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Building a Cone


Date: 10/28/2001 at 23:32:24
From: Gary Schwarze
Subject: Building a cone

I am trying to find a formula for building a cone. This particular 
one is for a chimney flashing. It would have to be 21 inches tall with 
a top opening of 8 inches and a bottom opening of 20 inches. I would 
need a vertical seam overlap of 2 inches. Is there a formula that 
would provide me the proper arcs for top and bottom? 

I built one by using a hit-and-miss method of drawing on a huge piece 
of cardboard and folding and cutting until it sat perfectly vertical 
on the floor.

Thank you.


Date: 10/29/2001 at 03:34:08
From: Doctor Jeremiah
Subject: Re: Building a cone

Hi Gary,

First of all I am going to assume that the 8 inches and 20 inches are 
the diameters of the openings (all the way across)  That means the 
radius of one openings is 4 inches and the other is 10 inches (because
the radius goes only from the center to the edge).

I am also going to assume that the 21 inches is the real vertical
height of just the flashing, and not the measurement of the length of 
the side. That is what you said, but the difference is important.

Look at the diagrams and if I didn't understand completely what you 
were asking, please let me know so I can change my assumptions.

I will use * to mean multiplication (I could use the letter x but it 
might get confusing). I will use ^ to mean exponentiation (so ^2 will
mean squared).

A whole cone (without the hole in the top) is a huge circle with a 
radius of the total side length:

           +++++
     ++++         +++
      \                +++
       \                    +
        \                     +
         \                     +   \
          \                     +   \
           \                     + 2*Pi*10 around outside
            \                     +  |
             \                    +  }
              +-------------------+ ---
              |---side length-----|

The length of the outside is 2*Pi*10, but it doesn't go all the way 
around. The radius is the total side length of a full cone (another 
value we don't have).

You mentioned that 21 inches is only from the 8-inch opening to the 
20-inch opening. If we look at the rolled-up cone from the side:

              |---21---|------L----|
              |        |           |
  --+-------- +        |           |
    |         |  +     |           |
    |         |     +  |           |
    |   --+-- |        +           |
   10     |   |        |  +        |
    |     4   |        |     +     |
    |     |   |        |        +  |
  --+-----+-- |        |           +
    |     |   |        |        +   \
    |     4   |        |     +       \
   10     |   |        |  +
    |   --+-- |        +       B
    |         |     +   \
    |         |  +       \
  --+-------- +
               \    A
                \

The total side length is A+B but we don't need to worry about that 
yet. First we need to find L. What we have there is TWO triangles.  
Here is just one of them:

              |---21---|------L----|
              |        |           |
  --+-----+-- +--------+-----------+
    |     |   |        |      C +
    |     4   |        |     +
   10     |   |        |  +
    |   --+-- |        +
    |         |     +
    |         |  +
  --+-------- +

I have labelled the angle C in the triangle. There are really two 
triangles in this picture as well, a small one and a big one. Let's 
look at the small one first:

        |------L----|
        |           |
  --+-- +-----------+
    |   |      C +
    4   |     +
    |   |  +
  --+-- +

Using trigonometry we can say that the tangent of C is equal to the 
opposite length (4) divided by the adjacent length (L), or:

  tan(C) = 4/L

Now let's look at the big triangle:

        |-------21+L---------|
        |                    |
  --+-- +--------------------+
    |   |              C  +
    |   |              +
   10   |           +
    |   |        +
    |   |     +
    |   |  +
  --+-- +

And this one gives us:

  tan(C) = 10/(21+L)

They are both equal to tan(C), so we can make them equal to each 
other:

  tan(C) = 4/L = 10/(21+L)
  4/L = 10/(21+L)
  4(21+L) = 10L
  4*21 + 4L = 10L
  4*21 = 10L-4L
  4*21 = (10-4)L
  4*21/(10-4) = L
  L = 4(21)/(10-4)

Which gives us:

  L = 84/6
  L = 14

Aha! Now we have L. L is 14, which means that we can calculate A and 
B, get the side length, and know how big the circle must be.

Go back to the small triangle:

        |-----14----|
        |           |
  --+-- +-----------+
    |   |        +   \
    4   |     +       \
    |   |  +
  --+-- +       B
         \
          \

Perhaps you have heard of Pythagoras and his theorem? The theorem says 
that if we want to know the long side of a right triangle, the square 
of the long side is equal to the sum of the squares of the other 
sides:

  B^2 = 4^2 + 14^2
  B^2 = 16 + 196
  B^2 = 212
  B = sqrt(212)    (I use sqrt for the square root)
  B = 14.56

To find A we need to look at the large triangle:

        |-------21+14--------|
        |                    |
  --+-- +--------------------+
    |   |                 +   \
    |   |              +       \
   10   |           +
    |   |        +
    |   |     +
    |   |  +        A+B
  --+-- +
         \
          \

And again we look to Pythagoras:

      (A+B)^2 = 10^2 + (21+14)^2
  (A+14.56)^2 = 10^2 + (21+14)^2
  (A+14.56)^2 = 10^2 + 35^2
  (A+14.56)^2 = 100 + 1225
  (A+14.56)^2 = 1325
    A + 14.56 = sqrt(1325)
    A + 14.56 = 36.4
            A = 21.84

So our total side length (A+B) equals 36.4, and that is the radius of 
the circle you need to make to get a cone the right height and width.

So say you draw a circle that has a radius of 36.4  Then what? Well, 
you need a smaller circle too (for the top hole). That one will have a 
radius of 14.56, and the only thing left is to figure out where to cut 
into it to make the circle into a cone:

           +++++
     ++++         +++
      \                +++
       \                    +
        \                     +
         \                     +   \
          \  ++++               +   \
           \       +             + 62.83 around
            \        +            +  |
             \        +           +  }
              +-------+-----------+ ---
              |       |           |
              |-14.56-|---21.84---|
              |                   |
              |-------36.4--------|

We need to know where to cut to make the cone shape. We know the 
distance around is 2*Pi*10, which is equal to 62.83, so if you were to 
cut a string that long and measure around your big circle, you would 
know how much needed to be cut out.

Don't cut yet. You need to add the 2-inch overlap. But the overlap 
doesn't get smaller as you go toward the center, so it would look like 
this:

    \      +++++
  2  ++++         +++
 \  / \                +++
   \   \                    +
    \   \                     +
     \   \                     +   \
      \   \  ++++               +   \
       \   \       +             + 64.83 around (including overlap)
        \   \        +            +  |
         \   \        +           +  }
          \---+-------+-----------+ ---
              |       |           |
              |-14.56-|---21.84---|
              |                   |
              |-------36.4--------|

Measuring around with a string may seem silly, but it is the easiest 
way to measure it. If you want a more mathematical answer, then in 
percent it is the distance around the outside (62.83) divided by the 
total circumference of the circle (2*Pi*36.4) or:

  % of circle = 62.83 / 228.71 = 27.5 percent.

Since circles have 360 degrees, our cone will have 27.5 percent of 
that or:

  360 * .275 = 98.9 degrees

You could measure that with a protractor, but it would need to be a 
really big protractor or else it wouldn't be very accurate. The string 
idea might even be more accurate, especially if you are very careful.

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry
High School Triangles and Other Polygons

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