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Volume of a Soccer Ball

Date: 05/31/2003 at 01:06:18
From: K C
Subject: Volume of a Soccer Ball

What is the simplest way to find the volume of a truncated 
icosahedron (aka soccer ball or buckyball)?  Is it possible to do 
using only skills from Algebra I and Geometry?

I haven't been able to "split" the polyhedron into less complicated 
shapes to calculate the volume (as you can do with a dodecahedron).


Date: 05/31/2003 at 11:10:38
From: Doctor Jeremiah
Subject: Re: Volume of a Soccer Ball

Hi K C, thanks for writing.

To calculate the volume of a soccer ball or Buckminster Fullerine 
molecule you have to first realize that the shape is actually a 
truncated icosahedron. The easiest way, therefore, to calculate the 
volume is to first calculate the volume of the icosahedron and 
afterward calculate the volume of the bits that were cut off. 

An icosahedron is a 20-sided polyhedron made entirely of equilateral 
triangles, with five such triangles meeting at every vertex. So if 
you visualize an icosahedron with one vertex straight up at the top, 
then the top bit will look like a five-sided pyramid. Each vertex of 
the pyramid will be "R" distance from the center of the icosahedron.  
So really there are two pyramids: one pointing up and one pointing 
down to the center. Here is a picture of the top of the icosahedron 
and the five-sided pyramid that is formed at the topmost point:



Before we start we need to know that:

    1 - cos(72) = 1 - cos(36+36)
                = 1 - [cos(36)^2 - sin(36)^2]
                = 1 - cos(36)^2 + sin(36)^2
                = sin(36)^2 + sin(36)^2
                = 2 sin(36)^2

This isn't "needed" but it should make the equations smaller and more 
manageable.

Now, in the base of the pyramids there are five triangles.  Each 
triangle in the base of the pyramid looks like:

                   +
                  / \
                 /   \
                / 72  \
               /       \
            a /         \ a
             /           \
            /             \
           +---------------+
                   s

If we use the cosine law on this we get:

    s^2 = a^2 + a^2 - 2 a a cos(72)
    s^2 = 2a^2 - 2a^2 cos(72)
    s^2 = a^2 2(1 - cos(72))
    s^2 = a^2 2(1 - cos(36+36))
    s^2 = a^2 2(1 - (cos(36)^2 - sin(36)^2))
    s^2 = a^2 2(1 - cos(36)^2 + sin(36)^2)
    s^2 = a^2 2(sin(36)^2 + sin(36)^2)
    s^2 = 4 a^2 sin(36)^2
    a^2 = (s/2)^2 / sin(36)^2
      a = (s/2) / sin(36)

The area of that triangle can be found by finding b:

                   +          ---
                  / \          |
                 /   \         |
                /     \        |
               /       \       |b
            a /         \ a    |
             /           \     |
            /             \    |
           +---------------+  ---
                   s

    a^2 = b^2 + (s/2)^2
    b^2 = a^2 - (s/2)^2
    b^2 = s^2/[ 4 sin(36)^2 ] - s^2/4
    b^2 = s^2 [1 -  sin(36)^2]/[ 4 sin(36)^2 ]
    b^2 = s^2 [ cos(36)^2 ]/[ 4 sin(36)^2 ]
    b^2 = (s/2)^2 / tan(36)^2
      b = (s/2) / tan(36)

And knowing b we can get:

    Base_Triangle_Area = sb/2
    Base_Triangle_Area = (s/2)^2 / tan(36)

Now that we have the information about the base of the pyramids we 
can calulate the area of the base for the pyramids:

    Pyramid_Base = 5 Base_Triangle_Area
    Pyramid_Base = 5 (s/2)^2 / tan(36)

The height of the upward-pointing pyramid can be found from this 
other vertical triangle:

                          +
                      +   |
               s  +       |
              +           | h
          +               |
      +-------------------+
                a

    s^2 = a^2 + h^2
    h^2 = s^2 - a^2
    h^2 = s^2 - s^2/[ 4 sin(36)^2 ]
    h^2 = (s/2)^2 ( 4 - 1/sin(36)^2 )
      h = (s/2) sqrt( 4 - 1/sin(36)^2 )

The height of the downward pyramid is r-h. The value of r can be 
determined the same way:

                a
      +-------------------+
        +                 |
          +               |
            +             |
              +           |
                +         |(r-h)
                r +       |
                    +     |
                      +   |
                        + |
                          +

    r^2 = (r-h)^2 + a^2
    r^2 = r^2 - 2rh + h^2 + a^2
      0 = -2rh + h^2 + a^2
    2rh = h^2 + a^2
    2rh = s^2
      r = s^2 / (2h)

