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The Telescope Nut
by Jeff Baldwin

Photo Framing and Exposing

Ever plan to shoot a photo and wonder ahead of time how big the film's field of view will be?
Ever wonder how long to expose the film?  Read on ...

Image Scale:

To visualize how large a field of view an object requires, you can visualize your objective as having no aperture, that is, think of it as a point.  After all, if you put an aperture mask over the front of your scope, the size of the image doesn't change, just the intensity.  So let's keep decreasing the size of the aperture until it is only a point in the center of the lens.  This way you can see that, as light passes through the aperture from the stars in the sky, they pass through the lens, do a little twist as they pass through the lens, and basically keep going in the same direction afterward.  If I also do this at an area of the lens close to the side, the light will change direction and cross that previous path at the focus of the telescope.  But we're only going to use the hole in the middle right now.

Now imagine that we have two stars in the field of view.  One is dead center with the optical axis, the other is 1° off to the side.  After the light passes through the little hole in the middle of the lens, and down the telescope to the focal plane, there will be a star image dead center, and another 1° off to the side.  How many inches off to the side is this 1° ?

If you remember any trigonometry, you will recognize that the size that this 1° will have on the film represents the sine of the angle 1° times the length of the hypotenuse of the triangle.  If the hypotenuse is close to the focal length, f.l. (close enough), the size on the film is f.l. x sin(1°).  Let's try one.  A C-8 has a focal length of 80", so 1° will be 80" times sin(1°) = 80" times 0.01745 = about 1.4 ".  That means that a C-8 has 1.4 inches per degree on the film.  The Moon is a half degree wide, so it will be 0.7 inches wide on the film.  Our image scale rule is now focal length in inches times sin(1°), and the result is in inches per degree.  This simplifies to 0.01745 times focal length.  If you like metrics, then it is focal length in millimeters times 0.01745 millimeters per degree.  Working these backwards, you have degrees per inch = 57.3 / f.l.

Eyepiece projection is a challenge, but not impossible.  There is a computation for it which requires you to know where the geometric center of the eyepiece is.  Or, you can do this: let an object drift from side to side in your telescope using your camera with the clock drive off.  Time it.  Do the same using your eyepiece projection arrangement.  Divide the two numbers and multiply the result by your telescope's f ratio.  If you have a C-8, which is f/10, and an object takes 50 seconds to drift across the field, and with an eyepiece it takes 8 seconds, then you have 50/8 = 6.25, and 6.25 times 10 is 62.5.  Your set-up is now f/62.5.  For barlow projection, simply do the computation with the new focal length.  If you have a C-8 (f.l. = 80"), and a two power barlow, then do the arithmetic with an f.l. of 160".

So, you want to shoot the Andromeda galaxy, which is about 3° long, and you want to know what focal length will place it diagonally in a 35mm negative.  A 35mm negative is 24mm by 36mm, and its diagonal is 43mm.  You want a little room around it to make sure it's well composed, so let's try to have M31 about 36mm long on the negative.  This means that 3° of sky needs to be 36mm long on the film:

mm/degree = .01745 times f.l., or
12 = 0.01745 x f.l., or
f.l. = 12/.01745 = 687mm (27"). 
A 6" f/4.5 would do it.

Exposures:

How long do I expose this object?  Yikes.  Ask thirty astrophotographers and you'll get thirty answers!  I have two ways to compute this: one is Jack Marling's method using hypered film, the other is an old exposure scale from Astronomy magazine from about a million years ago.  I like it, so I'll go with it for this article, but call me if you’d like to see Jack's.

t is the exposure time in seconds,
f is the f ratio of the optic,
ISO is the film speed,
and B is the brightness rating for the particular object or field you are shooting determined by the following table:

BRIGHTNESS RATING TABLE
Scene B value
Bright day 150 - 250
Hazy day 75 - 150
Cloudy day 38 - 75
Overcast day 20 - 38
Blue sky 1000
Inside a room at night 2
Earthshine .001
Thin crescent moon 10
Wide crescent moon 20
Quarter moon 40
Gibbous moon 80
Full moon 200
Edge of penumbra 05
Mid-totality (varies) .005
Sun's disk 1,000,000
Prominences 100
Inner corona 50
Middle corona 5
Outer corona .5
Eclipse sky .01
Horizon at eclipse .5
Mercury 60
Venus 2000
Mars 60
Jupiter 30
Saturn 10
Uranus 40
Core of Orion Nebula .001
Most nebulae .0001
Nucleus of M31 .0001
Most galaxies .00001
Milky Way .00001

Let's try one:  You want to shoot M51 with a 16" f/4.5 using Fuji HG400 film.  The exposure time is f2 / (iso * b ) = 4.52/(400*.00001) = 20.25 / .004 = 5000 seconds, or an hour and 24 minutes.  How about the Full Moon using technipan film (which developed in technidol is ASA 25), using a Takahashi 5" at f/8?  Exposure time will be 64 / (25 * 200) = 64 / 5000 = 1/78 second.  You would select 1/64 sec. and 1/125 sec. since these are standard exposure times on a camera.  Bracketing is always suggested to make sure you have a good exposure.

Film has reciprocity, meaning that if you double the exposure time, you have to cut in half the aperture, the area of the hole where light enters the system.  This doubling-halving cancels out.  If a picture is properly exposed at 1/500 second at f/4, then at f/8 (two stops slower) the exposure would need to be 1/125 second.  This reciprocity arithmetic works up to around 2 to 10 seconds, depending on the film.  Exposures longer than that fail in reciprocity and the films become less sensitive, requiring longer exposures.  This is called film reciprocity failure, and it has plagued astrophotographers for a hundred years.  The extra exposure times needed differ with different films, but lots of people either increase by 50% or double their exposures.  Practicing with your films and equipment will make you an expert and you can determine what works best for you.

Glen Youman mentioned that three-element refractors also rob enough light to warrant a slight increase in exposure duration.  If each interface robs 6%, then the first air-to-glass interface will give you 94%, the next glass-to-glass interface will give you 94% of 94%, which is 88%, the next glass-to-glass interface will give you 94% of 88%, or 83%, and the last glass-to-air interface will give you 94% of 83%, or 78%.  This means you need to increase your exposure time by the reciprocal of 0.78, which is 1.28.  If an exposure warranted 10 seconds, then you need to increase it to 12.8 seconds.

You now have the tools to properly frame and expose images in the sky.

Clear Skies...Jeff Baldwin
For more information on Telescope Making jump to the ATM page.

Copyright © 2001 by Jeff Baldwin
Last Updated: 9/22/2001
http://astro.sci.uop.edu/~sas/Newsletter/TTN_PhotoFraming.html