DSLR plus lens - rule of thumb to avoid star trails?

[JamesF]

reducing the FoV increases the magnification so the star trailing is more visible. It's the same effect as using a longer focal length.

I really don't understand why this is, unless at the same time you're changing the pixel sizes in the sensor or adding more optical surfaces.

Given a lens with a fixed focal length, it projects an image of a fixed size onto the image plane. Regardless of the size of the sensor at the image plane, the magnification will not change. If the magnification does change when you drop from a full-frame sensor to a crop sensor (which is effectively just "trimming the edges off") what else has changed to make the magnification different? I'm clearly missing something here.

James

[JamesF]

The rule of the thumb I come with is a little bit more complex but takes into account the main physics of lenses, sensors and light :

max exposure time (s) = [19.3 x pixel size (µm) + 13 x aperture + 0.1 x focal length (mm) ] / ( focal length (mm) x cos declination )

I'm definitely in agreement that the pixel size, aperture, focal length and declination should be factored into such a calculation. I don't think it really matters, but unpleasant things happen with your formula if you're pointing the camera at the celestial poles

James

[Fred_76]

I'm definitely in agreement that the pixel size, aperture, focal length and declination should be factored into such a calculation. I don't think it really matters, but unpleasant things happen with your formula if you're pointing the camera at the celestial poles

James

Yes, because "cos d" equals 0 when d = 90°. That is why you could in theory expose a very long time without star trail close to the celestial pole. You should take into consideration the lowest declination of the pictured area instead of the central declination. It is not easy to calculate, but to do it quickly, just use :

dmin (°) ~= Max [ 0 ; declination (°) - 15.3 x sensor's width (mm) / focal length (mm) ]

[JamesF]

Yes, because "cos d" equals 0 when d = 90°.

Quite. As I said though, I don't think it matters too much. I'd be interested in how you derived the formula in the first place, if you have time to write it up at some point. I was about to do the same myself until I saw that you'd already done the hard work

James

[Fred_76]

I'd be interested in how you derived the formula in the first place, if you have time to write it up at some point.

The article is here, in french :

http://www.astrosurf.com/fred76/pose.html

or in Googlish :

http://translate.google.fr/translate?sl=fr&tl=en&js=n&prev=_t&hl=fr&ie=UTF-8&eotf=1&u=http%3A%2F%2Fwww.astrosurf.com%2Ffred76%2Fpose.html

I hope the translation is correct enough...

[Fred_76]

I hope the translation is correct enough...

Nope, it is not ! Google mixes up the formula... use the translation for the text only and refer to the original article in french for the maths.

[Fred_76]

Hy!

I came back on my calcs and refined the formula in a more consistent matter.

Let start from scratch.

The light coming from the star is going through the atmosphere and is affected by the seeing "s". The diameter of the star patch is given by :

d seeing = s x f

where s is the seeing in rad and f the focal length of the lens, in m

Then the light gathered by the lens is distributed over the Airy pattern. 91% of the energy of the incoming light is gathered in the central dot, 1st and 2nd circles of the pattern. The diameter of the 2nd circle is given by :

d Airy = 4.47 x lambda x aperture

Where lambda is the wavelength of the visible light, i.e. 550 nm, and aperture,, the aperture of the lens.

Finaly, the photosites of the sensor are organized in a Bayer pattern. The color of one pixel is in fact calculated from the light gathered by the surrounding photosites. A star shall therefore blur over several pixels, approximately 2.

d Bayer = 2 x pixel width

In the mean time, the Earth is rotating at a speed of 360° per 23 hours 56 minutes and 8 seconds.

On the DSLR chip, that gives a speed of displacement, per second, of the star of :

v = ( f x cos delta ) / 13 714

If you assume that the star shall not move more than the radius of the sum of the 3 diameters calculated above, and using common units for the various data, you can then calculate the rounded values with :

t max = [ 14 x pixel width (µm) + 17 x aperture + focal length (mm) / 15 ] / ( focal length (mm) x cos declination )

This formula is slightly different from the one I calculated a few years ago and explained on my website. It should however give more consistent results from what I experienced while shooting the Panstarrs C/2011 L4 comet.

Fred

[Mysteron2]

Very well done Dara

[Mysteron2]

Equally well done Nebula Merci

[stargazerlily]

I've been taking a few star photos with my Canon 1000D with a nifty 50mm lens closed down to f4.0, mainly of Orion and M42.  My location is 53 degrees north.

