Fitting an Arc to a Point and Two Tangent LinesDate: 11/14/2004 at 23:11:39 From: Phill Subject: Arc from point and 2 tangents I'm wondering how to fit an arc to a point and two circles. I have a picture of what I am trying to work out but I will try to describe it. There is a vertical line segment of length 0.3 m, with a circle of radius 0.037 m coming from the midpoint of that line, and another circle of radius 0.104 m coming from the bottom of the segment. I want to fit a 3 point arc to the top of the segment and the tangents of the two circles. Only one curve can fit these points. Date: 11/16/2004 at 08:20:57 From: Doctor Rick Subject: Re: Arc from point and 2 tangents Hi, Phill. I'm not sure what kind of answer will fill your need. What tools do you want to use to find the radius (and/or center) of the new circle? Do you want to construct it physically, or calculate it with algebraic formulas, or could you use a computer tool such as AutoCAD? I work with AutoCAD, and it has a function to construct a circle tangent to three given circles. We can convert your problem into this one easily. Draw a circle of arbitrary radius (a) around the top endpoint, and increase the radii of the two given circles by the same quantity, a. The circle tangent to these three new circles has the same center as the circle you seek, and its radius is "a" less than the radius you seek. With your example numbers, I get a radius of 0.6154. If you want algebraic formulas, we could work one up. I wouldn't write a single formula; it looks to be rather complicated, but several formulas to generate intermediate results would be simpler. I'd start by defining the top endpoint as the origin (0,0), with the line along the positive x axis. The first circle has center at (x1,0) and radius r1; the second circle has center at (x2,0) and radius r2. (I am generalizing a bit here, not requiring that the first circle be on the midpoint.) Then I can develop a formula to find the center (x,y) of the circle through the origin and tangent to the two circles. Here are the details of this coordinate-geometry method. We've got a point at (0,0), a circle of radius r1 with center (x1,0), and a circle of radius r2 with center (x2,0). If the circle we seek has center (x,y) and radius r, then the center is distance r from (0,0), distance r+r1 from (x1,0), and distance r+r2 from (x2,0). The locus of points such that the difference between the distances to (0,0) and (x1,0) is r1 is a hyperbola. I'll skip the derivation of the equation of the hyperbola, but it is (x - x1/2)^2/r1^2 - y^2/(x1^2 - r1^2) = 1/4 Likewise the locus of points such that the difference between the distances to (0,0) and (x2,0) is r2 is another hyperbola: (x - x2/2)^2/r2^2 - y^2/(x2^2 - r2^2) = 1/4 The center of the circle we seek is at one of the intersections of these hyperbolas. Before I solve the equations simultaneously, I'm going to define some coefficients to simplify the work: b1^2 = x1^2 - r1^2 h1 = x1/2 b2^2 = x2^2 - r2^2 h2 = x2/2 Now the equations are (x - h1)^2/r1^2 - y^2/b1^2 = 1/4 (x - h2)^2/r2^2 - y^2/b2^2 = 1/4 I will solve each for y^2, then equate them: y^2 = b1^2((x - h1)^2/r1^2 - 1/4) y^2 = b2^2((x - h2)^2/r2^2 - 1/4) b1^2((x - h1)^2/r1^2 - 1/4) = b2^2((x - h2)^2/r2^2 - 1/4) Expand: (b1^2/r1^2)x^2 - (2h1*b1^2/r1^2)x + b1^2(h1^2/r1^2 - 1/4) = (b2^2/r2^2)x^2 - (2h2*b2^2/r2^2)x + b2^2(h2^2/r2^2 - 1/4) Put in standard form: (b1^2/r1^2 - b2^2/r2^2)x^2 + 2(h2*b2^2/r2^2 - h1*b1^2/r1^2)x + (b1^2(h1^2/r1^2 - 1/4) - b2^2(h2^2/r2^2 - 1/4)) = 0 I will define the coefficients of the quadratic as follows: A = b1^2/r1^2 - b2^2/r2^2 B = 2(h2*b2^2/r2^2 - h1*b1^2/r1^2) C = b1^2(h1^2/r1^2 - 1/4) - b2^2(h2^2/r2^2 - 1/4) Then the solutions are x = (-B + sqrt(B^2 - 4*A*C))/(2*A) or x = (-B - sqrt(B^2 - 4*A*C))/(2*A) For each value of x, we have y = b1*sqrt((x - h1)^2/r1^2 - 1/4) or y = -b1*sqrt((x - h1)^2/r1^2 - 1/4) and r = sqrt(x^2 + y^2) That's the algorithm. When I program it into Excel with your input data, I get x = -0.081353778 y = 0.609965325 r = 0.615366667 This checks with the r = 0.6154 that I got using AutoCAD. But this algorithm will give you the numbers directly, if you put it in a spreadsheet as I did. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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