N8PPQ

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

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.

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.

Flail

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