We can find r by substituting in the other values. And after a little 
algebra...:

      r = s^2 / ( s sqrt( 4 - 1/sin(36)^2 ) )
      r = s / sqrt( 4 - 1/sin(36)^2 )

Now we have all the info to calculate the volume of the icosahedron. 
We start by gethering all the information we know:

      r = s / sqrt( 4 - 1/sin(36)^2 )
      h = (s/2) sqrt( 4 - 1/sin(36)^2 )
      a = (s/2) / sin(36)
      b = (s/2) / tan(36)
      Pyramid_Base = 5 (s/2)^2 / tan(36)
      Base_Triangle_Area = (s/2)^2 / tan(36)

The entire upward-pointing pyramid's volume can be found like this:

 Upward_Pyramid_Volume = Pyramid_Base x Height/3
 Upward_Pyramid_Volume = 5 Base_Triangle_Area x Height/3
 Upward_Pyramid_Volume = 5 (sb/2) h/3
 Upward_Pyramid_Volume = (5/6) sbh

Now, the downward-pointing pyramid's height is r-h because the 
distance from the center to the vertex is r and h of it is inside the 
upward-pointing pyramid. So the entire downward-pointing pyramid's 
volume can be found like this:

 Downward_Pyramid_Volume = Pyramid_Base x Height/3
 Downward_Pyramid_Volume = 5 Base_Triangle_Area x Height/3
 Downward_Pyramid_Volume = 5 (sb/2) (r-h)/3
 Downward_Pyramid_Volume = (5/6) sb(r-h)

The total volume for both pyramids is:

 Double_Pyramid_Volume = Upward_Pyramid_Vol + Downward_Pyramid_Vol
 Double_Pyramid_Volume = (5/6) sbh + (5/6) sb(r-h)
 Double_Pyramid_Volume = (5/6) sbr

But we don't care about a five-sided pyramid. We want the volume 
associated with one equilateral triangular section. And there are 5 
in the pyramid so we get:

 Equilateral_Triangle_Section_Volume = Double_Pyramid_Volume/5
 Equilateral_Triangle_Section_Volume = sbr/6

But there are 20 equilateral triangles in an icosahedron so:

 Icosahedron_Volume = 20 Equilateral_Triangle_Section_Volume

 Icosahedron_Volume = (20/6) sbr

 Icosahedron_Volume
        = (20/6) s^3 / ( sqrt( 4 - 1/sin(36)^2 ) 2 tan(36) )

 Icosahedron_Volume
        = (5/3) s^3 / ( sqrt( 4 - 1/sin(36)^2 ) tan(36) )

 Icosahedron_Volume
        = (5/3) s^3 cos(36) / sqrt( 4 sin(36)^2 - 1 )

This is actually the volume, but there are other ways of calculating 
it. Eric Weisstein's World of Mathematics calculates the volume a 
different way and gets:

   Icosahedron
   http://mathworld.wolfram.com/Icosahedron.html 

   Icosahedron_Volume
        = (5/12) s^3 ( 3 + sqrt(5) )

Which, if you calculate it out, is the same number with just a simpler 
way to calculate it.


Now that we have the volume of the icosahedron all we need to do is 
calculate the volume of the bits that were cut off to make it into 
a truncated icosahedron. When you cut the vertex off, each equilateral 
triangle turns into a hexagon, and the cut off bit is a five-sided 
pyramid. The side length of the five-sided pyramid must be s/3, 
because when you change an equilateral triangle into a hexagon the 
side lengths become 1/3 as long:

                    +
                   / \
                  /   \
                 +-----+
                /       \
               /         \
              +           +
             / \         / \
            /   \       /   \
           +-----+-----+-----+

           |-s/3-|-s/3-|-s/3-|

           |--------s--------|

So we are back to finding the volume of that same five-sided upward-
pointing pyramid. The triangle in the base is almost the same except 
the distance s becomes s/3.  If we do the same thing we did before:

                   +          ---
                  / \          |
                 /   \         |
                / 72  \        |
               /       \       |e
            d /         \ d    |
             /           \     |
            /             \    |
           +---------------+  ---
                  s/3

    d^2 = (s/6)^2 / sin(36)^2
      d = (s/6) / sin(36)

    e^2 = (s/6)^2 / tan(36)^2
      e = (s/6) / tan(36)

    Base_Triangle_Area = (s/3) e/2
    Base_Triangle_Area = (s/6)^2 / tan(36)

    Pyramid_Base = 5 Base_Triangle_Area
    Pyramid_Base = 5 (s/6)^2 / tan(36)

