Best Settings For Star Pics?

[Miaow]

Just wondering what are the best settings for getting a good clear pic of the stars?

I was playing around last night (finally got a tripod for the Canon YAY ) but I'm getting some movement on the stars (star trails i think) which i don't actually want.

I was using ISO200, F5.6 at about 175mm (75-300mm Zooms lens) and a 30 sec exposure - thinking the long exposure would make them clearer but obviously as i mentioned I'm getting some movement in the stars so I gather thats too long or is it also cause of using the zoom as well?

So i thought I'd ask here what others find the best settings for a nice clear shot are.

Thanks.

 

[Tasmaster]

The zoom has nothing to do with movement. I don't know if 30 seconds is long enough for movement to have an effect, but it might also be the camera shaking because of the mirror. Are you using mirror lock-up?

Apart from your camera, the best setting is as far away from city lights as you can get!

 

[skipper34]

With a lens at 175mm you will see stars trail at about 15 seconds. The longer the focal length of the lens used for star images, the less time can be exposed because of the movement of the earth. For focal lengths like this you need some sort of clock-driven mount. If you use a normal lens or wide-angle you can expose for longer periods, up to about 30 seconds without trailing. To keep the stars as pinpoints, you need to stop the lens down 1 or 2 stops. Astrophotography will test a lens to its limit as far as how good the lens is. Chromatic abberations will show up on star images more than anything else you shoot. For images of the night sky which show lots of stars, nebulae, clusters, etc., the camera must be guided on an equitorial mount with a clock drive. This is why astrophotographers use a telescope on a clock-driven EQ mount and attach the camera onto the telescope tube via a piggy-back mount and let the mount guide the camera through the exposure. I have taken images of the night sky with this set-up with a camera and wide-angle lens(17-40 EF)stopped down to f5.6 or f8 for up to 2 minutes without trailing. The mount must be accurately polar aligned, however, for this to work, but that's another subject.

 

[Tasmaster]

Guess i stand corected. How does the small magnification factor matter given the extremely long distance from the stars? Any links that explain how it works?

 

[shorty6049]

it works because you're still zooming in, or basically cropping the rest of the sky smaller than at a wide angle, so any tiny movement in the stars gets magnified by that amount. So at a really wide angle, they barely move at all, but when you zoom way in, and only see a small portion of the sky, everything moves throught he view faster. Same deal with a telescope. You have to make adjustments more frequently than at a lesser focal length

 

[Tasmaster]

Got it, thanks for the info.

 

[Miaow]

Thanks for the info :thumbup:

Will try another pic using the usual 18-55 rather than the zoom and see how it works out

Should i have the ISO higher than the 200 it was on also?

 

[astrostu]

With a 56 mm equivalent lens, I got trails with exposures longer than 15 seconds. If you don't have access to a motorized mount nor a faster lens, you will have to up the ISO to get a brighter picture.

 

[skipper34]

Thanks for the info :thumbup:

Will try another pic using the usual 18-55 rather than the zoom and see how it works out

Should i have the ISO higher than the 200 it was on also?

Yeah, I start at ISO 400 as a minimum. It depends on how dark the site is for shooting also. High ISO will be more light sensitive but will also add noise. If your camera has noise reduction, turn it on. If you are at a dark site with little or no light pollution, you can run higher ISO and capture more stars, but you also run the risk of sky glow creeping into your frame. It's a trial and error game like no other. If you use the 18-55, stop it down 1 or 2 and crank up the ISO. Otherwise the lens will scream out with aberations, especially at the edges of the frame. If you are using a Canon XT or 20d or newer, it will have noise reduction.

 

[rzambranap]

I've shot up to two minutes with no discernable movement

 

[enigma67]

I have had some problems with photographing the stars in the night-sky, so I read this thread, and from what I can understand I have absolutely to little equipment to do this. Is it not possible to take "ok" photos of the stars with just my basic Canon EOS 1100D with a normal lens?

 

[Patrice]

The length of the trails in any one exposure is dependant on the length of the exposure. The stars in the sky appear to make a complete revolution around the North Star (Polaris) (not exactly right but close enough for this discussion) once in 24 hours. That works out to 15 degrees of arc in one hour. So if you want your trails to go 1/10 of the way around then your exposure needs to be 1/10 of 24 hours, or 2 hours and 24 minutes. You can try this with one exposure or with more exposures of a shorter period spaced really close together and then merged into one image. Do not move the camera even one tiny little bit during any of this.

Any DSLR and kit lens on a tripod can take star trail images. The effect is quite noticeable when the camera is aimed in a such a way that the North Star is somewhere in the frame as well as some terrestrial features. Other than the really bright stars up there, the light from stars is comparatively pretty dim. The faster apertures and the higher iso's are your friends.

Focus is really pretty hard to nail down with stars, because they are dim and they are small. Plus a lot of kit lenses do not have a hard stop at infinity. So you need to focus manually. Trial and error are the key words. Take a test shot and look for focus in on the image at max enlargement on the preview screen. Might take a few tries before you nail it. Once you have it do not touch the focus ring anymore, set your camera on the tripod, use a remote control for multiple exposures and most importantly have fun with it.

After a few trial and error attempts you'll be getting images just as nice as the pros get.

 

[Dikkie]

I've shot up to two minutes with no discernable movement

Best way to bump a thread !

Probably, you shot a comet flying in the direction of the earth, turning with a same curve as the earth turns around. That's why you don't have movement after 2 minutes


Below, my first attempt for Orion, shot few days ago.
Sigma 30mm, after 25 seconds the stars already started moving. Not long enough for a trail, but long enough for a bad photo...

