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Rectangle Needed To Make Cone

Date: 9/5/96 at 8:34:50
From: Dean Thompson
Subject: Rectangle Needed To Make Cone

This is practical problem. I want to make a lampshade (actually many 
of them). It is made by rolling a piece of paper into a non-closed 
cone shape. The base is wider than the top. 

Given: the length of a side, the top diameter, and the bottom 

I need a formula that will take those three variables and output the 
X and Y dimensions of the piece of paper which will just inscribe the 
pattern. I can't even figure out whether this is a geometry, trig, or 
calculus problem.


Date: 9/7/96 at 6:45:5
From: Doctor Pete
Subject: Re: Rectangle Needed To Make Cone

You probably don't need calculus for this.  Here are a few ideas that 
might help:

When you take the lateral surface of a cone (i.e., without the 
circular base), cut it down the side, and unroll it flat, you get a 
sector of a circle.  Furthermore, the radius of the circle which is 
formed by the edge of the flattened paper is the length of the side.

So picture your lampshade as part of a cone; the cone's apex is 
missing, and what we want to find out is the length of the cone's side 
in terms of the length of the lampshade's side and the two radii.  In 
pictures, we have

                 / | \ m
                /  |  \
              /    | x  \
             /     |     \ n
            /      |      \
           /       |       \
                   |    y

So we want m+n in terms of n, x, and y.  By similar triangles, we see 

       m      m + n
     ----- = -------
       x        y

and solving for m, we have m = nx/(y-x).  So m+n = ny/(y-x).  

Now, the second quantity we need to find is the angle t which sweeps 
out the sector of the flattened lampshade.  This is simple, since 
t(m+n) is the length of the arc of the sector, which corresponds to 
the circumference of the lampshade's bottom, 2*Pi*y.  

Thus t = (m+n)/(2*Pi*y) = n/(2*Pi*(y-x)) when t is in radians.

Now, the hard part is figuring out the smallest rectangle that fits 
such a shape.  If t is greater than Pi radians, it follows that the 
optimal rectangle is symmetrically arranged (this is really hard to 
describe, but the sector's axis of symmetry should coincide with one 
of the rectangle's axes of symmetry).  If t is less than Pi radians, 
the optimal way to draw the rectangle will depend on the value of x, 
y, and n.  In some cases, the optimal rectangle will be drawn such 
that one side will coincide with a radius of the sector.

-Doctor Pete,  The Math Forum
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Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry
High School Triangles and Other Polygons

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