?ÿ
I dug these out of my archives.?ÿ They aren't the same problem, but very much like it: given chord and arc find radius.
They may be available on Profsurv.com under the Problem Corner archives there, too.?ÿ (Problems 47 and 71)
I also dug these out of the archives. Almost the same problem (different crossing height).
With a few logical attempts at a width (5, 10, and 20) I could narrow down the range and then with a few linear interpolations narrow it to whatever precision I wanted.
Yes, that's similar to what Newton's Method does, but Newton uses the derivative of the function to speed up the convergence. I like linear interpolation because it doesn't use calculus. Both need a good starting value, so the overhead is similar. In the ladder problem the first derivative is not a piece of cake to get, so linear interpolation may actually be be better choice.
The graphing calculator is doing essentially the same thing. It can just do it a heck of a lot faster than humans, so I let it do its thing on otherwise difficult-to-solve equations. I prefer solving in the grapher to using the calculator's Solve function because the shape of the curve is a check on the solution.
As always, you pose great problems!
Glad to see Dave Lindell ?ÿpost his solution to the "Arc and Chord" problem that MathTeacher posted.
This is a common problem in surveying and I was hoping that somebody would post about other ways to solve
this problem. By that I mean the easy way.
Before programming and graphing calculators what would you do. There was a book that had tables in it.
" Smoley's Segmental Functions" with one of the entry being "arc/chord".
With the given problem arc= 5cm and chord= 4cm you have 5/4=1.2500000
You would look in the table and see on page 254 that 1.250000 falls between 129D 30m and 129D 40m.
By interpolation you can get 129-36-53 with very little work.
entry in table on page 254;?ÿ 129d30m-----1.2494822?ÿ and for 129d40m-----1.2502340
By simple interpolation (one minute at most in time) you get your answer to the nearest second of arc with ease.
?ÿ
JOHN NOLTON
That's pretty neat. Calculate arc to chord ratios given the central angle theta and put them in a table. You can derive his formula from a unit circle, a circle with radius equal to 1. The arc, s, is theta; ( s = r*theta = 1 * theta = theta). The chord length, from the law of cosines, is
c^2 = 1^2 + 1^2 -2*1*1*cos(theta), Thus?ÿc = sqrt(2 - 2*cos(theta)) and?ÿs/c = theta/sqrt(2 - 2*cos(theta)).
For 129d30m in your post, theta = 2.26020138133 radians, cos(theta) = -0.636078220278, and s/c = 1.2494823, which matches Constantine Smoley up to the 7th decimal place.
Those old guys were pretty resourceful.
?ÿ
?ÿ
