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The Einstein-Szilard Refrigerator

This article looks at the refrigeration system patented by Albert Einstein and Leo Szilard and explores the underlying scientific principles.

The refrigerator is depicted in the patent filed by Einstein and Szilard, a copy of which is available here:  http://opensourceecology.org/w/images/6/63/Einstein_Fridge.pdf.  It’s copied below, but you’ll want to go ahead and click that link so you’ll have the image as a reference.

Einstein-Szilard Refrigerator Patent Drawing

As a quick orientation, the little box on the right, labeled “1,” is the part that gets cold.  All the other things in the drawing are the parts that are necessary to make that little box get cold.  Also note: if you wanted to use this system to, say, keep your food cold, you wouldn’t put the food in that little “1” box.  No, the “1” box would go inside of a larger insulated box, and your food would go in there.  The “1” box would get cold, heat would conduct from the inside of your insulated box through the walls of box “1” into its interior, and that in turn would make your food cold.

The fundamental principle by which this device makes cold is this: when liquids turn into gasses, they suck up heat from their surroundings.  A mass of water vapor at 212 degrees Fahrenheit has more heat energy than the same mass of liquid water at 212 degrees Fahrenheit.  One term for this concept is “enthalphy of vaporization.”  It’s an interesting and powerful scientific principle.  Just for example: it takes 5.4 times as much heat energy to convert water at a given temperature to water vapor at the same temperature as it does to heat the same water from 32°F to 212°F.  This is why sweat is so effective at cooling you off: every ounce of sweat that evaporates draws in enough heat energy to cool three quarts of water by 10°F.  [Can’t help adding: the specific heat of flesh is around 3,470 J/kg*K, so if you weigh 200 pounds, every 2.7 ounces of evaporated sweat is enough to cool your body 1°F.]

The Einstein-Szilard refrigerator, instead of using water as the evaporating substance, uses butane.  Here’s how butane stacks up against water in some key properties:

  Water Butane
ΔHvaporization (kJ/mol) 40.66 21
ΔHvaporization (kJ/kg) 2257 320
Boiling Point 100֗֠ºC -1ºC
Density (liquid) 1,000 kg/m3 600 kg/m3
Condensation pressure at 100ºF ~38 psi / 2.6 atm ~1 psi / 0.068 atm
chemical formula H2O C4H10

So butane has lower enthalpy of vaporization than water, meaning you need to evaporate more of it to produce temperature change.  How much more?  Seven times as much by weight, and almost 12 times as much by volume.  However, it has a couple of other key properties that make it more useful than water for this purpose.  Mainly, its boiling point is much lower than water’s.  Back to box “1;” the liquid butane is evaporated by the addition of gaseous ammonia, which is bubbled through the liquid butane by the porous stone “31.”  The gaseous ammonia causes the butane to evaporate much like water will evaporate into air – even well below its boiling point – until it reaches 100% relative humidity.  As the butane evaporates, it pulls in from its environment the energy required to overcome its enthalpy of vaporization, which makes its environment colder.  Butane’s low boiling point and condensation pressure make it well suited for refrigeration temperatures.

As we’ve now established, all we need is a steady supply of liquid butane and a steady supply of gaseous ammonia, and we’ll have a steady supply of cold.  The purpose of all of the other components of the Einstein-Szilard refrigerator apart from box “1” is to recycle the butane and ammonia so that you don’t have to keep supplying your cooler with fresh butane and ammonia.  That would get expensive.  Einstein and Szilard figured out a series of chemical processes by which the butane and ammonia could be recycled, requiring only a heat source and an environmental heat sink (such as the atmosphere, a lake, or a well).

Separating the Butane Vapor from the Ammonia Vapor

After the ammonia gas evaporates the butane in box “1,” the ammonia and butane gasses are combined.  The first step in recycling them is to separate them.  This is done by exploiting the fact that ammonia dissolves readily in water, whereas butane does not.  This happens in box “6” on the diagram.  The ammonia-butane gas mixture flows to box “6” via pipe “5.”  [Note that pipe “30” passes through pipe “5;” we’ll get to that later.]  In box “6,” water is sprayed through the ammonia-butane gas mixture using sprayer head “35.”  The ammonia dissolves into the water and the ammonia-water solution falls to the bottom of the tank (“26″ on the drawing) because liquids are heavier than gasses.  Removing the ammonia from the butane has the helpful side effect of condensing the butane (re-liquefying it), so the butane also falls to the bottom of the box, but it’s less dense than water so it floats above the water.  This actually completes the recycling of the butane, and the butane is allowed to flow back into box “1” via pipe “11.”

