# N8PPQ

## My Personal Site

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

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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

and

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.

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 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   fini!

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.