Thu Jul 06, 2017 8:05 am

I found this from a 2008 sky at night article Shooting the stars:

Why does "cos (Declination)" also appear as a divisor?

Should the formula be?

Exposure time <= 1000 * cosD / FL

I can see that the longer "fl" requires a shorter exposure and thus appears as a divisor in the formula.Take out the star trails

The rotation of the Earth will make Orion appear to move slowly from east to west. The shorter the focal length of your camera lens, the longer the exposure you can get away with before you start getting star trails in your image. The formula for working out how long you’ve got is

Time (seconds) = 1000 / (fl x cosD)

where fl = lens’s focal length (mm) and D = declination (degrees). Assume D is 0° for Orion.

Why does "cos (Declination)" also appear as a divisor?

Should the formula be?

Exposure time <= 1000 * cosD / FL

Thu Jul 06, 2017 3:34 pm

The figure of '1000' is a short cut to allow a simple division by the focal length to give the exposure time at the equator. The cosD is a modifier to reflect the slower movement of the stars away from the celestrial equator. The same result can be had by modifying the focal length with cosD. However your suggested formula modifies the '1000' figure which has already been calculated to make a complex equation easier. Therefore your suggested equation won't give the correct figure.

Personally I think messing around with the focal length figure and then dividing as suggested in the article is confusing. The obvious way for me would be to express it as (1000/fl)*cosD since the 'cos D' as they have expressed it is a modifier of the time taken for a star on the celestial equator to trail. (1000/fl). In reality of course (1000/fl)*cosD & 1000 / (fl x cosD) is the same equation, so I can't argue with the mathematical veracity of it! I just think my suggestion is less confusing.

If you need to understand why the modifier, cosD, is multiplied with the figure from 1000/fl, I'll try to explain. The fastest movement of a star will be at the celestial equator. There is no movement at the celestial poles. The cosine of the declination accuately reflects the difference in apparant movement of stars located between poles and equator.

Since declination is '0' at the celestial equator, and cos(0) is of course '1', Multiplying by 1 doesn't change the figure. As you move away from the celestial equator ( declination 0) and towards the celestial pole (declination 90) that cosine figure decreases towards zero. Eventually when you reach a declination 90, the cosine is zero. Dividing (1000/fl) by that smaller figure increases the exposure time before trailing, as you move away from the celestial equator, which is what we expect to happen.

Note . . .

(Dividing by zero would of course give an infinite exposure time at the poles! Theoretically correct, but your calculator won't like it!)

(In the celestial south pole the negative declination figures will revert to positive cosine figures, so the formula still works there.)

Personally I think messing around with the focal length figure and then dividing as suggested in the article is confusing. The obvious way for me would be to express it as (1000/fl)*cosD since the 'cos D' as they have expressed it is a modifier of the time taken for a star on the celestial equator to trail. (1000/fl). In reality of course (1000/fl)*cosD & 1000 / (fl x cosD) is the same equation, so I can't argue with the mathematical veracity of it! I just think my suggestion is less confusing.

If you need to understand why the modifier, cosD, is multiplied with the figure from 1000/fl, I'll try to explain. The fastest movement of a star will be at the celestial equator. There is no movement at the celestial poles. The cosine of the declination accuately reflects the difference in apparant movement of stars located between poles and equator.

Since declination is '0' at the celestial equator, and cos(0) is of course '1', Multiplying by 1 doesn't change the figure. As you move away from the celestial equator ( declination 0) and towards the celestial pole (declination 90) that cosine figure decreases towards zero. Eventually when you reach a declination 90, the cosine is zero. Dividing (1000/fl) by that smaller figure increases the exposure time before trailing, as you move away from the celestial equator, which is what we expect to happen.

Note . . .

(Dividing by zero would of course give an infinite exposure time at the poles! Theoretically correct, but your calculator won't like it!)

(In the celestial south pole the negative declination figures will revert to positive cosine figures, so the formula still works there.)

Thu Jul 06, 2017 5:54 pm

Are you agreeing that the Sky at Night article from 2008 is wrong?

Following on from that, we could have an angle of view centred on the pole of around 20degrees, where out at the corners of the field of view (for that lens) one would use the equation with Cos D using 10degrees giving your modifier = 0.98, when the corners of the FoV are near the pole.

Following on from that, we could have an angle of view centred on the pole of around 20degrees, where out at the corners of the field of view (for that lens) one would use the equation with Cos D using 10degrees giving your modifier = 0.98, when the corners of the FoV are near the pole.

Thu Jul 06, 2017 8:52 pm

The formula in the article isn't wrong as such. The formula that was given will give you a general idea of the maximum exposure time before the image shows trails. However, it doesn't take into account the characteristics of the camera being used. etc I thought the way it was expressed wasn't helpful in understanding it either, but it is basically correct if your only interest was to get a figure.

When using a wide field, then it makes sense to use a lowest declination figure in the image otherwise you might get trailing in part of the image. For instance if one corner of your image will show stars with a declination of 75 degrees, then you should make 'D' in the formula = 75, even if the rest of the photo shows stars of a larger inclination.

Are you going to try to get some starfield images?

When using a wide field, then it makes sense to use a lowest declination figure in the image otherwise you might get trailing in part of the image. For instance if one corner of your image will show stars with a declination of 75 degrees, then you should make 'D' in the formula = 75, even if the rest of the photo shows stars of a larger inclination.

Are you going to try to get some starfield images?

Tue Jul 11, 2017 11:32 pm

andrewt wrote:Are you agreeing that the Sky at Night article from 2008 is wrong?

Following on from that, we could have an angle of view centred on the pole of around 20degrees, where out at the corners of the field of view (for that lens) one would use the equation with Cos D using 10degrees giving your modifier = 0.98, when the corners of the FoV are near the pole.

Hi, jumping in here to point out that the Declination is measured from the equator, so if you're centred on the pole the adjustment for D is Cos 80 which is 0.17 rather than 0.98.

Whether it should be

1000 / (fl x cos D)

or

500 / (fl x Cos D)

is a moot point.

Sun Jul 16, 2017 10:25 am

The formula given at the top was designed to reduce a complex calculation to a simple generalised one which only requires the user to put in the declination of the stars they wish to photograph without trailing. The number '1000' or indeed any other number like '500' is arrived at from the characteristics of the camera being used and other characteristics of the lens.

In a sense, this approach is a bit out-dated. It comes from a time when you used expensive film in a camera and you had to wait hours or days before you could see the results.

My own pragmatic approach in a digital age would be to try something like 20 seconds, and if there is no trailing then increase the exposure times until trailing is seen. I could then use the image with the maximum exposure without trailing. All the other images would then be deleted

In a sense, this approach is a bit out-dated. It comes from a time when you used expensive film in a camera and you had to wait hours or days before you could see the results.

My own pragmatic approach in a digital age would be to try something like 20 seconds, and if there is no trailing then increase the exposure times until trailing is seen. I could then use the image with the maximum exposure without trailing. All the other images would then be deleted

Thu Aug 03, 2017 10:46 pm

I find 500 works but 1000 is too long. Remember that you need to take the field of view of the camera into account, so you cannot assume an infinite exposure at he pole, so you have to consider that for a short focal length, the field of view is 30 degrees. If you centre the field of view on the pole, the declination is 75 degrees to avoid star trailing at the edge of the photo.

A table is on Page 17 of this:

http://www.philippughastronomer.com/Ast ... 20DSLR.pdf

A table is on Page 17 of this:

http://www.philippughastronomer.com/Ast ... 20DSLR.pdf