Re: [CR]Bolt Circle Diameters

(Example: Events:Eroica)

From: Donald Gillies <gillies@cs.ubc.ca>
Date: Wed, 23 Feb 2005 15:04:21 -0800 (PST)
To: classicrendezvous@bikelist.org
Subject: Re: [CR]Bolt Circle Diameters

Ok, I have to chime in here as before Steve Maas gets the jump on this important matter of national security.

BCD can be calculated based on the angle between crank spider arms. if you know the number of spider arms, then you know the base of the isosceles triangle and the angle at the peak of the ososceles triangle. You are looking for the length of one side. You split the isosceles triangle down the middle to get two right triangles. Now you can use trig to easily get all the sides.

First though let's work a few ad-hoc simple cases.

For example, on a 4-arm crank we have right triangles (90 degrees) at the spindle. Everyone knows that the diagonal line across a square is sqrt(2) times the side of the square. So we measure between the bolts X and we know X = sqrt(2) * R. We want D = 2R = X * sqrt(2) = X * 1.414.

For example, on a 6-arm crank we have 60-60-60 equilateral triangles. This is important to Colnago owners. We measure between the bolts and get X = R, We want D= 2R = X * 2.0.

Now it gets tricky and we work the 5-arm case in general.

X = measured distance between bolts A = 360/N, the angle between spider arms R = bolt circle radius D = bolt circle diameter = 2R

Right triangle has a base of X/2, top angle of A/2.

X/2 = R sin(A/2) R = (X/2) / sin(A/2) D = X / sin(A/2)

In particular,

3-arm, A = 120, sin(A/2)= .8660, D = X/.8660 = X*1.154 4-arm, A = 90, sin(A/2) = .7071, D = X/.7071 = X*1.414 5-arm, A = 72, sin(A/2) = .5878, D = X/.5878 = X*1.701 6-arm, A = 60, sin(A/2) = .5000, D = X/.5000 = X*2.000

And for http://www.classicrendezvous.com/images/Italian/Campagnolo/Cam6T-S.jpg

10-arm, A = 36, sin(A/2) = .3090, D = X/.3090 = X*3.236

- Don Gillies
San Diego, CA