Of course the butane warms back up as it condenses, releasing as much heat energy as it absorbed when it evaporated.  This heat has to be dissipated before the butane goes back into box “1,” and that’s what the heat exchanger “12” is for.

Recycling the Ammonia

After the process in box “6,” the butane has been recycled, but the ammonia is now in solution with water.  It must be returned to its gaseous state in order to be useful.  To achieve this, the Einstein-Szilard refrigerator exploits another property of ammonia: more ammonia can dissolve in cool water than in hot water.  In other words, heating the water-ammonia solution will force out the ammonia, which will return to its gaseous state.  This heating is done in box “29.”  The heat supplied to box “29” is the primary energy required to make this heat pump work.

The hot ammonia gas travels through pipe “30” to box “1.”  On the way, it passes through pipe “5,” where it exchanges heat with the chilled ammonia-butane mixture that is exiting box “1,” so it enters box “1” pre-chilled and does not introduce too much heat to box “1.”

Recycling the Water

The byproduct of the ammonia recovery is hot water, which must be cooled so that it can again be useful for absorbing ammonia.  In the original Einstein-Szilard design, this is done using the heat exchangers at 28 and 12.  Heat exchanger 28 uses this hot, low-ammonia water to pre-heat the high-ammonia water en route from box 6 to box 29 while simultaneously cooling the hot water.  Heat exchanger 12 requires an external heat sink such as a source of cool water – maybe water pumped to a geothermal heat sink.  Air cooling could work, too, with some coils, a pump to move the cooling water through the coils, and a fan to move air over the coils.

This water cooling step is where a couple of common myths often claimed about the Einstein-Szilard refrigerator break down:

Myth #1: It’s a device that turns a heat source into a cold source!

Not entirely true.  It’s impossible to make a system colder by adding heat.  What the Einstein-Szilard refrigerator actually achieves is a heat pump, which is a device that moves heat energy from one place to another.   Specifically, this one moves heat from the surroundings of box “1” into box “1,” then from box “1” to box “6,” then from box “6” into the environment via heat exchanger “12.”  It is very interesting and useful that it does this without needing a motor, just a heat source, but of course it does not defy the laws of conservation of energy.

Myth #2: It’s a refrigerator with no moving parts!

Not quite.  It’s true that it’s a heat pump with no moving (solid) parts, but heat pumps just move heat from point “A” to point “B” (or in this case from box “1” to heat exchanger “12”).  Unless you have some mechanism in place to dissipate the heat from point “B,” the heat pump will overheat the environment at point “B” and stop working.  In order to have what we think of as a household refrigerator, we need a fan or pump or some other way to keep the environment around heat exchanger “12” from getting hot.

Likewise on the cold side, some air circulation around box “1” would be needed so that it doesn’t get too cold right next to box “1” and not cold enough in the rest of your insulated box.

Once the hot water is cooled down, the cycle is complete and we are back to our starting products: gaseous ammonia, liquid butane, and unheated water.  The Einstein-Szilard refrigerator continuously runs this process, cooling the surroundings of box “1” and heating the heat sink around heat exchanger “12.”

 

I intend to come back later and take a longer look at this style of heat pump, performing a somewhat more in-depth analysis.  For now, I have enjoyed considering the key concepts that make the machine work.  It will be a useful thing to consider for using waste heat from industrial processes to cool other processes.  I hope I get a chance to use this concept.

Games, Puzzles, and the Real World 2a: Monty Hall’s Doors, Bertrand’s Boxes, and the Three Prisoners

This is part of a series of articles that solves popular puzzles and demonstrates the applicability of the puzzle-solving hobby to everyday life. If you love puzzles like I do, and you have a favorite that I haven’t written about yet, please send it to me! I haven’t seen a new one in a long time, and I’d appreciate your message.

Today’s post addresses three puzzles that are all similar to the recent “Three Slips of Paper” puzzle: Monty Hall, Bertrand’s Boxes, and The Three Prisoners.  They are as follows:

Monty Hall

This puzzle is named for an old American television game show called Let’s Make a Deal, and more particularly for its host, Monty Hall.  On that game show, contestants were sometimes presented with three doors, each with a prize behind it, and asked to pick a door (numbered 1, 2, or 3); the contestant would then receive whatever prize was behind the door.  It might be something really nice, like a car, or it might be something less desirable, like a cheap coupon.  The puzzle goes like this:

You’re a contestant on a game show.  The host presents you with three doors to choose from, and informs you that behind the three doors are a goat, another goat, and a new car – but of course he doesn’t tell you which is behind which door.  You are going to pick a door, but then before you see what’s behind that door, the host (who knows what’s behind all the doors) will open another of the doors to reveal a goat.  You will then have the option to either take whatever prize is behind the door you initially chose, or you can switch to the other still-unopened door and take whatever prize is behind that door.  How can you maximize your odds of getting the car?

