2-23-10 - Mass Counting

[Doc Giggles]

I own a bank grade coin counter so I never have to worry about anything like this. I can count 2700 coins in a minute and I don't have to pay any fees.

[MasterChef]

Hi, everyone

In your latest comic series about coin counting, I'm surprised nobody has suggested using a water displacement method to count coins. It gets around the 'variable weights issue' given the fact that coins have been standard in size for longer than any have been in circulation.

Anyways, love your comic, and have been an avid reader for 7 years running now =)
(And I'm such a dork that I actually had to register on the forums to post this! )

[ShadowDog]

That's ... kinda cool actually.

[barawn]

The shape of coins has been defined for a long time, but the volume has not. The engravings have changed like crazy especially recently, with the Lincoln memorial from 1959 on, the Lincoln bicentennial pennies in 2009, and the 2010 shield. Though granted, the volume difference is much less than the weight differences.

But it's also much, much harder to measure volume than weight - especially with liquids. For a 4 cm jar opening, to measure volume accurately to 1 penny, you'd have to measure distances of 0.3 mm, the jar would have to be flat to 0.1 degrees, and you'd have to somehow know the meniscus behavior as well. You could use a small cylinder, though... but of course, that's the same as stacking pennies and then all you need to do is just measure the height.

[MasterChef]

But it's also much, much harder to measure volume than weight - especially with liquids. For a 4 cm jar opening, to measure volume accurately to 1 penny, you'd have to measure distances of 0.3 mm, the jar would have to be flat to 0.1 degrees, and you'd have to somehow know the meniscus behavior as well. You could use a small cylinder, though... but of course, that's the same as stacking pennies and then all you need to do is just measure the height.

No no, I think you might be over-complicating the analysis (really, this is the standard way in most sciences). If you have a small glass graduated cylinder, there are standard ways to account for the meniscus (i.e. the standard is to measure the bottom of the meniscus in glass), so fill it up to say 5 ml/cc and drop the penny in, the new volume-5cc=penny volume. If you know the volume of 1 penny, you now how many pennies will then fill up any given volume. So you would take the pennies, put them in a container of known volume (say a 3L mason jar), and fill it (to the 3L mark) while noting how much it takes to fill, the difference between 3L and what it took to get to 3L is the volume of the object of interest. The REAL limitation of the water displacement method is that it doesn't work for porous materials (which is a moot point because we're dealing with solid metal objects) or objects with an overall UNKNOWN density that is less than that of water (because if we knew the density we would know how much % volume would be above water, and thus the total volume is still calculable , like an iceberg)

The shape of coins has been defined for a long time, but the volume has not. The engravings have changed like crazy especially recently, with the Lincoln memorial from 1959 on, the Lincoln bicentennial pennies in 2009, and the 2010 shield. Though granted, the volume difference is much less than the weight differences.

Now this something I hadn't considered, but again it doesn't matter, you just have to account for it in your displacement method equation. If, for example, the old pennies (which are rarer) were say 0.01% smaller (for sake of argument) in volume, then by volume, 10000 old pennies=9999 new pennies. Again, for sake of arguement, lets set the volume of a new penny at 1cc (old pennies thus being 0.9999cc). Now, if one were to measure, say, a coin volume 3000cc exactly (e.g. which is 3 liters, which could be done if you are measuring in a 4 liter jar; are there bigger jars?), we would need to solve for the orthoganal set of #new and#old (x,y) penny counts with the equation
1[cc]*#new+0.9999[cc]*#old=3000cc
where only integer solutions of #new and #old >0are valid (and new coin count, our x, is effectively bounded 0-3000, otherwise they would not fit in the specified volume . So we can recast the equation to a more standard format
(1/.9999)*(3000-#new)=#old
we only need to determine those values of x that results in integer values of y
A really quick analysis in excel (because I don't want to take the time to do this in MatLab) gives us that there can only be 3000 new coins, as (3000,0) is the only solution that satisfies the condition. Therefore you must have 30 dollars worse of pennies. Another example (complex with multiple solutions), where we measure a 2500cc coin volume and old pennies are 2.5% larger (1.025cc, although)
The solutions are as follow
(#new,#old, coin total)
2500,0,2500
2213,280,2493
1967,520,2487
1926,560,2486
1885,600,2485
1434,1040,2474
1393,1080,2473
1352,1120,2472

(plus several more, but this is for illustration)
Given the rarity of old coins, I expect one would really have one of the top 3 solutions (which means one could be off by pennies, not 10s of cents as tony suggested in wednesday's comic). But this last case is probably not really valid, because as volume differences, there become less and less simultaneous solutions. In practice, I expect real volume differences to be 1% or less between the penny types (given the fairly fixed mass and metal compositions/density post-1982 would mean pretty constant volumes regardless of shape) . Overall, I would actually expect measurement error of water volume to be larger than anything else.

