Maximum exposure to avoid trails

[Earl]

Using a static DSLR with a 10mm lens how do I work out the maximum duration I can expose and not get a trail?

[Vroobel]

It's called 'the 500 rule' or similar. You have to divide 500 by the FL expressed in mm, the result means the exposure time expressed in second. You can also find that the 500 is replaced with 600.

[vlaiv]

A little bit of math is in order ...

You need to know several things before you start:

1. what is sampling rate of the final image in arc seconds per pixel (you can work that out from pixel size and focal length of lens and any resampling / binning or otherwise used)

2. what is DEC of target in question

3. what is acceptable trailing you are willing to work with.

At DEC 0, sky moves (or rather earth, but no need to go into details :D) at 15.043"/s - this is sidereal rate. You can also use King rate which is 15.0369 which accounts for average refraction in the sky - or you can simply round it up to 15.04"/s if you wish as it makes no difference at this level.

At DEC 90 - right next to celestial pole - things don't actually move - they "rotate".

This means that movement on sensor is actually sidereal rate * cos(DEC) (as cosine goes from 1 to 0 in appropriate fashion).

Now that you have rate at which target moves and you have acceptable trailing - you divide the two to get the time.

Example:

Say that you have DEC 30 target, you are using lens with 10mm, your pixel size is 4.54um and you are willing to accept 3px of trailing in your image. What is the sub duration?

Sampling rate will be = 206.3 * 4.54 / 10 = ~93.66"/px

3px of trailing will therefore be 93.66"/px * 3px = ~281"

Cosine of 30 is sqrt(3)/2 (or use calculator to get exact figure).

Sky will be moving at 15.043"/s * sqrt(3)/2 = ~13.02762"/s

And finally we have

281" / 13.02762"/s = ~21.57s or about 22s exposure.

 

[Earl]

1 hour ago, vlaiv said:

A little bit of math is in order ...

Where does the 206.3 "Sampling rate will be = 206.3" come from?

[vlaiv]

Just now, Earl said:

Where does the 206.3 "Sampling rate will be = 206.3" come from?

I think it is conversion between radians and degrees and small angle approximation.

Here is correct trig expression for arc seconds per pixel with diagram - so we can see if we can derive used formula (which I just picked up online).

In above diagram it is easy to see that:

Length / focal_length = tan(angle)

For very small angles, there is mathematical approximation that goes sin(x) = tan(x) = x, so we can substitute in above formula to get:

length / focal_length = angle

Above is of course in radians and we are interested in arc seconds per pixel. In order to have expression per pixel - we equate length with length of one pixel so expression becomes:

angle_radians = pixel_size / focal_length

angle_degrees *  pi / 180 = pixel_size / focal_length

angle_arc_seconds *  pi / (180 * 3600) = pixel_size / focal_length

angle_arc_seconds = (3600 * 180 / pi) * pixel_size / focal_length =  ~206264.806 *  pixel_size / focal_length

Now this holds if pixel size and focal length are in millimeters or meters - in the same units, but if we put focal length in millimeters and pixel_size in micrometers - we are off by factor of 1000 so it gets

206264.806 / 1000 = 206.264806 or about 206.3

There you go.

 

[Earl]

Thanks, using the 7D pixel size of 4.3 it gives me 20.xxx so 20s it is

[vlaiv]

Just now, Earl said:

Thanks, using the 7D pixel size of 4.3 it gives me 20.xxx so 20s it is

By the way - to see what would acceptable trailing be - you can do motion blur in image manipulation software.

Take image that you like with nice round stars and play with motion blur where you enter length of trail in pixels. Stars will become streaks after this operation - and you will get the sense of how much pixels of trail you can tolerate.

[Earl]

8 minutes ago, vlaiv said:

By the way - to see what would acceptable trailing be - you can do motion blur in image manipulation software.

Take image that you like with nice round stars and play with motion blur where you enter length of trail in pixels. Stars will become streaks after this operation - and you will get the sense of how much pixels of trail you can tolerate.

Yes, it needs a bit of real-world testing, then getting the focus right is the next bit.

But I have a good starting point now.

[happy-kat]

Generally East and West lower the altitude the less movement is seen in the northern hemisphere. The divide 500 (or like me I use 400 as not full frame) by lens focal length is a rough start and where you focus can effect overall image. Dead centre might leave the outer edge of image poor. I like to use where the third intersects to spread focus more evenly.

[Elp]

This is obviously ideal conditions, theoretical. With a 10mm lens I couldn't do 5s due to a nearby lamppost causing lens flare. I can just about get away with 50mm, depends where in the sky I'm imaging. Lens hoods help but with wide lenses you're limited as you'll soon see them in the FOV if you extend them too far forward (custom ones).

[900SL]

This calculator by Lonely Speck is great, covers some of the points raised by Mr V

 

https://www.lonelyspeck.com/advanced-astrophotography-shutter-time-calculator/

[vlaiv]

3 minutes ago, 900SL said:

This calculator by Lonely Speck is great, covers some of the points raised by Mr V

 

https://www.lonelyspeck.com/advanced-astrophotography-shutter-time-calculator/

That is very handy tool.

[ollypenrice]

It might also be worth knowing this Photoshop trick:

Copy layer. Change blend mode to darken.

Top layer active.

Go to Filter - Other - Offset. By nudging one layer relative to another you can make stars look much rounder. The main tool only works in increments of a full pixel but if you go to Edit-Fade you can reduce the nudge to less than a pixel.

Olly

Edited June 14 by ollypenrice
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