It was popularized by Marilyn vos Savant.

Monty Hall Solution

First off, obviously when you pick a door the first time, your odds of picking the door with the car are one in three.  The “trick” of this puzzle comes when the host then subsequently reveals one of the goats.  A lot of people, upon first confronting this puzzle, think that the reveal increases the odds of finding the car to one-half; they figure that once the goat is revealed, each of the remaining doors has a one-in-two chance of hiding the car.  That reasoning, however, ignores the contestant’s act of selecting a door.  See, that first contestant-selected door had only a one-in-three chance of concealing the car, so when the host opens one of the other doors to reveal a goat, that collapses the remaining two-thirds probability that the car was not behind the contestant-selected door into the remaining unselected, unopened door.  To prove it, here’s the decision tree:

Monty Hall Decision Tree
Monty Hall Decision Tree

 

Bertrand’s Boxes

Bertrand’s Boxes is named for Joseph Bertrand, who published it in his book Calcul des probabilités, published in 1889.

You are confronted with three boxes: one that contains two gold coins, a second that contains one gold coin and one silver coin; and a third that contains two silver coins.  You select a box at random and draw a coin at random from that box.  The coin happens to be gold.  What are the odds that the remaining coin in the box is also gold?

Bertrands Boxes

Bertrand’s Boxes Solution

Having just seen the Monty Hall puzzle, you probably figured out pretty quickly that the answer is two-thirds.  However, most people seeing this puzzle for the first time figure that drawing a gold coin narrows the puzzle down to two boxes, one of which contains two gold coins, and they’ll figure the odds are one in two.  As with other similar puzzles, however, the answer is readily discernible using a decision tree:

Bertrand's Boxes Decision Tree
Bertrand’s Boxes Decision Tree

Just eliminate the possibilities that didn’t happen, then do the math on the remaining possibilities.

The Three Prisoners

The Three Prisoners puzzle was published in Martin Gardner’s Scientific American column in 1959.  I grew up on Martin Gardner’s books of puzzles, and thoroughly loved them.  Here’s his formulation:

“Three prisoners, A, B and C, are in separate cells and sentenced to death. The governor has selected one of them at random to be pardoned. The warden knows which one is pardoned, but is not allowed to tell. Prisoner A begs the warden to let him know the identity of one of the others who is going to be executed. ‘If B is to be pardoned, give me C’s name. If C is to be pardoned, give me B’s name. And if I’m to be pardoned, flip a coin to decide whether to name B or C.’

“The warden tells A that B is to be executed. Prisoner A is pleased because he believes that his probability of surviving has gone up from 1/3 to 1/2, as it is now between him and C. Prisoner A secretly tells C the news, who is also pleased, because he reasons that A still has a chance of 1/3 to be the pardoned one, but his chance has gone up to 2/3. What is the correct answer?”

I would add one more key piece of information: the warden agrees to Prisoner A’s proposal and does not lie.

The Three Prisoners Solution

Prisoner C is correct in his thinking; based on this new information, Prisoner C’s odds of being pardoned are two-thirds.

Three Prisoners Decision Tree
Three Prisoners Decision Tree

As in the Bertrand’s Boxes puzzle, just draw out the decision tree, then eliminate the possibilities that didn’t happen.  In the puzzle scenario, the warden said that Prisoner B would be executed, so this eliminates the scenarios wherein he said that Prisoner C would be executed.  That leaves two possibilities: one in which the warden says that Prisoner B will be executed and Prisoner A gets pardoned (one-in-six overall chance) and one in which the warden says that Prisoner B will be executed and Prisoner C gets pardoned (one-in-three overall chance).  Because the warden’s statement eliminated all other possibilities, add up the odds and divide out to determine that, where the warden says B will be executed, C has a two-thirds chance of being pardoned.

Of course the odds would be different if the warden had, entirely of his own volition and without any structure, simply announced that Prisoner B would be executed.  In that scenario, there would then be a 50% chance each for A or C to be pardoned.  The critical element of the puzzle is that the warden agrees to follow the system set out by Prisoner A.

The Point

It’s good to look at these three similar puzzles together because it emphasizes the fact that the solution to a seemingly esoteric and obscure situation can be applied to other challenges in life. Even though the decision trees for these three puzzles are drawn differently, they are mathematically the same.  The different decision trees represent different valid ways to approach the same basic puzzle.  Once you have internalized the way of solving these puzzles, you can be on the lookout for other opportunities to gain valuable information in life by carefully applying reason to the facts available to you.

