

Subject:
Calculating arc lengh on ellipse, calculating start and end points worked exampl
Category: Science > Math Asked by: craigangusga List Price: $30.00 
Posted:
08 Oct 2004 16:39 PDT
Expires: 07 Nov 2004 15:39 PST Question ID: 412271 
I need to work out the length of an arc on an ellipse. I know the formula but its been so long since i have done integration taht I cant remember how to do it and I need it done quick! what I am looking for is a worked example with actual numbers in it. say for an ellipse of 36 wide and 25 high I need to the following worked out. Length of arc on ellipse How to work out the coordinates start and end point of teh arc on ellipse from given co ordinate This is for a program that writes text along the circumference of an oval  
 


Subject:
Re: Calculating arc lengh on ellipse, calculating start and end points worked ex
Answered By: mathtalkga on 12 Oct 2004 05:34 PDT 
Hi, craigangusga: The application described (formatting a text string around the perimeter of an "oval") raises some interesting issues that go beyond that of finding the length of an elliptical arc. In a circle orienting the characters perpendicular to the circumference is equivalent to aligning them parallel with a radial line. Because an ellipse is essentially a stretched circle, this perpendicularity is not generally valid in the case of an ellipse (ie. placing the characters perpendicular to the perimeter of the ellipse is not simply a matter of drawing them aligned to a line from perimeter to the center of the ellipse). However if the height of characters is fairly small in relation to the distance to the center of the ellipse, then the tilting of the characters necessary to align them to the perimeter will not have a big effect on the length of the display string. The letters will of course be "crowded" somewhat at the ends closer to the center than at the ends furthest from the center. We will be focused in this answer on your original problem of finding the length of an elliptical arc and on the inverse problem of finding an elliptical arc of a given length. A general theory of arc length is outlined in the Comment posted below. We illustrate first the computation of length for an elliptical arc using the example you suggested, of an ellipse (centered at the origin, for simplicity) whose width is 36 units and whose height is 25 units. The equation of such an ellipse in "standard position" would then be: x² y²  +  = 1 A² B² with A = 18 and B = 12.5 (because A,B represent the halflengths of the axes of the ellipse). Using the parameterization (t is in radians, but t is not a "central angle" as explained in the Comment): x = A cos(t), y = B sin(t) we will begin with computing the length of the entire ellipse, or rather (because of symmetry) with computing the quarter of its length that lies in the second quadrant, ie. where t ranges from pi/2 to pi. Thus we have for this length an integral (explained below): INTEGRAL sqrt( (A sin(t))² + (B cos(t))² ) dt OVER [pi/2,pi] The midpoint rule approximates this integral by the length of the interval times the value of the integrand at the midpoint of the interval: (pi  pi/2) * sqrt( (A sin(3pi/4))² + (B cos(3pi/4))² ) I set up rows in an Excel spreadsheet, using varying values of t running down the first column and computing the lengths and midpoint values (and their products) across the corresponding rows. For the single midpoint above, I get: 1.570796 * 15.49597 = 24.34101 when rounded to the places shown. To assess the accuracy of this value, I then computed using a "composite" midpoint rule for three equal subintervals in the same range. That is, the interval [pi/2,pi] was broken into: [pi/2,2pi/3], [2pi/3,5pi/6], and [5pi/6,pi] and the midpoint rule applied to each subinterval separately. The values for the three subintervals are then summed to get a total for the overall range: 9.259898 + 8.113669 + 6.776249 = 24.14982 The "refined" approximation to the integral (and thus to the length of the elliptical arc) has changed only slightly. Since the computer is doing most of the work, with a bit of cutandpaste on my part, refining the subintervals once again by thirds produced the sum 24.14978 (using nine subintervals). So the length of one quarter of the ellipse is pretty clearly converged, at least as far as a computer graphic pixelization would require. Let's turn now to finding a pair of endpoints, symmetric with respect to the yaxis, which give an arc of some prescribed length. In the Comment I suggested that we might want a string/arc of length 19.8 and that by symmetry we would put half to the left and half to the right of the yaxis. Thus we would want to find T between pi/2 (the positive yaxis) and pi such that the arclength over [pi/2,T] is 9.9. As a starting point let me give the approximate lengths for the nine subintervals from the final tabulation above: [pi/2,5pi/9] 3.135409 [5pi/9,11pi/18] 3.086633 [11pi/18,2pi/3] 2.992814 [2pi/3,13pi/18] 2.861552 [13pi/18,7pi/9] 2.704556 [7pi/9,5pi/6] 2.537868 [5pi/6,8pi/9] 2.381663 [8pi/9,17pi/18] 2.258750 [17pi/18,pi] 2.190539 If we add up the "lengths" from the first three of these intervals, we get 9.214855, which is a bit less than the 9.9 we want. Therefore some but not all of the "fourth" interval will have to be included. We have narrowed down T to between 2pi/3 and 13pi/18. Splitting the interval [2pi/3,13pi/18] itself into three subintervals gives: [2pi/3,37pi/54] 0.969610 [37pi/54,19pi/27] 0.953850 [19pi/27,13pi/18] 0.937126 Because we only want to increment the arc's length by: 9.9  9.214855 = 0.685145 the first of the three new subintervals is more than enough. An adequate approximation can now be obtained by taking T to be proportional within that interval to the fraction 0.685145/0.969610 of its length which we require. T = (2pi/3) + (pi/54)*(0.685145/0.969610) which yields roughly T = 2.1355. If more than about two decimal places of accuracy were required, we could of course carry out a more refined computation. Please let me know if some part of this discussion needs Clarification. regards, mathtalkga 

