Calculating Angles Between Faces of a SolidDate: 09/15/2003 at 13:15:27 From: Dave Kiefer Subject: Geometry - Calculating angles between faces of a solid I am making a Great Rhombicosidodecahedron, http://mathworld.wolfram.com/GreatRhombicosidodecahedron.html and I need to calculate the angle between the square-hexagon, hexagon- decagon, and square-decagon faces, i.e., the dihedral angles. Would the angles between faces be the same? Or would I have to calculate the angles between the different types of faces? For example, if I were to make the faces of a flat material, with a "clip" adapter to hold them together to assemble the solid, would I need one type of clip, or three? I can't begin to address this mathematically - I've tried with models and tools to measure the angles, but the results aren't accurate enough! Date: 09/15/2003 at 16:27:34 From: Doctor Douglas Subject: Re: Geometry - Calculating angles between faces of a solid Hi Dave. The dihedral angles are not all the same, because the polygons are not all identical. You can perhaps convince yourself of this by imagining the vertex at which two squares and a hexagon meet. Suppose that you place the vertex at the origin: y O is the origin, and we are looking | downward from the top (along -z). D-----E The zipping angle (EOF) becomes zero | | .F. when E and F meet. A-----O-------G--x Square ABCO lies in the xy-plane. | | | Square ADEO is tilted upward by | | | "folding" on line AO. B-----C . . H Hexagon COFGHI is tilted upward by I "folding" on line OC. It seems reasonable that you will have to fold the hexagon upward by a greater amount than you will have to fold square ADEO. After all, when you are done, you will have formed a hexagonal prism, and clearly the dihedral angle between the two squares is 120 degrees, while the dihedral angle between the hexagon and a neighboring square is 90 degrees. Hence, dihedral angles between faces with fewer sides ought to be shallower (dihedral angle closer to 180 degrees). So, on to calculating the dihedral angles for your problem (square/ hexagon/decagon). Because you have a square, it is convenient to place it in the xy-plane as above, and find the vector V=(vx,vy,vz) that represents where the zipping angle is completely closed. This vector will point somewhere in the first octant (all of its components are positive). We also can refer to the point V as a point on the decagon/hexagon edge, exactly 1 unit from the origin. Then angle AOV must measure 120 degrees (square/hexagon edge folds on AO), and angle COV must measure 144 degrees (square/decagon edge folds on CO). Using the properties of dot products: (-1,0,0).(vx,vy,vz) = -vx = cos(120 deg) (0,-1,0).(vx,yz,vz) = -vy = cos(144 deg) So that vx = -cos(120 deg) = 1/2 and vy = -cos(144 deg) = 0.809017. We can obtain vz from the fact that V is on the unit sphere: vz = sqrt(1 - vx^2 - vy^2) = sqrt(1 - 0.25 - 0.6545085) = sqrt(0.095491) = 0.309017 Note that vy^2 + vz^2 = 1 - vx^2 = 1 - 0.5^2 = 0.75. On to computing the dihedral angle. Let's first try to find the angle between the hexagon and square. For this we need to take the dot product between the normal vectors to each face. The normal to the square is obviously (0,0,-1) and the normal to the hexagon can be found from the cross product between the vectors OA and OV: OA x OV = (-1,0,0) x (vx,vy,vz) = (0,vz,-vy) So that the angle between the planes is given by Q(square,hexagon) = arccos{(0,0,-1).(0,vz,-vy)/[sqrt(vy^2+vz^2)]} = arccos[vy/sqrt(0.75)] = arccos[0.809017/0.866025] = 20.905 degrees The dihedral angle is the supplementary angle to this, or D(square,hexagon) = 180 deg - 20.905 deg = 159.095 degrees. To obtain the dihedral angles involving the decagon, you'll need the normal to that face, which can be found from the cross product of OC with OV: OC x OV = (0,-1,0) x (vx,vy,vz) = (-vz,0,vx) Then the angle between the square and the decagon is Q(square,decagon) = arccos{(0,0,-1).(-vz,0,vx)/[sqrt(vz^2+vx^2)]} = arccos[-vx/sqrt(0.095491 + 0.25)] = arccos[-0.5/sqrt(0.345491)] = 148.28 degrees This is already the dihedral angle, because it is bigger than 90 degrees. Finally, we can compute the remaining dihedral angle (hexagon-decagon) by using the arccosine formula and the dot product. We already have the two required normals OA x OV and OC x OV, and we must remember to be careful to divide by the lengths L in the dot product: Q(hexagon/decagon) = arccos{(OAxOV).(OCxOV) / [L(OAxOV)L(OCxOV)]} = arccos{(0,vz,-vy).(-vz,0,vx) / [sqrt(vz^2+vy^2)sqrt(vx^2+vz^2)]} = arccos{(vx vy) / [sqrt(0.75)sqrt(0.345491)]} = arccos{0.5 x 0.809017 / [sqrt(0.75)sqrt(0.345491)]} = arccos(0.794655) = 37.377 degrees so that the dihedral angle between the hexagon and decagon is 180 degrees - 37.377 degrees = 142.62 degrees. So the angles we're looking for are hexagon-decagon: 142.62 degrees square-decagon: 148.28 degrees hexagon-square: 159.095 degrees It makes sense that the square/hexagon dihedral angle is is the largest, because the square and hexagon have the fewest sides, and that the hexagon/decagon dihedral angle is the smallest, by the reasoning above. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ |
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