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The Pattern for a
Spiked Flail

    This dodecahedron provides the pattern I want for the spikes on a flail. The sphere is circumscribed. One spike corresponds to each node, where three edges meet. Note that a great circle will transverse above the altitudes of four pentagons and two edges. The job of locating these nodes on a ball of known diameter involves four equations and four variables.

    This is a crummy drawing. I used MS Paint because it's convenient. We need to relate the edge e to altitude h.



                 Since the angles of a triangle (ABC) add to 180º



which may or may not have been obvious. 

               Now, to get the base e in terms of h.



                       Equation 1  






   Equation 1:                                                            

     Now to relate the edge and altitude of the pentagons to the radius of the ball.  We will need to relate the central angles u and w.



            Equation 2                                                                      

   Now using trigonometry of right triangles we see that


                          Equation 3



                            Equation 4

       So, we have four equations relating the four variables e, h, u, w
    We'll use substitution to put Equations 3 and 4 in terms of e and w and solve both for e.



      from Eq. 2

        from Eq. 1          
   (we cannot avoid a half angle or double angle calculation)

     Now, setting these two expressions for e equal we have an equation in one variable. This we (should be able to) solve.







   I was surprised to see the R's cancel, but of course   
   they should. The central angles are not related to the radius.

   Now we invoke the identies:   and


   Now sin 90º = 1, of course, and cos 90º = 0.

  Now square both sides to get rid of the radical.

   We'll have to check our results for extraneous solutions.

   I looked at this for a while. I expected the Pythagorean identity to help, i.e.
    , hence   ,   which factors thus:

    , neither of which cancels the denominator.




   So, I rearranged the terms to look like a quadratic equation in cos(w).

   Now I'll use the "completing the squares" technique.











   By taking the square root of both sides.

  Time to find a calculator.

   Not bad!
   Now we'll plug this in to the equation  
   to find our edge e. We're done!

   For the 3" ball, R = 1.5" and


     Now, we know the length of the pentagon edges. My thought was to layout a rectangular piece of sheet steel with five divisions this size, then to bend it into a sort of cookie-cutter. I considered laying this on the ball, in order to locate the nodes. I planned to undercut the sides of the cookie-cutter in order to bring its corners close to the ball's surface.

Click on the link below to see the real thing.