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Longest exposure possible for the moon without blurring?

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Anonymous
August 4, 2005 12:02:14 PM

Archived from groups: rec.photo.digital (More info?)

I was attempting to take some pictures of the moon yesterday, when it
was about 2% lit. It looked really cool, but I made the mistake of
taking long exposures, and the moons movement across the sky blurred
the image.

I was wondering what the longest exposure you could take of the moon
would be before the spinning of the earth moves the moon a noticible
amount and ruins the shot?


And, on a related note, what kind of exposure time you'd use for
getting some perseids meteors.


By the way, a great resource for tracking the moon is this:

http://aa.usno.navy.mil/data/docs/RS_OneDay.html
Anonymous
August 4, 2005 2:56:27 PM

Archived from groups: rec.photo.digital (More info?)

badchess wrote:

> I was attempting to take some pictures of the moon yesterday, when it
> was about 2% lit. It looked really cool, but I made the mistake of
> taking long exposures, and the moons movement across the sky blurred
> the image.
>
> I was wondering what the longest exposure you could take of the moon
> would be before the spinning of the earth moves the moon a noticible
> amount and ruins the shot?

It depends on your focal length.. You have less time
with longer lenses because the narrow field of view
exaggerates the movement.

I've managed clear shots of the new crescent as low as 1/8 second
using a 400mm lens at ISO 100
Anonymous
August 4, 2005 2:57:19 PM

Archived from groups: rec.photo.digital (More info?)

ASAAR wrote:
> On 4 Aug 2005 08:02:14 -0700, badchess wrote:
>
> > I was attempting to take some pictures of the moon yesterday, when it
> > was about 2% lit. It looked really cool, but I made the mistake of
> > taking long exposures, and the moons movement across the sky blurred
> > the image.
> >
> > I was wondering what the longest exposure you could take of the moon
> > would be before the spinning of the earth moves the moon a noticible
> > amount and ruins the shot?
>
> It depends on the amount of the sky that your lens sees in degrees
> (in other words, its focal length), the size of your camera's
> sensor, and the amount of blur (in pixels) you can tolerate. To
> give a very rough example in order for you to plug in more suitable
> values, you can assume that for a particular lens focal length, the
> amount of the sky that it covers is 10 degrees (out of 360). The
> number of seconds in a day is 60 * 60 * 24 (== 86,400). The number
> of seconds it would take for the moon to travel 10 degrees (using
> either the width or diagonal of the camera's sensor) would then be
> 86,400 * (10/360) == 2,400. Assuming a 3mp sensor that is 2000 x
> 1500 pixels, its diagonal would be equivalent to 2,500 pixels if the
> pixels had the same spacing on the diagonal. The moon would take
> 2,400 seconds to traverse 2,500 pixels, so if you'd tolerate a blur
> that's only two pixels wide, you'd be able to use an exposure of
> (2,400 * 2 / 2,500) == 1.92 seconds. If you'd prefer to use the
> sensor's width instead of its diagonal for the calculations, you'd
> use 2,000 instead of 2,500, which would increase the exposure to
> (2,400 * 2 / 2000) === 2.4 seconds.
>
> Using a camera with combinations of a higher resolution sensor,
> smaller angle (greater focal length) as well as tolerating blurs of
> no more than one pixel width would require shutter speeds of less
> than one second. This should be no problem with the portion of the
> moon that's lit by the sun's light. I guess that when you say that
> the moon "was about 2% lit", you're referring to the dark side of
> the moon that's illuminated by other sources, such as light
> reflected from the earth, etc., and that's what you want to capture,
> with necessarily longer shutter speeds?

Must be an engineer! :) 
Anonymous
August 4, 2005 3:51:45 PM

Archived from groups: rec.photo.digital (More info?)

Thanks for all the in depth and interesting answers.

I was using a EF 100-400mm f/4.5-5.6L IS USM zoomed all the way to 400
with a x2 extender on it attached to a canon 20D. I can't remember
what settings I used.

I'll try again next time I get the right combination of a sliver moon
rise/set at a evening/morning when it is not too light and I am awake,
but not at work...
Anonymous
August 4, 2005 4:43:22 PM

Archived from groups: rec.photo.digital (More info?)

