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Exposure time im space

 
 
ASAAR
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      07-02-2007
On Sun, 01 Jul 2007 22:14:07 -0600, Roger N. Clark (change username
to rnclark) wrote:

>>>> Just curious - assuming you were on a planet, Mars, Saturn, Venus,
>>>> Neptune etc., what exposure time would be necessary to take a photo
>>>> at ISO 100 and F4 or F8 ?
>>> Assuming you wanted to expose an 18% gray card, then
>>> the exposure would be the square of the relative distance
>>> of the planet from the sun compared to the earth.
>>> Don's answer is the only correct one I've seen so far in this
>>> thread.

>>
>> There's another correct answer, that there would be no difference
>> in exposure time. You of all people should know why. <g>

>
> With changing light levels you can do three things
> with a given camera:
> 1) change exposure time,
> 2) change the aperture, and
> 3) "change the ISO."


What I meant was that when Alfred asked :

> assuming you were on a planet, Mars, Saturn, Venus, Neptune etc.,
> what exposure time would be necessary to take a photo at ISO 100
> and F4 or F8 ?


I was considering that one option when on the planet (or actually,
in it, as you noted), the photographer might not want to take
pictures of the planet itself, or even take pictures illuminated by
the sun. The camera might be used to take pictures of other stars,
constellations, galaxies, etc., and so the exposure time, assuming
no local interference, should be about the same. Apologies if a
trick answer to a tricky question is not allowed.


>> http://opensourceschools.org/article...794&mode=print

>
> There is one mistake: 2 millibars. For that depth, it is probably
> like 2 megabars (a bar = 1 earth atmosphere in pressure).


I smelled a rat there, considering the immense size of those
planets, but I was more interested in light levels, temperature and
the extremely high winds. At any appreciable depth below the
uppermost edges of the atmosphere, the camera would probably quickly
vanish, and if not, even if lithium AA batteries were used, they
wouldn't be up to the job anyway.


> Is vims a 0.004 megapixel camera (64*64/1,000,000)
> or a 1.4 megapixel camera (64*64*352), or a 1/1,000,000
> (one millionth) megapixel camera? The foveon people
> would call it a 1.4 megapixel camera. I call it a
> 1/1,000,000 megapixel camera.


If there's any shimmer in the image, I'd call it either a floor
wax or a dessert topping.

 
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eclyma
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      07-02-2007
Thank You Gentlemen for a fun and informative thread and not the usual "who
has the biggest or the best".
ernie


 
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David J Taylor
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      07-02-2007
Roger N. Clark (change username to rnclark) wrote:
[]
> I forgot to answer the question in my last reply.
> Pluto is about 20 times
> further from the sun than the Earth, so exposures will be
> about 20*20 = 400 times longer than on Earth.
> So a 1/100 second Earth exposure becomes a 4 second
> exposure at Pluto.
>
> Roger


... and hence the need for a good tripod (or equivalent <G>).

David


 
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John Bean
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      07-02-2007
On Sun, 01 Jul 2007 21:09:03 -0600, "Roger N. Clark (change
username to rnclark)" <(E-Mail Removed)> wrote:

>John Bean wrote:
>> On Sun, 01 Jul 2007 14:04:51 -0600, "Roger N. Clark (change
>> username to rnclark)" <(E-Mail Removed)> wrote:

>
>>> and if
>>> it has an atmosphere, the absorption of sunlight through
>>> the atmosphere.

>>
>> So you think the planetary atmosphere wouldn't affect a 18%
>> grey card?

>
>That's why I said "absorption of sunlight through
> >> the atmosphere."


You didn't, Roger. You said:

"Assuming you wanted to expose an 18% gray card, then
the exposure would be the square of the relative distance
of the planet from the sun compared to the earth."

No mention of effect of atmosphere in the calculated
exposure, which is why I asked.

Since that wasn't what you meant it's a moot point.

