5 Gallons of Pennies.
Carl Mittler
It took 25 years and now my 5 gallon water cooler bottle is full. It
seemed like a fun idea back then. Anyone know approximatly how many
pennies I have, How much they weigh (I can't even move it), and are
there any suggestions on what to do with them.
Thanks
Carl (New York)
Tony Clayton
That is a dangerous question ;-) You might get told!!
I personally would carefully sort them by year and mint-mark to do a
statistical survey of the various quantities, then let us all know
the result.
I would then keep the best or the scarcest ones, and cash them in
at the bank, and have a holiday on the proceeds.
If it had been a 5 gallon jar of quarters you might have been able
to go on a longer holiday :-)
--
Tony Clayton ar...@charterhouse.surrey.sch.uk
ReedOnly
>5 gallon water cooler bottle is full.
>pennies I have, How much they weigh (I can't even move it), and are
>there any suggestions on what to do with them.
>
volume is less than 5 gallons because of the spaces left between the coins
inside the jar. One way you could measure volume by displacement is to
fill your jar with water while the coins are in it (I know, I know, this
is just for illustration purposes!). Measure how much water goes into the
"full" jar, and then subtract that number from 5 gallons, and you'll have
the displacement volume of your coins.
Of course, you could just weigh the jar on a scale, then dump out the
coins, weigh the jar empty. The full weight minus the empty weight gives
you the weight of the coins.
Now, the problem with calculation by weight is that the weight of a cent
has changed recently. In 1982, the coins were changed from a brass alloy
weighing 3.11 grams to a copper-plated zinc weighing 2.5 grams. If you're
talking one coin, that doesn't sound like much, but multiplied a thousand
times and it becomes significant. Since you began the project 25 years
ago, and the change occurred about halfway through your project (14 years
ago, in 1982) let's assume half of your coins are post-1982 (2.5 grams)
and half are pre-1982 (3.11 grams) for an average per-coin weight of 2.805
grams.
Now, 1 gram equals 0.0022 avoirdupois (I think I spelled that wrong, but
it's not the troy kind) pounds; or 1 pound contains 453.592 grams. At a
2.805 average per coin weight, we get 0.356 cents per gram, or $1.61 per
pound. A gallon of water weighs about 7 pounds. Pure copper has a specific
gravity (a measure of density compared to water) of 8.96; pure zinc is
7.13. Let's say the coins you have have an average density of 8 times that
of water. We would expect a gallon (displaced volume) of coins to weigh 8
times a gallon of water, or 56 pounds. Five gallons of coins, then, would
weigh approximately 280 pounds. By this measurement, we can predict a face
value of your coins to be $450.80.
Measured another way, we know the volume of a cent to be about 0.027 cubic
inch. (Don't ask me how I know this!) I calculate approximately 1,154
cubic inches for 5 gallons, and can then estimate about 42,000 coins, or
$420 face value.
I'm going to guess your face value to be about $400. You probably have
about 40,000 cents in your jar. Check with the major banks in your area
about thei policies on accepting bulk cents. DON'T ROLL THEM! We had a
story recently about someone who collected thousands of cents, rolled
them, then had to cut them out of the rolls before the bank would take
them!
Let me know how I did with my estimates!
---Brad Reed
---COIN WORLD "Enriching Coin Collecting Through Knowledge."
E.B.W.
In article <4qm5pc$a...@newsbf02.news.aol.com>,
ReedOnly <reed...@aol.com> wrote:
>>5 gallon water cooler bottle is full.
>>Anyone know approximatly how many
>>pennies I have, How much they weigh (I can't even move it), and are
>>there any suggestions on what to do with them.
>>
>First, these calculations are based on volume by displacement. Your actual
>volume is less than 5 gallons because of the spaces left between the coins
>inside the jar. One way you could measure volume by displacement is to
>fill your jar with water while the coins are in it (I know, I know, this
>is just for illustration purposes!). Measure how much water goes into the
>"full" jar, and then subtract that number from 5 gallons, and you'll have
>the displacement volume of your coins.
