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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.
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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.
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