Re: Fiveminute challenge Message #18 Posted by Gerson W. Barbosa on 21 Jan 2011, 12:25 p.m., in response to message #13 by Karl Schneider
Karl,
Quote:
(While I was busily writing, David Hayden, Gerson, Dave Shaffer were posting identical results using what appears to be the same approach...)
I don't remember the word you used in your first version of this text in place of busily. It was not diligently, but it had a close meaning and expressed better what you did. Congratulations for the nice work and your detailed and didactic explanations, as always. It takes time to provide a complete solution as you have done. Most here just don't have time to, anyway it is a good thing they occasionally take the time to solve problems and post the results so we can compare them to our own. As of me, I am just kind of lazy.
Quote:
I'd say that Martin's answer is correct, or very nearly so. I got 44.4984551 feet on my HP32SII.
I do agree. Martin is a Civil Engineer, if I am not wrong, so his twodigit answer is good enough for practical purposes. I remember myself solving transmission lines problems (involving grossbeak aluminum cables and other birds names) and the answers to six decimal places during examinations, when two would have sufficed. My professor found that strange but would never say anything :)
Quote:
Solving for the circulararc expanded track proved to be more complicated.
I am glad you have said this. So, I am not the only one :)
Here is my solution, using a similar approach. Yesterday, at first I had used an expression involving the ASIN function at first, then switched to SIN. I have decided to use ASIN again, only because it's similar to the answer I got through Calculus. Anyway, both are equivalent:
yaxis

r___
e ^  L L = half length of the arc (2,640.5 ft)
v ______\ r = radius of the circular arc
^  d /
 / h = height from cente of the arc to the track
 / e = r  h (vertical expansion)
h  /r
A / A = angle in radians
 /
/
v _______________ xaxis
0 r
< d > d = half lenght of track (2,640 ft)
Using
2L = r*2A (length of arc subtended by central arc)
and
d
sin(A) = 
r
L
We get A = 
r
/ d \
and A = arcsin  
\ r /
which implies
/ d \
L = r*arcsin  
\ r /
Replacing L and d with their known values, 2640.5 and 2640, respectively, we get
r*arcsin(2640/r) = 2640.5
when solved for r (initial estimate r = 50,000) on the HP32SII in radians mode yieds
r = 78,335.0808851 ft
We are now able to compute the vertical expansion e:
From the drawing (well, sort of),
h = sqrt(r^2  d^2)
h = sqrt(78,335.0808851^2  2640^2)
h = 78,290.5824303 ft
e = r  h
e = 78,335.0808851 ft  78,290.5824303 ft
e = 44.4984548 ft
This morning, hoping to avoid the solver, I tried Calculus. It turns out I got the same equation:
From the Length of an Arc Formula:
L = Integrate(x=a, b, sqrt(1+[f'(x)]^2)dx)
(SCHWARTZ Abraham, Calculus and Analytic Geometry. Holt, Rinehart and Winston, 3rd edition)
f(x) = sqrt(r^2  x^2)
f'(x) = x*sqrt(r^2  x^2)/(x^2  r^2)
[f'(x)]^2 = x^2/(x^2r^2)
L = Integrate(x=O, 2640, sqrt(1  x^2/(x^2  r^2)))
L = r*arcsin(2640/r)
or r*arcsin(2640/r) = 2640.5
On the HP50g:
'sqrt(R^2X^2' DERVX x^2 1 + sqrt > 'ABS(R)/ABS(R^2X^2)*SQRT(R^2X^2)'
INTVX > 'ASIN(X/R)*(R*SIGN(R)/SIGN(R^2X^2))'
Since R > 0 and X=2640 > 0, this simplifies to
'R*ASIN(X/R)')
Regards,
Gerson.
Edited: 21 Jan 2011, 1:26 p.m.