>
...snip...
>I'm going to guess your face value to be about $400. You probably have
>about 40,000 cents in your jar. Check with the major banks in your area
>about thei policies on accepting bulk cents. DON'T ROLL THEM! We had a
>story recently about someone who collected thousands of cents, rolled
>them, then had to cut them out of the rolls before the bank would take
>them!
>
>Let me know how I did with my estimates!
>
Hmmm.... I did this twice so far and came out with approx the same value.
1 gal = ~2900 pennies. I measured this with a 1 gallon juice jar and
a 64 oz of Hefenrefer beer bottle (no, I didn't drink it). Both times
it came to approx 2900. So, I calculate there to be $145 in there.
Let me know how close I am. I know it depends on the geometry of the
bottle, but these were packed in there real good.
It is a usefull tidbit of info in case you see a gallon jar of wheat
pennies at a flea market or yard sale.
-Ernie
Randy Flaming
In article <4qi2s2$3n...@usenetz1.news.prodigy.com>, DPC...@prodigy.com
says...
>
>It took 25 years and now my 5 gallon water cooler bottle is full. It
>there any suggestions on what to do with them.
>Carl (New York)
>
Be very careful moving those cents if the jar they are in is glass. My
stepfather cut his hand badly when he tried to move my mother's water jar full
of cents. The top of the jar broke off in his hand.
Norman Diamond
In article <4qi2s2$3n...@usenetz1.news.prodigy.com>, DPC...@prodigy.com (Carl Mittler) writes:
>It took 25 years and now my 5 gallon water cooler bottle is full. It
>seemed like a fun idea back then. Anyone know approximatly how many
>pennies I have,
At the university where I went to graduate school, one time an organization
put on display a one-gallon bottle (this was when gallons were still legally
acceptable in Canada) full of one-cent coins, and people entered their
guesses. Later the bottle disappeared, and I assumed that it had been given
to the winner. Later I read that I was wrong! I was the winner, but the
bottle had been stolen. In consolation, they gave me another one-gallon
bottle full of one-cent coins, slightly more than the first. Both quantities
were pretty close to 5,000. A five-gallon container would be expected to
contain five times as many, if they were Canadian cents, or slightly fewer
if they were American cents.
Pennies used to be much larger than Canadian one-cent coins. However,
if I recall correctly when Britain decimalized, it was when you started
collecting, so you have collected entirely new pennies. Sorry I don't
know the exact proportion of sizes and can't really estimate how many
you have.
--
<< If this were the company's opinion, I would not be allowed to post it. >>
"I paid money for this car, I pay taxes for vehicle registration and a driver's
license, so I can drive in any lane I want, and no innocent victim gets to call
the cops just 'cause the lane's not goin' the same direction as me" - J Spammer
Andrew Tumber
Norman Diamond (dia...@tbj.dec.com) writes:
> In article <4qi2s2$3n...@usenetz1.news.prodigy.com>, DPC...@prodigy.com (Carl Mittler) writes:
>>It took 25 years and now my 5 gallon water cooler bottle is full. It
>>seemed like a fun idea back then. Anyone know approximatly how many
>>pennies I have,
>
> At the university where I went to graduate school, one time an organization
> put on display a one-gallon bottle (this was when gallons were still legally
> acceptable in Canada) full of one-cent coins, and people entered their
And a Canadian (Imperial) Gallon is larger than an American gallon...
Imperial terms are still used on some products in Canada, Norman
butter still says 1lb (as well as 454g) on it and supermarkets
still price meat and produce in both pounds and kg. It is quite legally
acceptable.
Last time I was in the US I was amazed at how metric the groceries had become.
2 litre pops, ect...
Andrew
OnIineFIN
actually it is quite easy to estimate how many pennies you have.
Assume 95% (or whatever you wish) of the volume is filled with pennies,
and the pennies are copper (pre 1982). Well, since the density of copper
is known, and the weight of the average penny is known, it is simple
mathematics to determine the number of pennies in a 5 gallon container.
