N is determined by
the number of photons collected, and that's the same regardless of how
much amplification is applied between the sensor and the A/D (which is
what the ISO setting controls). A few cameras apply the gain inside the
sensor, which means you do get a slight edge on readout noise (which in
that case is constant regardless of ISO setting). But usually, for
longer exposures, readout noise isn't the major noise source.

The situation is further confused by the sort of image processing that
goes on inside the camera. Nearly all cameras (all Canon's for sure)
subtract a computed dark frame, really a scaled bias, from the image.
They do this even when shooting in raw mode. That changes the noise
characteristics. Also, many cameras (but not Canon's in raw mode) apply
filters to the data to make noise less visible, even though doing this
also reduces the resolution of the image.

_________________________________________________

Chris L Peterson
Cloudbait Observatory
http://www.cloudbait.com

Date: 11 Jun 2007 21:06:25
From: Peter Webb
Subject: Re: A beginner question

"Chris L Peterson" <clp@alumni.caltech.edu > wrote in message
news:u01p63leoukjm0m7rsmdq6fb051md0fb5n@4ax.com...
> On Mon, 11 Jun 2007 01:13:18 +1000, "Peter Webb"
> <webbfamily@DIESPAMDIEoptusnet.com.au> wrote:
>
>>Whilst I have not taken astronomical photographs, I have a pretty good
>>understanding of optics. So I was surprised to find that I could find no
>>way
>>of calculating exposure times of astronomical objects (other than the
>>moon)
>>by comparison to terrestrial ISOs, f-stops etc. I note above that this is
>>not the route you took to your solution. Can it be done that way?
>
> For the most part, no. The ISO setting on a digital camera is at best
> loosely related to the concept of film ISO values. In fact, using high
> ISO values with dim astronomical targets is rarely advantageous. That's
> because the important components of noise are amplified at the same rate
> as the signal.

Essentially then digital ISOs are analogous to push-processing on film. To
somewhat extend the analogy, its like a "digital zoom" - no extra
information is actually added.

> There can be some advantage to shooting at high ISO if
> you save to JPEG, but if you save to RAW (which you should always do
> with astronomical objects if your camera supports it) there is almost no
> difference between high and low ISO settings. Or to put it a little
> differently, you can shoot at low ISO and multiply all the pixel values
> by 2, 4, 8 etc in your image processing program, and get the same result
> as if you had shot at high ISO. That's _not_ the case with film.
>
> You can consider the effect of f-stops to be quite similar to ordinary
> terrestrial photography or imaging (assuming you are using the camera's
> lens). Each stop doubles the time required to get the same amount of
> signal, and more than doubles the time required to get the same S/N.
>

Sure. My issue was that I had two different "scales" in use in my head, and
now way to calibrate between them. I know its much the same principles for
astro and terrestrial photography (of course, no depth of field issues,
point sources only, some others), but I had no way "in my head" of equating
the two scales. When I looked at the question concerning a magnitude 5.8
object on a terrestrial camera, I had no scale to equate this to daytime
photography. Unlike, for example, photographing the moon which is directly
illuminated by the Sun. I don't even know how to work out the exposure etc
for photographing the Sun directly (and nor would I try!); if I did, I could
pull out an inverse square law and work out what would be needed for the Sun
as a distant point source to be "resolved".

>
>>Anyway, that's a useful computation, and I now know that a 3mm aperture
>>aperture will see the same objects as a human observer with a 2 second
>>exposure. Did you assume an equivalent ISO rating for the CCD, or are you
>>just assuming it is cranked up to the maximum in the calculation?
>
> As noted, the ISO is pretty meaningless. The basic S/N is determined by
> the number of photons collected, and that's the same regardless of how
> much amplification is applied between the sensor and the A/D (which is
> what the ISO setting controls). A few cameras apply the gain inside the
> sensor, which means you do get a slight edge on readout noise (which in
> that case is constant regardless of ISO setting). But usually, for
> longer exposures, readout noise isn't the major noise source.
>
> The situation is further confused by the sort of image processing that
> goes on inside the camera. Nearly all cameras (all Canon's for sure)
> subtract a computed dark frame, really a scaled bias, from the image.
> They do this even when shooting in raw mode. That changes the noise
> characteristics. Also, many cameras (but not Canon's in raw mode) apply
> filters to the data to make noise less visible, even though doing this
> also reduces the resolution of the image.

Presumably the camera optics are not much better than are needed to resolve
to physical pixel size on the CCD array (what would be the point)? So
presumably also the point source can easily spill over into adjacent pixels.
Most digital cameras will let you reduce resolution - I wonder whether this
improves sensitivity (as it does for film grain size). This is presumably
whether they sum the baseline data from adjacent pixels to produce the lower
resolution image. If they do, halving the total image size should provide a
massive 1.5 dB gain.

Of course, this can as easily be done in post-processing, I guess?

Finally, and thankyou for your answers so far, a somewhat more technical
one.

I am something of an expert on digital imaging as applied to business
documents. I therefore know that JPEG compression produces bad artefacts on
sharp edges ("ringing", effectively the digital equivalent of diffraction
lines). Not a good look.

Is the intent to capture in each JPEG frame the "dark background" in
sufficient resolution to add multiple images and decrease noise by summing?
So is the JPEG really better considered as a map of the background noise
rather than the foreground point sources, and the digital artefacts don't
matter as much?

And if the JPEG contains the data to exactly reconstruct the original
bitmap, then its hard to see how it could be much smaller than the bitmap.
You have got to be chucking information away somewhere.

