Date: 03/30/99 From: Anonymous Subject: Catenary Curve I know what the general equation for a catenary curve is, but a friend is trying to find where the vertex of the curve is if the poles on each side are at a different height because they lie on a slope. Is there a variation of the formula that would allow us to know where the vertex is (making sure it is far enough above the ground)?
Date: 03/30/99 From: Doctor Anthony Subject: Re: Catenary Curve We have supports at different heights. Let one support be at (x1,y1) and the other at (x2,y2) with y1 > y2. To avoid negatives I am measuring x1 positive to the right (from the lowest point on the curve) and x2 positive to the left. This is okay provided we are consistent. It means that x1+x2 is the distance between the posts and can be measured. We also know s1+s2, the length of the cable L. The equation of a catenary curve is: y = a.cosh(x/a) and the equation for arc length from the lowest point is: s = a.sinh(x/a), where a is the distance below the lowest point to the origin of coordinates. When x = 0, y = a and this is the lowest point. We can combine the following equations y1 - y2 = a[cosh(x1/a) - cosh(x2/a)] = 2a.sinh((x1 + x2)/2a).sinh((x1 - x2)/2a) ....(1) s1 + s2 = a.sinh(x1/a) + a.sinh(x2/a) = 2a.sinh((x1 + x2)/2a).cosh((x1 - x2)/2a) to get y1-y2 ----- = tanh[(x1-x2)/2a] s1+s2 We know all the values on the left, which means we can find the value of (x1-x2)/2a. So we can find sinh[(x1-x2)/2a] = k1 (say) With x1+x2 = L, from equation (1) we have y1-y2 = 2a.sinh(L/2a).k1 So 2a.sinh(L/2a) = (y1-y2)/k1 This equation has 'a' as the only unknown and can be solved by an iterative method. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
Date: 03/30/99 From: Doctor Rob Subject: Re: Catenary curve If the catenary is y = a*(e^x + e^[-x])/2 and the line is y = m*x + b, then the point on the catenary of minimum distance from the line is (x0, y0), where x0 = ln([m + sqrt(m^2 + a^2)]/a) y0 = sqrt(m^2 + a^2) The catenary is above the line if and only if y0 > m*x0 + b This was found by using the formula for the distance from the point (x,y) to the line y - m*x - b = 0: d = (y - m*x - b)/sqrt(1 + m^2) I substituted for y the catenary formula, differentiated with respect to x0, set the result equal to zero, and solved for x = x0. Then, I used the catenary formula to find y0. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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