So get out your calculator and you have your answer.
Kaleb KEITHLEY
bi...@FreeNet.Carleton.CA (Andrew Tumber) writes:
>And a Canadian (Imperial) Gallon is larger than an American gallon...
>Imperial terms are still used on some products in Canada, Norman
>butter still says 1lb (as well as 454g) on it and supermarkets
Hmmh. A pound of butter here in the U.S. weighs 454g too. Are you sure
the butter you're looking at isn't weighed in U.S. pounds. :-)
--
Kaleb KEITHLEY
Jon Hamkins
If you did know the average weight of the cents in the jar, it would be
much easier to just weigh the full jar and divide the average cent
weight to obtain the number of cents.
However, you may not have realized that the real beauty of your method
is that you don't have to assume that the cents are all pre-1982 to get
the right answer. Not convinced? Follow me. Your method can be
summarized by following algorithm:
A. volume of 1 cent = (average weight)/(density of copper)
B. volume of all cents in jar = 95% of volume of 5 gallons
C. number of cents = (volume of all cents in jar)/(volume of 1 cent)
At first, I thought there were two problems with this method:
1. The average weight of the cent in the jar is not known. As the
original author stated, he has put cents in this jar over the years
and he just now finished filling it. Thus, some of the cents in
this jar are 3.11 g (pre-1982) and some are 2.5 g (post-1982), and
there isn't an accurate way to estimate the fraction of either.
Note that 3.11 is 24.4% higher than 2.5.
2. Using the density of copper is OK for pre-1982 cents, which are 95%
copper, but not for post-1982 cents, which are 97.6% zinc.
Fortunately, problems 1 and 2 cancel out! That's because the volume of
a cent is probably exactly the same before 1982 as after 1982. Thus,
weight of copper cent weight of zinc cent
--------------------- =~ -------------------- = volume of cent
density of copper density of zinc
Looking the specific gravties in my dictionary, I found that this is
the case:
3.11 2.5
---------- =~ ---------
8.92 7.14
So, in step A, the answer is the same for all cents in the jar. (My
take on specific gravity is that 8.92 means copper is 8.92 times
as dense as water, which has about 1000 kg/m^3, if I am not mistaken.
Some chemist or home brewer please correct me if I am wrong.)
Let's crunch the numbers:
Volume of cent = 3.11 g / (8.92 x 1000 kg/m^3)
= 3.5 x 10^-7 m^3
95% of Volume of 5 gallons = 0.95 x 5 x 3.7854 liters
= 0.95 x 5 x 0.0037854 m^3
= 0.01798 m^3
Number of cents = (95% of volume of 5 gallons)/(volume of cent)
= 0.01798 m^3 / 0.00000035 m^3
=~ 51,000
There is one problem with this analysis. 95% is way too high! To
illustrate how hard guessing the density of the cents in the jar is,
consider this even simpler example: if you just laid all of the cents
flat on a big table, with none of them overlapping, what would you say
is the biggest fraction of the table surface that you can cover? It
turns out the answer is pi/sqrt(12) or about 90.7%, which is what you
get when you put the cents in a hexagonal pattern, with each cent
touching six others. If you imagine these cents packed in three
dimensions, no matter how arranged, I don't think you'd get a much
higher density. In fact, it is known that the density of a sphere
packing in three dimensions--- by which I mean the percentage of space
occupied by spherical objects, say a bunch of marbles piled up---is
less than 77.84%.
[Digression: in fact, the usual stack of oranges at the grocery store,
called the face-centered cubic packing by mathematicians, has a density
of only 74.05%. In the year 1611, Kepler conjectured that this is the
densest packing in three dimensions, but to this day no proof of this
has ever gained widespread acceptance.]
Not only is 95% higher than the best 2D packing of 90.7%, but my guess
is that when cents stack up randomly, there is a lot of space between
them. I would think 50% is a much better ballpark estimate, which
would give about 27,000 cents in the jar, or 5,400 per gallon. But
even this is much higher than another poster, who claimed 2,900
cents/gallon.
