Maximum exposure time without trailing?

[dlp]

I was just looking at another post http://stargazerslounge.com/imaging-deep-sky/125165-my-first-photo-night-sky.html, and was amazed to see an 8s exposure without star trailing. I seem to get trailing after only a couple of seconds.

Seriously though, I know the earth's rotation speed is ≈465m/s, or 7.2921150 ×10−5 radians per SI second, which I think equates to about .004 deg/s (7.2921150 ×10−5 * 180/pi), or 14.4 arcseconds . Given that for example Betelgeuse is only ≈0.050 arcseconds across, in 1 sec it surely must have moved a lot in relation to the pixel detector on your camera sensor?

Just wondered what other people's experience is, and indeed whether my maths is right?

David

[jgs001]

David, I've got decent results with 30 seconds at 18mm down to 8 seconds at 50mm with my 450d. I checked by taking a shot and zooming all the way in on preview, looking for trails. If the star looked elongated, I reduced the exposure time and tried again.

[barkis]

At what focal length are you imaging at ? Also, Betegeuse is only some 7 degrees above the Celestial equator, where trailing arcs are a lot longer than when at higher latitudes.

[jgs001]

I found this chart that Ron posted a few years back...

[dlp]

Thanks guys - most useful. I'm using an 80 - 300mm lens so that adds a further variable to the equation doesn't it...

PS Betelgeuse was just an example for star size, although my last 'trailed' image was Orion.

[themos]

Star size doesn't enter into it. What matters is how closely squished the pixels are. At 300mm focal length, a 5 micron pixel pitch sensor has a resolution of about 3.5 arcseconds per pixel. On the celestial equator (stars go 15 arc-seconds per second, 15=360/24) the star goes across 4 pixels in a second. Multiply by the cosine of the declination for stars away from the equator.

[brianb]

A small amount of trailing simply doesn't show up. The conventional limit for a lens of focal length F is 500/F seconds. So 10 sec for 50mm, 2.5 sec for 200mm etc.

[dlp]

Ahh - brianb - thanks, I can deal with rule of thumb and that clearly explains why 300mm doesn't work at 2s!

themos - thanks! How do you work out the res in arcseconds per pixel, related to the lens focal length? My sensor is apparently 4.78µm, which means little to me!

David

[themos]

I googled "arcsecond per pixel calculator"....

But I also have the formula (648/π)*(p/f) (comes from the geometry of a very thin right angle triangle with the pixel size on one side and the focal length on the other side of the right angle)

In your case (648/3.14)*(4.78/300) = 3.3 arcseconds per pixel

[FraserClarke]

The more classical "plate scale", in arcseconds per millimeter, is 206265/FocalLength(mm) -- which is exactly the same a themos' formula above just with slightly different units.

[dlp]

Thanks for that themos - a handy formula to keep around. I also found, for anyone still interested ) some useful formulae and calculators on Wilmslow Astro site - Useful Formulae

- even down to calculating length of star trails, which with my 300mm lens at 2s would be 7.9 px, or at the 80mm end, 2.1px, at 30deg declination.