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Q: Applied Geometry ( Answered ,   0 Comments ) Question
 Subject: Applied Geometry Category: Science > Math Asked by: zkane-ga List Price: \$2.00 Posted: 02 Jan 2003 12:09 PST Expires: 01 Feb 2003 12:09 PST Question ID: 136558
 ```I am working on the design of a curved wall in a building. The horizontal length of the wall is exactly 34'-0". The actual length of the wall along the curve needs to be exactly 35'-0". This curve is a portion of a circle and is symmetrical. What is the radius of the circle and the degree of the arc needed to meet these two requirements. I would also like a general equation for solving this problem with different spacings. Thanks``` Subject: Re: Applied Geometry Answered By: mcfly-ga on 02 Jan 2003 15:32 PST Rated: ```Hi zkane, A seemingly simple problem, this question is actually surprisingly complex. However a useful site that I found gives a guide as to how to proceed: Ask Dr. Math FAQ: Segments of Circles http://mathforum.org/dr.math/faq/faq.circle.segment.html Using the parameters defined by this site, we know the arc length 's' to be 35, chord length 'c' to be 34, and we wish to determine the circle radius 'r' and the angle subtended by the arc, 'theta'. Using rules of right-angled triangles, we can say that sin(theta/2) = c / 2r (1) If we initially measure angle theta in radians (converting into degrees can be done later), then theta = s / r and therefore r = s / theta (2) Substituting equation (2) into (1) we get sin(theta/2) = (c*theta)/2s Renaming theta/2 to be 'x' and rearranging gives the formula quoted at http://mathforum.org/dr.math/faq/faq.circle.segment.html#1 c / s = sin(x) / x (3) We know that theta = 2x and r = s / theta so we just need to solve equation 3. Unfortunately the solution to sin(x)/x comprises of an infinite series but an accurate approximation can be achieved using Newton's Method. This uses the repeated application (interation) of the formula x(n+1) = x(n) - (sin[x(n)]-kx(n))/(cos[x(n)]-k) where k = c /s . This may seem rather overwhelmingly complex, but working with our numeric example (c=34, s=35) simplifies things somewhat. k = 34/35 = 0.9714 A suggested starting value for the iteration is x(0) = sqrt(6-6k) = 0.4140, then applying the formula repeatedly gives: x(1) = 0.4140 - [sin(0.4140) - (0.9714*0.4140)]/[cos(0.4140) - 0.9714] = 0.4167 x(2) = 0.4167 - [sin(0.4167) - (0.9714*0.4167)]/[cos(0.4167) - 0.9714] = 0.4174 (NOTE: These calculations must be done with a calculator in 'rad' mode, not 'deg') This should be enough for us to say with a high level of certainty that x = 0.42 to two decimal places. Therefore, in radians, angle theta = 2*0.42 = 0.84 and radius r = 35/0.84 = 41.7' = 41' 8'' To convert theta into degrees we can multiply by 180/pi to give theta = 48.1 degrees. In answer to your question then, an arc of 48.1 degrees of a circle of radius 41'-8'' will give the wall you describe. The above process can of course be applied for any values of 's' and 'c'. I hope this answer has been both interesting and informative. If you would like any clarifications please request them before rating my answer. Thanks, mcfly-ga :-) Search strategy: circle theorems maths length chord length chord arc``` Request for Answer Clarification by zkane-ga on 03 Jan 2003 06:06 PST ```Wow, now I don't feel so bad about not being able to solve this on my own. The numbers do work and I am happy with the answer. My only question is how to go about increasing the accuracy of the numbers. When I construct the arc with the given numbers I am getting errors between .25" and .5". Regardless I am really impressed with the answer and the whole -Google Answers setup.``` Clarification of Answer by mcfly-ga on 03 Jan 2003 10:21 PST ```Hi zkane, Thanks for the great rating! The accuracy of the figures for radius and subtended angle is limited by the number of iterations used when applying the Newton Method. In my answer I applied the formula twice to get a precision of 2 decimal places. To improve on a tolerance of 0.5 inches out of 41'-8'' requires precision of 4 decimal places. Therefore substituting the value of x(2) into the formula x(n+1) = x(n) - (sin[x(n)]-kx(n))/(cos[x(n)]-k) and repeating until x(n+1) varies from x(n) by less than 0.00005 will reduce the error even further. Good luck in your wall building :) mcfly-ga```
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