[MetalZilla]

I have to calculate the number of pins in the diameter of a circle based on the circle radius (diameter) where the pins are tangent to each other and lying inside the circle diameter. Any help would be most appreciated. Thank you.

[deputydawg]

.965" pins, using 13 pins as pictured in a 10" diameter hole.

[Anniepoo]

If I've done my sums correctly -

let R be the cavity radius
let r be the pin radius

draw a triangle abc between two adjacent pins, ab, and the center c

bisect ab to make point d.
adc is now a right triangle.

let theta be the angle acd (the one at the center).
let k be the length of the hypotenuse

Now, since the pin must be centered on a and it's surface touch d,

len(ad) = r

Since the pin must touch the outside R,

A) k = R - r

and by trig definition of sin

B) k = r sin(theta)

substituting A into B

r = (R - r) sin(theta)

collecting r terms on left side

r(1 + sin(theta)) = R sin(theta)

divide by (1 + sin(theta))

C) r = (R sin(theta)) / (1 + sin(theta))

Now, let n be the numer of pins. Then 2n is the number of triangles similar to adc, so

2 pi = 2n theta

so

D) pi / n = theta

substituting D into C

r = (R sin(pi/n)) / (1 + sin(pi/n))

This solution is exact.

The seemingly simpler solution is just to take a circle of radius (R-r) , take it's circumference, 2 pi (R - r), and divide by n,
but this has an error, as it doesn't account for the difference in length along the arc and along ad for two triangles.

As the number of pins rise the solutions converge, using two approximations

sin(pi/n) ~= pi/n (the so-called 'small angle approximation')

and

1 + sin(pi/n) ~= 1

Hope that helps

[JobShopJeff]

Thank you I will try it out.