Now my only question is then why would Tony complain about drying the penny's to repeat the measure, i.e. why would one 'do it again'? But Liz's idea works too

[Mr. Xero]

This whole argument is Moot as i kow for a fact that Weighing is exactly how banks check rolls of coins

[teddy]

This whole argument is Moot as i kow for a fact that Weighing is exactly how banks check rolls of coins

Way to fail at imagination.

[collegestudent22]

This whole argument is Moot as i kow for a fact that Weighing is exactly how banks check rolls of coins

That isn't exact and they only use it to make sure that you didn't just wrap the thing around a short steel rod with pennies on the end or something crazy like that. Basically, if the rolls are the right coins it should weigh between X and Y, and so if it does, it passes the check.

[ampersand]

How much does a bank-grade coin counter cost?

[adciv]

Free! Use your hands. Just like the banks did 60 years ago.

[Doc Giggles]

How much does a bank-grade coin counter cost?

$350 for a low use one to over $5000 for a high use. Mine is a portable unit that sells new for $1500. I picked up it up for $200 at a car wash that was closing. It was new in box because no one there knew how to work it. Lucky for me because it's the same type my company uses.

[barawn]

No no, I think you might be over-complicating the analysis (really, this is the standard way in most sciences). If you have a small glass graduated cylinder, there are standard ways to account for the meniscus (i.e. the standard is to measure the bottom of the meniscus in glass),

The reason you measure the bottom is because that's where the guys who put the markings on the cylinder calibrated it to. If you were trying to put markings on your own custom container, you'd have to compensate for it somehow yourself. Which you could do... but I doubt you could do it to the precision needed.

And if you've got a small glass graduated cylinder, and you're dropping pennies in... they're going to be stacking one by one. In which case you can just measure the height, and skip the water.

container of known volume (say a 3L mason jar), and fill it (to the 3L mark) while noting how much it takes to fill

Emphasis mine. I don't believe it's possible to say that a 3L mason jar is "full" to 0.5 ml precision, which was the point (I also doubt that a 3L mason jar is 3L to 0.5 ml precision, either). Assuming that the mason jar is ~4 inches across, you've got to stop within a fraction of a millimeter, and the jar has to be virtually completely flat.

You could certainly get within a few cents, though. But it'd probably do the same as weighing.

A really quick analysis in excel (because I don't want to take the time to do this in MatLab) gives us that there can only be 3000 new coins, as (3000,0) is the only solution that satisfies the condition.

Ah, see, this clued me in to the "real" solution, which satisfies both Tony's (precise, repeatable) and Greg's (simple, basic) requirements, and it's not as brute force as Liz's. It's also ludicrous, which satisfies both of the guys.

1: Weigh the empty jar, weigh the full jar, difference the two. That weight is X.
2: Weigh a sample (or Google) of old coins (call that weight A) and new coins (call that weight B)
4: Pour half (exact value is unimportant) the coins out. Repeat the process (call that weight Y)
5: Take the other half. Repeat the process (call that weight Z)
6: You now have pA+qB=Y, rA+sB=Z, and (p+r)A+(q+s)B=X, which is much more strongly constrained than just the single measurement case. Repeat from 4 using the sub-pots created above until only one solution remains.

As a silly example, if you had coins of two types A and B, A weighing 3 and B weighing 2, and the coins totalled 26, you could have:
13 B
10 B 2 A
7 B 4 A
4 B 6 A
1 B 8 A

If, when you split them up, the first half weighs 13 and the other 13, now only
10 B 2 A (5B+1A, 5B+1A)
4 B 6 A (2B+3A, 2B+3A)

remain. In this case you have to divide the sub-pots up (say, getting 6 and 7). This still doesn't do it (you could have 3B/2B+1A or 2A/2B+1A). and then divide them up *again* (say, divide the 6 and get 4 and 2), which finally breaks the degeneracy, leaving only 10 B, 2 A.

[adciv]

Except that you've added an additional constraint, the total coin count. If the total count is unknown, then your formula is not solvable.

[barawn]

It only looks unconstrained (4 unknowns, 3 equations) if you neglect the fact that the p, q, r, s have to be integers. The entire process is guaranteed to converge once the weights get below the least-common multiple of the two coin weights A and B, but there's a good chance that it would get down to one possibility well before that.

[adciv]

Ok, upon second reading your solution is even more idiotic than I thought, as it requires multiple recursive measurements and precision to 1/100th of a gram. You still have the issue of not having a garuanteed unique solution under real world conditions.