At any rate, it's not easy without some experimental data!
----Jon Hamkins
OnIineFIN
Let's look at your assumptions:
A. volume of 1 cent = (average weight)/(density of copper)
B. volume of all cents in jar = 95% of volume of 5 gallons
C. number of cents = (volume of all cents in jar)/(volume of 1 cent)
Well, c'mon, the size (volume) of the penny hasn't changed since 1856, so
forget A and B as they are meaningless, that is (C) can be calculated
directly, since the dimensions of a penny are known -
Now, there are 2 unknowns, the packing efficiency (as recognized) and
another unknown, which I am surprised you haven;t taken into account; the
wear on the penny (e.g. what is the volume due to wear -- it might be
higher due to contaminants so don't assume it is less than 100%). NOW
RE-CALCULATE your answer!!
ReedOnly
>another unknown, which I am surprised you haven;t taken into account; the
>wear on the penny
>
ten-thousands of an inch (or less!) and even multiplied by the full volume
of the jar will result in a barely 1 percent variation.
Packing density can be determined by pouring measured water into the jar
full of pennies. Let it "percolate" through, and continue until no more
water will go into the jar. The total amount of water, subtracted from the
total volume of the jar, results in the pennies' displacement volume.
The jar is just to heavy to practically weigh it as a whole.
pk...@pica.army.mil
Prior analytic discussion omitted.
>I would think 50% is a much better ballpark estimate, which
>would give about 27,000 cents in the jar, or 5,400 per gallon.
>
>At any rate, it's not easy without some experimental data!
>
> ----Jon Hamkins
Thankyou for the analysis. Several years back my grandfather had
given me a one gallon apple juice bottle filled with pennies. When
counted the amount was approximately 5200 one cent pieces.
Jon Hamkins
>Actually, you made this problem much harder than you had to.
>Let's look at your assumptions:
>A. volume of 1 cent = (average weight)/(density of copper)
>B. volume of all cents in jar = 95% of volume of 5 gallons
>C. number of cents = (volume of all cents in jar)/(volume of 1 cent)
>Well, c'mon, the size (volume) of the penny hasn't changed since 1856
Just to nit-pick, the volume of a cent was significantly different
before 1864, as the small cents before this year were significantly
thicker and weighed 4.67 g, not 3.11g or 2.5 g. But yes, you are
essentially correct--- I doubt there is much of anything in the jar
before 1930, let alone 1864.
>, so
>forget A and B as they are meaningless, that is (C) can be calculated
>directly, since the dimensions of a penny are known -
Hey, this is your method, not mine! :) Let me remind you of what
you said:
"Well, since the density of copper is known, and
the weight of the average penny is known, it is simple mathematics to
determine the number of pennies in a 5 gallon container."
If you are going use the dimensions of the cent, why bother with the
weight and density of the cent, as you say? The only thing you
can calculate from weight and density is the volume of the cent, so I
made the steps of your suggestion explicit in (A), (B), and (C).
In any case, your original idea of using weight and density was a
pretty good one. That is because the volume of a cent is NOT readily
available in any references, at least not the ones I have, and neither
is it calculable from readily available dimensions, of say, diameter
and thickness. This is because "thickness" refers to the thickness of
the rim, but the coin is not uniformly this thick. If you lie a coin
flat on the table, the only part that touches the table is the rim, and
there is some amount of empty space between the table top and the field
which is facing the table. This empty space is not part of the volume
of the coin, and would significantly affect any calculations
performed.
This empty space further contributes to the lousiness of the packing
density, and makes it even harder to estimate.
>Now, there are 2 unknowns, the packing efficiency (as recognized)
Yes, I see this as the major stumbling block. It's just too hard to
analyze theoretically. I just took a stab in the dark saying 50%,
which may or may not be as bad as your guess of 95%. Does anyone want
to do an experiment on what the packing density is? (Just take a
"full" jar of cents, and fill it with water. Then remove and measure
the water. The cents take up the remaining space of the jar.)
