Exact dimensions of a 3-sided coin?

Factual Questions
1

I have always said that a tuna can is the perfect three-sided coin (or d3). Truth is, there are too many shapes of tuna cans and I have no idea of what the actual proportions would be for a can to have equal chances of landing on either side or on edge.

Can anyone calculate this? The ratio of diameter to height that would make a homogeneous cylinder equally likely to fall on any of its 3 surfaces?

Bonus point for an answer in the format: “a 6 oz can of Del Monte whole kernel corn”

(And yes, I know there are better ways of randomly deciding among 3 choices)

2

I know this probably isn’t what you’re looking for, but a quick google search basically said that it’s impossible to make a thick coin a fair 3-sided die.

From: http://www.maa.org/editorial/mathgames/mathgames_05_16_05.html

To expand one what he’s saying, if you flip the thick going, it would have to be pretty thick to guarantee an equal chance of landing on any of the 3 sides. However, if it’s much closer to being rolled along the round edge, it would need to be thinner to guarantee an equal chance. To be fair, there shouldn’t be so much dependency on the methodology of rolling/flipping.
The way a D3 is generally rolled is by rolling a D6 and dividing by 2 and rounding up.
If you’re really interested in a 3-sided die, I can envision one that would guarantee all sides are the same shape. It would be like a triangular prism, but rather than having a triangular base, it would taper to a point, sort of like a football. I haven’t found any examples of something like yet, but I imagine one is out there.

3

I missed the edit window, but I found a US Patent for exactly the sort of 3-sided die I envisioned.

http://www.google.com/patents?id=EZYZAAAAEBAJ&pg=PA1&lpg=PA1&dq=United+States+Patent+D412537&source=web&ots=OuHKc1j2WT&sig=ApjiNhDKHnXNOjq6iEy36q0IeUA&hl=en&sa=X&oi=book_result&resnum=2&ct=result#PPA1,M1

4

A three-sided pencil-shape (triangular prism) sharpened at both ends would work just fine. It would just need to be long enough not to be stable apart from on the three sides.

5

Brazil nut.

6

That’s a great read.

Yeah, we always did d3 as a d6/2. I was thinking more of a daily solution to the problem (plus the purely academic angle to it). D6 are pretty common anyways.

I can see how throwing a fat coin rolling on its edge would bias the chances.

Still, how would I go about estimating the height of a coin for a more or less fair d3, assuming it will be tossed to flip as a coin and with it falling on a level surface where it will not roll but stay on the surface it landed?

7

Thought of that. I actually was eating one once and thought it would make a great d3. Not available at all times, though. And if the decision to be made is “who is going to eat the last Brazil nut?” Then it is not so useful, unless you have a clean surface to roll it.

It is also pretty much the shape described by Blaster Master and then shown in his patent link.

I like the shape on the link as it has some details such as the wavy edge, probably to aid rolling. Shows some thought and not just some guy patenting the obvious.

8

You take a standard Die 6, and number it 1,2,3, 1,2,3. I own several.

9

Is it really that hard to see that (s)he’s not looking for a fair method of picking from three?

To work on the problem, start with a very simple, and untrue form, and then you can see how complicated it will get.

Assume that the outside of our cylinder is smooth, and as you said, the density is constant. Although constant density is not strictly necessary for this ultra-simplified form, it’s a good simplifying assumption anyway.
Call the radius of the cylinder r and the height h.

Next, ignore all rotations. Assume it is just falling through the air without turning. Also ignore (this is kind of a big one) any rotational momentum once it hits the ground.

What our assumptions mean is that when it touches the ground, it will always land whichever way it is ‘leaning’. It’s the same as if you just tilted it randomly and then set it down. The only variable of concern is the angle at which it hits the ground.

The problem reduces to two dimensions (and half a circle of angles since it’s symmetric); we can just look at the can in cross section from the side. In other words, a rectangle with sides of length 2r and h.

The only point of interest is the angle at which it will fall either flat on a face or on its side. This is when the center of gravity is directly over the corner of the can, which touches the ground first (outside of the isolated cases of landing perfectly flat; those don’t change our answer).

It’d be nice if I could draw a diagram here, but imagine it thusly: A rectangle drawn so that the center point is directly above one corner, and that corner is on a flat line (the ground). The vertical line from the center to ground is the hypotenuse of a triangle with sides r and h/2. Call the angle the can makes with the ground (along the ‘flat’ face) phi.
Hopefully it’s apparent that the angle phi is equal to the angle inside the triangle at the center point (opposite the radius side). In other words,

tan phi = (r / (h/2) )

So that’s all we need at this point to get an answer to this problem. Since it should fall on the three faces with equal likelihood, the angle we want is π/3 (60º).

