As old Senator Everett Dirkson once famously and sarcastically said, “A billion here, a billion there, and first thing you know, you’re talking real money!”
In these times of budgetary crisis, I wondered if there’s a better way for us to get our heads around the quantities involved when we talk of thousands, millions, billions and trillions. I mean, when we lapse into talk of such sums it’s easy to simply say “jillions” or “gazillions” or some such, but those starting letters ARE meaningful.
So, I decided to do an example using coins rather than paper money. Paper is so thin and you can print any number on it, but a coin is more tangible. You can feel the weight of a coin and 5 coins are worth 5 times as much and weigh 5 times more than one. Good. So, I picked the silver dollar for my example.
Wikipedia gives the specifications of the “Peace Dollar” as follows:
- weight (w): 26.73g, or 0.058 lbs.
- diameter (d): 38.1 mm, or 0.125 ft. (or, 1.5 inches)
- thickness (t): 2.0 mm, or 0.08 inches
Then I set out to determine what size coin would result if we made a thousand-dollar coin, a million-dollar coin, and so forth, assuming the same alloy composition as for the Peace dollar, i.e., 90% silver and 10% copper. OK, here we go.
The volume (V) of the coin is its area times its thickness
Equation 1. V = pi r^2 t = (pi d^2 t)/4; for the Peace dollar, this comes to 8.05276 10^-5 cu. ft.
also, d/t for the Peace dollar = 38.1/2.0 = 19.05, so t = d / 19.05 (the ratio of thickness to diameter)
solving Equation 1. for d then, we get Equation 2., d = cube root of [ 76.2V/pi ]
Now we can fill in the below table. Note that volume and weight naturally go up by multiples of 1,000 as we go from 1 to 1,000 to 1,000,000, etc. In other words, if one Peace dollar weighs 0.059 lbs, then 1,000 Peace dollars weigh 59 lbs., and so on. Also, knowing V, we can solve for d (diameter), and knowing d we can then use the ratio to get t (thickness) because we want each coin to be in the same proportion as the original silver dollar.
Coin V(cu.ft.) w (lbs.) d (ft.) t (ft.) t (in.)
$1 8.05276 10^-5 0.059 0.125 6.562 10^-3 0.078
$1K 8.05276 10^-2 59 1.25 6.562 10^-2 0.78
$1M 80.5 59,000 12.5 6.562 10^-1 7.8
$1B 80,527 59 million 125 6.5
$1T 80,527,000 59 billion 1,250 65
Now we see something interesting. The diameter and thickness are each ten times that of the previous coin. Does this make sense? Yep. Volume is a cube function and the value-multiple of a thousand to a million is a factor of one thousand, the cube root of which is, viola’, ten. So our coin gets ten times bigger in diameter and ten times thicker each time we multiply the value (or weight) by 1,000. Isn’t that neat?
As you can see, a silver alloy coin worth, nominally, $1 Billion, or $1,000,000,000, would be 125 feet in diameter, six and a half feet thick and weigh 59,000,000 pounds, or almost 30 tons. Of course, this overstates the case somewhat because the price of actual silver has shot up, but I think the calculation makes the point. I think even Everett Dirkson would be impressed, don’t you?