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Angles in a Hexagonal Pyramid

Date: 07/25/2002 at 09:34:14
From: Chris
Subject: Angles of a hexagonal cone

Dr. Math:

The pitch of the roof of a gazebo is 45 degrees, and the entire roof
can described as a hexagonal pyramid. My question is this: What is the
interior angle that the planes of the pyramid make with each other?
Or, what is the interior angle of the intersection of two roof panels
and how does this angle change with the pitch of the roof? 

Obviously, if the pitch of the roof were 0 degrees, the angle of
intersection would be 180 degrees. And if the pitch of the roof were 
90 degrees (vertical roof), the angle of intersection of the roof 
panels would be equal to 60 degrees (the interior angle of a 2D
hexagon).

For a roof pitch between 0 and 90 degrees, can you assume a linear 
relation for the angles of intersection? Also, how do these angles 
change along the length of the roof (or height of the pyramid)?  

I hope that this is not too involved and I really appreciate your 
time.

Thanks,
Chris


Date: 08/01/2002 at 02:16:58
From: Doctor Douglas
Subject: Re: Angles of a hexagonal cone

Hi, Chris,

If you have a pyramid whose base is a regular hexagon (all sides and
all angles equal), and the apex is located directly above the hexagon
center, then the dihedral angle (the angles that the faces make with
each other) is given below.  

If the hexagon has sides of length A and has height H, then the ratio
z = H/A can be thought of as a sort of "slope" (although it is
actually the slope of the edge from base vertex to the point).

The angle that you are seeking is

             [sqrt(3)/2] sqrt(1+z^2) z
  x = arcsin ------------------------- 
                     z^2 + 3/4

  where 90 deg <= x <= 180 deg
 
This is not a linear relation, but you can check that it has the 
proper "limiting" behavior. If z is very large (i.e., you have a very 
pointy pyramid), then the sides are like those of a hexagonal prism, 
and x becomes arcsin(sqrt(3)/2). A calculator yields the result of 60 
degrees here, but we have to take the supplementary angle, or 
180 - 60 = 120 degrees. This answer agrees with our reasoning from the 
hexagonal prism, which has faces that meet at 120 degrees.

If z is very small, or approaches zero, then x approaches 180 degrees
(recall that x is in the range 90 deg to 180 deg), which again agrees 
with our intuition of a squat pyramid whose faces are nearly parallel.

The computation of the formula for x above is found from good old
three-dimensional geometry, and the use of the three-dimensional
vector cross product. I'll sketch out where it comes from, and you
can work through it if you're interested in the details.

1.  z = H/A is the only parameter we need. The angles are 
    independent of the actual size of the pyramid, so we might
    as well choose a pyramid with side A = 1, and height H = z.

2.  Let the base of the pyramid be centered on the (x,y) plane,
    and let one of the vertices be at (x=0,y=1).  We need to
    compute the direction of two normals (directions perpendicular
    to the faces). We can compute these normals at any points on
    two adjacent faces.

3.  Let the first such point be at (x=sqrt(3)/2, y=0, z=0), at the
    midpoint of one side. The normal to the plane here is in the
    direction (z, 0, sqrt(3)/2).  The length of this normal is
    sqrt(z^2 + 3/4).

4.  The normal to the adjacent plane will be essentially the same,
    but rotated in the xy plane by +60 degrees counterclockwise. 
    This direction is (z/2, z*sqrt(3)/2, sqrt(3)/2).

5.  The vector cross product between these two directions is
    (-3z/4, sqrt(3)*z/4, z^2*sqrt(3)/2). The length of the 
    vector cross product is the square root of the sum of the
    squares of the 3 components, or 

         sqrt(9z^2 + 3z^2 + 12z^4)/4

       = sqrt(12)sqrt(z^2 + z^4)/4 

       = 2*sqrt(3)*z*sqrt(1 + z^2)/4

       = sqrt(3)*z*sqrt(1+z^2)/2

6.  The vector cross product length is also equal to the 
    product of the lengths of the two original vectors times
    the sine of the angle that we seek:

        sin(x)*sqrt(z^2 + 3/4)*sqrt(z^2 + 3/4) 

      = sin(x)*(z^2 + 3/4)

7.  Equating these last two expressions allows us to solve for x
    to get the equation that I gave above. We need to take account
    of the fact that the cross product gives us the smallest angle
    between the two normals, and in your application, we're typically
    looking for the supplementary angle to this.

I hope this helps answer your question!  

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Polyhedra

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