These are also interesting puzzles to look at together with the preceding “Three Slips of Paper” puzzle, even though they are not identical to that one, because the fundamental thought process is the same. Once you know how to solve these information-probability puzzles, they’re a breeze.

Games, Puzzles, and the Real World 2: What Do Slips of Paper Have to do With Dating?

This is part of a series of articles that solves popular puzzles and demonstrates the applicability of the puzzle-solving hobby to everyday life. If you love puzzles like I do, and you have a favorite that I haven’t written about yet, please send it to me! I haven’t seen a new one in a long time, and I’d appreciate your message.

The Puzzle

I have seen this puzzle in many different places, but this particular phrasing is taken (with minor edits for clarification) from the Car Talk radio show.

“Three different numbers are chosen at random, and each is written on one of three slips of paper. The slips are then placed face down on the table. The objective is to choose the slip upon which is written the largest number.

Here are the rules: You can turn over any slip of paper and look at the amount written on it. If for any reason you think this is the largest, you’re done; you keep it. Otherwise you discard it and turn over a second slip. Again, if you think this is the one with the biggest number, you keep that one and the game is over. If you don’t, you discard that one too, and you’re stuck with the third.”

At first glance, you might think the odds of getting the highest number are one in three. But I wouldn’t be bothering to write this down if it were that simple, would I?

Solving Discussion

Any time you’re trying to maximize an outcome, your go-to tool should be a decision tree. They cover these things in business classes. It’s like a flow chart that reads left-to-right. Here’s what the decision tree looks like if you keep the first piece of paper you pull:

Figure 1: Decision Tree 1

Assigning a value of “1” to a “win” where you found the highest number, and a value of “0” to a “loss” where you are stuck with one of the lower numbers, the expected value of this strategy is 0.333…. Which is just a fancy way of saying you have a one in three chance of getting the highest number this way.

Now, this becomes a pretty simple puzzle because there are really only three other strategies possible within the framework of the rules:

  1. Discard the first number you pull and keep the second.
  2. Discard the first and second numbers you pull and keep the third.
  3. Discard the first number you pull, look at the second number, and then decide whether to discard the second number and pull the third.

You can reason out pretty readily that options 1 and 2 are not different from just keeping the first piece of paper you pull. Option 3, however, is interesting because there’s a choice involved. Does that choice make a difference? If you want to work it out yourself, go ahead; this page will still be here. If not, read on…

The winning strategy for solving this puzzle is to pull a number at random, look at it, discard it, and pull a second number. Then, if the second number is higher than the first, keep it. If the second number is lower, discard it and keep the third and final number. Here’s the decision tree:

Figure 2: Decision Tree 2

So if you discard the first number you pull, there is a one-in-three chance that it was the biggest number and you lose. There is also a one-in-three chance that it was the middle number. If it was the middle number, and the second number you pull is the highest number, then by following the strategy you will keep it and you will win. If the second number you pull in this situation is the lowest number, then by following the strategy you will discard it and pull the third number, which will be the highest number, and you will win. So this strategy means you always win if the first number you pull is the middle number. This means that so far, we have a one-in-three chance of losing and a one-in-three chance of winning. What happens in the remaining one-in-three scenario where the first number pulled is the lowest number? Well, this means that the second number pulled will definitely be higher than the first, so we will keep it, and there is a one-in-two chance that it was the highest number and a one-in-two chance that it was the middle number. So on the one-in-three chance that the first number pulled is the lowest number, we have a 50% chance of winning and a 50% chance of losing. Adding all these percentages up, we have a one-in-three chance of losing, a one-in-three chance of winning, another one-in-six (33.33% times 50%) chance of losing, and another one-in-six chance of winning. That comes out to a one-half chance of winning and a one-half chance of losing. That’s better than the one-in-three chance given by random selection, so this is a superior strategy.

Beyond the Puzzle

It’s useful to work out this sort of small-scale puzzle to learn basic mathematical patterns, but in order to really wring every bit of value out of a puzzle and internalize the underlying concepts, it’s important to expand it and see how it can apply in other contexts. In order to solve the puzzle, we worked out what happens with a set of three numbers, but what happens with a set of ten? Should we just examine and discard the first selection? Examine and discard one-third of the numbers? Discard all but three of the numbers and then follow the strategy from the first puzzle? Follow some other strategy? Let’s find out…

First off, the odds are pretty easy to calculate if we decide to discard seven of the ten and then follow the strategy from the first puzzle. There is a 30% chance that the highest number remains undiscovered after discarding 70% of the numbers, and the first puzzle solution gives us a 50% chance of finding the highest of the three remaining numbers, so our overall odds of winning are 15%. This is a good check; it proves that the first puzzle still has value here. The odds of 15% beat the odds of 10% that we get if we choose at random. But can we beat 15%? Obviously, we can: if we just slightly modify the original strategy to say that we will discard the ninth number not only if it is smaller than the eighth, but if it is smaller than any of the first eight numbers, this of course slightly betters the odds. These first couple of intuitive tests having borne fruit, we are motivated to proceed to look for the best solution by working through the decision trees.

