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Volume of a Soccer BallDate: 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: 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 |
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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
