geek bodyart

Sometimes, a little geometry is useful. It's even more useful when someone has already done it for you. Here are the answers to these questions:

The derivations are scanned, and my writing isn't the clearest. If you can't make something out, mail me.

How big a ring can I thread through my eyelet?

Well, you'll need to know the inner diameter of your eyelet, the length of the eyelet, and make a choice about either the inner diameter of the ring OR its gauge. Then....

The inside *radius* of the CBR that goes through an eyelet of length L and inner *diameter* H is: 

  R = (H-G)/2 + L^2/[4(H-G)]
where G is the gauge of the ring. R, G, H and L must all be in the SAME UNITS.

Conversely, for a ring of inner radius R, the gauge would be: 

  G = H - R + sqrt[R^2 - (L/2)^2]

These produce a fit with ZERO clearance. You probably won't be able to get it through, and if you do, it'll be scratched as all hell. For clearance, you need a ring of inner radius larger than R (less curvature) or a ring thinner than gauge G.

Here's the geometry & derivation:


What's the effective gauge of two rings worn through the same hole?

First, keep in mind that it's the perimeter around the rings, not the area they comprise, which is important here: that's what your skin needs to accomodate (going from round to square jewelry with the same xsectional area would be a stretch).

The correct formula for the perimeter around a "rounded teardrop" type shape formed by two touching circles of radius r1 and r2 is 

   P = 4*sqrt(r1*r2) + 2*(pi-a)*r1 + 2*a*r2 ,
and the correct formula for the area is 
   A = (r1+r2)*sqrt(r1*r2) + (pi-a)*r1^2 + a*r2^2,
where a is the angle in *radians* given by a = arccos[(r1-r2)/(r1+r2)] .

Here's the geometry & derivation. The teardrop is the yellow (both dark and light) region:

This is what the text says, since it's not very clear:

The total area is comprised of two identical trapezoids [like the one in dark yellow] and two partial circles [the "pacman" shaped part of the big circle (c1) and the non-pacman part of the small circle (c2)]:

A = 2Atrap + Ac1 + Ac2
Atrap = 1/2 (r1+r2) h = (r1+r2) * sqrt (r1r2)
Ac1 = ([pi] - [alpha]) r12 , where [alpha] is in radians
Ac2 = [alpha] r22

Similarly, the perimeter is comprised of two lengths h and the circumference of two partial circles:

P = 2h + Cc1 + Cc2
= 4 sqrt (r1r2) + 2 ([pi] - [alpha]) r1 + 2 [alpha] r2

Off to the side, the derivation for h and [alpha]:

h = sqrt ( (r1+r2)2 - (r1-r2)2 ) = 2 sqrt (r1r2)
[alpha] = arccos [ (r1-r2) / (r1+r2) ]

Content last modified Fri Mar 9 12:40:45 CDT 2001 by myrrh

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