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Old 08-19-2003, 10:07 AM   #11
Thoran
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Well... I don't have the dimensions of a bottle of Dom handy so I'll make one assumption that Dom comes out with a new bottle that is hexagonal in shape (and the hexagon just happens to be some multiple of the coin diameter) but contains the same 750ml. of good stuff. Given that assumption the container would hold:

Quarters (3.5cm x .155cm) -

Volume of the bounding hexagonal column - .94 cc
Number of coins in a 750ml (750cc) hexagonal container - 798
Value of coins - $199.00 - too much

Penny (1.905cm x .155cm) -

Volume of the bounding hexagonal column - 0.51 cc
Number of coins in a 750ml (750cc) hexagonal container - 1466
Value of coins - $14.66 - not enough

Those were the only dimensions for us coins I could find in a quick search... but frome these two you can extrapolate a conclusion:

A nickel is a bit larger than a penny but much smaller than a quarter... so lets guestimate that 1000 could fit in our container (high side estimate)... this would be worth $50.00... not enough

A dime is smaller than a penny... but lets say it's the same size (high side estimate)... so 1466 dimes would be worth $146.00... so the dime is our winner since more than 146 dollars worth would fit into our ideal container.

With a profile of a Dom bottle these calculations could be run taking into account bottle diameter... but it's unlikely the result would change.

EDIT... just looked at previous posts... the only problem you guys have is that coins stacked efficiently form hexagonal shapes... not square shapes (try it out yourself). What you get is a bunch of columns of coins, each of which occupies a hexagonal volume that is:

(((Coin Radius / Sin(60degrees)) ^ 2) * 2.598) * Column Height

[ 08-19-2003, 10:13 AM: Message edited by: Thoran ]
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Old 08-19-2003, 12:48 PM   #12
Chewbacca
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I wanna know who's doing all this math after drinking a bottle of Dom? Not me! [img]tongue.gif[/img]
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Old 08-19-2003, 04:28 PM   #13
Faceman
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Hey guys! Great job. [img]graemlins/thumbsup.gif[/img]
I actually assumed you'd leave out the space-between-coins issue as we are looking for an approximation not an exact result (which we can't get anyway lacking the knowledge of the exact shape of our champagne bottle)

I used US coins in my example because I figured most of the members here would be from the US but what the heck. I'm from Euroland myself so the answers quite suit me.

It is with US coins the same as with EU coins (they are quite similar in weight and shape). The 10-cent (dime) is the correct answer.

calculations have been outlined quite fabulously in the above posts

disregarding the space-between issue it comes down to a price/volume question
which is easily calculated for the champagne by price/volume (duh!)
and not anymore difficult for the coins (where volume may be unknown) by
(diameter/2)^2*pi*thickness = volume
face value/volume

I refer to diameter/2 instead of radius (which would be the common term in the formula) and thickness instead of height in this calculation as these are the standard specifications by the Fed and the EZB.

[ 08-19-2003, 04:56 PM: Message edited by: Faceman ]
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Old 08-19-2003, 05:55 PM   #14
Faceman
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Now for another easy and quick one.

And again I leave fact collection, approximations and interpretations to you.

Assuming you are driving in a very nice sports car (any of the Italian V12s fore example) at top speed.
What would the cinetic energy of the car be?
How long could you as a human sustain on that energy? (You may approximate or use your actual values).

And another one I always fancied:

How big (and filled with what gas) should a baloon or Zeppelin so that upon his explosion (and after a cooldown) there'd be a noticeable rainfall over a major city (e.g. New York).
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Old 08-19-2003, 08:24 PM   #15
Azred
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[img]graemlins/erm.gif[/img] Why not do it the easy way?