Let’s next look at the other obvious permutation of the strategy from the original puzzle: pick a number, look at it, discard it, then proceed to draw additional numbers until we find one that is higher than the first number. This gives us:

  • a 10% chance of drawing the highest number first, in which case we lose;
  • a 10% chance of drawing the second-highest number first, in which case the next number we draw that is higher than any preceding number will necessarily be the highest number in the set, which means we will win;
  • a 10% chance of drawing the third-highest number first, in which case there is a 50-50 chance that the next number we draw that is higher than any preceding number will be the highest number;
  • a 10% chance of drawing the fourth-highest number first, in which case there is a one-in-three chance that the next number we draw that is higher than any preceding number will be the highest number;

… and so on. You see the pattern. The odds of selecting the highest number via this method are (.1 × 0) + (.1 × 1) + (.1 × .5) + (.1 × .333) + (.1 × .25) + (.1 × .2) + (.1 × .1667) + (.1 × .1429) + (.1 × .125) + (.1 × .111), totaling about 28.3%. Already this is way better than any prior answer. But can we improve on it further?

Now let’s try drawing three numbers, discarding them all, and proceeding from there. The first thing you’ll notice is that your decision tree becomes a LOT more complex if you try to map out all the possibilities of which three numbers you might draw. In fact, there are 720 (ten times nine times eight) possibilities for what those first three numbers could be. We need to collapse this down some. So let’s think: which of these possibilities are alike? It turns out that the only thing that matters about those first three numbers is the highest number. Think about it: if the highest of the first three numbers is, let’s say, the second-highest number in the set, it doesn’t matter what the other two are. The rest of the puzzle proceeds the same. So, collapsing things down: in 216 out of the 720 possibilities, the highest of the three numbers is the highest number in the set; in 168 of the 720 possibilities, the highest of the three numbers is the second-highest number in the set; in 126 of the 720, the highest of the three is the third-highest in the set; in 90 of the 720, the highest of the three is the fourth-highest in the set; 60 of 720, fifth-highest; 36 of 720, sixth-highest; 18 of 720, seventh-highest; 6 of 720, eighth-highest; and of course if you have three different numbers out of a set of ten, the highest of the three can be no smaller than the third-smallest, which is also the eighth-highest. Add ‘em up and you get 720.

So what are the odds of victory in each of these scenarios? Well, if the highest number got drawn as one of the first three, we lose. And the rest of the odds track what we just worked out in the preceding scenario: 100% if the highest among the three is the second-highest of the set; 50% if the highest among the three is the third-highest of the set; one-in-three if the highest among the three is the fourth-highest of the set; and so on. Multiplying it out, our total comes to almost a 40% chance of finding the highest card! (39.75% to be precise). So by reviewing and discarding almost a third of the numbers, thereby accepting an automatic 30% chance of failure, we actually boost our overall odds of success!

Now, just for the sake of confirmation, let’s run the math if you discard four. … Actually, let’s not do that part together. If you want to, I hope you will, but I’m just going to give you the answer: 39.83%. And then if you run the numbers on discarding five, you get 37.28%. So we’ve identified a peak at around three or four cards discarded. The key seems to be to discard a third of the numbers. For example, if you have 12 numbers, the peak odds are 39.6% if you discard four cards.

The Point

Can you think of a parallel scenario from real life? A situation where you have to get “involved,” let’s say, with an unknown in order to learn its value, and then before you can learn the value of another unknown, you have to discard the first one? How about dating? After you’ve narrowed the field to a set of likely possibilities, you have to date one of your prospects in order to find out how compatible you are. But before you can investigate another prospect, you are typically going to have to discard the first one… and by the time you’ve investigated your second prospect, the first will probably be involved with someone else and therefore no longer an option.

The metaphor isn’t perfect, but it’s pretty good. As a young, single person looking for a relationship, you typically have no idea what the range of possible values are for relationship bliss. Even after you’re in a relationship, it’s impossible to tell how your relationship stacks up in terms of objective quality. You need to experience a few relationships in order to have some confidence that you’ve found a good match.