>and
>another unknown, which I am surprised you haven;t taken into account; the
>wear on the penny (e.g. what is the volume due to wear -- it might be
>higher due to contaminants so don't assume it is less than 100%).
Good point. I wonder if anyone has statistics on what wear does to
volume or weight. I bet the mint has performed reasearch on this when
fiddling with new alloys, because they need to know how long a coin
will last in circulation and still work in vending machines. If you've
ever had a lot of junk silver coins, you know that the wear starts
playing a significant part. For example, some of these junk coins
don't even stay properly in a Whitman folder because they have been
worn to a smaller size. (Perhaps that's just as well, preservationists
might say!)
>NOW RE-CALCULATE your answer!!
No thanks!
----Jon Hamkins
Skip Floyd
********************
Volume
Density
Weight
Mass
These terms are not interchangeable...
You don't determine the Volume of an object from its Weight...
A good Bathroom scale will weigh the coins...
But, in the end... The formula is:::
One roll equals 50 coins...
Skip
<o> <o>
/\
\__/
.
OnIineFIN
however, then you would have a 5 gallon container full of water and
pennies -- not exactly very eye appealing!!
I have a new method, hire a schoolkid to count and roll the pennies for
you. Then if you are really cruel, make him sort the '82 pennies by type.
Tell him for his payment he can have all the' 95 DD's he finds - or if
you are a cheapskate, the '60 SD's!.. That outta keep him busy.
Jon Hamkins
Yes, good idea. :) Actually, I think more than a few of us would be
interested in the statistical breakdown of all of the dates and mint
marks.
----Jon Hamkins
Kevin Redden
Kevin
Skip Floyd
That's Too Easy.!!!
NO Fun at All.!!!
Carl A Fischer
the volume of a roll of pennies ( i.e. cylender) and taking 1/50th
Michael Dworetsky
>actually it is quite easy to estimate how many pennies you have.
>Assume 95% (or whatever you wish) of the volume is filled with pennies,
>is known, and the weight of the average penny is known, it is simple
>mathematics to determine the number of pennies in a 5 gallon container.
>
Even better--if you know the weight of the filled container, and can
estimate the weight of the empty container (by weighing a similar one)
why not just divide the total weight in kg of the cents by the weight of one
cent, 3.11 grammes up to 1982? Even a very worn coin would not lose more
than 10 percent of its weight, and most of your coins would weight nearly
the full minted weight.
--
Mike Dworetsky, Department of Physics | Haiku: Nine men ogle gnats
& Astronomy, University College London | all lit
Gower Street, London WC1E 6BT UK | till last angel gone.
email: m...@star.ucl.ac.uk | Men in Ukiah.
Leonard Augsburger
OnIineFIN <onii...@aol.com> wrote:
>The concept of deteremining packing efficiency by adding water is valid,
>however, then you would have a 5 gallon container full of water and
>pennies -- not exactly very eye appealing!!
OK, how about taking a sample of the pennies and putting them in a
smaller glass, then filling with water.........from this you can
get the "packing density" or whatever you want to call it and then
extrapolate the calculation to a large container.......
Regards,
Len.
--
---------------------------------------------------------------------
Leonard D. Augsburger Motorola Inc, CIG
1501 W Shure Dr. IL27-3227AR
Mail: augs...@cig.mot.com Arlington Heights, IL 60004 USA
Skip Floyd
>Mike Dworetsky, Department of Physics | Haiku: Nine men ogle gnats
>& Astronomy, University College London | all lit
>Gower Street, London WC1E 6BT UK | till last angel gone.
> email: m...@star.ucl.ac.uk | Men in Ukiah.
We used to burn ants with a magnifier on summer afternoons when I was a
kid...
But, why would 9 physicists, in a University, find it interesting to do
the same to Gnats.???
TTFN
Skip
just an ignorant RedNeck, with more Q's than A's...