Anybody that thinks the D70 is not *fast*..

Bob wrote:

On Tue, 13 Jul 2004 17:49:13 -0700, Crownfield wrote:

leo wrote:

"Crownfield" wrote in message
...
leo wrote:

"Crownfield" wrote in message
...
leo wrote:

Hummingbird is fast, about 50 - 80 beats a second. Many P&S can
manage
at
least 1/1000 of a second shutter so what're you talking about?

thinking aloud...
at 80 bps amd 60mm motion, that is almost 5,000 mm per second.
1/1000 flash will give about 5mm motion during the exposure.
focal plane shuters can not do it.
hummers take about 1/50,000 flash to really freeze their wingtips.

Alright, maybe 1/1000 is a bit slow. The D70 can do 1/8000 or 1/500 with
flash at best, so your 1/50,000 figure is a bit off.

and...

1: the wingtips were blurred.
2:did you notice that they were flash pictures?


a flash, at low power, is usually very short duration.
if the flash was 1/10,000, and the shutter was 1/2 second,
what effect would the shutter have on the exposure?

um? I am talking the D70 can do at optimal 1/8000 sec.

um? you do know that at 1/8000th,
the whole image is not being exposed at the same time?

check up on the traveling of focal plane curtains,
and the time the first shutter opens
and the time that the second shutter starts to close.


Perhaps under ideal
condition, without the need of flash, that the motion can be stopped. At
1/500. It's damn good, just a tiny bit blur, so I don't think it needs
1/50,000 of a sec as you calculated.

you are blithely assuming that the 1/500 was the controlling variable,
and the secret to the picture. think about the picture,
the numbers, and the problem.

again,
thinking aloud...
at 80 bps amd 60mm motion, that is almost 5,000 mm per second.
1/1000 flash will give about 5mm motion during the exposure.
focal plane shuters can not do it.
hummers take about 1/50,000 flash to really freeze their wingtips.


if the shutter was 1/500, and the flash was 1/10,000
and was 24 inches away from the subject...

As far as I know, the max flash sync shutter speed is determined by the max
speed that the shutter is fully open, right? So in the D70, one curtain is open
and the other has not started up until you demand faster then 1/500 sec, then
the shutters become a traveling slit.

perfect!

i guess that at 1/5,000th of a second,
the slit width is about 1/10 of the frame.

I would also imagine that the 1/8000 limit
of the camera has to do with the max speed the camera can scan the sensor into
memory. We have to remember that at 1/500 or faster, the shutters move at
constant speed, only the slit-width-duration changes. Therefore the process
speed of the camera has to do with the minimum size of the shutter slit opening
as well.

the process time available is at least 1/500th of a second.
that is determined by the minimum shutter curtain travel time. (~1/500)

think of the faster time / smaller slit
as painting with a thinner brush (slit width).
the time is still the 1/500th, the curtain travel time.

Too much math for me!!

The CMOS chips process each pixel inside each cell. Is it any faster than
CCD?

Bob wrote:

On Tue, 13 Jul 2004 04:14:20 GMT, Gary Eickmeier
wrote:



leo wrote:
"Bob" wrote in message
news

Sure, but the camera has to scan it's sensor into memory at least as fast!



Hummingbird is fast, about 50 - 80 beats a second. Many P&S can manage at
least 1/1000 of a second shutter so what're you talking about?

It's just a test of person's skill, long lens and external flash (if needed)
of SLR over compact cameras.

He is thinking that the camera has to scan in the info from the CCD just
as fast as the flash duration, but I don't think that is right.

Gary Eickmeier


The sensor has to be scanned before the image is gone,

how long does the sensor hold the image before it decays?


so you would need to know
the persistance level of the sensor as well as the light duration. If the
persistance is too long you wouldn't be able to capture fast action or move the
camera, since you would get blurring as in a 1960s video camera!

You would also have to know how the sensor is scanned in what parallel mode,
since I discussed this a while back with other techs and we determined that it
isn't possible to series scan a 5m sensor in 1/8000 sec with present day
computers!

