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Old 2005-12-11, 23:39   #34
drew
 
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Quote:
Originally Posted by nibble4bits
Another shape I've considered is interesting... A sphere with three slices taken out of it. If you had tetrahedron circumscribed by the sphere and cut segments out of the sphere in parallel with 3 of the 4 sides of the tetrahedron, then you could argue that it has the effects of a three-sided coin. LOL I'd love to actually see some coins made using these designs just to see how well they work in practice.

It would be interesting to have people come up with odd types of dice. BTW Odd numbers are naturally difficult in 3D using a 2D surface/polyhedron/combination. Think about why this is so... and the effects for other numbers of dimensions.
Your tetrahedral shape would be more effective if you had more curvature in the remaining spherical part...sort of if it was a hemisphere. It will be less stable, and tend to topple onto one of the flat surfaces.

As far as arbitrary numbers go, you can have a regular n-gon prism or pyramid, and create any n-sided die. You just need to make it unstable to land on the remaining ends.

Drew
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Old 2005-12-13, 16:32   #35
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I kind of figured to make the "fourth side" curved in a way to make it unlikely to be settled on. An added consequence is that it will not roll as far. There's also the issue of weight and balance but it's possible to tweak one side to make it more balanced.

LOL nice tweak of my design. It would be obvious if I had actually used one with a simple prism shape. Adding the curves would be stupid not to add (unless patented).

Last fiddled with by nibble4bits on 2005-12-13 at 16:36
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Old 2005-12-13, 17:10   #36
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I'm not certain if you are describing the same thing, but here is a design that will work to make any n-sided "dice".

Take a regular n-agon (for 3, it is an equilateral triangle)
Extrude it along the axis perpendicular to the face for a distance in excess of twice (it may not need to go that far, but the concept will work at that length) the diameter of the circle circumscribing the polygon.
Add end caps to this prism by adding an n-sided pyramid to each end. The pyramid needs to be tall enough to assure that the CG of the object would never lie above an end facet if it were to attempt to rest on said facet.

Since the ends are designed to be unstable, the object will always come to rest on one of the major faces. By symmetry, they are each equally likely.

One problem with using any kind of dice that has an odd number of faces is that it will be difficult to read. The only clearly designated face is the one on the bottom. Its value is obscured by the object itself.

With the regular solids such as the cube (and higher), the designated face is the one "on top", opposite the face upon which the object is resting. This provides a ready surface upon which the designation can be easily marked.

Last fiddled with by Wacky on 2005-12-13 at 17:12
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Old 2005-12-13, 17:26   #37
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Wacky,

While it is hard to read the "bottom" face of a die, what I've seen done (with a tetrahedron) is write three numbers on each face, one on each edge, in such a way that when the die lands, the number along the bottom edge on each face next to the floor is the same number, and this is the number rolled.
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Old 2005-12-13, 17:48   #38
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Zeta,

yes, for the tetrahedron, that is a reasonable approach. However, it is significant that the angle between the faces is acute. When you go to more faces than 6, the faces adjacent to the floor are also point "down" and therefore, they, too are going to be difficult to read. If you attempt to use any other face, I fear that it would be difficult to determine which ones to "read" because they are removed from any distinct reference plane.

The same can be said for determining which face is the "top" one when the number of faces grows beyond a very small number.
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Old 2007-01-10, 13:36   #39
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Quote:
Originally Posted by drew View Post
This is an invalid approach. As you revolve this circle around the coin's axis to create the 3-d sphere, you'll see that the 'edge' region sweeps out a much larger area per projected length than the face, due to it's larger displacement from the axis of revolution. Therefore your result gives more weight to the faces than it should. mfgoode made the same mistake of trying to turn it into a 2-d problem without making the appropriate adjustment.

The assumption about circumscribing a sphere is reasonable, though, for the static case. Let's see if someone can corroborate a value of sqrt(2)/4 for the thickness/diameter ratio.

Drew
Coming late to this thread, I gather that the argument was
about whether the angles subtended by the three faces of the
cylinder shoud involve the arc length or the area.

Drew plumps for area and considers a sphere radius R into
which the cylinder fits. For the areas to be equal, the height
of the cylinder h = 2R/3

If r is the radius of the cylinder then
r2 + h2/4 = R2 = 9h2/4

so h/r = sqr(2)/4 as Drew says.

However, if the "coin" is flipped as usual, it spins about a
horizontal diameter. Freezing this motion at a random instant
suggests the 2D arc length approach giving h=R=2r/sqr(3)

David

Last fiddled with by davieddy on 2007-01-10 at 14:11
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Old 2007-01-10, 15:30   #40
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Quote:
Originally Posted by Wacky View Post
If we circumscribe a sphere about the cylinder, then this means that 1/3 of the total spherical surface must be the projection of one of the faces.
Following this principle, experiments have been made. For example, one so proportioned yielded 32.6% edges in a sample of 3000 tosses, which is in fairly good agreement with the model.
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Old 2007-01-10, 16:46   #41
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Quote:
Originally Posted by fatphil View Post
Following this principle, experiments have been made. For example, one so proportioned yielded 32.6% edges in a sample of 3000 tosses, which is in fairly good agreement with the model.
I find this surprising. Drew (in post #20) points out that the C of G
is lower when the coin lies on a face, which should favour this in
a dynamic situation.
As I said, it also depends on how the coin is spun.
Finally, think about my "stick problem".
The "good agreement" of experiment with naive theory is,
I suggest, fortuitous.

David
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Old 2007-01-10, 17:56   #42
xilman
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Quote:
Originally Posted by davieddy View Post
I find this surprising. Drew (in post #20) points out that the C of G
is lower when the coin lies on a face, which should favour this in
a dynamic situation.
As I said, it also depends on how the coin is spun.
Finally, think about my "stick problem".
The "good agreement" of experiment with naive theory is,
I suggest, fortuitous.

David
As I said in post #12, this probability question has been known for a very long time as being insoluble without making a number of assumptions explicit. There are many assumptions as to what constitutes "tossing" the coin which agree with common notions of "tossing" and there are many different geometries which meet the constraints so imposed.

Going over to a static model doesn't help much, as I pointed out in a subsequent post.

Paul
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Old 2007-01-11, 07:47   #43
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Quote:
Originally Posted by xilman View Post

Going over to a static model doesn't help much, as I pointed out in a subsequent post.

Paul
But it at least sets a soluble well defined problem,
apparently tricky enough to confuse certain frequent posters
(who shall remain nameless :)

A sphere with differently coloured areas is a good start
for the design of a general dice.

David

Last fiddled with by davieddy on 2007-01-11 at 07:51
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Old 2007-01-12, 08:55   #44
davieddy
 
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Default Static Model

Everyone is in agreement that the question of
what happens to a spinning coin AFTER it hits
the surface is too complex for informal discussion.

The "static model" consists of choosing an orientation
of the coin "at random" and deeming that the coin lands
on the surface vertically below the centre of gravity.

Choosing a random orientation in 3D is like choosing a
random location on earth. If we choose latitude and
longitude at random, our choice is biased towards locations
near the poles. We should instead choose longitude and the
distance from the equatorial plane.

However, in the context of a coin spun about a horizontal
axis, it is not unreasonable to take the angle at which it
strikes the ground as the orientation chosen for the static
model. There is a marked tendency for the coin to strike the
ground with the leading rim rather than the trailing one, which
IMO favours ending on the edge rather than a face in a dynamic
scenario.

Perhaps this explains the experimental coincidence I noted
in an earlier post.

David

Last fiddled with by davieddy on 2007-01-12 at 08:59
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