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Old 09-11-2008, 11:43 AM
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Location: Tennessee
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Default Re: Multi-Dimensional Sterling Bracelet

Here's a re-post from the comments section. Anyone that comments here in this thread will be answered in the comments thread in the regular forum ... thank-you...


ok Roger ... you triggered an obsessive compulsive moment with the word "calculate".

Please remember everybody, that I am not a mathematician or geometry wizard nor have I ever played one on television.

Let's consider this to be a hollow cylinder in which a hollow cylinder is a solid bounded by two co-axial cylinders of the same height.

C = outer circumference
c = inner circumference
D = outer diameter
d = inner diameter
R = outer radius
r = inner radius

Pi = 3.14
C = 6 inches (or standard bracelet length)

Formula: D = C/Pi or D = 6/3.14 = 1.9

R = 1.9/2 = .95

We also know this:

1. Open-work/top sterling thickness = 20Ga = .032"
2. Square wire separator = 12Ga = .081"
3. Western/bottom sterling thickness = 22Ga = .025"

for a total thickness of 0.138"

Our inner radius (r) = .95-0.138 = 0.812

Our inner diameter (d) = 0.812 X 2 = 1.624

inner circumference (c) = pi X 1.624 = 5.1" Basically, I need to remove less than .5 inches from each end of the bottom layer.

Technically I should split the difference between each of the thicknesses of the top and bottom pieces (average). In order to keep this simple :smilie5: for our fellow readers (and Conservative) I will just exclude the thickness of the bottom layer in calculating the inner radius.

That inner radius, excluding bottom thickness of the western bright cut, is 0.95 - 0.113 = 0.837

our conservative inner circumference is now: c = pi X (0.837 X 2) = 5.256 ..... let's call it 5.25 inches.

That's 3/4" less than our standard bracelet length of 6 inches .... So, I'll take about 3/8ths of an inch off each end of the bottom layer.

QED (and just in time for lunch)