# How long does the exposure have to be for a specific astrophotography object with a given lense, iso and aperture?

how do I know how many frames I need (overall exposure) for a particular object at a given ISO setting? (200mm, f2.8, canon 6D)

I will be shooting on a barndoor tracker, at a very dark place about 40 km from the nearest city. Is there a "scientific" way of calculating how long the total amount of exposures has to be in order to gather enough light?

Thanks

You can estimate this for a particular object you want to photograph. E.g. suppose that you want to take a picture of the Andromeda galaxy such that none of the foreground stars are going to be blown. The brightness of the foreground stars and the viewing conditions determine the longest exposure for a single shot. The seeing condition determines the maximum sharpness you could in theory reach. But with your 200 mm focal length, you are limited to 4 arc seconds of resolution anyway, while seeing conditions need to get pretty bad before this will lead to seeing disks larger than 4 arc seconds diameter.

So, when focusing perfectly you should in theory be able to get each star to affect only one pixel. However, in practice a star will still be spread out over a few pixels (which is good because if each star would only be one pixel, the information about the colors of the stars would be missing). So, the light of each star will be spread out over a small disk, in your case the size of this disk doesn't depend strongly on the seeing conditions, it's pretty much the same every time.

Then given the information about the brightness of the brightest star, you can estimate the maximum exposure time as follows. You can take pictures of stars or use older pictures you've taken in the past for which you can look up the apparent magnitudes. Transform these pictures to linear color space, measure the value of the brightest pixels in the center of the disk in case the value is not clipped and also measure the diameter of the disk. Also record the exposure time, and ISO.

You then try to find a good fit for the constant of proportionality A in the formula:

center pixel value = ATISO/D^2 * 10^(-2*magnitude/5)

where T is exposure time, and D the diameter of the blur. A simple way is to just calculate A for each picture and take the average. Note that stars have different colors, a red star of the same magnitude as a blue star won't lead to the same pixel value as the magnitude scale. So, you won't get it exactly right, but it's good enough for this purpose. With this measurement of A, you can calculate how long you can expose before some foreground star will get blown.

The next step is to determine how long you would need to expose to make e.g. the Andromeda visible. This depends on the surface brightness of the object and the noise level per picture. The ratio between the two exposure times is then the minimum number of pictures you need to take. In case of Andromeda, you can look up that the brightness is m = 3.44 and it's 190 arc minutes by 60 arc minutes. From your camera sensor dimensions and the number of megapixels, you can calculate the size of a pixel, a typical value is around 4 micrometers. Then that pixel size of, say, 4 micrometers divided by the focal length of 200 mm gives you the angle in the sky that one pixel covers. So, this could be 2*10^(-5) radians = 0.07 arc minutes. This means that the galaxy will be roughly 2700 by 900 pixels in size.

The formula ATISO/D^2 * 10^(-2*magnitude/5) for m = 3.44 and D roughly 2000 pixels will give a reasonable estimate of the center pixel brightness due to the light from Andromeda, it will underestimate this as the surface brightness of a galaxy diminishes rapidly from the center, so we've underestimated the center surface brightness.

If you set T equal to the maximum value ensuring that bright stars don't get blown, you get the pixel values due to the light of Andromeda in a single exposure. The next step is to measure the noise level. Yo can do that from older pictures taken at the same ISO, or you can just put the lens cap on and take some pictures. The brightness values you measure will have a standard deviation, and this will be proportional to the square root of the exposure time. From that you can calculate the noise in each picture, but you can just as well simply measure this in the pictures you take.

Then if you add up N pictures, the pixel brightness due to the light from Andromeda gets multiplied by N, while the noise increases by a factor of sqrt(N). There is also a noise floor which increases linearly as a function of N, but the light from Andromeda is added to that noise floor, so this effect isn't relevant.

Example: Suppose that you've calculated that the light from Andromeda contributes to 3 units of brightness while the standard deviation of the noise is 5. If you add up N pictures, the brightness becomes 3 N while the noise will get a standard deviation of 5*sqrt(N). The average of N pictures will then have a brightness of 3 due to Andromeda, the noise will be 5/sqrt(N). If you want the noise to be 10% or less of the brightness of Andromeda, then that means that 5/sqrt(N) = 0.3, so you need to average over about 280 pictures.

• upvote for your answer - I started it several times, but boy, is hard or what to put it to writing. Great answer – Abdul N Quraishi Jul 12 '15 at 17:40
• now i have to copy and paste this over on the math site to get it explained.. :D thank you very much! was hoping for something more along the lines of the 500 rule, something simple but still scientific. But it is indeed a great answer and answers my question even tough i dont fully understand it. but thats my limitation. thanks! – sharkyenergy Jul 12 '15 at 17:47
• Your link to `"seeing condition"` is broken. – shrx Aug 2 '15 at 15:47