So to give a real-world example: let’s say you’re in college. Out of your fellow students, numbering 20,000, you winnow them down based on criteria such as age, religion, sex, socio-economic background, geography, similar interests, physical attractiveness, and maybe other factors to ten viable dating prospects. You could pick the one that, based on your best guess, seems best to you, but let’s face it: you’re not that far along in your life. There’s a pretty sizeable margin of error on your guess. Your best move is really to get some experience by dating a few people, and then start looking in earnest.  If you have ten prospects, this puzzle suggests that your optimum strategy would be to try out four of them to maximize the value of your experiences.  Of course this puzzle is just one equation among many that would inform an actual decision, but it does suggest a base experience level is necessary to learn enough about yourself and relationships generally to make further decisions with some degree of confidence.  It also suggests that, if a person has a high number of potential mates, it makes sense to test out more of them before beginning to look in earnest.

The most obvious way in which this puzzle differs from reality is that it is not actually a loss scenario if you fail to end up with the best possible mate, and of course this fact changes the decision tree significantly.  However, it does perhaps describe that persistent part of the human psyche that does feel like anything short of the “best” is failure.

This is a wonderful puzzle because it provides an example of how to think through decisions in a limited-information environment.  Sometimes people lock up when presented with a decision and incomplete information, and will sometimes throw up their hands and make a decision at random; especially in a situation like this, where losing is more likely than winning.  However, you can frequently increase your odds in an apparently hopeless situation by just carefully thinking through what you can deduce from the information you do have, and what else you can learn efficiently in the time you have.

Games, Puzzles, and Reality: Glass Marbles and Googling

This is part of a series of articles that solves popular puzzles and demonstrates the applicability of the puzzle-solving hobby to everyday life. If you love puzzles like I do, and you have a favorite that I haven’t written about yet, please send it to me! I haven’t seen a new one in a long time, and I’d appreciate your message.

The Puzzle

There is a building 100 stories tall.  You have a collection of glass marbles. You want to know what is the structural integrity of these marbles: from what height (from which floor) you would need to drop a marble in order to break it.  The marbles are substantially identical, so you can assume that if one would break when dropped from a given height, so would any other.  You may also assume, of course, that if a marble would break if dropped from a given floor, it would also break if dropped from any floor above that floor.  Finally, there is an elevator that, due to its particular design, takes about as much time to travel to or from any given floor as to or from any other.

You don’t have enough information about the annealing process, etc., used to make the marbles to calculate a solution, so you are going to have to find an empirical answer.  Fortunately, you can afford to break up to two of your marbles in pursuit of an answer.  What is the most efficient way to find your answer?

Solving Discussion (Spoilers!)

If you had only one marble, the solution would be easy: just start at the first floor, drop the marble, retrieve it, then drop it from the second floor, then the third, and so on.  In this scenario, you would expect to drop the marble an average of 50.5 times in order to find the right floor.  (=(100+1)/2)  See that equation?  That’s a basic probability equation, used to calculate the expected value of (for example) a die roll.  To calculate the average outcome of a die roll, you would add all the potential numbers and then divide by the number of numbers.  So for a standard six-sided die, you would add 1+2+3+4+5+6 and then divide by 6.  The result would be 3.5.  The cool thing is that you can shortcut the process by leaving out all the numbers except the lowest and the highest.  See, one plus six equals seven, and two plus five equals seven, and three plus four also equals seven.  So just leave out all the stuff in the middle, reduce your divisor to two, and you’ll get the same answer.  A mathematician named Friedrich Gauss is often credited for noticing this.  The story goes that when Friedrich was a child, his teacher was frustrated by the task of occupying Friedrich’s mind, and assigned him to add together all numbers from 1 to 100, hoping that this would occupy him for a long time.  However, Friedrich came back seconds later with an answer.  Friedrich, having observed that 1+100=2+99=3+98=4+97=… and so on, simply multiplied 101 by 50, immediately coming up with the answer: 5,050.  So anytime you need to average a long series of consecutive numbers, you can just do what Friedrich did and shortcut the process: add the highest number to the lowest, and divide by two.

Moving on: the presence of the second marble triggers a realization. You can break up the number of floors.  For example, you might drop the first marble from the 50th floor, and then if it breaks start dropping the second marble at floor one, or if it doesn’t, start dropping the second marble at floor 51.  This would be a better solution.  How do we know it’s better?  Under this revised solution, we now expect to drop the marble an average of 26.49 times.  That’s a 1/100 chance that floor 50 will be the breaking floor, in which case we will drop the marble 50 times (first marble breaks at floor 50, second marble gets dropped at each floor 1 to 49 to confirm); a 49/100 chance that one of floors 1 to 49 will be the breaking floor, in which case we will drop the marble 2 to 50 times, with an average of 26; and a 50/100 chance that one of floors 51-100 will be the breaking floor, in which case we will drop the marble 2 to 51 times, with an average of 26.5.  Do the math, and you get an expected drop count of 26.49.