I imagine you could test this by spinning around and taking a strobe picture in
the dark with the strobe set to a known high speed.

A CCD responds to light by creating an electron-hole pair
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Sure, but the camera has to scan it's sensor into memory at least as fast!

A CCD detects light (a photon) by creating an electron-hole pair in the
silicon. The pair will remain there as long as the voltage reamins on
the CCD; exposures of many hours can be taken with a CCD. The readout
of the device consists of transferring the charge in each pixel to a
final output well, where it is measured. The speed with which this is
done is up to the designer. Astronomical CCDs can take as long a minute
to read out a single image, using this slow speed to minimize noise per
pixel; the noise can be under two electrons per pixel. In digicams, the
CCDs are generally designed so that the charge is first trasferred to
another area of the chip, where it is then read out, but this doesn't
chage the principle.

Now to the case of photographing a hummingbird's wings. The exposure
must be short. This can be done with a short shutter speed or with a
strobe/flash. Suppose it is 1/1000 of a second. In the short interval,
all of photons that will be detected are detected, and the CCD now has
its pixels with varying amounts of charge per pixel. How long it takes
to read it out is (almost) irrelevant. That 1/1000 sec exposure is
frozen in the CCD. You could spend 1/1000 second or one minute reading
it out (in principle- digicams can't really do this). The "almost"
connected with readout time above is there for several reasons. One,
digital camera CCDs have signifacant dark current, and if you wait too
long to read out your image, it will be swamped by the dark current.
This holds for any image, not just short exposure images. Second, if
you try to read out the CCD too fast, the noise will go up. Again, this
has nothing to do with exposure length.

In short, that fact that a hummingbird's wings can be frozen in a
digital camera image only means that a short exposure was taken. You
could do this with a large number of cameras. It has absolutely nothing
to do with the readout time of the CCD.

Sure, but the camera has to scan it's sensor into memory at least as fast!

A CCD detects light (a photon) by creating an electron-hole pair in the
silicon. The pair will remain there as long as the voltage reamins on
the CCD; exposures of many hours can be taken with a CCD. The readout
of the device consists of transferring the charge in each pixel to a
final output well, where it is measured. The speed with which this is
done is up to the designer. Astronomical CCDs can take as long a minute
to read out a single image, using this slow speed to minimize noise per
pixel; the noise can be under two electrons per pixel. In digicams, the
CCDs are generally designed so that the charge is first trasferred to
another area of the chip, where it is then read out, but this doesn't
chage the principle.

Now to the case of photographing a hummingbird's wings. The exposure
must be short. This can be done with a short shutter speed or with a
strobe/flash. Suppose it is 1/1000 of a second. In the short interval,
all of photons that will be detected are detected, and the CCD now has
its pixels with varying amounts of charge per pixel. How long it takes
to read it out is (almost) irrelevant. That 1/1000 sec exposure is
frozen in the CCD. You could spend 1/1000 second or one minute reading
it out (in principle- digicams can't really do this). The "almost"
connected with readout time above is there for several reasons. One,
digital camera CCDs have signifacant dark current, and if you wait too
long to read out your image, it will be swamped by the dark current.
This holds for any image, not just short exposure images. Second, if
you try to read out the CCD too fast, the noise will go up. Again, this
has nothing to do with exposure length.

In short, that fact that a hummingbird's wings can be frozen in a
digital camera image only means that a short exposure was taken. You
could do this with a large number of cameras. It has absolutely nothing
to do with the readout time of the CCD.

On Tue, 13 Jul 2004 20:09:59 -0700, Crownfield wrote:


The sensor has to be scanned before the image is gone,

how long does the sensor hold the image before it decays?


I suppose thats the important thing to know - and does the camera erase it to
make the next picture or does it just fade away?

I don't know about differances between the DSLRs and the P&Ss but the small cams
can do video so that means the sensor has to be clear in 1/60 second.

On Tue, 13 Jul 2004 20:09:59 -0700, Crownfield wrote:


The sensor has to be scanned before the image is gone,

how long does the sensor hold the image before it decays?


