Frustum of a Cone

Date: 12/09/96 at 00:55:37
From: Bill Blackburn
Subject: Cones ?

Hi,

My question is about cones.  I am trying to find the formula for the
following:

If you take a cone of a given size and cut it from the small end to 
the large end, then lay it out on a flat surface, what will the inside 
and outside sizes of this flattened-out cone be? 

Example:
   x = diameter of small end of cone   x is 3"
   y = diameter of large end of cone   y is 6"
   z = length of cone                  z is 2"

Any help or suggestions would be greatly appreciated.  Thank you.

Bill Blackburn

Date: 12/11/96 at 21:12:03
From: Doctor Rob
Subject: Re: Cones ?

Interesting problem!

Actually, you seem to be dealing with what is called a "frustum" of a 
cone.  A cone usually refers to a figure formed as follows: Start 
with a plane and a circle lying in it. The circle is called the base 
of the cone. Construct the line perpendicular to the plane and 
passing through the center of the circle. Choose a point, called the 
vertex, on that line but not in the plane. Now consider all the lines 
connecting the vertex to points on the starting circle. The distance 
from the vertex to the center of the base is called the height of the 
cone. The distance from the vertex to the points on the circle is 
called the slant-height of the cone.

A frustum of the cone is gotten by cutting the cone with a plane 
parallel to the base and lying between the vertex and the base, and 
discarding everything on the same side of this cutting plane as the 
vertex.

If I understand your question properly, you get a figure in the plane 
which looks like this:

           ______________
          /-.    d     .-\
         /   `--....--'   \
        /       Pi*x       \
      s/                    \s
      /                      \
     /________________________\
     `-.         D          .-'
        `--...________...--'
                Pi*y

The upper arc length is Pi*x, the lower arc length is Pi*y.  

I have a small problem with your term "length" here.  I will assume 
you mean the height of the frustum, which is the perpendicular 
distance between the two bases.  This is not the same as the slant-
height, which would be the length s of the two sides in the above 
diagram.

If you extend the frustum to a full cone, then unroll it in the plane, 
you would have the same effect as extending the two sides in the above 
diagram until they meet and point P.  Call the angle formed by these 
two extended lines A.  Then the upper and lower arcs are arcs of 
circles centered at P.  

If we look at a cross-section of the cone taken through the center 
line, we would get a diagram like this:

                  P
                 /|\
                / | \ 
               /  |  \
              /   |   \
             /    |    \
            /     |     \
           /      |h-z   \ S
          /       |       \
         /________|________\
        /   x/2   |   x/2   \
       /          |z        s\
      /           |           \
     /____________|____________\
           y/2         y/2

If we let h be the height of the entire cone, then using similar
triangles, we can get the ratios h/(y/2) = (h-z)/(x/2), which we can 
solve for h = y*z/(y-x) and h-z = x*z/(y-x).  The slant-height of the 
frustum is then s = Sqrt[z^2 + (y-x)^2/4], which is the length of the 
two sides in the first diagram.

If you want to know the length D of the horizontal line in the first
diagram, that will tell you the width of the figure.  It can be 
computed as the slant-height of the full cone times 2*sin(A/2).  The 
slant-height of the full cone from the second diagram is
S = Sqrt[h^2 + y^2/4], and the angle A is given by Pi*y/S = Pi*x/(S-s).
D = 2*S*sin(Pi*y/(2*S)). Similarly, d = 2*(S-s)*sin(Pi*y/(2*S)).

The height of the first diagram, from the line marked d to the arc
marked Pi*y, would be given by S - (S-s)*cos(Pi*y/(2*S)).

If you don't understand this, or if I have misinterpreted your 
question, please feel free to write back and ask for further 
clarification.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/