Packing 4 Spheres Into a TetrahedronDate: 09/03/99 at 16:11:06 From: Daniel Lam Subject: Packing 4 spheres into a tetrahedron I have 4 spheres. I need to pack them as snugly as possible, and then find the dimensions of the smallest tetrahedron that can serve as a container for the spheres, so the spheres are in a tetrahedral arrangement (3 at the bottom in a triangular fashion, with one on top.) The method I have been trying involves the angle between the faces (I believe it's 73.5 degrees) but I can't show conclusively that this is correct. I have been advised to construct a tetrahedron with vertices at the center of each sphere, so that its edge length is 2r (where r is the radius of the sphere), but I don't know where to go from there. Date: 09/04/99 at 17:48:12 From: Doctor Mitteldorf Subject: Re: Packing 4 spheres into a tetrahedron Dear Dan, This is a sophisticated problem, which requires many steps and a good ability to visualize in 3 dimensions. The latter has always been difficult for me, and requires all the concentration I can muster. There are many ways to go about it. I'll suggest one. The Rosetta stone for a great number of 3-D geometry problems is a relation I call the "hinge formula." It can be applied at each of the corners of the tetrahedron, and also at the middle. Draw two angles on a hinge, with the spine of the hinge for their common edge. What is the angle between the two lines that you drew? The hinge formula says cos(c) = cos(a)cos(b) + sin(a)sin(b)cos(t) where angles a and b are painted on the hinge with the axis of the hinge for a common side, the hinge is open to a plane angle t, and c is the angle formed by the outer limbs of angles a and b. If the hinge is closed, then the angle c is the difference of a and b, and the hinge formula reduces to the formula for the cosine of the difference of two angles. If the hinge is open flat, then angle c is the sum of a and b, and the hinge formula reduces to the formula for the cosine of the sum. (Remember that the cosine of the sum has a minus sign in it, and the cosine of the difference has a plus sign.) But the interesting cases are the ones in between. You can apply the hinge formula at any vertex of the tetrahedron, where a, b and c are all 60 degrees, and solve for t, which is the angle formed by any two faces of the tetrahedron. Now draw 4 lines from each vertex of the tetrahedron that meet at the solid center. Imagine these lines on 4 fins, planes that meet at 120 degrees. Use the hinge formula again with t = 120 degrees to solve for a, b and c, which are all the same in this case. The angle you've found is the "tetrahedral angle," the angle formed by any two lines from the center of a tetrahedron to two different vertices. What angle do these lines make when they meet the edges of the tetrahedron at a vertex? Again, use the hinge formula, this time imagining the hinge open to 120 degrees and the line to the center on the spine of the hinge. The angle you want is either a or b (they are equal) and the angle t is 120 degrees. The angle between two of the edges that meet at the vertex is c, which is 60 degrees. Now you have all the angles you need. Do a little plane geometry to locate the center of one face. Use the angles you have and trigonometry definitions to find the vertical height of the tetrahedron: the distance between any vertex and the center of the opposite face. Also, use the angles you've found to compute the distance between the solid center and the center of one face, and subtract these two to give the distance between the solid center and any vertex. If you get this far, you've calculated all the dimensions and angles of the tetrahedron, and there's not far to go to get your answer. Take the center of each ball as a vertex to make a small tetrahedron. You know that the edge of that small tetrahedron has length 2r, and you can use the above calculation to find the distance between the solid center and the 3-ball center, i.e. the distance between the midpoint of all 4 balls and the midpoint of any three balls. Imagine drawing that line and extending it past the plane of the three-ball-centers by an additional distance r. Now it just reaches the surface of the large tetrahedron, the one that can enclose all 4 balls. You know then, the distance between any face of this large tetrahedron and its center, and you can use all the calculations you did at the beginning to work backwards and find the length of one edge of the large tetrahedron in terms of r. Why don't you get started with this calculation? If you get all the way to the end, please send me your answer and I'll check it against what I get. If you get stuck, be sure to write again and I'll try to fill in some more details of the calculation. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 09/11/99 at 09:45:50 From: Daniel Lam Subject: Re: Packing 4 spheres into a tetrahedron Dear Dr. Mitteldorf, Thanks very much for replying to my question. Your reply gave me a lot of insight into the problem, and I now have a good idea of what has to be done, but I haven't arrived at a correct answer (the balls don't fit into the box) The reason could be, in order of likelihood: 1) I made a mistake in the calculation 2) The method is incorrect 3) The solution doesn't apply to reality. Here are my results for the calculation. Maybe you can tell me where I went wrong. Angle between two faces of a tetrahedron: Approx. 70.528 degrees Tetrahedral Angle: Approx. 109.471 degrees Angle where line from center meets vertex: Approx. 35.264 degrees Center of one face (where e is the length of an edge): e/[2 * sqrt(3)] from a perpendicular to an edge Height of a tetrahedron (where f is the angle between two faces): (e tan f)/[2 * sqrt(3)] Distance from solid center to center of a face (where d is half the tetrahedral angle): 2e / [2 * sqrt(3) * tan d] Distance between solid center and vertex: e(tan f tan d - 2) / [2 * sqrt(3) * tan d] Distance from 4 ball center to 3 ball center: (e = 2r) 4r / [2 * sqrt(3) * tan d] Distance from center of large tetrahedron to face: 4r / [2 * sqrt(3) * tan d] + r and 2e / [2 * sqrt(3) * tan d] therefore 4r / [2 * sqrt(3) * tan d] + r = 2e / [2 * sqrt(3) * tan d] therefore e = r[2 + sqrt(3) * tan d) = approx. 4.45r Measuring the radius of a tennis ball to approx. 3.28 cm, the length of an edge of the box would be approx. 14.594 cm. But, I built a box of this size and 4 tennis balls definitely don't fit. By trial and error, I found that the edge needs to be about 21 cm, so my answer is quite a long way off. Date: 10/07/1999 at 16:51:55 From: Doctor Mitteldorf Subject: Re: [Re: Packing 4 Spheres into a tetrahedron] Dear Dan, You're doing great - you clearly understand the whole process, and you've done all the little algebra problems. I think the only reason the answer didn't come out right is that you lost a factor of 2 in the following place. I get 2r*[1+sqrt(6)] = 6.899 r as an answer. >Distance from solid center to center of a face (where d is half the >tetrahedral angle): > > 2e / [2 * sqrt(3) * tan d] ***************************** Check this ***************************** ************* I think you're a factor of 2 too big here ************** One suggestion: you don't need to express any of this in terms of trig functions, because all the trig functions derive from geometry, so they are all fractions or radicals. For example, tan(d) = sqrt(2) and tan(f) = 2sqrt(2). If you want to stretch a bit further, here's a challenge: I realized there's a way to derive the tetrahedral angle without the "hinge formula." An edge and an adjacent altitude are part of a right triangle whose base is the line between the center of a triangle and its vertex, which we know from plane geometry to be e/sqrt(3). This gives us a simpler way to calculate the 35.26-degree angle. Then consider the triangle formed by two lines center-to-vertex and one edge. The fact that it's an isosceles triangle gives us a way to calculate the tetrahedral angle in terms of the 35.26-degree angle. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/