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Old 08-19-2003, 04:36 AM   #1
Faceman
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Join Date: February 18, 2002
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Okay after I did some calculation on "Operation Pennydrop" in the
Chemical Weapons topic I decided to continue these silly calculations.
If the topic doesn't get too boring I'm going to post silly math problems (like: If all the oceans were made of booze and mankind drank them up, how hammered would everybody be?)and solve them after enough people have started cursing me.

So for starters an easy one.

Which is the smalles standard US coin you would need to fill an empty bottle of good champagne (e.g. Dom Perignon) so that the coin-filled bottle would be worth equal or more than the champagne?
Or in other words: Which US coins are worth their volume in champagne?
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Old 08-19-2003, 05:42 AM   #2
shadowhound
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Argh american questions [img]tongue.gif[/img]
It doesn't matter either way, all I know is that its worth more then a bottle full of australian coins...
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Old 08-19-2003, 05:48 AM   #3
Deejax
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Since I lack the knowledge of the american monetary system (and champagne prices) I calculated it with the dutch values.

One bottle of champagne: 111,- euro for 0,75 liter.

The champagne has a value density of 0.15 euro/ml

The dutch (well, european) coin values are: 2, 1, 0.5, 0.2, 0.1, 0.05, 0.02 and 0.01 euro. The volumes of these coins are 1.15, 0.99, 1.10, 0.83, 0.59, 0.59, 0.46 and 0.35 respectively.

This means that the bottle could best be filled with the 10 cents piece. It has a value density of 0.17 euro/ml. (the next closest values are 0.24 for the 20 cent piece and 0.08 for the 5 cnet piece)

If you fill the bottle with these coins you end up with a little bit overpriced (127 euro) bottle containing 1268 (and a half) coins which weigh 5.20 kilo.
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Old 08-19-2003, 06:02 AM   #4
daan
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Though keeping in mind that you wont be able to fill the entire volume with coins as they cant be aligned so that they leave no holes .. and the fact that the coins are relatively small and therefore the holes they leave when put next to eachother are relatively small too, I think the price might acutally be very close to the real thing since the theoretical value is a little higher than the practical [img]tongue.gif[/img]
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Old 08-19-2003, 06:22 AM   #5
Deejax
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Quote:
Originally posted by daan:
Though keeping in mind that you wont be able to fill the entire volume with coins as they cant be aligned so that they leave no holes .. and the fact that the coins are relatively small and therefore the holes they leave when put next to eachother are relatively small too, I think the price might acutally be very close to the real thing since the theoretical value is a little higher than the practical [img]tongue.gif[/img]
You're absolutely right. I just made the simplest calculation.

Let's take it one step further.
Stack the coins in neat columns and arrange the columns next to each other in a triangular way. The mass you get is filled for 78.5% (the fraction is Pi/4) with coin and 21.5% air.

This means that, continuing the euro example, the bottle contains 996 (and a quarter) 10 cents coins, weighing 4.08 kilo, and costing you 99.6 euro. A bit less then the champagne price. [img]smile.gif[/img]

This is still not the actual amount of coins you can fit in a bottle. You will probably lose more space due to its shape.
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Old 08-19-2003, 06:44 AM   #6
daan
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Im intrigued .. how did you come up with pi/4 ?
And do you mean that just the space left in between ( the middle of ) the triangle of coins is 21.5 % ?
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Old 08-19-2003, 07:15 AM   #7
Deejax
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It is a bit difficult to explain without drawings.

Look at a layer of coins. The closest you can get them together is by making rows of coins and placing them next to each other. Start with a horizontal row of coins. Make a second row above it, displaced half a coin to the right (or left). And so on. You get a layer of coins in alternating rows, each row displaced half a coin compared to the previous. If you were to draw lines between the centers of neighbouring coins you would get triangles. (i do not know the correct term, but the are triangles with 60 degree angles and equal side lengths.).