Now let me introduce one more wrinkle: if we drop the marble from floor 50 and it doesn’t break, then we have 50 floors left and still two marbles.  So rather than starting at floor 51 and continuing upward, we might repeat our method and drop a marble from floor 75.  Then if it didn’t break, we might repeat our puzzle again and drop a marble from floor 87, and so on.  At this point, the probability calculation is getting more complex, and it looks like this:

Drop # Floors Remaining Floors to Skip Drop Floor # Expected Drops Likelihood Total
1 100 50 50 26.48 0.5 13.24
2 50 25 75 14.96 0.25 3.74
3 25 12 87 9.42 0.12 1.13
4 13 6 93 7.33 0.06 0.44
5 7 3 96 6.67 0.03 0.2
6 4 2 98 7.00 0.02 0.14
7 2 1 99 7.00 0.01 0.07
8 1 1 100 8.00 0.01 0.08
19.04

So instead of 26.49, the average number of drops for this iterative process is 19.04.  It’s better.

Now that you have a framework for analysis, the question therefore becomes, what is the most efficient way to split up 100 floors with two marbles?  One solution I’ve seen goes like this: realize that the two marbles are basically two factors in a multiplication problem which must have as its product a number greater than or equal to 100, and the goal is to have the sum of the two numbers be as low as possible.  If you look at a few possibilities, you’ll notice a pattern… 2 times 50 equals 100 and sums to 52; 4 times 25 equals 100 and sums to 29; 5 times 20 equals 100 and sums to 25… as the two numbers get closer together, their sum goes down.  So in order to get the two numbers as low as possible, let’s make them the same.  Find the square root of 100, which is 10.  So 10 times 10 equals 100 and sums to 20 – your answer is to drop the first marble at floors 10, 20, 30, 40, 50, 60, 70, 80, 90, and 100, see where it breaks, and then run the marble up through the 9 floors below that floor.  This solution has a pleasing simplicity and gives you an expected process run time of 10.9, calculated as follows:

Drop # Floors Remaining Floors to Skip Drop Floor # Expected Drops to Find Result Likelihood Weighting
1 100 10 10 6.4 0.1 0.64
2 90 10 20 7.4 0.1 0.74
3 80 10 30 8.4 0.1 0.84
4 70 10 40 9.4 0.1 0.94
5 60 10 50 10.4 0.1 1.04
6 50 10 60 11.4 0.1 1.14
7 40 10 70 12.4 0.1 1.24
8 30 10 80 13.4 0.1 1.34
9 20 10 90 14.4 0.1 1.44
10 10 10 100 15.4 0.1 1.54
10.9

… or, if you follow the iterative strategy, 10.81:

Drop # Floors Remaining Floors to Skip Drop Floor # Expected Drops to Find Result Likelihood Weighting
1 100 10 10 6.4 0.1 0.64
2 90 9 19 6.9 0.09 0.62
3 81 9 28 7.9 0.09 0.71
4 72 8 36 8.4 0.08 0.67
5 64 8 44 9.4 0.08 0.75
6 56 7 51 9.9 0.07 0.69
7 49 7 58 10.9 0.07 0.76
8 42 6 64 11.3 0.06 0.68
9 36 6 70 12.3 0.06 0.74
10 30 5 75 12.8 0.05 0.64
11 25 5 80 13.8 0.05 0.69
12 20 4 84 14.3 0.04 0.57
13 16 4 88 15.3 0.04 0.61
14 12 3 91 15.7 0.03 0.47
15 9 3 94 16.7 0.03 0.5
16 6 2 96 17.0 0.02 0.34
17 4 2 98 18.0 0.02 0.36
18 2 1 99 18.0 0.01 0.18
19 1 1 100 19.0 0.01 0.19
10.81

The thing is, you can immediately improve this solution to 10.74 by just changing the first drop floor to 11.  In fact, by just experimenting a little, changing the first 6 drops to floors 11, 22, 33, 44, 55, and 66, you can get down to 10.44 drops.  So there has to be a better answer.

