Finding the Arc Length of a Hanging CatenaryDate: 07/23/97 at 19:19:43 From: Anonymous Subject: Finding the arc length of a hanging catenary I have been assigned a project in my class to find the arc length of a catenary suspended between two equal poles at equal height. The distance between the poles is 400 feet, and it sags in the center 40 feet. Can anyone out there help me? Date: 07/23/97 at 21:12:50 From: Doctor Anthony Subject: Re: Finding the arc length of a hanging catenary Take the x axis horizontal and the y axis vertical, through (but not at) the lowest point. Let s be the length of arc measured from C, the lowest point, to any point, P, and let w be the weight per unit length of the chain. We let 'phi' be the angle the tangent at P makes with the horizontal. Consider the equilibrium of the portion CP. The forces on it are the tension T at P along the tangent at P, the horizontal tension T' at C, and the weight ws. Resolving horizontally and vertically, T cos(phi) = T' and T sin(phi) = ws By division ws/T' = tan(phi) = dy/dx If we write T' in the form wa, i.e. if the tension at the lowest point is equal to the weight of a length 'a' of the chain, we have dy/dx = ws/wa = s/a Now ds/dx = sqrt[1 + (dy/dx)^2] = sqrt[1 + s^2/a^2] sqrt(a^2 + s^2) = --------------- a ds dx -------------- = ------- sqrt(a^2 + s^2) a Integrating: sinh^(-1)(s/a) = x/a + const at x = 0, s = 0, so const = 0 So we get s/a = sinh(x/a) But we had dy/dx = s/a = sinh(x/a) so integrating y = a cosh(x/a) + const. Now when x = 0, y = a cosh(0) + const y = a + const, and it is convenient to choose the origin so that the const. is zero. So if y = a when x = 0, we get y = a cosh(x/a) and 'a' is the distance of the origin of coordinates below the lowest point. Now, considering the problem as set in the question, when x = 200, the value of y is a+40, so we have a+40 = a cosh(200/a) Trial and error shows a = 507 (approx.). With this value, the left hand side is 547, and the right hand side 546.962 So using this value in the equation s = a sinh(x/a) for x = 200 we get s = 507 sinh(200/507) = 507 sinh(.39447) = 507 x 0.40479 = 205.22 This is for half the span, so the total length of chain is 2 x 205.22 Length of chain = 410.455 feet. If we carry forward the solution of the equation for a to 3 places of decimals, we get a = 506.489 So we get s = 506.489 sinh(200/506.489) = 205.238, so the total length = 410.476 feet. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/