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calling all clayart mathematicians (long)

updated wed 16 feb 05


D Bouchette on tue 15 feb 05

If you read all this, you'll understand why most
potters get a chain, a blank wall, a coupla nails, a
tape measure & a level, and use those to define a

The formula for a generic catenary arch is

e^x + e^(-x) = y

where the ^ symbol means "raised to the power of"
(since I'm not using superscript formatting).

"e" is an irrational constant with the approximate
value of 2.718281828.

This gives you a catenary arch "upside down" (concave
up) with its apex at (0,2).

(I.e., substituting 0 for each x, we get
e^0 + e^(-0) = 2
--remember, anything raised to the 0 power gives the
value 1.)

But that formula isn't very useful as it stands.

For any given horizontal span, there is an infinite
number of vertical spans. (Pick your horizontal span,
and you could drop any length of chain to create an
arch of different vertical spans.)

For any given vertical span, there is an infinite
number of horizontal spans. (Pick your height, and
you could find a length of chain that makes an arch of
any horizontal span for that height.)

So it's hard to answer your question about "would the
arch always be the same?". If you mean, given a
defined catenary arch and hack it off to a shorter
height, would the remaining curve still be a catenary
arch? As long as you accept the reduced width at its
base (and all the remaining points on the curve), Yes.

What we really want is to define the height "h" and
width "w" of a catenary arch and pop out x,y pairs to
plot the rest of the arch. One of us (Prune) derived
this very close approximation (with the arch rightside

y = [-1 * {e^(a*x) + e^(-a*x)}] + (h-2)
where a = [2 * {ln(h-2)}] / w

and "ln" is the "natural log"

(We're using ( ), { }, and [ ] as parentheses pairs so
you can better see the required order of operations.)

For a kiln that is 5 feet wide at its base and 5-1/2
feet high at its peak, let's find the height of the
side above the point at the base halfway from the
middle to the outside. (At the base, the middle point
is 2-1/2 feet in from the side, and half of that is
1-1/4 feet in from the side.)

Use inches or centimeters to get better accuracy (than
feet or meters).
w = 60 inches (5 feet)
h = 66 inches (5-1/2 feet)
x = 15 inches (1-1/4 feet)

First we figure "a"
a = [2 * {ln(66-2)}] / 60
a = 0.138629436
then we figure "y"
y = [-1 * {2.718281828^(0.138629436*15) +
2.718281828^(-0.138629436*15)}] + (66-2)
y = 55.875 inches

Figure other points the same way.

Deb Bouchette and Prune Wickart
Hillsboro, Oregon

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