Shooting Stars: Shutter speed vs. focal length

[briancmo]

I know there's a mathematical formula that related the focal length one uses and the maximum exposure time before stars start to 'egg' or streak, but I can't find it for the life of me.

Does anyone know the formula for calculating this?

[Bernoulli]

Interesting question. I think I can tell you the answer for a 50D, which has an APS-C sensor with 15.5M pixels:

61.2/(focal length in mm) = seconds for the sky to move one pixel

That's based on a pixel being 1.02 arc seconds at 900 mm in the 50D, which I know from shooting the Moon at that FL. It's also based on the assumption that my algebra is right!

But I see from your signature that you have a 5D so it would be different. It also depends on how many pixels of smear is acceptable.

[chris.bailey]

http://www.wilmslowastro.com/software/formulae.htm

Go down to star trail length. Easy to plug different values into the formula and make a judgement call.

[Bernoulli]

http://www.wilmslowastro.com/software/formulae.htm

Go down to star trail length. Easy to plug different values into the formula and make a judgement call.

To use that website you'll need the pixel size for your camera, you can find it here:

http://www.the-digital-picture.com/Reviews/Canon-EOS-50D-Digital-SLR-Camera-Review.aspx

[henleg]

So for me, having the EOS300D at 6.1mp, I could comfortably cook a roast before moving 1 pixel... *sigh*

Interesting question. I think I can tell you the answer for a 50D, which has an APS-C sensor with 15.5M pixels:

61.2/(focal length in mm) = seconds for the sky to move one pixel

That's based on a pixel being 1.02 arc seconds at 900 mm in the 50D, which I know from shooting the Moon at that FL. It's also based on the assumption that my algebra is right!

But I see from your signature that you have a 5D so it would be different. It also depends on how many pixels of smear is acceptable.

[chris.bailey]

Only if you are cooking your roast on some form of thermonuclear device :-)

[jjbaulikki]

http://www.wilmslowastro.com/software/formulae.htm

Go down to star trail length. Easy to plug different values into the formula and make a judgement call.

Thanks

[briancmo]

Wow, the formula I read a long time ago was much simpler, though this makes more sense and is far more accurate. I never even thought about the angle of the camera in relation to the speed of smear...ness.

[briancmo]

My physics friend sent me this email just now...hilarious!

First of all, a star will occupy more than just one pixel no matter what. It's light will spread out due to atmospheric turbulence (seen as twinkling) and also as light disperses when it enters and exits the glass of your lens. So, you'll never get completely crisp images if you really look closely.

Here's my "ballpark analysis" method. You aren't going to use a calculator or computer when shooting in the middle of a field (I'm guessing) so unless there's some fatal flaw in my reasoning, I hope this will guide you to happy results:

- ballpark estimate #1: stars travel through the sky in circles at 360 degrees per 24 hours (i.e. once around, each day), which is 1/240 of a degree per second, or 1/4 degree per minute.
- ballpark estimate #2: what is the field of view of your lens? -> Depends on the focal length, and whether you're talking about horizontal or vertical or diagonal, but for a ballpark estimate involving diagonals (source=Wikipedia...Gasp!): 90 degrees at 20mm; 45 degrees at 50mm; 30 degrees at 85mm. All assume full frame, and remember: you usually shoot star trails at wide angle.
- ballpark estimate #3: determine over what percentage of the frame you are willing to accept any star trails. 1 percent? 1/2 percent? 1/10 percent?
- ballpark estimate #4: calculate your max. exposure time based on estimates #2, and #3.

Sample Calculation:
- 20mm lens gives 90 degree coverage.
- Stars move at 1/4 degree per minute, so you've got 90x4=360 minutes till they cross the entire frame (of course, this ignores the fact that they move in curved paths, but it's just a ballpark estimate).
- at 360 minutes crossing time, you will get 1 percent coverage in 3.6 minutes, and 1/10 percent coverage in 0.36 minutes.
- 0.36 minutes in around 30 seconds, which is what I told you a week ago when you asked me how long you can expose, before star trails become noticeable.

[evorgsumaf]

My physics friend sent me this email just now...hilarious!

First of all, a star will occupy more than just one pixel no matter what. It's light will spread out due to atmospheric turbulence (seen as twinkling) and also as light disperses when it enters and exits the glass of your lens. So, you'll never get completely crisp images if you really look closely.

Here's my "ballpark analysis" method. You aren't going to use a calculator or computer when shooting in the middle of a field (I'm guessing) so unless there's some fatal flaw in my reasoning, I hope this will guide you to happy results:

- ballpark estimate #1: stars travel through the sky in circles at 360 degrees per 24 hours (i.e. once around, each day), which is 1/240 of a degree per second, or 1/4 degree per minute.
- ballpark estimate #2: what is the field of view of your lens? -> Depends on the focal length, and whether you're talking about horizontal or vertical or diagonal, but for a ballpark estimate involving diagonals (source=Wikipedia...Gasp!): 90 degrees at 20mm; 45 degrees at 50mm; 30 degrees at 85mm. All assume full frame, and remember: you usually shoot star trails at wide angle.
- ballpark estimate #3: determine over what percentage of the frame you are willing to accept any star trails. 1 percent? 1/2 percent? 1/10 percent?
- ballpark estimate #4: calculate your max. exposure time based on estimates #2, and #3.

Sample Calculation:
- 20mm lens gives 90 degree coverage.
- Stars move at 1/4 degree per minute, so you've got 90x4=360 minutes till they cross the entire frame (of course, this ignores the fact that they move in curved paths, but it's just a ballpark estimate).
- at 360 minutes crossing time, you will get 1 percent coverage in 3.6 minutes, and 1/10 percent coverage in 0.36 minutes.
- 0.36 minutes in around 30 seconds, which is what I told you a week ago when you asked me how long you can expose, before star trails become noticeable.

HA HA I read this whole things and all I got was 30 seconds is what I have got before streaking happens.

[Bill Boehme]

..... First of all, a star will occupy more than just one pixel no matter what. It's light will spread out due to atmospheric turbulence (seen as twinkling) and also as light disperses when it enters and exits the glass of your lens. So, you'll never get completely crisp images if you really look closely. .........

That is correct. Becuse of diffraction in the lens, a point of light becomes a pattern that is somewhat like that which is created when a pebble is dropped in water.

And ... it is actually a good thing that the light from a star strikes more than a single pixel (we need to remember that a sensor "pixel" a.k.a. "element" or "site" or "sensel" is considerably different from an image pixel -- while each image pixel contains full color information, the individual sites on an image sensor are arranged in a Bayer pattern where the colors are physically separated. Part of the demosaicing process which convrts the data into a viewble image involves interpolating the "missing" colors at each location. Many, if not most, of the interpolation algorithms used in RAW converters include logic that will discard anomolous data from a single site if the brightness is significntly different from all of the neighboring sites with the assumption that it may be giving a false reading (i.e., hot pixel).

[Karl Johnston]

This is a great read...