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Distance of Chord from Circumference

Date: 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


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!   

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 

Can you, help?

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)


  (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 

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:


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!   
Associated Topics:
High School Conic Sections/Circles
High School Geometry

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