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Calculating Angles Between Faces of a Solid

Date: 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, 

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 

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

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:


  = 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 
Associated Topics:
College Linear Algebra
College Polyhedra
High School Linear Algebra
High School Polyhedra

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