| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1093.1 | | ALLVAX::ROTH | If you plant ice you'll harvest wind | Fri Jun 30 1989 15:31 | 5 |
| � Give answer to the nearest foot.
... using a slide rule :-)
- Jim
|
| 1093.2 | Slide rule if you wish. But.... | PSYCHE::ROOS | | Fri Jun 30 1989 15:48 | 4 |
|
OK. Set up the solution. If necessary write a computer program
to solve it from there.
|
| 1093.3 | 44 | DEC25::ROBERTS | Reason, Purpose, Self-esteem | Fri Jun 30 1989 18:43 | 6 |
| To the nearest foot, I find the answer to be 44. However, the answer
MAPLE gives me is 44.498 so only a slight inaccuracy might push the
answer to 45.
/Dwayne
|
| 1093.4 | Same Answer - 44 feet. | PSYCHE::ROOS | | Wed Jul 05 1989 13:23 | 6 |
|
That is the same answer I got. 44.4922 feet.
The accuracy depends on what your calculating device does with
solving for the solution to 2640.5(sin X) = 2640(X).
|
| 1093.5 | | ALLVAX::ROTH | If you plant ice you'll harvest wind | Thu Jul 06 1989 08:07 | 50 |
| � That is the same answer I got. 44.4922 feet.
� The accuracy depends on what your calculating device does with
� solving for the solution to 2640.5(sin X) = 2640(X).
Which is actually the point of the problem - to analytically process
the equations and eliminate the massive loss of significance that
results when subtracting nearly equal quantities.
Let l = 5280 be the track length, d = 1 be the expansion, t = 1/2 the
angle subtended by the arc of length (l+d), r be the radius of the
circular arc, and h be the height of the bulge.
Then
r*sin(t) = l/2
r*t = (l+d)/2
sin(t)/t = l/(l+d)
1 - t^2/3! + t^4/5! - t^6/7! + ... = l/(l+d)
t^2/6 = d/(l+d)/(1 - t^2/(4*5)*(1 - t^2/(6*7)*(1 - ...)...)
Since t is small, this forms a rapidly converging sequence of
approximations for t^2, and even neglegting the tail of the series
gives better accuracy than the answer quoted above.
Also
h = r*(1-cos(t)) = r*2*sin(t/2)^2
Using a first approximation to t^2:
t^2 = 0.001136148456731680
r = 78337.305672595523
h = 44.4971912816139739
Iterating the expression for t^2 gives:
t^2 = 0.001136148456731680
t^2 = 0.001136213000318023
t^2 = 0.001136213003984796
t^2 = 0.001136213003985004
t^2 = 0.001136213003985004
r = 78335.080504554166
h = 44.4984550191007981
- Jim
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