

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 "cutable" material, and you could use the templates to serve as a guide for
matching the pots and their lids. Cost is ~$60. Their email is: www.altosezmats.com and phone number:
18002252497. 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