On 4 Aug 2005 08:02:14 -0700, badchess wrote:

> I was attempting to take some pictures of the moon yesterday, when it
> was about 2% lit. It looked really cool, but I made the mistake of
> taking long exposures, and the moons movement across the sky blurred
> the image.
>
> I was wondering what the longest exposure you could take of the moon
> would be before the spinning of the earth moves the moon a noticible
> amount and ruins the shot?

It depends on the amount of the sky that your lens sees in degrees
(in other words, its focal length), the size of your camera's
sensor, and the amount of blur (in pixels) you can tolerate. To
give a very rough example in order for you to plug in more suitable
values, you can assume that for a particular lens focal length, the
amount of the sky that it covers is 10 degrees (out of 360). The
number of seconds in a day is 60 * 60 * 24 (== 86,400). The number
of seconds it would take for the moon to travel 10 degrees (using
either the width or diagonal of the camera's sensor) would then be
86,400 * (10/360) == 2,400. Assuming a 3mp sensor that is 2000 x
1500 pixels, its diagonal would be equivalent to 2,500 pixels if the
pixels had the same spacing on the diagonal. The moon would take
2,400 seconds to traverse 2,500 pixels, so if you'd tolerate a blur
that's only two pixels wide, you'd be able to use an exposure of
(2,400 * 2 / 2,500) == 1.92 seconds. If you'd prefer to use the
sensor's width instead of its diagonal for the calculations, you'd
use 2,000 instead of 2,500, which would increase the exposure to
(2,400 * 2 / 2000) === 2.4 seconds.

Using a camera with combinations of a higher resolution sensor,
smaller angle (greater focal length) as well as tolerating blurs of
no more than one pixel width would require shutter speeds of less
than one second. This should be no problem with the portion of the
moon that's lit by the sun's light. I guess that when you say that
the moon "was about 2% lit", you're referring to the dark side of
the moon that's illuminated by other sources, such as light
reflected from the earth, etc., and that's what you want to capture,
with necessarily longer shutter speeds?
Anonymous
August 4, 2005 7:11:24 PM

Archived from groups: rec.photo.digital (More info?)

salgud wrote:
> ASAAR wrote:
> > On 4 Aug 2005 08:02:14 -0700, badchess wrote:
> >
> > > I was attempting to take some pictures of the moon yesterday, when it
> > > was about 2% lit. It looked really cool, but I made the mistake of
> > > taking long exposures, and the moons movement across the sky blurred
> > > the image.
> > >
> > > I was wondering what the longest exposure you could take of the moon
> > > would be before the spinning of the earth moves the moon a noticible
> > > amount and ruins the shot?
> >
> > It depends on the amount of the sky that your lens sees in degrees
> > (in other words, its focal length), the size of your camera's
> > sensor, and the amount of blur (in pixels) you can tolerate. To
> > give a very rough example in order for you to plug in more suitable
> > values, you can assume that for a particular lens focal length, the
> > amount of the sky that it covers is 10 degrees (out of 360). The
> > number of seconds in a day is 60 * 60 * 24 (== 86,400). The number
> > of seconds it would take for the moon to travel 10 degrees (using
> > either the width or diagonal of the camera's sensor) would then be
> > 86,400 * (10/360) == 2,400. Assuming a 3mp sensor that is 2000 x
> > 1500 pixels, its diagonal would be equivalent to 2,500 pixels if the
> > pixels had the same spacing on the diagonal. The moon would take
> > 2,400 seconds to traverse 2,500 pixels, so if you'd tolerate a blur
> > that's only two pixels wide, you'd be able to use an exposure of
> > (2,400 * 2 / 2,500) == 1.92 seconds. If you'd prefer to use the
> > sensor's width instead of its diagonal for the calculations, you'd
> > use 2,000 instead of 2,500, which would increase the exposure to
> > (2,400 * 2 / 2000) === 2.4 seconds.
> >
> > Using a camera with combinations of a higher resolution sensor,
> > smaller angle (greater focal length) as well as tolerating blurs of
> > no more than one pixel width would require shutter speeds of less
> > than one second. This should be no problem with the portion of the
> > moon that's lit by the sun's light. I guess that when you say that
> > the moon "was about 2% lit", you're referring to the dark side of
> > the moon that's illuminated by other sources, such as light
> > reflected from the earth, etc., and that's what you want to capture,
> > with necessarily longer shutter speeds?
>
> Must be an engineer! :) 

And a good one too!