Obtain the volume of each US coin via water displacement, from which you can calculate each coin's value in units $/mL; measure the mass of the coin and you can also calcuate its density. Take any bottle of champagne and use the volume on the front, as well as the price, to calculate its value in $/mL. Now you can compare base value/volume directly.
Once you have these, take any glass and calculate the volume it holds by filling it with water and compare to mass to the empty glass. Multiply glass volume * coin density, divide by the mass of one coin and round down to the largest integer to deduce the maximum number of coins the glass may contain; from this you can figure the value of the coins in the glass. Finally, it is a simple calculation to figure out how much a glassful of champagne would cost (volume * value in $/mL).
In the end, you know how much a glassful of each coin in worth and you may compare these against the value of a glassful of champagne.

Don't forget to use half-dollar, "silver" dollar, Susan B. Anthony dollar, and Sacajawea dollar coins.

After all this, you'll most likely want to drink what is left of the champagne. [img]graemlins/awesomework.gif[/img] [img]graemlins/petard.gif[/img]
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Old 08-19-2003, 08:44 PM   #16
Azred
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Quote:
Originally posted by Faceman:
Assuming you are driving in a very nice sports car (any of the Italian V12s fore example) at top speed.
What would the cinetic energy of the car be?
How long could you as a human sustain on that energy? (You may approximate or use your actual values).
Let's use 1750 kg for the mass of the car travelling at 200 kph. Kinetic energy is (1/2)*mass*(velocity^2). 200 kph = 55.556 m/s, so the energy is 2,700,660.49 (kg * (m/s)^2); 1 joule = 1 (kg * (m/s)^2). 2,700,660.49 joules = 645.46 kcal. A good diet is 2500 cal/day, so that many calories would equal about 258 days. [img]graemlins/saywhat.gif[/img] A new diet, perhaps? [img]graemlins/laugh3.gif[/img]

Quote:
Originally posted by Faceman:
How big (and filled with what gas) should a baloon or Zeppelin so that upon his explosion (and after a cooldown) there'd be a noticeable rainfall over a major city (e.g. New York).
[img]graemlins/erm.gif[/img] I'll have to browse the CRC to look for gases that are both lighter than air and have condensation points below...say, what is the ambient air temperature at the time of the explosion, anyway? We would need temperature and pressure readings every 1000 ft (300 m), since these have a direct effect on the gas being used; of course, we could assume a simple gradient between readings using linear interpolation.
Many, many variables there....
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Old 08-20-2003, 12:57 AM   #17
Faceman
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Quote:
Originally posted by Azred:
2,700,660.49 joules = 645.46 kcal. A good diet is 2500 cal/day, so that many calories would equal about 258 days.
Your calcs are correct [img]graemlins/thumbsup.gif[/img]
but there are some flaws.
1. I said nice car at top speed and was therefore thinking about something in the area of 200mph, 320kph, 90m/s
2. A good diet is 2500kcal/day so that energy won't last very long. In fact it's about the energy value of a litre of milk.

Quote:
Originally posted by Azred:

[img]graemlins/erm.gif[/img] I'll have to browse the CRC to look for gases that are both lighter than air and have condensation points below...say, what is the ambient air temperature at the time of the explosion, anyway? We would need temperature and pressure readings every 1000 ft (300 m), since these have a direct effect on the gas being used; of course, we could assume a simple gradient between readings using linear interpolation.
Many, many variables there....
Okay, I'll give you a hint: I'm looking for REAL rainfall, i.e. Water.

[ 08-20-2003, 01:06 AM: Message edited by: Faceman ]
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Old 08-20-2003, 03:59 AM   #18
Deejax
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Quote:
Originally posted by Faceman:
Okay, I'll give you a hint: I'm looking for REAL rainfall, i.e. Water.
Combustion of compounds containing hydrogen (e.g. hydrocarbons) produces water. Water contains 2 hydrogen atoms and 1 oxygen atom. The oxygen is present in the air surrounding the balloon, so we need a source of hydrogen. The simplest source is hydrogen gas, which btw was used in zeppelins.

Now to calculate:
What is 'noticable rainfall'? Maybe 1 mm of rain? If somebody can inform me of the approximate surface of a city like New York, a volume of water can be calculated. Let's assume a nice round 100 square kilometers.

