Jeff the Green
2013-07-03, 04:56 AM
I was listening to a radio report about the Middle East yesterday and one of the guests said something to the effect that a Bill of Rights without rule of law "isn't worth a hill of beans."
Which of course made me wonder what a hill of beans would be worth.
Problem 1: What is a hill, anyway?
This one's relatively easy (http://www.howitworksdaily.com/environment/when-does-a-hill-become-a-mountain/), actually. At least according to America, a hill is anything less than 300 meters high. On the other hand, the OED defines it as anything less than 600 meters high. We'll use both.
Problem 2: How big is a hill of beans?
For this we need to know the angle of repose (http://en.wikipedia.org/wiki/Angle_of_repose) for beans. This is the angle at the base of a cone formed by granules. This is a nice picture explaining it.
http://upload.wikimedia.org/wikipedia/commons/thumb/b/ba/Angleofrepose.png/220px-Angleofrepose.png
Thankfully, I do not have to measure this, as Benedito C. Benedetti and José T. Jorge have done it for me (http://spiru.cgahr.ksu.edu/proj/iwcspp/pdf2/5/1777.pdf). Depending on moisture content, the angle is about 33.8°.
The equation for volume (v) of a cone is v = ⅓πr2h. We know h (300-600 m), but what about r? Time for some trigonometry.
...SOH-CAH-TOA (www.mathwords.com/s/sohcahtoa.htm). Right.
So h/r = tan(33.8°). Rearranged, r = h/tan(33.8°). Plug that into our volume equation and we get v = πh3/tan2(33.8°). When h=300, this is 63,069,448 m3, and when h=600 this is 504,555,580 m3.
Problem 3: How much does this hill weigh?
Mercifully, this part involves no trigonometry, just a quick conversion from volume to mass. So we need the density of beans. Again, I don't have to measure this because someone already did (http://www.tacwebdesign.com/clients/oe/materials.shtml): 45 lbs./ft., or (since I despise Imperial units) 720.83 kg/m3.
So 63,069,448 m3 of beans weighs 45,462,000,000 kg = 45,462,000 tonnes and 504,555,580 m3 weighs 363,700,000,000 kg = 363,700,000 tonnes
Looked at another way, this is a quarter to two million adult blue whales.
Problem 4: How much is this hill worth?
Another quick conversion. The cheapest bulk pinto beans (http://www.alibaba.com/trade/search?fsb=y&IndexArea=product_en&CatId=&SearchText=pinto+bean) I could find were $600/tonne and the most expensive were $1700/tonne. So our 300 m hill is worth $27,277,000,000 to $77,285,000,000 and our 600 m hill is worth $218,220,000,000 to $618,290,000,000.
This is in the range of GDP of countries ranging from Bolivia to Saudi Arabia. Fewer than fifty countries could buy our larger hill at the lowest price with their entire GDP for a year, and only nineteen could buy it at the highest.
Conclusion
This is a stupid metaphor.
Which of course made me wonder what a hill of beans would be worth.
Problem 1: What is a hill, anyway?
This one's relatively easy (http://www.howitworksdaily.com/environment/when-does-a-hill-become-a-mountain/), actually. At least according to America, a hill is anything less than 300 meters high. On the other hand, the OED defines it as anything less than 600 meters high. We'll use both.
Problem 2: How big is a hill of beans?
For this we need to know the angle of repose (http://en.wikipedia.org/wiki/Angle_of_repose) for beans. This is the angle at the base of a cone formed by granules. This is a nice picture explaining it.
http://upload.wikimedia.org/wikipedia/commons/thumb/b/ba/Angleofrepose.png/220px-Angleofrepose.png
Thankfully, I do not have to measure this, as Benedito C. Benedetti and José T. Jorge have done it for me (http://spiru.cgahr.ksu.edu/proj/iwcspp/pdf2/5/1777.pdf). Depending on moisture content, the angle is about 33.8°.
The equation for volume (v) of a cone is v = ⅓πr2h. We know h (300-600 m), but what about r? Time for some trigonometry.
...SOH-CAH-TOA (www.mathwords.com/s/sohcahtoa.htm). Right.
So h/r = tan(33.8°). Rearranged, r = h/tan(33.8°). Plug that into our volume equation and we get v = πh3/tan2(33.8°). When h=300, this is 63,069,448 m3, and when h=600 this is 504,555,580 m3.
Problem 3: How much does this hill weigh?
Mercifully, this part involves no trigonometry, just a quick conversion from volume to mass. So we need the density of beans. Again, I don't have to measure this because someone already did (http://www.tacwebdesign.com/clients/oe/materials.shtml): 45 lbs./ft., or (since I despise Imperial units) 720.83 kg/m3.
So 63,069,448 m3 of beans weighs 45,462,000,000 kg = 45,462,000 tonnes and 504,555,580 m3 weighs 363,700,000,000 kg = 363,700,000 tonnes
Looked at another way, this is a quarter to two million adult blue whales.
Problem 4: How much is this hill worth?
Another quick conversion. The cheapest bulk pinto beans (http://www.alibaba.com/trade/search?fsb=y&IndexArea=product_en&CatId=&SearchText=pinto+bean) I could find were $600/tonne and the most expensive were $1700/tonne. So our 300 m hill is worth $27,277,000,000 to $77,285,000,000 and our 600 m hill is worth $218,220,000,000 to $618,290,000,000.
This is in the range of GDP of countries ranging from Bolivia to Saudi Arabia. Fewer than fifty countries could buy our larger hill at the lowest price with their entire GDP for a year, and only nineteen could buy it at the highest.
Conclusion
This is a stupid metaphor.