                          +
                      +   |
             s/3  +       |
              +           | f
          +               |
      +-------------------+
                d

    (s/3)^2 = d^2 + f^2
    f^2 = (s/3)^2 - d^2
    f^2 = (s/6)^2 ( 4 - 1/sin(36)^2 )
      f = (s/6) sqrt( 4 - 1/sin(36)^2 )

 Truncated_Pyramid_Volume = Pyramid_Base x Height/3

 Truncated_Pyramid_Volume
        = (5/3) (s/6)^3 sqrt( 4 - 1/sin(36)^2 )/tan(36)

So far this is exactly the same as above except the s is now s/3. But 
this time we have 12 full five-sided pyramids that need subtracting 
(one for each vertex):

 Truncated_Volume
        = 12 Truncated_Pyramid_Volume

 Truncated_Volume
        = 12 (5/3) (s/6)^3 sqrt( 4 - 1/sin(36)^2 )/tan(36)

 Truncated_Volume
        = (60/3) (s/6)^3 sqrt( 4 - 1/sin(36)^2 )/tan(36)

And the total volume of the truncated icosahedron is:

 Total_volume = Icosahedron_Volume - Truncated_Volume

 Total_volume
        = (5/3) s^3 cos(36) / sqrt( 4 sin(36)^2 - 1 )
          - (60/3) (s/6)^3 sqrt( 4 - 1/sin(36)^2 ) / tan(36)

 Total_volume
        = (1080/648) s^3 / ( sqrt( 4 - 1/sin(36)^2 ) tan(36) )
          - (60/648) s^3 sqrt( 4 - 1/sin(36)^2 ) / tan(36)

 Total_volume
        = Icosahedron_Volume (84 + 6/sin(36)^2 )/108


That's the volume of the whole shape. But we can simplify things a 
bit more. Using these two solutions for the volume of an icosahedron:

 Icosahedron_Volume
        = (5/3) s^3 cos(36) / sqrt( 4 sin(36)^2 - 1 )
        = (5/12) s^3 ( 3 + sqrt(5) )

We can derive the following values for the cosine and sine of 36 and 
72 degrees (the others are just there for reference):

   cos 90    =    sin  0    =    (1/2) sqrt( 0 )

   cos 75    =    sin 15    =    (1/2) sqrt( 2 - sqrt(3) )

                                              3 + sqrt(5)
   cos 72    =    sin 18    =    (1/2) sqrt( ------------- )
                                             7 + 3 sqrt(5)

                                             5 + sqrt(5)
   cos 54    =    sin 36    =    (1/2) sqrt( ----------- )
                                             3 + sqrt(5)

   cos 60    =    sin 30    =    (1/2) sqrt( 1 )

   cos 45    =    sin 45    =    (1/2) sqrt( 2 )

   cos 30    =    sin 60    =    (1/2) sqrt( 3 )

                                             7 + 3 sqrt(5)
   cos 36    =    sin 54    =    (1/2) sqrt( ------------- ) 
                                              3 + sqrt(5)

                                             25 + 11 sqrt(5)
   cos 18    =    sin 72    =    (1/2) sqrt( --------------- )
                                              7 + 3 sqrt(5)

   cos 15    =    sin 75    =    (1/2) sqrt( 2 + sqrt(3) )

   cos  0    =    sin 90    =    (1/2) sqrt( 4 )


It's interesting that all of them are the square root of something 
times one half ...

Anyway, inserting the algebraic values of the trig values gives us:

 Total_volume
        = Icosahedron_Volume (84 + 6/sin(36)^2 )/108

 Total_volume
        = (5/12) s^3 ( 3 + sqrt(5) ) (84 + 6/sin(36)^2 )/108

 Total_volume
        = (210/648) s^3 ( 3 + sqrt(5) )
          + (120/648) s^3 ( 7 + 3 sqrt(5) )/( 5 + sqrt(5) )

 Total_volume
        = s^3 ( 4200 + 1680 sqrt(5) )/( 3240 + 648 sqrt(5) )
          + s^3 ( 840 + 360 sqrt(5) )/( 3240 + 648 sqrt(5) )

 Total_volume
        = s^3 ( 210 + 85 sqrt(5) )/( 135 + 27 sqrt(5) )

And that will be the volume of any truncated icosahedron (e.g. a 
soccer ball or Buckminster Fullerine molecule) where s is the side 
length of the equilateral triangles. 

