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Square Pyramid Stack

05/31/2011 4:59 AM

Can somebody please show me calculations for the effective weight on cannonballs at the base of a spherical arrangement of cannonballs.

To clarify further... it is a square base pyramid arrangement of cannonballs...

What is the weight felt by each cannonball?

What pressure is felt by the brass monkey?

What is the pressure multiplication factor as we go down towards the base layer?

I am writing a software and would really appreciate it if someone could explain to me the simple math involved... i am already familiar with the projections of the weights but am miscalculating something which is screwing the way they add up.

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#1

Re: Square Pyramid Stack

05/31/2011 7:58 AM

If it's a square stack of cannonballs, then each layer is a square number n2, when n is the layer number going from the top to the bottom. There is 1 cannonball in the top layer, 22 = 4 balls in the second layer and so forth. Each ball has a weight, W.

The top layer (1 ball) has no balls pressing on it, so it feels no weight from above.

The second layer has 1 ball above it and its weight, W, is equally shared by all 4 balls, so each feels a weight of W/4 (or 0.25W).

There are 9 balls in the 3rd layer and there are 5 balls above them, weighing a total of 5W, pressing on them; so each ball in the 3rd layer feels a weight of 5W/9.

The math is fairly simple, so from this discussion you should be able to work out the equation (or algorithm) on your own. Check back once you've worked it out.

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#2
In reply to #1

Re: Square Pyramid Stack

05/31/2011 9:49 AM

What about the brass monkey ? What weight does he feel ?

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#3

Re: Square Pyramid Stack

05/31/2011 10:20 AM

In naval terms there's no such thing as a Brass Monkey.

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#5
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Re: Square Pyramid Stack

05/31/2011 5:21 PM

Term: brass monkey - Naval History and Heritage Command:

It has often been claimed that the "brass monkey" was a holder or storage rack in which cannon balls (or shot) were stacked on a ship. Supposedly when the "monkey" with its stack of cannon ball became cold, the contraction of iron cannon balls led to the balls falling through or off of the "monkey." This explanation appears to be a legend of the sea without historical justification. In actuality, ready service shot was kept on the gun or spar decks in shot racks (also known as shot garlands in the Royal Navy) which consisted of longitudinal wooden planks with holes bored into them, into which round shot (cannon balls) were inserted for ready use by the gun crew.

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#8
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Re: Square Pyramid Stack

06/01/2011 5:56 AM

The coefficient of thermal expansion of brass is 18.7m/m°C, whereas the coefficient of thermal expansion of iron is between 10 and 12 m/m°C. the balls didn't "fall through" the rack they jumped off/out of the rack when the tightening brass base overcame the static friction.

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#13
In reply to #8

Re: Square Pyramid Stack

06/01/2011 11:20 AM

Ooops x10-6

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#11
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Re: Square Pyramid Stack

06/01/2011 8:46 AM

.............and you knew it was cold, when it was cold enough to "freeze the balls off a brass monkey"

Although not the correct "brass monkey" it suitably illustrates the problem...........it's bloody cold...........or freezing.

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#4

Re: Square Pyramid Stack

05/31/2011 11:25 AM

I've thought some more about my initial reply, and I think I'll revise it.

In the 3rd layer, where there are 9 balls bearing the weight of the 5 balls above them, the 4 balls in the 2nd layer are in contact with the 3rd layer at 16 points. The corner balls in the 3rd layer each have 1 contact point; the balls on the centers of each side have 2 contact points and the middle ball has 4 contact points.

So the corner balls each carry 5/16*W, the side balls each carry 2*5/16*W, and the middle ball carries 4*5/16*W. The net weight on the 3rd layer remains 5*W, of course:

4*5/16*W + 4*2*5/16*W + 1*4*5/16*W = 5*W.