I suppose thats the important thing to know - and does the camera erase it to
make the next picture or does it just fade away?

I don't know about differances between the DSLRs and the P&Ss but the small cams
can do video so that means the sensor has to be clear in 1/60 second.

On Wed, 14 Jul 2004 12:57:29 -0700, Joseph Miller wrote:


Sure, but the camera has to scan it's sensor into memory at least as fast!

A CCD detects light (a photon) by creating an electron-hole pair in the
silicon. The pair will remain there as long as the voltage reamins on
the CCD; exposures of many hours can be taken with a CCD. The readout
of the device consists of transferring the charge in each pixel to a
final output well, where it is measured. The speed with which this is
done is up to the designer. Astronomical CCDs can take as long a minute
to read out a single image, using this slow speed to minimize noise per
pixel; the noise can be under two electrons per pixel. In digicams, the
CCDs are generally designed so that the charge is first trasferred to
another area of the chip, where it is then read out, but this doesn't
chage the principle.

Now to the case of photographing a hummingbird's wings. The exposure
must be short. This can be done with a short shutter speed or with a
strobe/flash. Suppose it is 1/1000 of a second. In the short interval,
all of photons that will be detected are detected, and the CCD now has
its pixels with varying amounts of charge per pixel. How long it takes
to read it out is (almost) irrelevant. That 1/1000 sec exposure is
frozen in the CCD. You could spend 1/1000 second or one minute reading
it out (in principle- digicams can't really do this). The "almost"
connected with readout time above is there for several reasons. One,
digital camera CCDs have signifacant dark current, and if you wait too
long to read out your image, it will be swamped by the dark current.
This holds for any image, not just short exposure images. Second, if
you try to read out the CCD too fast, the noise will go up. Again, this
has nothing to do with exposure length.

In short, that fact that a hummingbird's wings can be frozen in a
digital camera image only means that a short exposure was taken. You
could do this with a large number of cameras. It has absolutely nothing
to do with the readout time of the CCD.


Thanks for clearing that up!

I had already figured out that reading a 5m sensor in 1/8000 sec was kind of
hard to do with our processor speeds!

On Wed, 14 Jul 2004 12:57:29 -0700, Joseph Miller wrote:


Sure, but the camera has to scan it's sensor into memory at least as fast!

A CCD detects light (a photon) by creating an electron-hole pair in the
silicon. The pair will remain there as long as the voltage reamins on
the CCD; exposures of many hours can be taken with a CCD. The readout
of the device consists of transferring the charge in each pixel to a
final output well, where it is measured. The speed with which this is
done is up to the designer. Astronomical CCDs can take as long a minute
to read out a single image, using this slow speed to minimize noise per
pixel; the noise can be under two electrons per pixel. In digicams, the
CCDs are generally designed so that the charge is first trasferred to
another area of the chip, where it is then read out, but this doesn't
chage the principle.

Now to the case of photographing a hummingbird's wings. The exposure
must be short. This can be done with a short shutter speed or with a
strobe/flash. Suppose it is 1/1000 of a second. In the short interval,
all of photons that will be detected are detected, and the CCD now has
its pixels with varying amounts of charge per pixel. How long it takes
to read it out is (almost) irrelevant. That 1/1000 sec exposure is
frozen in the CCD. You could spend 1/1000 second or one minute reading
it out (in principle- digicams can't really do this). The "almost"
connected with readout time above is there for several reasons. One,
digital camera CCDs have signifacant dark current, and if you wait too
long to read out your image, it will be swamped by the dark current.
This holds for any image, not just short exposure images. Second, if
you try to read out the CCD too fast, the noise will go up. Again, this
has nothing to do with exposure length.

In short, that fact that a hummingbird's wings can be frozen in a
digital camera image only means that a short exposure was taken. You
could do this with a large number of cameras. It has absolutely nothing
to do with the readout time of the CCD.


Thanks for clearing that up!

I had already figured out that reading a 5m sensor in 1/8000 sec was kind of
hard to do with our processor speeds!