Through trial and lots of errors, I was able to get stars without trails when the exposure time was 4 seconds or less.  This is in agreement with the maximum time predicted by the equation when I enter the camera stats and a declination of 10 degrees.

Pete

[Photosbykev]

I really don't understand why this is, unless at the same time you're changing the pixel sizes in the sensor or adding more optical surfaces.

Given a lens with a fixed focal length, it projects an image of a fixed size onto the image plane. Regardless of the size of the sensor at the image plane, the magnification will not change. If the magnification does change when you drop from a full-frame sensor to a crop sensor (which is effectively just "trimming the edges off") what else has changed to make the magnification different? I'm clearly missing something here.

James

I've come back to this a, because I didn't answer the question and b, the thread came back onto the mian topics

James, let's call it apparent magnification or reduced FoV, rather than actual. If I go out with my cameras, one a full frame and the other a crop sensor (say a Canon with x1.6 factor) and just a 24mm lens.

If I set up to photograph a star trail with the ff camera and I have 10 fence posts in the foreground and during a 1 hour exposure the stars trail across the top of 6 of those fence posts if you print the image the stars cover 60% of the width of the print. Same lens with the crop body and same settings, the crop camera will only see 6 of the fence posts in the foreground (an apparent increase in magnification, or reduction of FoV) and the stars will trail over all six of the posts in one hour and the print will have the star trail right across it.

[Dara]

Fred, can I clarify a couple of things about your equation, please? I'm trying to implement it and I just want to make sure I'm understanding the order of operations correctly...

If we use the following variables for shorthand -

P : pixelwidth

a : aperture

f : focal length

d : declination

This gives

t = (14p + 17a + f/15) / (f cos[d])

Is that right?

Secondly - if someone was still using film what sort of value should be used in place of pixel width?

[Pint Of Stellar]

I've come back to this a, because I didn't answer the question and b, the thread came back onto the mian topics

James, let's call it apparent magnification or reduced FoV, rather than actual. If I go out with my cameras, one a full frame and the other a crop sensor (say a Canon with x1.6 factor) and just a 24mm lens.

If I set up to photograph a star trail with the ff camera and I have 10 fence posts in the foreground and during a 1 hour exposure the stars trail across the top of 6 of those fence posts if you print the image the stars cover 60% of the width of the print. Same lens with the crop body and same settings, the crop camera will only see 6 of the fence posts in the foreground (an apparent increase in magnification, or reduction of FoV) and the stars will trail over all six of the posts in one hour and the print will have the star trail right across it.

A great visual explanation to go with the one above:

[JamesF]

A great visual explanation to go with the one above:

It is however necessary to account for the fact that the pixel sizes in the cameras are not the same and that the images have (as far as I can tell) been resized to different scales to fit the video frame, so it isn't just the change of sensor size that is being compared in the video.

James

[andrea1984]

I  want to give a contribution to the calculation, this is the calculation that I use to know how many seconds of exposure can be set on the camera before the trail is noticeable, based on the calculations made by the site (http://www.sceneplanner.com/tool3.php):

Tha calculation does not take care of:

- seeing

- resolution of the optics (optical resolution = 116 / focal lenght)

Example when the orientation is portrait:

Input parameters:
 

HorSize = Horizontal Sensor size (mm)

HorPx = Horizontal resolution of the sensor (pixel)

FL = Focal length of the optics (mm)

Dec = Declination (degree)

TrailPx = Maximum number of pixel we accept before before assuming the photo has the trail

Calculated parameters:

FOVHor = Field of view orizontal (rad)

Res = Resolution (rad / pixel)

AngSpeed = Angular speed (rad / s)

Seconds = number of seconds that a pixel takes to move of TrailPx pixels

FOVHor = 2 * atan( HorSize / (2* FL))

Res = FOVHor / HorPx

AngSpeed = (2 * Pi) / 24 * abs(cos(Dec * Pi / 180))

-> Seconds = Res * TrailPx / AngSpeed * 3600

To calculate the maximum exposure for the top or bottom frame add or subtract FOVHor / 2 when calculating the AngSpeed:

AngSpeed = 2 * Pi / 24 * abs(cos((Dec + FOVHor / 2)* Pi / 180))

For landscape orientation use the vertical resolution and size.

Setting the TrailPx to 12.5 pixel this calculation gives the same results as the site previously published