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Maclaurin Solution for Angle Subtending Arc and Chord

02/09/2014 - 12:41 von humbled survivor | Report spam
The result shows a close match on the interval -0.2 < B/A < 0.25

Since n=0,1,2,3,.. , the result was obtained by using only n=1, but can also
be n=2, n=3 and so on.

http://www.stonetabernacle.com/angl...chord.html

B=chord A=arc x=angle subtending arc and chord on a circle

Bx/(2A) = sin(x/2)

Let B/A=x and x/2=f(x)

Then

x f(x)=sin f(x)

f(0)=n*pi

Differentiating implicitly,

f(x) + x f'(x) = cos f(x) f'(x)

f'(0) = [(-1)^n]*n*pi

.
.

There doesn't appear to be any pattern in later terms of the series, which I
carried out to 6 terms.

"What is a non-iterative solution for the unknown angle subtending known arc
and known cord on a circle?"

Notice that A may have many revolutions while B is limited to twice the
radius, so as B/A -> 0 there are infinite solutions to the series.

Jon Giffen
intrepid@bellaire.tv
 

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#1 humbled survivor
02/09/2014 - 19:20 | Warnen spam
I added some more information to my web page on the matter:

http://www.stonetabernacle.com/angl...chord.html

When theta = 0, B/A=1, also derived by L'Hospital's rule.

Other than that, referring to the graph, when n=1 it is the poorest fit and
is
way off 0.25<B/A<1. The series is only harmonic when n is odd.

I carried out the Maclaurin series to 5 derivatives. Probably 10
derivatives would result in a closer fit.
This tedious calculation only needs done once and the formula applicable to
any B/A.

f(0)=y(0) reads, "the function of x=0 equals y of x=0"

"humbled survivor" wrote in message
news:c9hNv.137050$
The result shows a close match on the interval -0.2 < B/A < 0.25

Since n=0,1,2,3,.. , the result was obtained by using only n=1, but can
also
be n=2, n=3 and so on.

http://www.stonetabernacle.com/angl...chord.html

B=chord A=arc x=angle subtending arc and chord on a circle

Bx/(2A) = sin(x/2)

Let B/A=x and x/2=f(x)

Then

x f(x)=sin f(x)

f(0)=n*pi

Differentiating implicitly,

f(x) + x f'(x) = cos f(x) f'(x)

f'(0) = [(-1)^n]*n*pi

.
.

There doesn't appear to be any pattern in later terms of the series, which
I
carried out to 6 terms.

"What is a non-iterative solution for the unknown angle subtending known
arc
and known cord on a circle?"

Notice that A may have many revolutions while B is limited to twice the
radius, so as B/A -> 0 there are infinite solutions to the series.

Jon Giffen



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