20051211, 23:39  #34  
Jun 2005
2×191 Posts 
Quote:
As far as arbitrary numbers go, you can have a regular ngon prism or pyramid, and create any nsided die. You just need to make it unstable to land on the remaining ends. Drew 

20051213, 16:32  #35 
Nov 2005
2×7×13 Posts 
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 20051213 at 16:36 
20051213, 17:10  #36 
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
I'm not certain if you are describing the same thing, but here is a design that will work to make any nsided "dice".
Take a regular nagon (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 nsided 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 20051213 at 17:12 
20051213, 17:26  #37 
May 2003
1547_{10} Posts 
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. 
20051213, 17:48  #38 
Jun 2003
The Texas Hill Country
3^{2}×11^{2} Posts 
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. 
20070110, 13:36  #39  
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Quote:
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 r^{2} + h^{2}/4 = R^{2} = 9h^{2}/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 20070110 at 14:11 

20070110, 15:30  #40 
May 2003
E7_{16} Posts 
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.

20070110, 16:46  #41  
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Quote:
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 

20070110, 17:56  #42  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
280E_{16} Posts 
Quote:
Going over to a static model doesn't help much, as I pointed out in a subsequent post. Paul 

20070111, 07:47  #43  
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Quote:
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 20070111 at 07:51 

20070112, 08:55  #44 
"Lucan"
Dec 2006
England
194A_{16} Posts 
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 20070112 at 08:59 
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