 

[TCampbell]

You can use some math to work this out and it's all fairly basic math.

Stars move at "sidereal" (yes, I know it looks like "side-real" but it's actually pronounced "sid-EAR-ee-al") speed. It's technically not the stars that move, but the ground beneath our feet. A "day" is 24 hours because that's how long it takes for the sun to go from directly above the meridian one day (true local noon) to the same position in the sky the following day. Except there are 365 days in a year and 360º in a circle. That means each day the Earth moves just fractionally less than 1º in our orbit around the sun. So technically the Earth does a full rotation every 23 hours, 56 minutes, and 4 seconds. It takes us 3 minutes and 54 seconds to catch up to the point where the Sun will appear to have moved (really due to our movement around the sun.)

To make the math easier... it takes the Earth about 1444 minutes to do a complete rotations. Divide 360º by 1444 and you get a value that says we move just about 1/4º (14.95 arc-minutes) per minute of real time. That's how far a star will appear to travel in one minute if the star is roughly above Earth's equator (declination 0º). Stars nearer to the Earth's poles will not appear to move so much. Expressed as "degrees" it works out to .249 degrees.

Hop over to a website such as http://www.tawbaware.com/maxlyons/calc.htm and scroll down to find their "Angular Field of View Calculator", then punch in the focal length of your lens AND the crop-factor of your camera.

For example, if you were your 175mm lens and you have a Canon 400D (that's what your profile says) then your crop-factor is 1.6.

When I feed in these two values, it tells me the angular area of your sensor is:
Horizontal: 7.4º
Vertical: 4.9º

We've already established that the sidereal speed is about 1/4º per minute. That means that in just a 60 second exposure, a star will have moved 1/30th of the distance across your horizontal field of view. If you rotate the camera it will have moved about 1/20th of the distance across your vertical field of view. To get this, I'm dividing 7.4 by .249 for the horizontal dimension and I'm dividing the 4.9 by .249 to get the vertical dimension.

Your sensor is 3888 x 2592 pixels. That means that in the time it takes for the star to drift 1/30th of the distance across the

You would _definitely_ see star trails in that amount of time.

Suppose we switched to a wide angle lens such as Canon's 10-22mm and we zoom wide to a 10mm focal length with it. This changes things:
Horizontal: 96.7º
Vertical: 73.7º

NOW when we take the 60 second photo, a star near the equator would only have traveled 1/388th of the distance across the horizontal field of view. Based on the horizontal sensor resolution of 3888 pixels, 3888 ÷ 388 = about 10 pixels. That means in an enlarged size version of the image with close inspection you may notice the stars are elongated, but it'd be much harder to notice.

The wider the lens, the longer the exposure can be before you notice elongated stars.

 

[BrickHouse]

You can use some math to work this out and it's all fairly basic math.

Stars move at "sidereal" (yes, I know it looks like "side-real" but it's actually pronounced "sid-EAR-ee-al") speed. It's technically not the stars that move, but the ground beneath our feet. A "day" is 24 hours because that's how long it takes for the sun to go from directly above the meridian one day (true local noon) to the same position in the sky the following day. Except there are 365 days in a year and 360º in a circle. That means each day the Earth moves just fractionally less than 1º in our orbit around the sun. So technically the Earth does a full rotation every 23 hours, 56 minutes, and 4 seconds. It takes us 3 minutes and 54 seconds to catch up to the point where the Sun will appear to have moved (really due to our movement around the sun.)

To make the math easier... it takes the Earth about 1444 minutes to do a complete rotations. Divide 360º by 1444 and you get a value that says we move just about 1/4º (14.95 arc-minutes) per minute of real time. That's how far a star will appear to travel in one minute if the star is roughly above Earth's equator (declination 0º). Stars nearer to the Earth's poles will not appear to move so much. Expressed as "degrees" it works out to .249 degrees.

Hop over to a website such as http://www.tawbaware.com/maxlyons/calc.htm and scroll down to find their "Angular Field of View Calculator", then punch in the focal length of your lens AND the crop-factor of your camera.

For example, if you were your 175mm lens and you have a Canon 400D (that's what your profile says) then your crop-factor is 1.6.

When I feed in these two values, it tells me the angular area of your sensor is:
Horizontal: 7.4º
Vertical: 4.9º

We've already established that the sidereal speed is about 1/4º per minute. That means that in just a 60 second exposure, a star will have moved 1/30th of the distance across your horizontal field of view. If you rotate the camera it will have moved about 1/20th of the distance across your vertical field of view. To get this, I'm dividing 7.4 by .249 for the horizontal dimension and I'm dividing the 4.9 by .249 to get the vertical dimension.

Your sensor is 3888 x 2592 pixels. That means that in the time it takes for the star to drift 1/30th of the distance across the

You would _definitely_ see star trails in that amount of time.

Suppose we switched to a wide angle lens such as Canon's 10-22mm and we zoom wide to a 10mm focal length with it. This changes things:
Horizontal: 96.7º
Vertical: 73.7º

NOW when we take the 60 second photo, a star near the equator would only have traveled 1/388th of the distance across the horizontal field of view. Based on the horizontal sensor resolution of 3888 pixels, 3888 ÷ 388 = about 10 pixels. That means in an enlarged size version of the image with close inspection you may notice the stars are elongated, but it'd be much harder to notice.

The wider the lens, the longer the exposure can be before you notice elongated stars.

MIND BLOWN!!!!