I assume that RAW is just that, the pixel values. Every camera I have owned
in the last five years has it, wouldn't that be heaps better? (Although I
have never used it, and until recently was prohibitively expensive in
storage).

Finally, and again, thanks for your input thus far. This is a fascinating
area - the interplay between optics and information theory.

Date: 11 Jun 2007 16:00:35
From: Chris L Peterson
Subject: Re: A beginner question

On Mon, 11 Jun 2007 21:06:25 +1000, "Peter Webb"
<webbfamily@DIESPAMDIEoptusnet.com.au > wrote:

>Essentially then digital ISOs are analogous to push-processing on film. To
>somewhat extend the analogy, its like a "digital zoom" - no extra
>information is actually added.

I don't think the analogy is precise, but I'd agree that digital ISO is
more like push processing than native film ISO. However, as I noted
earlier, if the gain is applied in the sensor, before the readout stage,
it's possible to see a slight improvement in S/N because the readout
noise contribution isn't amplified (or isn't fully amplified).

>Sure. My issue was that I had two different "scales" in use in my head, and
>now way to calibrate between them. I know its much the same principles for
>astro and terrestrial photography (of course, no depth of field issues,
>point sources only, some others), but I had no way "in my head" of equating
>the two scales. When I looked at the question concerning a magnitude 5.8
>object on a terrestrial camera, I had no scale to equate this to daytime
>photography. Unlike, for example, photographing the moon which is directly
>illuminated by the Sun. I don't even know how to work out the exposure etc
>for photographing the Sun directly (and nor would I try!); if I did, I could
>pull out an inverse square law and work out what would be needed for the Sun
>as a distant point source to be "resolved".

Yeah, I don't think there's any way to really equate the calculation of
exposure for terrestrial imaging with the calculation for astronomical
imaging.

>Presumably the camera optics are not much better than are needed to resolve
>to physical pixel size on the CCD array (what would be the point)? So
>presumably also the point source can easily spill over into adjacent pixels.
>Most digital cameras will let you reduce resolution - I wonder whether this
>improves sensitivity (as it does for film grain size). This is presumably
>whether they sum the baseline data from adjacent pixels to produce the lower
>resolution image. If they do, halving the total image size should provide a
>massive 1.5 dB gain.
>
>Of course, this can as easily be done in post-processing, I guess?

There is only one noise source that is pixel dependent, and that's
readout noise. For short exposures (typically less than a few seconds
with ordinary digital cameras, and up to many minutes with cooled
cameras under dark skies) the readout noise is dominant, so there is a
value to having less pixels involved. However, this advantage is only
realized when the camera is capable of binning the pixels within the
sensor, so a single readout operation is involved. If you reduce the
resolution after the sensor is read, either in the camera or during post
processing, there is no reduced noise.

For exposures long enough that readout noise is fairly insignificant, it
makes no difference how many pixels are involved.

>Is the intent to capture in each JPEG frame the "dark background" in
>sufficient resolution to add multiple images and decrease noise by summing?
>So is the JPEG really better considered as a map of the background noise
>rather than the foreground point sources, and the digital artefacts don't
>matter as much?
>
>And if the JPEG contains the data to exactly reconstruct the original
>bitmap, then its hard to see how it could be much smaller than the bitmap.
>You have got to be chucking information away somewhere.

Well, a lossless JPEG compression (which isn't an option with any camera
I've seen, although the standard supports it) could still be highly
compressed with respect to a bitmap for many astronomical images, given
that much of the background may be essentially zero. Naturally, if you
fully encode the noise, the JPEG will have to be the same size as the
bitmap.

>I assume that RAW is just that, the pixel values. Every camera I have owned
>in the last five years has it, wouldn't that be heaps better? (Although I
>have never used it, and until recently was prohibitively expensive in
>storage).

That's what you'd like RAW to be, although with most cameras there's
still a certain degree of processing that has occurred before the raw
image is saved. But it's close enough to raw pixel values in most cases
that typical image processing operations such as stacking can be
performed. Besides the lack of compression artifacts, what's very
important with RAW is that the original dynamic range of the sensor is
largely preserved. The data will be 10-12 bits deep (60dB-72dB),
compared with at best 8 bits (48dB) with JPEG encoding.

_________________________________________________

Chris L Peterson
Cloudbait Observatory
http://www.cloudbait.com

Date: 10 Jun 2007 15:45:49
From: William Hamblen
Subject: Re: A beginner question

Date: 09 Jun 2007 14:11:55
From: RMOLLISE
Subject: Re: A beginner question

On Jun 9, 2:56 pm, "actu...@mchsi.com" <actu...@mchsi.com > wrote:
> To All:
>
> I reently purchased a Canon PowerShot SD1000 for strictly home use. I
> was out the other night trying to locate Vesta and I did but then I
> wondered if I could capture Vesta (magnitude 5.8) on a 2 second
> exposure (maximum exposure time) using the Canon 1000SD camera.
>
> I don't have a tripod so I didn't try. What is the applicable
> mathematics to answer the question?
>
> Thanks.
>
> Larry

Hi:

Two seconds won't be enough. About 15 - 30 seconds might do it, but
you'll need a tripod or other suppot for the camera (even at 2
seconds), and you'll need to crank your ISO up as high as it will go
(400 or 800 or 1600) and you'll want the slowest shutter speed and
widest f/strop your camera can deliver.

Unk Rod