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Creating Ellipses in Pottery
Name: Mark S.
Status: Student
Age: 30s
Location: N/A
Country: N/A
Date: May 2003
Question:
I am an art student and need help with a geometry
issue. I am making teapots which are oval so the openings are
ellipses. I need to create lids which are also then reshaped as ellipses
to fit.
What I need to figure out is with a given ellipse is there a way to
determine it's circumference to use as a guide for creating the same
circumference of a circle lid?
The idea is that I use a wheel to throw the lids which makes them round
then I compress them a bit to reshape them oval. It seems like as long as
the two circumferences match they will fit?
Replies:
According to "http://home.att.net/~numericana/answer/ellipse.htm", there is no
exact solution in closed form, but this approximation is very good:
PI * (3*(a+b) - sqrt((3*a+b)*(a+3*b)))
where 2*a is the length of the major axis (the largest measurement through
the center) and 2*b is the length of the minor axis (the smallest measurement
through the center).
Tim Mooney
The perimeter,P, of an ellipse that has a major axis, the long way, equal to: 2A, and a minor axis, the
short way, equal to 2B is approximately: pi*(2(A^2 +B^2))^1/2 where pi = 3.14... And this is an
approximation. The exact analytical answer is much more messy algebraically. I am not sure this is going
to be of much practical value.
I have another suggestion of how you might approach the problem. A company, Alto Inc., sells supplies for
matting photographs, art work, etc. They make elliptical templates that you could cut out of matting board,
or cardboard, or some other "cut-able" material, and you could use the templates to serve as a guide for
matching the pots and their lids. Cost is ~$60. Their e-mail is: www.altosezmats.com and phone number:
1-800-225-2497. I do not do pottery, but this seems like a more practical and simpler solution. Since you
could save the template(s), all the tea pots could be the same shape.
Vince Calder
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Update: June 2012
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