Alan
Anonymous
August 4, 2005 8:10:37 PM

Archived from groups: rec.photo.digital (More info?)

On 4 Aug 2005 10:57:19 -0700, salgud wrote:

> Must be an engineer! :) 

Not now, but I played one in school. :)  This is more the type of
problem that might be covered in an early physics course, though.
Anonymous
August 4, 2005 8:27:23 PM

Archived from groups: rec.photo.digital (More info?)

In article <1123164896.749285.121700@o13g2000cwo.googlegroups.com>,
badchess <badchess@msn.com> wrote:
>I was attempting to take some pictures of the moon yesterday, when it
>was about 2% lit. It looked really cool, but I made the mistake of
>taking long exposures, and the moons movement across the sky blurred
>the image.
>
>I was wondering what the longest exposure you could take of the moon
>would be before the spinning of the earth moves the moon a noticible
>amount and ruins the shot?

Depends on the field of view.

I have used a 400mm lens on a 10D to take photographs of a lunar eclipse. I
found that I generally needed to keep the exposure time down to around a
second or less. With a wide-angle lens, one could get away with more.

Having a really fast lens, and a camera with low noise at high ISO is going
to help a lot here.
Anonymous
August 4, 2005 9:59:38 PM

Archived from groups: rec.photo.digital (More info?)

"badchess" <badchess@msn.com> writes:

>I was wondering what the longest exposure you could take of the moon
>would be before the spinning of the earth moves the moon a noticible
>amount and ruins the shot?

It's pretty easy to calculate this for yourself.

The earth rotates 360 degrees in almost 24 hours. Divide one by the
other to determine how many degrees it moves per second. (4.2e-3)
(This ignores the difference between solar time and sidereal time, and
the moon's own rotation around the earth, but those have only a small
effect on the answer).

Now you want to know how much movement that translates to on the sensor.
The focal length of the lens times the tangent of the angle you
calculated above (in degrees/sec) gives you the image motion in mm/sec.
For example, if you're using a 500 mm lens, the motion is 0.036 mm/sec.

Then divide by the sensor pixel pitch in mm to convert the motion into
pixels/sec. Using a DSLR with a 8 um = 0.008 mm pixel pitch, this is
4.5 pixels/sec.

Decide how many pixels of blur you can tolerate, and you know your
maximum exposure time in seconds. If you want at most 2 pixels of blur,
your exposure needs to be less than 1/2 sec - with this particular lens
focal length.

Dave
Anonymous
August 5, 2005 2:49:33 PM

Archived from groups: rec.photo.digital (More info?)

There is no dark side of the moon, really
as a matter of fact it's all dark...
Anonymous
August 5, 2005 8:12:56 PM

Archived from groups: rec.photo.digital (More info?)

In article <1123164896.749285.121700@o13g2000cwo.googlegroups.com>,
badchess <badchess@msn.com> writes
>I was attempting to take some pictures of the moon yesterday, when it
>was about 2% lit. It looked really cool, but I made the mistake of
>taking long exposures, and the moons movement across the sky blurred
>the image.
>
>I was wondering what the longest exposure you could take of the moon
>would be before the spinning of the earth moves the moon a noticible
>amount and ruins the shot?
>
>
>And, on a related note, what kind of exposure time you'd use for
>getting some perseids meteors.
>
>
>By the way, a great resource for tracking the moon is this:
>
>http://aa.usno.navy.mil/data/docs/RS_OneDay.html
>
I won't repeat the excellent stuff others have written, but just add
that the solution used by astrophotographers is to use a motorised
equatorial mount - an electric or clockwork mechanism which moves the
camera in synch with the star, planet or whatever you are observing.
Some commercial designs are available IIRC for a few hundred
pounds/dollars, and there are instructions around for making your own.
Either way, they do require some practice to get set up, but once you
have it, exposures can be in hours rather than seconds.

Of course, for the bright part of the moon, fractions of a second are
all you need, but for the dark side - ah, there's always the dark side!

David
--
David Littlewood
Anonymous
August 5, 2005 11:26:44 PM

Archived from groups: rec.photo.digital (More info?)

badchess wrote:
>
> There is no dark side of the moon, really
> as a matter of fact it's all dark...

damn, must tell Floyd to rewrite! :o )

--
Paul (And I'm, like, "yeah, whatever!")
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