| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 687.1 | | ENGINE::ROTH | | Fri Apr 03 1987 08:04 | 13 |
| This is a good kind of problem to give to beginning numerical
analysis students; the variant that was posed to me in high school
(by a physics substute teacher of all things) was like this:
You have a mile long straight railroad track. At night someone
comes in and welds in one extra foot of track. How much does
the track bulge to one side? (assuming it now makes a circular
arc).
After much scribbling on paper, I had managed a few decimal digits on
a slide rule... (anyone remember those?)
- Jim
|
| 687.2 | similar idea on note 495.* | THEBUS::KOSTAS | Wisdom is the child of experience. | Fri Apr 03 1987 10:57 | 7 |
| re. .0
Note 495.{0,1,2} titled: "The equator hoop problem" is similar
to this one, take a look.
-K
|
| 687.3 | This one is different, as stated | MODEL::YARBROUGH | | Fri Apr 03 1987 14:20 | 15 |
| > Note 495.{0,1,2} titled: "The equator hoop problem" is similar
> to this one, take a look.
Not really. If you read them carefully, the three problems mentioned in
{.0,.1,.2} are all different in essential ways. The present problem boils
down to
{h = R*(secO-1),
R*O + .5
cotO = ------------}
R
where R is the earth radius, O is the angle at the center of the earth,
between the vertical at the lift point and the end point of the tangent
created by lifting the rope by the distance h.
Which doesn't look like it has an analytical solution.
|
| 687.4 | Wish we had graphics | TOOK::APPELLOF | Carl J. Appellof | Fri Apr 03 1987 17:18 | 11 |
| I think it should be
R*O + .5
tanO = ----------
R
rather than cotO
regardless, it looks like a mess to me.
I got a little over 121 meters by brute force? I don't think I
can believe DECALC's roundoff at this point.
|
| 687.5 | no analytic solution as far as I know | ENGINE::ROTH | | Sat Apr 04 1987 11:07 | 37 |
| The crux of this type of problem is to avoid roundoff error
by not subracting nearly equal quantities.
Let r be the radius of the earth, nominally 4.0E7/2/pi meters,
dc be half the increase in length of rope, 0.5 meters, t
be half the angle subtended by the rope when you lift it, and
h be the height you're after.
By elementry trigonometry you have dc/r = tan(t) - t
Expanding tan(t) in a power series near zero and cancelling t we have
dc/r = t^3/3 + 2*t^5/15 + 17*t^7/315 + ...
So to a first approx t[0] = cubrt(3*dc/r)
You can improve this by simple iteration:
t[k+1] = cubrt(3*dc/r - 2*t[k]^5/15 - 17*t[k]^7/315 - ...)
t[0] = 0.006176423196527
t[1] = 0.006176391779955
t[2] = 0.006176391780754
Finally, trigonometry also gives
h/r = (1-cos(t))/cos(t)
Again, to avoid roundoff, since cos(t) ~= 1 rewrite this as
h/r = 2*sin(t/2)^2/cos(t)
and plug in to get
h = 121.430197997747698
- Jim
|