--
John Bean
 
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AustinMN
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      07-02-2007
On Jul 1, 8:05 am, Don Stauffer in Minnesota <(E-Mail Removed)>
wrote:
> On Jul 1, 6:20 am, Alfred Molon <(E-Mail Removed)> wrote:
>
> > Just curious - assuming you were on a planet, Mars, Saturn, Venus,
> > Neptune etc., what exposure time would be necessary to take a photo at
> > ISO 100 and F4 or F8 ?
> > --

>
> > Alfred Molon
> > ------------------------------
> > Olympus 50X0, 7070, 8080, E3X0, E4X0 and E5X0 forum athttp://tech.groups.yahoo.com/group/MyOlympus/http://myolympus.org/photosharing site

>
> The light falls off as the square of the distance.


This is only true of a point source. At our distance (and Mars'
disance), the sun is *not* a point source.

Austin

> Find a chart of
> planetary distances in AUs (one AU is radius of earth orbit). Or,
> convert from miles (1 AU = 93,000,000 miles). Now, square the number
> of AUs, this is the factor the earth exposure time must be multiplied
> by.
>
> As an example, I believe mars is about 2 AU. Thus one must make an
> exposure four times what it would be on earth.


In reality, it requires a little less than one stop more exposure than
on Earth.

Austin

 
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jpc
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      07-02-2007
On Sun, 01 Jul 2007 14:04:51 -0600, "Roger N. Clark (change username
to rnclark)" <(E-Mail Removed)> wrote:


>JPC's idea of using infrared doesn't work. Sunlight peaks in the
>visible part of the spectrum. Going into the infrared, there
>are less photons. In the outer solar system, the bodies are colder
>than earth, so emit less thermal radiation than the earth,
>so exposure times are long in the infrared too.


I'll dispute that. Maybe my off the cuff estimate of three was a bit
optimistic, but according to my handy-dandy GE Radiation Calculator 50
% of the sun's radiation (watts per sq cm per micron) is in the
visible and UV with the rest in the IR. And the energy carried by a
photon drops linearly with increasing wavelength So at 2.4 microns
you need 4 photons to carry the same amount of energy 1 photon carries
at .6 microns--the sun's spectral peak. Since the number of photon per
micron drops far more slowly in the infrared than in the UV extending
the spectal range in that direction will increase the camera's
effective ISO.

And don't forget the night shots of of the glowing sulfur volcanos.
Fot that you'll need a MCT sensor

jpc



>


 
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Roger N. Clark (change username to rnclark)
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      07-03-2007
AustinMN wrote:
> On Jul 1, 8:05 am, Don Stauffer in Minnesota <(E-Mail Removed)>
> wrote:
>> On Jul 1, 6:20 am, Alfred Molon <(E-Mail Removed)> wrote:
>>
>>> Just curious - assuming you were on a planet, Mars, Saturn, Venus,
>>> Neptune etc., what exposure time would be necessary to take a photo at
>>> ISO 100 and F4 or F8 ?
>>> --
>>> Alfred Molon
>>> ------------------------------
>>> Olympus 50X0, 7070, 8080, E3X0, E4X0 and E5X0 forum athttp://tech.groups.yahoo.com/group/MyOlympus/http://myolympus.org/photosharing site

>> The light falls off as the square of the distance.

>
> This is only true of a point source. At our distance (and Mars'
> disance), the sun is *not* a point source.


While technically true, the deviation from an inverse square
rule is negligible and that deviation decreases as you move
out in the solar system. Just compute the fractional area of
the sun compared to a hemisphere. The Sun at the earth's
distance is a very small fraction of that hemisphere.
>
> Austin
>
>> Find a chart of
>> planetary distances in AUs (one AU is radius of earth orbit). Or,
>> convert from miles (1 AU = 93,000,000 miles). Now, square the number
>> of AUs, this is the factor the earth exposure time must be multiplied
>> by.
>>
>> As an example, I believe mars is about 2 AU. Thus one must make an
>> exposure four times what it would be on earth.

>
> In reality, it requires a little less than one stop more exposure than
> on Earth.


The mean distance of Mars from the sun is 1.52 times the
earth's distance. So the sunlight is 1.52*1.52 times
less intense = 1.2 stops fainter.

Roger
 
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Roger N. Clark (change username to rnclark)
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      07-03-2007
ASAAR wrote:

> I was considering that one option when on the planet (or actually,
> in it, as you noted), the photographer might not want to take
> pictures of the planet itself, or even take pictures illuminated by
> the sun. The camera might be used to take pictures of other stars,
> constellations, galaxies, etc., and so the exposure time, assuming
> no local interference, should be about the same. Apologies if a
> trick answer to a tricky question is not allowed.


