Hi everyone,

I was just wondering what exposures are used when the Milky Way is photographed. I would have thought that long shutter speeds would have been necessary since in order to get a clear view of this, there can't be any extraneous light. But long SS would lead to star trails. I have never taken an image of the MW before but at some point would like to do so.

TIA

The shutter speed calculations for star images that don't turn to trails is a matter of focal length used. The rule of thumb I was told is at 500mm, you can only shoot for 1 second before the stars start trailing. If you half the focal length, you double the shutter speed. So a 250mm lens lets you shoot for 2 seconds. A 125mm lens about 4 seconds etc. Going the rest of the way, 60mm is about 8 seconds, 30 mm about 15 seconds and 15 mm about 30.

So essentially, a very wide lens can shoot between 15 and 30 seconds before you need to worry about trails. I've found that 1600 ISO or higher gets you a lot of stars in dark sky conditions.

Martin,

here's an example you can do from a fixed tripod:

viewtopic.php?f=4&t=210105&hilit=comet+lovejoy

The secret is to stack images to get detail. you shoot multiple images with the time set to avoid trailing, a fast prime lens is the ideal candidate, something like a 50mm f1.8/f1.4 and a good high ISO camera. The image above was 12 images of 15 and 25s. Taken at f2 and ISO800. I have a 1DMkIIn which is not the greatest high iso camera, you can see some abberations in the corner due to coma withe 50mm lens.

To stack you can use freeware Deep sky stacker, it wil automatically align frames, which can be quite difficult in photoshop due to field rotation. It will handle RAW or tif files to stack. There is some argument for processing your RAW files through PS to take advantage of the lens corrections prior to feeding them as tifs to the stacker. Removing distortions helps as the sky move sthrough the frame for each consecutive exposure and removing distortions helps with alignment of teh stars. You bascially have to crop the image to common material to take account of sky movements. With a modern high ISO capable camera look at doing 10-15 frames of 25-30 sec each (on a 50mm lens) at ISO1600 and f2.8.

I've found 20-30 sec is fine for a 50mm lens, it varies depending on where in the sky you point.

I have found the following to work really well:

Add a zero to your latitude and then divide by the focal length. This leaves quite a bit of safety margin so you can cheat this a bit.

So for example, me living at 33 degrees north latitude and shooting with a 50mm lens would yield a maximum shutter speed of 330/50 or about 7 seconds. If I shot it with a 28mm lens it would be about 12 seconds.

On the other hand if I were around the northern border of the US or about 48 degrees the same lenses would produce a maximum shutter speed of 10 sec and 17 seconds.

In Fairbanks at 65 degree shooting aurora I could shoot the same lenses at 13 and 23 seconds respectively.

Why the difference? Check out the globe, the angular velocity of the earth at the equator and at the poles are a whole lot different. The farther north you go, the lower the velocity the ground you are standing on is moving relative to the universe since the circumference is much less and goes from 40,000km (25,000 miles) to zero at the pole.

So just add a zero to your latitude and divide by the focal length for a safe time. If you need to push this 10-15% or so you usually can.

I did a test the other night between a D700 and D800 to see the difference and as expected I could see a star trail at 100% pixel view on the D800 before I could on the D700. That's of course due to the smaller pixels on the D800.

In fact the variable time limit on exposures applies anywhere you are, because the sky moves at different linear rates but constant angular rate depending on declination.( which is angular degrees from the celestial equator). So the time you can expose varies depending on where you point the camera. The formula is

time (sec) = 1000/(F*cos d)

F - focal length mm, d - declination in degrees

This is from an older book for 35mm film cameras so you may want to reduce 1000 by a bit based on trials and use the crop factor with your focal length as it assumes a certain size enlargement.

Here's some samples of times based on the formula :

Declination: 0° +-30° +-45° +-60° +-75°
18mm lens 55 65 80 110 220
50mm lens 20 23 28 40 75
400mm lens 2.5 3 3.5 5 10

wide angle lenses you may need to be a little careful due to their wide angular coverage, they will cover multiple declinations.

As I said declination is the angular distance from the celestial equator and the north celestial pole is at +90°. The belt of Orion is at 0°, Vega is +40°, Cassiopeia is + 60°. Antares in Scorpius is -30° , the southern Cross is at -60° and the magellenic clouds are at -70°.

Do some trial exposures, work out what enlargement you want then view the image at that size and work out if th trailing is acceptable to you.

There some relatively inexpensive devices available that you can use that will track the stars as they move across the sky. They allow for longer exposures and more images which really opens up detail in the MW.

Yup Chris, that's the old formula but I can't do that in my head where I can do latitude times 10 divided by focal length in my head and know it will be good with absolutely no star trail Plus with today's tiny pixels, at the pixel level you will see a star trail sooner. I guarantee that even on a 12 megapixel camera at 30 degrees with a 50mm lens you will absolutely see star trail formation at 20 seconds (I just proved that the other day ). I live at 33 degrees and I can see star trail formation even on a 28mm lens at 20 seconds and that formula says I should be good to 45 seconds.

Scott, those devices only work if you don't include any earthbound foreground objects.

that's right but the key point is that as you approach the poles you can expose for longer. Stacking exposures allows you to do similar things to tracking but again including the FG is difficult as when you stack the stars the FG moves each time, so you end up having to do a composite. For me a slight amount of trailing is acceptable if you get more diffuse detail from the Milky Way, but then I have 8MP on the 1DMkIIn and I'm not so sensitive to trailing.

Exactly and that's the point most people I run into don't seem to understand. At the equator, you are traveling at 1000MPH due to the earth's rotation. At 60 degrees latitude you are only moving 500MPH. At the pole you are not moving forward at all, just spinning around your own vertical axis once every 24 hours.

I must be at the pole then, as I seem to be just going in circles some days.

Guys, are you figuring in the crop factor of your camera when doing your calculations? Ie... my 24mm lens on my D300 would effectively be 36mm. Just curious, thx.

Thanks for the answers everyone - this is really helpful. Unfortunately, some of you may remember an earlier post I made regarding my D2x. I just received a modified estimate for repair from Nikon and it seems my sensor needs to be replaced at a price far greater than the camera is worth ... so right now, I am SOL ... and not happy!

Thanks for your help!

For an illustration of the effect of the difference in angular velocity as you move away from the poles, just take a long exposure shot of a celestial pole and look at the length of trails near and away from the pole.
Here is a rather average shot of the South Celestial Pole over 20 min.

Hey Neil, y'all need to get a South Star

That would make things easier

We may not have a pole star, but the the Southern Milky way directly overhead in winter and the magellenic clouds makes up for it 1000x over.

Good point, the Milky Way seems much more intense south of the equator.

It casts shadows on a good night!

I was told last year about the 600 guideline; 600 divide by the focal length gives the length of time in seconds you can record stars without streaks. example; 600 divided by 20mm = 30 sec . Of course aperture and ISO need to be considered.

I like the add a zero to latitude as suggested by EJ above. And, will have to give this a try when shooting in the Arctic again this winter.

Chas