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BigDon
2015-Jun-19, 07:33 PM
(Some of you enjoy word problems as much as I enjoy anacrostics.)

I have to ask an odd question.

My piggy bank, when full, consistently counts out to \$103, plus or minus two dollars, despite being a random collection of pennies, nickels and dimes. (Quarters were set aside as laundry tokens.)

I think I may have asked this before and somebody came up with a formula as to why this was so. If that could be re-explained I would appreciate it.

The reason I bring this up again is...

I've moved to a wonderful new place and no longer need to set quarters aside to do laundry. The piggy bank is getting near full again and I was wondering if somebody here could predict the probable count out now that quarters are included in the mix.

grapes
2015-Jun-19, 07:42 PM
TNSTAAFWP

Gimme a sec...

NEOWatcher
2015-Jun-19, 08:24 PM
For a start, Here's the dimensions of the coins into mm^3 and then cents/cm^3. Math mistakes. See later posts
1: 19.05 mm x 1.52 = 45.5: 0.022
5: 21.21 mm x 1.95 = 65: 0.77
10: 17.91 mm x 1.35 = 38: 0.26
25: 24.26 mm x 1.75 = 66.7: 0.37

It looks nickels and pennies would overwhelm/underwhelm the balance if widely off of random, but overwhelming or underwhelming dimes wouldn't.
So if the prior 2 are somewhat consistent, the entire bank would be somewhat consistent.

PS; I'm glad to see you're back... safe and settled in.

Grey
2015-Jun-19, 09:15 PM
Here (http://www.csirik.net/coins.html)'s a good site that gives typical numerical distributions of coins (and here (http://www.notmartha.org/archives/2008/10/10/how-much-is-a-pound-of-coins-worth/)'s an empirical test, rather than a theoretical prediction). Unfortunately, it doesn't do it by volume, but I'll bet that random coins are similar enough in their packing that you could use the by weight calculations and still be close. So you could work out ratio of value to weight both with and without quarters, and see how they compare.

grapes
2015-Jun-19, 10:12 PM
For a start, Here's the dimensions of the coins into mm^3 and then cents/cm^3.
1: 19.05 mm x 1.52 = 45.5: 0.022
5: 21.21 mm x 1.95 = 65: 0.77
10: 17.91 mm x 1.35 = 38: 0.26
25: 24.26 mm x 1.75 = 66.7: 0.37

Somethings off, the quarter and the nickel have almost the same volume, but their ¢/cm3 ratio is only 2? O, nickel is greater?

NEOWatcher
2015-Jun-19, 11:07 PM
Somethings off, the quarter and the nickel have almost the same volume, but their ¢/cm3 ratio is only 2? O, nickel is greater?
I have no clue what I did wrong. :doh: Maybe not recording my intermediate steps.
I must have lost my way ....Recalculating....

Here's the revised list. mm^3, ¢/mm^3

1
285.02295699234248518029736010015
0.0035084893180266398995627478647298

5
353.32243791844602610177061231
0.014151379769303248316003323081127

10
251.93067661648903912362725975803
0.039693459067007058561966420818657

25
808.9273571959580853096196981372
0.030905123652461528931585630869233

Still, without quarters, the dimes are still prevalent.

Solfe
2015-Jun-20, 02:39 AM
My best shot is

Pennies =\$3.40, Nickels =\$10.65, Dimes =\$43.20, and Quarters =\$45.75 to total \$103.00.

There is no real methodology in my answer. I took three random pulls from my coin jar, counted the coins, added the value and extrapolated what \$103.00 handfuls of coins could be. I have no idea how many coins are in my jar, nor the distribution. In each pull, I had to remove dollar coins, gum wrappers, a couple of screws, and wire nuts.

grapes
2015-Jun-20, 04:01 AM
My best shot is

Pennies =\$3.40, Nickels =\$10.65, Dimes =\$43.20, and Quarters =\$45.75 to total \$103.00.

There is no real methodology in my answer. I took three random pulls from my coin jar, counted the coins, added the value and extrapolated what \$103.00 handfuls of coins could be.
I like that methodology! :)

Let's see, the value and volume of those PNDs are 3.40+10.65+43.20 and 3.40*.00351/.01+10.65*.01415/.05+43.20*.03969/.10, using NEOWatcher's new results, or \$57.25 and 21.3534, uh, cubic cm? For there to be \$103 to be in Don's piggy bank, in those proportions, we need a volume of 21.3534*103.00/57.25, about 38.42

Now. To fit those quarters in there...oops, got distracted.

