2-23-10 - Mass Counting

[MasterChef]

(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.

Ok fine then use a 3L Beaker properly calibrated: http://www.capitolscientific.com/estyle ... m=B3190-3L

Also: the 0.75 inch diameter and 1.5mm thickness seems to have been the standard since before 1982, only mass (again precise to 0.01g) has changes. So again I believe the engraving changes would account for less than a fraction of a percent difference in volume. Thus volume analysis the most scientifically correct method, other than pure empirical methods (i.e. direct counting), and leaves very rare instances of no unique solution to the equation (which is even easier now than I previously supposed since there is no significant difference between coins of different eras)

precision to 1/100th of a grams.

Thats actually a pretty standard level of precision (to 0.01) for something that measures mass in the 'gram' order of magnitude. I've used devices calibrated to measure mass to 1/1000th mg (so really to microgram precision, but its sensitive to wind, so you shut the doors and let the sensor settle ) Dont know why that seems unreasonable, its even the precision level the US mint reports.

[adciv]

It's unreasonable for two reasons.

1) When measuring 100s of pennies at a time, you'd need a scale capable of reading to at least five significant digits accurately. While I have seen scales accurate to 1/100th of a gram, they generally do not go past 100 grams. To get measurements for a reasonable count of pennies, you're going to need a scale that will go to at least 500g with the same resolution.

2) Corrosion or foreign material on the pennies will throw off the weight by more than 1/100th of a gram over your sample, throwing off the count. I've seen paper stuck to coins before, as well as oxidation and other things.

[barawn]

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.

Multiple recursive measurements: Right, so what? Any goofball method that you want to be precise will require multiple measurements. If you want one measurement, easy, done, and perfect, count them by hand or use a coin counter.

Precision to 1/100th of a gram: Uh, no. Pennies weigh 2.5 g or 3.1 g - that is, separations on the order of tenths of a gram. So you'd need precision to a tenth of a gram, which is pretty easy (The 2.7 g pennies from 1943 can probably be neglected since you can separate them by sight).

No guaranteed solution: No. Let me say it again - you are guaranteed to get a unique solution once you get the weights down below the LCM of the individual weights, and quite likely you'll end up with a solution earlier than that.

[barawn]

Ok fine then use a 3L Beaker properly calibrated: http://www.capitolscientific.com/estyle ... m=B3190-3L

Check the image more carefully. It only has demarcations of several hundred milliliters. You need 0.5 ml precision, which just isn't happening easily.

[MasterChef]

Ok fine then use a 3L Beaker properly calibrated: http://www.capitolscientific.com/estyle ... m=B3190-3L

Check the image more carefully. It only has demarcations of several hundred milliliters. You need 0.5 ml precision, which just isn't happening easily.

No no, I think you misunderstood my suggestion. The 3L canister for holding the pennies (so the only marking you'd need on it is the 3L mark) You would ALWAYS fill this container to 3L

What you'd need then is a smaller (with more precise graduation) cylinder to fill, say a 250ml beaker to get close, and then pipette gun the rest (I have used both 50ml for quicker transfers, and long 10ml pippets with 0.1ml demarcations for precision)

Basically
Step 1: Put pennies in canister with known volume mark
Step 2: Fill canister with 250ml beaker, repeat till <250ml to known mark remains
Step 3: Fill canister with 50ml beaker
Step 4: Use 10mL pipette gun to correct volume to 3L, precise to 0.1mL
Done
Steps 2/3 can be altered as needed (i mean, in reality, step 2 isn't needed, it just might make things faster)

And remember if you have more or less pennies, just pick a bigger (so it fits) or less (so it doesn't take too many fills) canister to hold them in

[adciv]

Multiple recursive measurements: Right, so what? Any goofball method that you want to be precise will require multiple measurements. If you want one measurement, easy, done, and perfect, count them by hand or use a coin counter.

At which point there is no difference between the recursion method and counting it by hand.

Precision to 1/100th of a gram: Uh, no. Pennies weigh 2.5 g or 3.1 g - that is, separations on the order of tenths of a gram. So you'd need precision to a tenth of a gram, which is pretty easy (The 2.7 g pennies from 1943 can probably be neglected since you can separate them by sight).

3.11[/] grams

No guaranteed solution: No. Let me say it again - you are guaranteed to get a unique solution once you get the weights down below the LCM of the individual weights, and quite likely you'll end up with a solution earlier than that.

And again, it is unrealistic when LCM>10.

[barawn]

Doesn't matter if it's 3.11 grams or 3.1100018581 grams. All that matters is that "old penny weight - new penny weight = O(0.1) grams", which it is - well, it's actually bigger, so 0.1 g accuracy is easily enough. That's all you have to measure to. Of course it's best if you have the penny weights to as high precision as possible for larger numbers of coins, but the measurements themselves only need to be accurate to of order the difference of the two.

If you want to be precise, you look for integer solutions of coins which have the least overall error compared to the total weight. By the time you repeat the procedure a few times, one solution will likely have much, much lower error than the others. When the others all have error greater than the difference between the two coins, you've got your solution.

And it becomes impractical when the LCM gets *small*, not large. Large is better - in the example I gave, the LCM was small - 3 coins (the LCM was 6, which was 3 of one type, 2 of the other). But you only have to go down to the LCM if you're really unlucky. Assuming you're capable of measuring to 0.1 g, it means at worst you would have to get down to below ~30 coins. (To see what I mean, imagine if one penny weighed 2.0 g, and the other weighed 4.0 g. You'd have to get down to a single coin.)

Geez, you're making me want to actually simulate the whole thing. Or do it.

[adciv]

Well, I've got about 11,000 lying around if you want to try it.