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Re: Help on Radius of Arc



brian whatcott wrote:

At 22:53 11/16/97 -0500, you wrote:
...from the geometry, I get the two
equations:
h/(2r)=sin(T/2) and s/r=T, where h=chord length, s=arc length,
T=subtended angle.

Bob Sciamanda

By inspection, I reckon the half chord divided by the radius
is the tangent (half central angle)

i.e
h/(2r) = tan(T/2)

Sincerely

brian whatcott <inet@intellisys.net>
Altus OK


Nope, look again, Brian; the radius is the hypotenuse of the right
triangle!

While I'm here, my final equation: {Sin(T/2)}/(T/2) = h/s just looks
bad because we don't usually have tabulated (or calculator stored)
values for the LHS. In fact it is an oft occuring function and has been
defined and given a name: Sinc(x)={Sin(x)}/x . Perhaps your students
should tabulate (or plot) Sinc(x) and then simply read the value of
(T/2) as that which is paired with the known ratio h/s. (Then, of
course, r=s/T). It is a good lesson for these students that there is
nothing magic about those functions which we have chosen to define and
tabulate; these choices are perhaps more a statement about "us" than
about any objective reality.
--


Bob Sciamanda sciamanda@edinboro.edu
Dept of Physics sciamanda@worldnet.att.net
Edinboro Univ of PA http://www.edinboro.edu/~sciamanda/home.html
Edinboro, PA (814)838-7185

You never do anything until the first time you do it.
-Me, here, now