But after it's been truncated the shape doesn't have equilateral 
triangles. The side length of the hexagons and pentagons is s/3, so if 
we call the side length of the soccerball "x" then x=s/3 and we get 
this:

 Total_volume
        = (3x)^3 ( 210 + 85 sqrt(5) )/( 135 + 27 sqrt(5) )

 Total_volume
        = 27 x^3 ( 210 + 85 sqrt(5) )/( 135 + 27 sqrt(5) )

 Total_volume
        = 5 x^3 ( 42 + 17 sqrt(5) )/( 5 + sqrt(5) )

Eric Weisstein's World of Mathematics calculates the volume of a 
truncated icosahedron a different way and gets:

   Truncated Icosahedron
   http://mathworld.wolfram.com/TruncatedIcosahedron.html 

    Total_volume
        = (1/4) x^3 ( 125 + 43 sqrt(5) )

Which, if you calculate it out, is the same number with just a simpler 
way to calculate it.

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 05/31/2003 at 12:07:04
From: Doctor Rob
Subject: Re: Volume of a Soccer Ball

Thanks for writing to Ask Dr. Math, K.C.

The simplest way I can see is to split it into pyramids whose bases
are the faces of the solid and whose vertices coincide at the center.
The volume of each pyramid is 1/3 the area of the base times the 
height. Then there are 20 regular hexagon bases and 12 regular
pentagon bases. The heights of the hexagonal pyramids and the
pentagonal pyramids are the hard numbers to calculate.

Let's assume that all the face edges have length 1. If we can do this 
case, we can scale to get the general case.

Start with a hexagonal pyramid with base edges of length 1. The other 
edges will have length R, the circumradius of the solid. The distance 
from each vertex of the base to the center of the base is 1 unit. The 
area of the base is 3*sqrt(3)/2. For these quantities, see the regular 
hexagon formulas in the Dr. Math Geometric Formulas FAQ:

   Regular Polygons - Regular Hexagon
   http://www.mathforum.org/dr.math/faq/formulas/faq.regpoly.html#6 

   Then the height of the hexagonal pyramid is given by sqrt(R^2-1).

Similarly for the pentagonal pyramid with base edges of length 1, the 
other edges will have length R. The distance from each vertex of the 
base to the center of the base is sqrt(2+2/sqrt[5])/2. The area of the 
base is 5*sqrt(1+2/sqrt[5])/4. For these quantities, see the regular 
pentagon formulas in the Dr. Math Geometric Formulas FAQ:

   Regular Polygons - Regular Pentagon
  http://www.mathforum.org/dr.math/faq/formulas/faq.regpoly.html#5 

   Then the height of the pentagonal pyramid is given by
   sqrt[R^2-(2+2/sqrt[5])/4].

This will give you an answer in terms of R, the circumradius. It 
remains to compute this quantity. That can be found from the
circumradius of the icosahedron that was truncated, call it R_I, as
follows. 

Draw lines from the vertices of the icosahedron to its center, all of 
length R_I. The edges of the icosahedron will all have length 3 units, 
because the truncation leaves the edges of the faces 1/3 the lengths 
of the edges of the original icosahedral faces. Then the height of the 
triangles with sides R_I, R_I, and 3 will be sqrt([R_I]^2-[3/2]^2), by 
the Pythagorean Theorem, and then

   R = sqrt([R_I]^2-[3/2]^2+[1/2]^2) = sqrt([R_I]^2-2),

also by the Pythagorean Theorem:

               o
              /|\
             /,|.\
            / ,|. \
           / , | . \
          /  , | .  \
      R_I/   , | .   \R_I
        /  R,  |  .R  \
       /    ,  |  .    \
      /     ,  |  .     \
     /     ,   |   .     \
    /  1   ,1/2|1/2.   1  \
   o-------o---o---o-------o
               3

Now it is known that R_I = 3*sqrt([5+sqrt(5)]/8) units if the edge
length is 3 units. For the derivation of this quantity, see Eric 
Weisstein's World of Mathematics:

   Icosahedron
   http://mathworld.wolfram.com/Icosahedron.html 

Thus

   R = sqrt(9*[5+sqrt(5)]/8-2) = sqrt(58+18*sqrt[5])/4.

This agrees with the quantity at:

   Truncated Icosahedron - Weisstein
   http://mathworld.wolfram.com/TruncatedIcosahedron.html 

This is the best I can do.  It uses properties of the truncation,
properties of hexagons and pentagons, together with the Pythagorean
Theorem and the volume formula for a pyramid, but that's all.

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 06/01/2003 at 01:13:31
From: K C
Subject: Thank you (Volume of a Soccer Ball)

To Doctors Jeremiah and Rob,

Thanks so much for answering my question and helping to solve my 
problem. I appreciate your time and effort immensely. THANK YOU!

Sincerely,
KC
Associated Topics:
College Polyhedra
High School Polyhedra

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