2 tan (π/3) = r / h

r/h = 2 sqrt(3) ~= 3.46

(I just measured a can of tuna which came out at 2.28, by the way.)

This post is already too long but there’s a lot of complicating conditions, most of which will tend toward requiring a lower value. So you can almost think of this as an upper bound on that ratio.

10

He. And yes, your assumptions are about right. Thanks.

So we have around a diameter of around 7 times the height, right? Or flatter if you let reality kick in a bit more. That’s pretty much what I was looking for. It is a LOT flatter than I would have thought.

Now if the can hits the ground flipping flat over flat, that would need a taller can, right?

11

I wish I could have summed up in words for my high school math teachers that this is exactly the reason I did not want to go on to higher studies in math.

People who are really into math would get all hotted up by this kind of stuff.

I find myself wondering just exactly how much free time you people must have.

:wink:

12

Oops; I realized my ratio for the tuna can I measured was diameter, not radius. So that tuna can was 1.14 radius to height.

Yes, complicating assumptions will decrease the ratio of radius to height. Once you include angular momentum, the can will definitely tend to tip over to flat more often, so you need more relative height to compensate.

The number I came up with, while not rigorously proven, is effectively a practical upper bound - it almost certainly couldn’t be flatter than that and still be fair.

About the only condition pushing it down would depend on throwing style - if it fell for a very long time, the effects of air resistance might tend toward a side landing. But I imagine the best style is to spin very quickly from a very low height.

The next complicating step would be to work on angular momentum that occurs just as it hits the ground - but that immediately does require a lot more time and effort to solve.

13

I don’t think this is quite correct. You’re right that it essentially reduces to a two-dimensional problem, but the problem is that there’s a lot more probability for a randomly oriented sphere to have “up” be near the “equator” than there is for “up” to be near a “pole.”

Rather, you need to think about inscribing your cylinder inside a sphere. The two edges of the can divide the sphere into three regions, two “caps” and a “belt”. A moment’s thought tells us that if the can lands with one of its faces “up”, that’s equivalent to having the top point of the sphere be on one of the caps; if the can lands with the side up, then the top point of the circumscribed sphere will be in the “belt”. We want to figure out the angle phi (as defined above) such that each of the caps and the belt have equal areas. Going through the arithmetic, it turns out that you get this when the ratio of the can’s diameter to its height is sqrt(8), or about 2.828.

Not as much as I’d like, but that doesn’t stop me from solving interesting problems when they’re presented to me. :slight_smile:

14

It is for me, but that is probably because I am verra verra stupid.

How is that not most plausibly construed as asking for “a fair method of picking from three”?

-FrL-

15

I did a computer simulation of this once. The answer is, given sufficient distance to fall, the height needs to be ~1.33 * the radius.

16

Oops, you’re right. It doesn’t exactly reduce to two dimensions the way I did it. I also really like the simple explanation using the point on the sphere to make it obvious how it will ‘balance’ on landing. No trigonometry needed for that.

edit in response to Frylock : The parenthetical at the end makes it clear that the OP wasn’t looking for a general method, but for how to figure this specific method. I was only responding to the replies that weren’t helping with the can flipping. Yes, if you take the words literally the question is to find a fair three-sided method, but only using the can.

17

Or, maybe, just maybe, you could allow the OP to speak for themselves if that’s not what they want?

18

Regardless of the points made later on, there is an error in the calculation here.
You have
tan phi = (r / (h/2) )
which is quite right, but then you shuffle stuff around, and end up with
2 tan (π/3) = r / h
but that’s wrong of course - just a simple algebraic manipulation bungle. You want:
(tan (π/3))/2 = r / h
i.e., your end result was off by a factor of 4. The real r/h should be ~0.866. This is intuitively a much more reasonable figure. As you said, this can be an upper bound on the ratio, and this fits quite well with the figure that Shalmanese gives of h/r=1.33, or, inversely, r/h= 0.75.

19

If I ask “how can I kill a bear with a spoon” and somebody responds with “shoot it with a gun” then they aren’t answering my question. The OP clearly states the coin is supposed to be a cylinder.

20

Maybe, just maybe, the OP already did just that.