Hi Alan,
>> * Or you just get used to the fact that some numbers are not exact in
>> floating point.
>
> This got me thinking. How many decimal places do you need to
> accurately, say, aim a laser somewhere in a 180 degree arc accurately
> enough to hit a dime on the surface of the moon?
Here is a quick back of the envelope estimate for you. (While I am still
learning the Python, I can answer this one!)
The angle subtended by a dime on the earth is (approximately) given by
sin( theta ) = d / sqrt( R^2 + d^2 )
where
d = 1 cm (the diameter of a dime)
R = 384,403 km (the average distance from the center of the earth to the center
of the moon - the moon traverses an elliptical path about the earth)
To make the approximation simple, take advantage of the series expansion for
sin (theta) and 1 / sqrt(R^2 + d^2)
first
d / sqrt( R^2 + d^2 ) = d / R * 1 / sqrt(1 + d^2 / R^2 )
~= d / R * (1 - 1/2 * d^2 / R^2 + ...)
now
d / R = 1 * e-2 / (384403 * e3)
~= 3 * e-11
so the d^2 / R^2 correction will be very small, and won't effect the
determination. So we now have
sin (theta) ~= d / R
This will be a very small angle. The next approximation to make is for small
angles
sin (theta) ~= theta + ...
leaving us with
theta ~= d / R
To be approximate, assume the precision you need is equal to the size of the
dime. This means you need an precision of
d theta ~= d/R ~= 3 * e-11 ( = 3 * 10^{-11} if you aren't familiar with the "e"
notation)
this is the minimum precision you would need in both the "x" and "y" direction
to accurately hit the dime on the moon with your laser (at its average
distance).
Corrections to this estimate will come from the fact that the moon's radius is
~1737 km and the earth's radius is ~6370 km, so you are actually this much
closer (R is this much smaller).
Of course both the earth is spinning and the moon is moving relative to us, so
you would have to account for those extra corrections as well.
Hope that wasn't too much info,
André
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