??????????Guess-Big cube of gold???????????

XpipedreamR · October 23, 2003 12:41AM

According to my Redbook, the mint produced 445,500 double eagles in 1933. For the most part, these were all then melted down. Once melted, how big of a pure gold cube(equal sides) could be produced with the recovered metal?

Hint: It's between 1 and 10 feet per side

I'll post the answer later

(Yes, I was bored this morning!)

relayer · October 23, 2003 12:44AM

6.35185421

RiffRaff1967 · October 23, 2003 1:36AM

24.68 cubic feet, 2.9115 feet per side.

mgoodm3 · October 23, 2003 2:55AM

I think the second guess is correct. I just estimated based on about an ounce of gold per coin, 12 troy oz/ pound, 19 g/cm3 for gold. I didn't bother to look up the correct numbers. I got about 3 feet.

cladking · October 23, 2003 3:14AM

Off the top of my head, I got about 5 and a half or six feet.

PhillyJoe · October 23, 2003 3:22AM

How big? More like where are you going to put it. You could put this 27,800 lb. cube on your front lawn and watch the people try to steal it!
(I have it on good authority that if you put this cube outside Russ' shop it would disappear though.)

Joe

coinlieutenant · October 23, 2003 4:05AM

Xpipe,

What is the thickness in millimeters of a double eagle?

JOhn

mgoodm3 · October 23, 2003 4:08AM

2nd reply: I look up the weight of a double eagle = 1.07 troy oz = 11.2 eagles/lb = 13328/ ft3 (based on 1188lb/ft3 gold) = 33 cubic feet for 445,000 eagles. which is a cube 3.22 feet on a side. the only problem is that 100% gold is 1188 lb/ft3, 90%, is probably slightly less, I don't know the density of coinage gold.

coinlieutenant · October 23, 2003 4:21AM

My point is that in order to solve the problem you need one of two things.

1. You need the height of the planchet that produced the double eagle in which the problem is fairly simple.

The formula for that is:
2*pi*(17mm)^2*height=volume in cubic mm per coin of total metal. To get gold volume, multiply by 90% or .9 . Multiply that number by 445,500 and you will get the total volume in Cubic mm. That would be the answer unless you wanted it specifically in cubic feet. I'm a metric guy myself....

OR

2. You would need a mass/volume ratio, i.e. density of pure gold. I dont have the height of a planchet and I dont know the density of gold off the top of my head so....
This would be easy enough, just multiply:

1/density of gold*30.06grams/coin would give you cubic meters per coin with the 90% figured in (30.06 is 90% of 33.4grams). Multiply cubic meters per coin times the number of coins melted and you should get the answer. Does anyone have either the height or the density of gold????

John

coinlieutenant · October 23, 2003 4:23AM

GOT IT!!

1 cubic centimer of gold weighs 19.3 grams...

Nysoto · October 23, 2003 4:23AM

The cube would measure 2 feet, 9 inches on each side, give or take a fraction.

coinlieutenant · October 23, 2003 4:28AM

(.0518cm^3 per gram gold)*(30.06grams gold/coin)*445,000coins=692913.06 cubic centimeters

Let me convert to cubic feet...

coinlieutenant · October 23, 2003 4:31AM

2.903 feet per side

I guess Nysoto got it first, but I will concur

John

baccaruda · October 23, 2003 4:42AM

I don't know - let me go down to the basement and measure it.

Nysoto · October 23, 2003 4:42AM

Formula is: .9675x445,500/14.58/density of gold/cu.ft=23.46cu.ft3

RiffRaff1967 · October 23, 2003 5:02AM

<< <i> guess Nysoto got it first, but I will concur >>

Wow - it's like I didn't even post an answer......

nwcs · October 23, 2003 5:06AM

Not very big overall. If I remember right, all the gold on and in the earth would amount to a 100 by 100 by 100 yard area. It's so frickin' dense. If I'm off, it ain't by much. So I'd guess that amount, roughly 430,000 troy oz. would be between 1 and 8 ft^3. (the diff between 1 ft. per side and 2 ft per side).

Neil

Edit: nice to see that my 2 second slightly educated guess wasn't too far off. Meaning if I chose 2 and 3 ft per side, then I'd be right in that range. Instead I opted conservative and went 1 to 2 ft per side. Oh well...

coinlieutenant · October 23, 2003 5:24AM

Sorry RiffRaff,

didnt see your post....

John

Riff wins!!

Conder101 · October 23, 2003 7:56AM

3 ft 1.33 inches on a side

Oops just saw you want the size of the pure gold cube not the cube made from just the melted coins.

2 ft 10 3/4 inches

mozeppa · October 23, 2003 8:07AM

i'm all squishy inside just thinking about it

mgoodm3 · October 23, 2003 8:40AM

Wait a minute.

Tell me what's wrong with this.

1 double eagle = .9675 oz troy pure gold (weighs 1.07 oz overall). *445000 = 430538 oz gold /12 troy oz/lb = 35878 lb gold.

gold = 19.3 g/cm3 = 316.3 g/in3 = 546515 g/ft3 = 546.515 kg/ft3 = 1202.3 lbs/ft3

Therefore, 35878/1202.3 = 29.84 ft3 gold, take cube root and = 3.101 ft per side. or 3ft 1.2 inches.

XpipedreamR · October 23, 2003 8:45AM

Congratulations to everyone, you're all winners in my book!

The correct answer is (to one decimal place) 2.9 feet, or a hair under 2 ft 11 in.

Did you know that you could draw this amound of gold into a cable 1cm in diameter and almost 5.5 miles long?

To all the winners, knowledge is its own reward!

XpipedreamR · October 23, 2003 8:57AM

mgoodm3

In calculating the mass of the gold, you arrived and a value in troy pounds, and your density value is in avoirdupois pounds.

One pound troy =12 oz troy=0.377 kg

One pound avoirdupois =16 oz avoirdupois =0.453 kg

The two pounds are not equal, that's why your calculation is off(by a factor of the cube root of 0.453/0.377)

Conder101 · October 23, 2003 11:03PM

mgoodm3,

Second error was that the number of double eagles was 445500 so you were short 500 double eagles. (Ok what did you do with them?)

So proper troy pounds weight of the cube would be 35918.44

And as was pointed out earlier your density weight was not in troy pounds /ft3. Proper figure is 1464.40 troy pounds/ft3

or 24.528 ft3 of gold, take cube root to get the size of the side of the cube = 2.9 ft = 2 ft 10 13/16 inches. So your cube is a 16th inch larger than mine. Probably do to various round off errors/values chosen.

??????????Guess-Big cube of gold???????????

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