Math Class

[RErender]

Please help me with this problem, I would like to be able to solve for each of the segments given x and the number of sides of a regular polygon. Any ideas?



Thanks

[jujubee]

x*(a-b)² ???

[Dusan_Pavlicek]

It's simple:

a = x * [ sin( pi/(2*n) * 1 ) - sin( pi/(2*n) * 0 ) ] = x * sin( pi/(2*n) * 1 ) - 0
b = x * [ sin( pi/(2*n) * 2 ) - sin( pi/(2*n) * 1 ) ] = x * sin( pi/(2*n) * 2 ) - a
c = x * [ sin( pi/(2*n) * 3 ) - sin( pi/(2*n) * 2 ) ] = x * sin( pi/(2*n) * 3 ) - a - b
d = x * [ sin( pi/(2*n) * 4 ) - sin( pi/(2*n) * 3 ) ] = x * sin( pi/(2*n) * 4 ) - a - b - c
e = x * [ sin( pi/(2*n) * 5 ) - sin( pi/(2*n) * 4 ) ] = x * sin( pi/(2*n) * 5 ) - a - b - c - d
etc.

n ... number of segments (5 in the example above)

I hope you see the general rule in it...

[jujubee]

I dont know if I believe it but Ill take your word for it lol. Now how are you getting pi from a shape that obviously does not have any smooth radius?

[Dusan_Pavlicek]

Now how are you getting pi from a shape that obviously does not have any smooth radius?

"pi/2" is the same as 90 degrees
For it is the angles that matters here...

[RErender]

Thanks Dusan! Makes sense now, I was close but couldn't put it all together. Soon we will have a script to extrude polys with a nice round bevel.

[pog]

"pi/2" is the same as 90 degrees

not trying to through a wrench into the works, I admit my math skills are rough, but doesn't that mean a+b+c+d+e = pi/2 or, in other words, the length adjacent(perpendicular) to x would always have to be less than or equal to pi/2?

[RErender]

I seems I am missing something, all of the expressions come out equal for me . I think it is on the right track though. I have an idea of another way to approach it. Have to see if I can make it work

[RErender]

Well here is what I came up with:

a = 2*x * sin(180/n) * cos(180/n)
b = 2*x * sin(180/n) * cos(3*(180/n))
c = 2*x * sin(180/n) * cos(5*(180/n))
d = 2*x * sin(180/n) * cos(7*(180/n))
e = 2*x * sin(180/n) * cos(9*(180/n))

It's in max script so it is a bit ugly

[Dusan_Pavlicek]

RErender:

I double-checked and tested my equations again and everything works perfect!
Consider your image above:

x = approx. 376 pixels
n = 5
then
a = x * sin( pi/(2*n) * 1 ) - 0 = 376 * sin(pi/10 * 1) - 0 = 116 (ok)
b = x * sin( pi/(2*n) * 2 ) - a = 376 * sin(pi/10 * 2) - 116 = 105 (ok)
c = x * sin( pi/(2*n) * 3 ) - a - b = 376 * sin(pi/10 * 3) - 116 - 105 = 83 (ok)
etc.

or (the other form)

a = x * [ sin( pi/(2*n) * 1 ) - sin( pi/(2*n) * 0 ) ] = 376 * [ sin( pi/10 * 1 ) - sin( pi/10 * 0 ) ] = 116 (ok)
b = x * [ sin( pi/(2*n) * 2 ) - sin( pi/(2*n) * 1 ) ] = 376 * [ sin( pi/10 * 2 ) - sin( pi/10 * 1 ) ] = 105 (ok)
etc.

The equations you mentioned give incorrect results:
a = 2*x * sin(180/n) * cos(180/n) = 2 * 376 * sin(180/5) * cos(180/5) = 358 (incorrect - that's almost as much as x !)
b = 2*x * sin(180/n) * cos(3*(180/n)) = 2 * 376 * sin(180/5) * cos(3*(180/5)) = -137 (obviously incorrect )
etc.

pog:
a + b + c + d + e = x, not pi/2
also: a, b, c, d, e, x are lengths (in pixels, meters etc.) while pi/2 is an angle (in radians) - so the units differ and you can neither add them together nor compare them

[RErender]

Ok, I'm a believer, something must of gone wrong when I converted your equations to script.

Woops, My equation is set where n = all the sides of the polygon, works if n= n/4


I think my method will be easier to include in a scripted loop, and I apologize for saying you were wrong

[Dusan_Pavlicek]