Subject:
Re: Calculating arc lengh on ellipse, calculating start and end points worked ex
From: mathtalkga on 10 Oct 2004 22:38 PDT 
The length of an elliptical arc is surprisingly hard to compute, at least in comparison to that of a circular arc. To avoid "singularities" introduced by points of a curve where x or y may fail to define a function, e.g. for the circle x² + y² = r², it's often simplest to parameterize the curve. Suppose x = x(t) and y = y(t) for some range of values t. Calculus gives us the general formula for arclength: s(t) = INTEGRAL Sqrt( x'(t)² + y'(t)² ) dt [For the sake of brevity we will write definite integrals using an indefinite integral notation, with the convention that the constant of integration is determined so that the integral is zero "at" t = 0.] The circle has a "uniform" parameterization: x(t) = r cos(t) y(t) = r sin(t) where s(t) = rt and t may be considered as a central angle sweeping from the positive xaxis counterclockwise around the circle. The ellipse has a similar parameterization in that: x(t) = A cos(t) y(t) = B sin(t) periodically generate all points on the ellipse centered at the origin: x² y²  +  = 1 A² B² as t ranges from 0 to 2pi. Note however that t cannot be identified with the corresponding central angles, and s(t) is no longer a simple constant times t. It is nonetheless possible to evaluate the resulting integral for arclength: s(t) = INTEGRAL Sqrt( (A sin(t))² + (B cos(t))² ) dt in a variety of ways: 1) Analytically Although no "simple" expression for this integral can be given in terms of composition of elementary transcendental functions and usual arithmetic operations, it is given readily in terms of the incomplete elliptic integral of the second kind, implemented in Mathematica as EllipticE[t,m]: [Elliptic integral of the second kind  MathWorld] http://mathworld.wolfram.com/EllipticIntegraloftheSecondKind.html 2) Interpolation In a broad sense approximating the values through a tabulation or using a precomputed formula whose error is known to be small. 3) Numerical integration The formula above is very smooth, so any good quadrature scheme succeeds. 4) Arithmeticgeometric mean An algorithm discovered by Gauss for computing values of hypergeometric functions, to which elliptic integrals are easily related. For the sake of a concrete example, let's suppose the ellipse is like the one suggested by craigangusga. To be 36 units wide means the semimajor axis is: A = 18 and to be 25 units wide means the semiminor axis is: B = 12.5 Perhaps the piece of text to be printed along (say) the upper rim of the ellipse symmetrically about the yaxis is to have a length of 19.8 units. Clearly the ellipse is wide enough (in this case) to allow printing such a string, so we will not dwell (unless asked!) on detecting when a string is so long that it begins to "wrap around". The positive yaxis is where t = pi/2 on the parameterized ellipse. So we are asked to find t > pi/2 such that: s(t)  s(pi  t) = 19.8 or by symmetry: s(t)  s(pi/2) = 9.9 This sort of problem is easily handled by the numerical integration approach, as is suggested by this thread: [Elliptical Arc Length  Ask Dr. Math] http://mathforum.org/library/drmath/view/51945.html Perhaps the simplest decent quadrature rule for such integrals is the midpoint rule: [Numerical integration] http://www.damtp.cam.ac.uk/user/fdl/people/sd/lectures/nummeth98/integration.htm It will suffice in any case for our illustration here, though a "quality" canned routine might estimate the error of the midpoint rule using a more accurate Simpson's rule approximation as a comparison. An Excel spreadsheet can be used to provide such an illustration. I'll go off and set that up, then come back and (unless there are objections) post the results as an Answer. regards, mathtalkga 
Subject:
Re: Calculating arc lengh on ellipse, calculating start and end points worked exampl
From: craigangusga on 12 Oct 2004 00:39 PDT 
An excel spreadsheet illustration would be excellent. I dont think the results have to be too accurate. An approximation would be best if it can be calcutlated easier. 
Subject:
Re: Calculating arc lengh on ellipse, calculating start and end points worked exampl
From: craigangusga on 14 Oct 2004 04:15 PDT 
Answer looks good, but I'll need a bit of time to review, if I need any clarification I will get back to you. Thanks 
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