Associated Topics || Dr. Math Home || Search Dr. Math

### 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
diameter.

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.

Thanks.
```

```
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
that

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
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry
High School Triangles and Other Polygons

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search