[ExI] tumbling pyramids again

[Adrian Tymes]

I'm not sure, but I *think* this reduces to "keep the CG equidistant from
all vertexes".

If the CG is h/4 from the square face, then it is 3h/4 from the "top"
vertex (as per your diagram).  So 3h/4 needs to be the distance from the
other vertexes.

The 1 cm square face has a diagonal of sqrt(2), so it is sqrt(2)/2 from any
of its corners to its center.

We can use that corner-to-center line as one side of a right triangle, with
another side being from said center of the square to the CG.  Per your
calculation, this is h/4.

The hypotenuse of this triangle is the distance from the CG to the square's
corners.  It is sqrt( (sqrt(2)/2)^2 + (h/4)^2 ) = sqrt( 2/4 + h^2/16 ).  We
need this to equal 3h/4.

3h/4 = sqrt( 2/4 + h^2/16 )
square:
9h^2/16 = 2/4 + h^2/16
multiply by 16:
9h^2 = 8 + h^2
subtract h^2 then divide by 8:
h^2 = 1
square root:
h = 1

So if I have my math and assumption right, that comes to a height of 1 cm
if the square base is 1 cm on a side.

On Wed, Feb 15, 2017 at 10:25 PM, spike <spike66 at att.net> wrote:

>
>
> A question I posed a long time ago, but I have new insights.
>
>
>
> Imagine a cube a cm on a side, tossed like a gaming die.  By symmetry, any
> face has the same probability of coming up (or down): one in six.
> Likewise, a tetrahedron could be tossed and each face would have a one in
> four chance of landing face down.  The same symmetry argument could be made
> for any of the five platonic solids.
>
>
>
> Cool but what if we wanted to create an unsymmetrical solid, such as a
> square based pyramid one cm on a side for the square, and we wanted to make
> it such that the probability of landing face down on any of the five
> surfaces was the same.  How tall do we need to make it?
>
>
>
>
>
>
>
> I get a CG height of h/4.
>
>
>
>
>
> So if we want to make a fair pyramid shaped five sided die, what is h?
>
>
>
> spike
>
> _______________________________________________
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>
>
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