ETA: Oops, made a mistake there. With the quarters, the ("usable") volume is
3.40/.00351+10.65/.01415+43.20/.03969+45.74/.03091 or about 4290 cubic mm. Without quarters, it's 2810.

EETA: so the new volume is 2810*103.00/57.25, the value with quarters (predicted) is 2810*103.00/57.25 * 103.00/4290, or

\$121.38

NEOWatcher
2015-Jun-20, 01:55 PM
uh, cubic cm?
For my latter post, I stuck with mm. Bigger numbers, but no conversion distraction.

ETA: Oops, made a mistake there. With the quarters, the ("usable") volume is
3.40/.00351+10.65/.01415+43.20/.03969+45.74/.03091 or about 4290 cubic mm. Without quarters, it's 2810.
By "usable", I'm assuming you mean no air gaps.
The next step would be to put that random pile of coins in a container and fill it with water to determine the ratio of empty space.

Without quarters, my SWAG would be that there is less wasted space than with quarters. I don't know if that would change Solfe's ratios.

grapes
2015-Jun-20, 04:06 PM
By "usable", I'm assuming you mean no air gaps.

Yeah, the actual volume may be different. But maybe approx proportionate.

The next step would be to put that random pile of coins in a container and fill it with water to determine the ratio of empty space.

The amount of empty space would be important, if the ratio was different for with quarters vs without. So I'm going to assume its close--go with an extra twenty bucks in the bank. It'll vary.

Without quarters, my SWAG would be that there is less wasted space than with quarters. I don't know if that would change Solfe's ratios.
The percent empty space for just quarters is probably close to that of just pennies. For instance, it's the same for small spheres as it is for large spheres--just a matter of scale. But combining large spheres with much smaller spheres allows the density to go up, because the small ones can pack in the interstices of the large ones.

profloater
2015-Jun-20, 04:42 PM
I go for 111.25 dollars for no good reason, you might get a good steak dinner with pudding (as we say in England) and coffee for that.

JohnD
2015-Jun-20, 10:17 PM
Big Don,
In the UK, coins are made so that they weigh the same if they represent the same value.
For instance two penny coins weigh the same as one 2p coin.
This is a convenience for banks, who need not separate them onto the same denominations, just weigh them.
As all the coins are made of the same metal (copper-plated steel), with the same density, they will occupy the same volume, more or less, even if they are composed of a random number of the two denominations, as long as the collecting jar is large enough to let them pack together.
I don't know if the above is true of US coins, but I think it likely, again for the banks' convenience, so this would explain the standard value of your full cookie jar.

This is also true of 'silver' coins (nickel-plated steel), but as a ten-penny piece is nothing like the weight of ten penny coins, a mixture of 'silver' and 'copper' will not have a standard weight.

I suspect that this will mean that if you add a random proportion of quarters, 'silver' coins that weigh a lot less than 25 one cent coins, the value of your cookie jar will not be constant.

JOhn

Hornblower
2015-Jun-21, 01:10 AM
Big Don,
In the UK, coins are made so that they weigh the same if they represent the same value.
For instance two penny coins weigh the same as one 2p coin.
This is a convenience for banks, who need not separate them onto the same denominations, just weigh them.
As all the coins are made of the same metal (copper-plated steel), with the same density, they will occupy the same volume, more or less, even if they are composed of a random number of the two denominations, as long as the collecting jar is large enough to let them pack together.
I don't know if the above is true of US coins, but I think it likely, again for the banks' convenience, so this would explain the standard value of your full cookie jar.

This is also true of 'silver' coins (nickel-plated steel), but as a ten-penny piece is nothing like the weight of ten penny coins, a mixture of 'silver' and 'copper' will not have a standard weight.

I suspect that this will mean that if you add a random proportion of quarters, 'silver' coins that weigh a lot less than 25 one cent coins, the value of your cookie jar will not be constant.

JOhnUntil the mid-1960s, when the USA still used silver coins from 10 cents up, the weight of the coin was exactly proportional to the denomination. When we switched to nickel-alloy-clad copper coins the weights were reduced to make them interchangeable with the silver coins in vending machines, so mixing them would make weighing unreliable. My guess is that the weights of the new coins were in proportion to their denominations, so after the silver coins disappeared from circulation weighing would become reliable again. Pennies and nickels had different weight to denomination ratios, and the nickel was nowhere near five times the weight of the penny.

JohnD
2015-Jun-21, 09:32 AM
Thank you, Hornblower!
Sorry, Don, I fear that you will need to separate out the quarters - perhaps a second cookie jar?
JOhn