[Tutor] [OT] Re: Floating Point Craziness

[Andre' Walker-Loud]

On Jun 13, 2011, at 1:44 PM, Emile van Sebille <emile at fenx.com> wrote:

> On 6/12/2011 1:55 PM Andre' Walker-Loud said...
>> 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,
>> 
> 
> 
> Of course not.  I enjoyed it.  However, don't you need to work divergence in, as per wikipedia, "...At the Moon's surface, the beam is only about 6.5 kilometers (four miles) wide[6] and scientists liken the task of aiming the beam to using a rifle to hit a moving dime 3 kilometers (two miles) away."
> 
> (http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment)


Of course, I was performing just the 'theoretical' calculation.  It is up to others to dtermine if it is actually practical :)

Andre



> 
> Emile
> 
> _______________________________________________
> Tutor maillist  -  Tutor at python.org
> To unsubscribe or change subscription options:
> http://mail.python.org/mailman/listinfo/tutor