Acer Victoria <siddim01@xxxxxxxxxxxxxxx> moved upon the face of the 'Net and
spake thusly:
> OK, I get it that stars are a point source; no focal length will render
> them anything other than points. But how dow focal length affect how
> fast the trails form? How is the motion magnified when the point isn't?
Focal length is inversely proportional to field-of-view.
The stars are aparrently moving at a constant angular velocity, so
they will traverse a narrow field faster than a wide one.
It's the field of view that affects aparrent movement, not
magnification. It just happens that both factors depend on focal
length.
You can calculate the field-of-view of a (non pathological) lens like this:
fov = 2 * arctangent ( film-size / ( 2 * focal-length ) )
eg, for 50mm lens on 36mm wide frame
fov = 2 * arctangent ( 36 / ( 2 * 50 ) )
= 0.6911111611634243 radian
= 0.69111 * 180 / pi degree
= 39.59 degrees
Remember to keep your units the same, and to be aware of whether
you're working in radians or degrees.
Now the earth rotates 360 degrees in 24 hours, or 15 degrees per hour,
or one degree per four minutes, so (to a first approximation), a star
will traverse your 50mm lens in under three hours.
In reality, only a star on the celestial equator will appear to move
at the same rate as the earth turns; a star closer to the celestial
poles will appear to move more slowly.
cjb.
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