Yes, I should have thought of that!

Roger
 
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Roger N. Clark (change username to rnclark)
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      07-03-2007
jpc wrote:
> On Sun, 01 Jul 2007 14:04:51 -0600, "Roger N. Clark (change username
> to rnclark)" <(E-Mail Removed)> wrote:
>
>> JPC's idea of using infrared doesn't work. Sunlight peaks in the
>> visible part of the spectrum. Going into the infrared, there
>> are less photons. In the outer solar system, the bodies are colder
>> than earth, so emit less thermal radiation than the earth,
>> so exposure times are long in the infrared too.

>
> I'll dispute that. Maybe my off the cuff estimate of three was a bit
> optimistic, but according to my handy-dandy GE Radiation Calculator 50
> % of the sun's radiation (watts per sq cm per micron) is in the
> visible and UV with the rest in the IR. And the energy carried by a
> photon drops linearly with increasing wavelength So at 2.4 microns
> you need 4 photons to carry the same amount of energy 1 photon carries
> at .6 microns--the sun's spectral peak. Since the number of photon per
> micron drops far more slowly in the infrared than in the UV extending
> the spectal range in that direction will increase the camera's
> effective ISO.
>
> And don't forget the night shots of of the glowing sulfur volcanos.
> Fot that you'll need a MCT sensor
>
> jpc


See Figure 1 at
http://speclab.cr.usgs.gov/PAPERS.calibration.tutorial
where the solar spectrum is shown. You'll see the solar
spectrum as seen on the earth's surface peaks near
0.6 microns (6000 angstroms) but the intensity is down
by more than 10x by 2 microns. That is a range of
about 0.4 to 2.5 microns, or just over 2 microns.

Now look at Figure 3b at:
http://speclab.cr.usgs.gov/PAPERS.refl-mrs/refl4.html
where thermal infrared radiation is shown peaking at
10 microns and distributed from 4 to beyond 20 microns.
That is a range of over 16 microns.

So near infrared solar radiation is much lower than
visible light, and thermal radiation is distributed
over such a large range (a larger range than most thermal
infrared detectors), that detectable signals are weaker.

Next look at detectors. Read noise from CCD and CMOS
sensors is a few electrons, and the devices have
high quantum efficiencies. Read noise in near and
thermal infrared detectors is on the order of hundreds
to thousands. Broader range thermal detectors tend to have
lower quantum efficiencies.

So you have lower signals, lower detection capabilities,
and higher read noise in the thermal or near-infrared.

So, at 2.4 microns, the solar response is down by a factor
of about 20, and each photon is 4 times lower energy
than 0.6 micron photons. Your losing by a factor
of ~4*20 ~ 80 plus your sensors aren't as good.

Roger
(I work in the infrared; I wish what you said was true;
it would make my work much easier .
 
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Roger N. Clark (change username to rnclark)
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      07-03-2007
John Bean wrote:
> On Sun, 01 Jul 2007 21:09:03 -0600, "Roger N. Clark (change
> username to rnclark)" <(E-Mail Removed)> wrote:
>
>> John Bean wrote:
>>> On Sun, 01 Jul 2007 14:04:51 -0600, "Roger N. Clark (change
>>> username to rnclark)" <(E-Mail Removed)> wrote:
>>>> and if
>>>> it has an atmosphere, the absorption of sunlight through
>>>> the atmosphere.
>>> So you think the planetary atmosphere wouldn't affect a 18%
>>> grey card?

>> That's why I said "absorption of sunlight through
>>>> the atmosphere."

>
> You didn't, Roger. You said:
>
> "Assuming you wanted to expose an 18% gray card, then
> the exposure would be the square of the relative distance
> of the planet from the sun compared to the earth."
>
> No mention of effect of atmosphere in the calculated
> exposure, which is why I asked.


You didn't read the next paragraph in the same post,
where I said:

"If you want to expose for the average scene brightness
on that planet (or moon), then you also need to take into
account the reflectivity of the planetary surface, and if
it has an atmosphere, the absorption of sunlight through
the atmosphere."

Please read the complete post before jumping to conclusions.

Roger
 
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