Distance of Chord from CircumferenceDate: 12/31/97 at 19:39:06 From: David Simpson Subject: Trig, Distance of Chord from Circumference Is it possible to calculate the vertical distance, at a right angle, from a chord to the circumference of a circle? The following details are known: Radius of Circle Area of Minor Segment Dave Date: 01/01/98 at 08:40:16 From: Doctor Jerry Subject: Re: Trig, Distance of Chord from Circumference Hi Dave, See the figure: I'm assuming that the shaded part is a minor segment. It is known that the area of such a segment is A = (a^2/2)(t-sin(t)) where t is in radians. If you know A and a, then the angle can be calculated with a numerical procedure. So, t is known (approximately, as accurately as you like) and you can calculate t/2. Then, cos(t/2) = a/(a-h), where h is the distance from the chord to the circumference. Now you can calcuate h. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 01/02/98 at 18:15:19 From: Dave Subject: Re: Trig, Distance of Chord from Circumference Many thanks for your prompt help it is very good of you. However I am going to have to ask you to indulge me a little further if you will (it is a long time since I did any of this stuff). Given that A and a are known we have the lefthand side of the equation, but I have no idea how to calculate t-sin(t) from this information. Can you, help? Dave Date: 01/06/98 at 08:50:12 From: Doctor Jerry Subject: Re: Trig, Distance of Chord from Circumference Hi Dave, Okay, we have the equation A = (a^2/2)(t-sin(t)), where we know A and a. As a sample problem, suppose we take a = 3 and t = pi/3. We find A = 0.81527... As the sample problem, take a = 3 and A = 0.81527. When we solve the equation A = (a^2/2)(t-sin(t)) for t we should find t = pi/3=1.047... The equation is 2*A/a^2 = t-sin(t) or (1) 0.18117 = t-sin(t). I don't know if you have a calculator and can set it in radians, but I'll assume that this is a possibility for you. Of course, if you have a scientific calculator with a "solver" on it, or if you can graph and zoom, then your problems are over. There are many procedures for locating a value of t that solves (1). Some of these are quite sophisticated, others are more or less informed guessing. I'm not at all sure which to choose. Perhaps the "bisection method." First, you must locate an interval containing the t we are looking for. How to do this? First, let g(t) = 0.18117-t+sin(t). We want to find a value of t (called a zero of g) for which g(t) = 0. Usually, g(t)>0 on one side of the zero and g(t)<0 on the other side. We start with the interval [1,1.5]. Note that g(1)=0.02... and g(1.5) = -0.32... So, the graph of g crosses the axis somewhere between 1 and 1.5. We calculate the midpoint 1.25 of the interval and say: the zero must be in either the left half or right half. To decide, calculate g(1.25)=-0.11... So the zero is in the left half. Our new interval is [1,1.25]. We now repeat this procedure. The midpoint is 1.125; g(1.125)<0 so the new interval is [1,1.125]. Repeat until you obtain the accuracy you need. Here are successive midpoints: 1.25 1.125 1.0625 1.03125 1.046875 1.0546875 1.05078125 1.048828125 1.0478515625 Recall that pi/3 = 1.047... You can see that the Bisection method isn't fast. It's sure, but not fast. Newton's method is much faster. This require calculus. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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