The comes the question of how much hydrogen is needed to produces that volume of water. Water contains 1/8 part hydrogen by mass. For 1 kg (= 1 liter) of water we need 125 grams of hydrogen.

The density of hydrogen is (I believe) approximately 90 grams per cubic meter.
Thus for 1 liter of rain we need 125 grams or 1.4 m3 of hydrogen.

100 km2 with 1 mm of rain equals 105 m3 of water. This is 108 or 100,000,000 kg (or liter) of water!
For 108 kg water we need 1.25*107 kg of hydrogen. This equals a volume of 1.4*108 m3 of hydrogen.

To visualise: This means a hydrogen filled balloon with a diameter of almost 800 meter (yes half a mile [img]smile.gif[/img] )

I hope I did all the calculations correctly
It sounds like a lot of hydrogen!

Final note: In my calculation I ignored the pressure of the hydrogen inside the balloon. If, for example, the pressure inside is 10 times the air pressure, the diameter of the balloon decrease by a factor 3. But I have no idea at what pressure you should fill a balloon with hydrogen.
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Old 08-20-2003, 06:05 AM   #19
daan
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round square kilometers .. you english peeps are weird.
And though I suppose you mean that if you get the pressure inside the balloon 10 times the pressure outside, it would "explode" covering a wider area, we are talking about a piece of cloth and not a metal container which is unexpandable. Therefor I doubt that raising the pressure inside the piece of cloth to 10 times the pressure outside the piece of cloth would acutally make that balloon shrink
But the point is still valid if the balloon can hold the pressure though [img]smile.gif[/img]

I'm going to mimick your calculations, but in Mol.
2 H2 + O2 = 2 H2O
H2 + 0.5 O2 = H20
We want 10^8 kg H20 --> 1 Mol H20 = 18.016 gram
So we need 1*10^11 gram / 18.016 gram = 5550621670 Mol H20 and thus also that much Mol H2.
Im gonna assume I'm letting my Balloon up to 5 km height. Air Pressure there is 5*10^4 Pascal, assuming that the balloon is flexible and thus adjusts, we can use the Gay-Lusac : ((p*V)/ T ) = n*R
We know everything but V:
((5*10^4*V)/ +-250 K ) = 5550621670 * 8,3145
V = 230753219,4 m^3 = 2,3 * 10^8 m^3

This difference can probably be contributed to the fact that I used the lower pressure and temperature of 5 km height.

EDIT:
Assuming for ease we're putting it in a perfecty sphere-shaped balloon and it stays that way:
Volume = 4/3 Pi*r^3 = 230753219,4
r = 380,4985913 m ( A span of about 761 meters )

[ 08-20-2003, 06:14 AM: Message edited by: daan ]
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Old 08-20-2003, 06:22 AM   #20
Deejax
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Quote:
Originally posted by daan:
round square kilometers .. you english peeps are weird.
Hey man, I'm dutch!
I was just saying 100 is a nice round number. I did not mean a circular square kilometer, although that is also possible.
Quote:
I'm going to mimick your calculations, but in Mol.
2 H2 + O2 = 2 H2O
H2 + 0.5 O2 = H20
We want 10^8 kg H20 --> 1 Mol H20 = 18.016 gram
So we need 1*10^11 gram / 18.016 gram = 5550621670 Mol H20 and thus also that much Mol H2.
Im gonna assume I'm letting my Balloon up to 5 km height. Air Pressure there is 5*10^4 Pascal, assuming that the balloon is flexible and thus adjusts, we can use the Gay-Lusac : ((p*V)/ T ) = n*R
We know everything but V:
((5*10^4*V)/ +-250 K ) = 5550621670 * 8,3145
V = 230753219,4 m^3 = 2,3 * 10^8 m^3

This difference can probably be contributed to the fact that I used the lower pressure and temperature of 5 km height.
Yep, works too. I'm glad to see the difference is quite small.
But still, a balloon that's half a mile wide....

Faceman, you care to reply?
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