Now we need to draw a grid over the coins. Do this by drawing paralel horizontal lines through the centers of the coins. To draw the vertical lines take a single row and draw vertical line through the centers of every coin in that row. If you've done this correctly the vertical lines will not run through coins in the rows directly above and below this row.

If you look at the grid. every square should contain two quarters and one half coin, totalling one coin.

If we set the radius of the coins to 1, the calculation becomes easy.

Every square contains one coin.

The surface of a coin is Pi*Radius2 = Pi*12 = Pi

The side of a square is two times the radius of a coin = 2. The total surface of a square is then 2*2 = 4

This means that the fraction of the surface covered with coin is Pi/4 = approx 78.5%


I hope you understand this. I don't know of a way to draw at IW. Would make it a LOT easier.
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Old 08-19-2003, 08:13 AM   #8
daan
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Yeah, I got it again. I did the calculations at school once ( math -analyses ), but the holidays diminished my intelligence to a level where I couldnt figure it out anymore.

Though I'd like to know what the diameter of a champagne bottle is, and the diameter of its neck. As the above calculations require the bottle to be a square with the coins touching the sides at convenient points.
I'm assuming that what you do to know the proportions of the "big square" that draws around all the traingular-placed coins is move each side of the "little square" a radius to the left,right, up and down ( the left side to the left, right side to the right etc. ). This would make the "Little square" match the "big square" ( which is the smallest possible sqaure to house all 3 coins ).

The only drawbacks this has is:
1. The big square has an extra 4 spots of aire in the square's corners that werent there in the calculations of the small square.

2. Both sqaures arent truly squares, as the width is indeed 1 coin for the small and 2 coins for the big square, but the lenght isnt 1, respectively 2 coins. They are squares if you use your method, but in reality the "top-coin" would slide into the gap that the 2 "base-coins" have because of their curvature.

Did I get that right ( I doubt it since the whole holiday_not_intelligent_enough_anymore is still true ) ?

[ 08-19-2003, 08:20 AM: Message edited by: daan ]
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Old 08-19-2003, 09:07 AM   #9
Deejax
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Whoa, discovered a major flaw in my calculation. [img]graemlins/blush.gif[/img]
And when I tried to correct it my internet connection went down. [img]graemlins/madhell.gif[/img]

But here it is: The square isn’t square!
One side has length 2*radius = 2, but the other side is shorter. It is equal in length to the height of the triangle I described earlier.
Via pythagoras, the height of the triangle equals the square root (22-12) = sqrt(3) = approx 1.73.

Thus the surface of one rectangle (see, not square [img]smile.gif[/img] ) equals 3.46

This means that 91% of the layer is filled with coin. This results in a bottle filled with 1150 coins, weighing 4.72 kilo and costing 115 euro.


Ah, I see, you already discovered my mistake.

This calculation isn't just for square bottles. It calculates the average of the air in a layer. You use the fact that there is a repeating pattern. The fraction of air in one unit (one repetition, for example the rectangle) is the same as the fraction of air in the total (discounting what happens at the edges).
The major problem with this method is that you ingore any effects of the shape of the container. But it gives a good estimate if the size of the container is significantly larger then the size of the coins.
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Old 08-19-2003, 09:42 AM   #10
daan
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Aye, the flaws resulting through the shape are relatively unsubstantial.
Though I'd still like to know how many of those "squares" fit inside a champagne bottle's base. We've got the numbers assuming exactly one triangle fits in there .. but how many do actually fit in there ?
And arent there other shapes thinkable that make better use of the circular space offered by the bottles base ?
Actually, I guess the triangle is the best shape if you lay the coin flat, but more coins would fit in if you put them on their side, as they'll be able to approach the curves more.
Anyway, putting that aside .. your calculations with the asumption you'll put a triangle of coins in there, got you very close to the champagne-price.

[img]graemlins/dance2.gif[/img] Score one for the dutch team [img]graemlins/dance2.gif[/img] ( Didnt do too much, but I'm dutch too [img]tongue.gif[/img] )
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