Here a cut-away sketch showing the contact points between the 2nd and 3rd layers:

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#7
In reply to #4

Re: Square Pyramid Stack

05/31/2011 11:25 PM

You are, of course, talking about the vertical component of the weight of the cannon balls. Since the weight of the balls in the next layer up is not centered over the balls in the current layer, I suspect the distribution to the layer below will not be equal, but I don't have time to study it. You may get into messy vector diagrams; maybe things will neatly cancel out, but my intuition says "no comment!"

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#10
In reply to #4

Re: Square Pyramid Stack

06/01/2011 6:46 AM

This is a nice problem that seems trivial until you think about it. From your diagram, I see that each ball is supported in four places. You cannot assume perfect inelastic spheres and use force vectors to get a solution because you have more unknowns than parameters. A tetrahedral stack would have a unique solution.

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#12
In reply to #4

Re: Square Pyramid Stack

06/01/2011 10:55 AM

Thanks for the revision, because I immediately saw the first analysis was oversimplified once you hit the 3rd row and the points of contact differed for different balls.

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#6

Re: Square Pyramid Stack

05/31/2011 11:14 PM

Start with a vector diagram. You are after the resultant forces.

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#9

Re: Square Pyramid Stack

06/01/2011 6:13 AM

I agree with Lehman 57 and Osborne83 GAs.

In Usbport's picture the top ball is applying an outward force to the four balls beneath it: that means that the force between the "corner balls" in the second and third rows is greater than the other forces (between the second and third rows).

The answer is going to depend a lot on the coefficient of friction: between the balls, and, between the balls and the monkey. How deep do you want the analysis to be?

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#14
In reply to #9

Re: Square Pyramid Stack

06/01/2011 11:37 AM

The OP just asked for the weight. It doesn't matter what the lateral forces are because they ultimately cancel; the weight of the cannonballs will yield a downward vector force equal to their weight.

If you were to stack the cannonballs in the normal manner or with a thin ('massless') sheet between each layer, the net weight on the layer below is the same, either way they are stacked.

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#15
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Re: Square Pyramid Stack

06/01/2011 12:01 PM

The OP also said

What pressure is felt by the brass monkey?

In the simplest case take five rusty cannonballs: arrange 4 touching each other in a square, and, place one on top; they will probably stay where you put them. Now take five large oily ball bearings and try the same thing. The monkeys holding the two sets of balls will "feel" two different types of load.

You may well have answered the OPs question adequately: it depends on what he is really trying to calculate.

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#17
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Re: Square Pyramid Stack

06/01/2011 2:30 PM

An experiment most everyone can try:

Stack billiard balls similarly and see how high they'll stack...

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#16

Re: Square Pyramid Stack

06/01/2011 1:57 PM

The old problem of supporting a weight on more than three supports, it's indeterminate as a slight dimensional difference changes everything. Or we assume some flexibilty as in a steel bridge, suppose the balls were rubber then the weight would spread equally, but wobbly iron cannon balls could go any way. But for this problem we assume the weight is equally spread four ways.

Next, the bottom layer would need a solid barrier to hold the balls in the formation, is this the monkey? It isn't possible to put one pool ball on top of four because the horizontal component forces them apart.

If this is a square base then what is the reference to a "spherical arrangement of cannonballs"

The effective weight on cannonballs at the base depends on the row they are in from the outside and if on a corner. Consider a 2D pyramid, and look at the weight distribution.

If all that is required is the total weight of balls then it seems to go 1 squared +2 squared + 3 squared + 4 squared + 5 squared and so on, until x = the number of layers, multiplied by the weight of a ball.

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#18
In reply to #16

Re: Square Pyramid Stack

06/01/2011 3:51 PM

GA:-

is this the monkey? Yes: 30 cannon balls on a brass monkey:-

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Anonymous Poster (1); engineertony (1); Lehman57 (1); lyn (1); MOBI (1); osborne83 (1); Randall (5); Rixter (1); RufusVS (2); TonyS (1); Usbport (3)

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