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Flattening the Frustum of a Cone


Date: 10/15/2000 at 23:21:06
From: Jonathan Scholl
Subject: Calculating the flat layout from the section of a cone

We are building a reception desk that has a curved and canted front, 
i.e. it forms a section of a cone. We know the radius of this section 
at its top and at the floor and we know the chord length at each of 
those locations.

Our question is; how do we generate a flat layout of this section? We 
want to lay out veneer so that when wrapped around the section it will 
fit the shape exactly.

Any help would be much appreciated.

Thanks,
Jon Scholl


Date: 10/16/2000 at 17:10:25
From: Doctor Peterson
Subject: Re: Calculating the flat layout from the section of a cone

Hi, Jonathan.

We receive similar questions from time to time for practical problems, 
and you can find several answers to this sort of thing in our 
archives; but not quite exactly what you want. I think the shape you 
are looking for is the shaded part of the lateral surface of a frustum 
of a cone, cut radially as shown in the top view below; I'll draw it 
with the smaller end at the top for convenience, though I think you 
want it upside-down:

Side view:
                           R1
                        |----->|
     ---         +------+--+---+
      ^         /|      |  |...|\
      |        / |      |   |..|.\
     H|       /  |      |   |..|..\ L
      |      /   |      |    |.|...\
      |     /    |      |    |.|....\
      v    /     |      |     ||.....\
     ---  +------+------+-----++------+
                        |------------>|
                             R2

Top view:

                   ***********
                ***          /****
             ***            /.....**
            *         *****/........*
           *       ***    /***.......*
          *      **      /    **......*
          *      *      + a    *......*
          *      **      \    **......*
           *       ***    \***.......*
            *         *****\........*
             ***            \.....**
                ***          \****
                   ***********

The circumferences of the top and bottom circles are

     C1 = 2 pi R1
     C2 = 2 pi R2

We'll also want the slant height L, which we can find from the actual 
height H using the Pythagorean theorem as

     L = sqrt[H^2 + (R2 - R1)^2]

Now if we take the surface of the WHOLE frustum (I'll reduce this to 
the section you want shortly), it will look like this when we flatten 
it out:

                        ***********
                  ******...........******  C2
               *** \.....................***
            ***     \.......................***
           *         \.........................*
          *          L\.........................*
        **             \.........................**
       *                \..........................*
      *                  \.*****.C1.................*
      *                 ***     ***.................*
     *                 *   \       *.................*
     *                *     \ A     *................*
     *                *      +------*----------------*
     *                *      |  r1  *        L       *
     *                 *     |     *                 *
      *                 ***  |  ***                 *
      *                    **|**                    *
       *                     |                     *
        **                   |r2                 **
          *                  |                  *
           **                |                **
             **              |              **
               ***           |           ***
                  *******    |    *******
                         *********

Given two arc lengths C1 and C2 (which are the top and bottom 
circumferences of our whole frustum), and a segment length L (our 
slant height), we need to find the two radii r1 and r2 and the angle A 
in order to draw the right shape.

Let's write equations for the numbers we have in terms of the numbers 
we want:

     C1 = A * r1    (A is the angle in radians)
     C2 = A * r2
     L  = r2 - r1

Now we can solve this for A, r1, and r2:

     C2 - C1 = A(r2 - r1) = A*L

so
          C2 - C1
     A  = ------- radians
             L

          C1   C1 * L
     r1 = -- = -------
          A    C2 - C1

          C2   C2 * L
     r2 = -- = -------
          A    C2 - C1

Plugging in the formulas for C1, C2, and L in terms of R1, R2, and H, 
and changing the angle to degrees by multiplying it by 180/pi, we get

     L  = sqrt[H^2 + (R2 - R1)^2]

          R2 - R1
     A  = ------- * 360 degrees
             L

          R1 * L
     r1 = -------
          R2 - R1

          R2 * L
     r2 = -------
          R2 - R1

I can simplify this a bit more by first calculating

            L               H
     X = ------- = sqrt[(-------)^2 + 1]
         R2 - R1         R2 - R1

and using that to get

     A  = 360/X degrees
 
     r1 = R1 * X

     r2 = R2 * X

Now, that gives you the whole frustum. To get just one part, you have 
to determine the angle I called "a" on the original top view. You can 
find this from the chord D2, and the corresponding radius R2 (or the 
corresponding pair at the other end):

              ***********+         ---
           ***          /****       ^
        ***          R2/.|...**     |
       *         *****/..|D2/2.*    |
      *       ***    /***|......*   |
     *      **      /a/2 |*......*  |
     *      *      +-----+*------*  | D2
     *      **      \    **......*  |
      *       ***    \***.......*   |
       *         *****\........*    |
        ***            \.....**     |
           ***          \****       v
              ***********          ---

Then
                D2/2
     sin(a/2) = ----
                 R2
and
     a = 2 arcsin(D2/(2 R2))

Now you just have to take a/360 (if you found A in degrees) of the 
angle in our solution, so that we multiply the angle A by a/360:

     A  = 360/X * a/360 = 2 arcsin(D2/(2 R2)) / X degrees

     r1 = R1 * X

     r2 = R2 * X

Use these radii and this central angle, and you have a pattern for the 
surface you need. Note, however, that under my assumptions of the 
shape, you will find that

     D1/R1 = D2/R2

If this is not true, then the shape is not what I assumed; in 
particular, the side edges will not be straight.

Let me know if I can help further.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry

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