So far, the best result I have found is to drop the first marble at floors 15, 28, 40, 50, 59, 67, 74, 80, 85, 89, 92, 95, 97, 99, and 100; yielding an expected drop count of 10.33.  The calculation for which floor to drop it from is to divide the number of floors remaining by the log2 of the number of floors remaining.  So if you have 100 floors remaining, the log2 of 100 is 6.64, 100 / 6.64 is 15, so you make your first drop from the fifteenth floor.  Here’s how it breaks down:

Drop # Floors Remaining Floors to Skip Drop Floor # Expected Drops Likelihood Weighting
1 100 15 15 8.9 0.15 1.34
2 85 13 28 8.9 0.13 1.16
3 72 12 40 9.4 0.12 1.13
4 60 10 50 9.4 0.1 0.94
5 50 9 59 9.9 0.09 0.89
6 41 8 67 10.4 0.08 0.83
7 33 7 74 10.9 0.07 0.76
8 26 6 80 11.3 0.06 0.68
9 20 5 85 11.8 0.05 0.59
10 15 4 89 12.3 0.04 0.49
11 11 3 92 12.7 0.03 0.38
12 8 3 95 13.7 0.03 0.41
13 5 2 97 14.0 0.02 0.28
14 3 2 99 15.0 0.02 0.3
15 1 1 100 15.0 0.01 0.15
10.33

What I haven’t done yet is prove mathematically that this is the best possible solution.  One day, when I feel like wringing more fun out of this puzzle, I’ll do that.

The Point

The hobby of puzzle-solving has a lot of real-world practicality.  When you discover a solution for yourself, it’s like adding a tool to your mental toolbox.  When you later encounter similar problems, the tool is there waiting to help you solve it.

This a search efficiency problem, and it deals more specifically with the concept of an index.  If you’ve ever opened a dictionary or used a library, you’ve manually operated an index.  If you’ve ever run a Google search, then you’ve benefited from the automatic indexing that Google’s computers are always doing.  Google has not only solved this puzzle, it has solved the next several levels of this puzzle (i.e., what is the optimal number of marbles to use?  what if there are 10,000,000,000 floors?  what if you don’t have the elevator, but you have to walk up and down the stairs instead?) and trained massive server farms to execute the solutions and probably also to solve this type of puzzle on the fly.

Each marble is a metaphor for a level of indexing.  Let’s say instead of floors on a building, you are looking for a word in a dictionary.  Your first index is the tabs on the side of the dictionary that separate the words by their first letter.  Your second index consists of the two words at the top of each page (e.g. “apple – art”) that tell you the ranges of words that appear on each page.  Then your third index is the fact that the words on the page are in an order you understand, so you can quickly look over the page and find your word.

This puzzle, like others, has valuable real-world applications.  In my computing career, I readily recognized the value of indexing when I saw process run times shrink from 28 hours to 7 minutes thanks to the addition of an index to a data set.  Later, in my legal career, I was confronted with a stack of a few hundred unsorted documents, and separately a lengthy and unsorted list of long ID numbers.  I needed to pull the documents that bore the listed ID numbers.  I was able to realize that I could save myself a bunch of time by first sorting the documents by ID number, thereby allowing me to index the documents and vastly improve my average search time.

I hope this puzzle helps you as it has helped me.

Hydroelectric Power – Part One: Is your stream viable?

This Part One discusses how to assess the viability of a site for hydroelectric power, and simultaneously provides a basic understanding of the physics and math behind hydroelectric power generation. Future articles are planned to address other aspects, such as turbine and generator design, but it may be a while. This here is the fun part.

The first thing to understand is that virtually all of the electricity generated in the world comes from using some physical process to create spin. If you can spin a turbine, that turbine can turn a generator, and that creates electricity. Spin the turbine against greater resistance, and you get more electricity. Hydroelectric power is about using water to spin the turbine. Pretty easy; just drop the water through the turbine. If you have a lot of water or if the water is under pressure – for example, if it has a lot of water above it pushing downward – then it will spin the turbine against greater resistance and generate more power. So hydroelectric power is fundamentally about harvesting the energy difference between water held at different elevations under the force of gravity. If you have a source of water at some elevation and you can control the water’s ability to flow to a lower elevation, then you can steal that energy and make it do work for you.

How much energy can you get? That depends on how much water you have and how much of an elevation change you can produce. Let’s walk through the equations:

  1. The equation we’re looking for defines the relationship between water, elevation, and energy production in familiar units we can use to understand the relationship. As Americans, we understand energy in terms of kilowatt-hours because that’s how our energy utility companies bill us for them, and we understand water in terms of gallons because that’s what we use when we cook, shower, and flush. So let’s try to build an equation that relates Watts to gallons.
    1. A kilowatt-hour (kWh) is a kilowatt (kW) (1,000 watts) of power exerted for one hour. Or half a kW (500 watts) exerted for two hours. Or two kW (2,000 watts) exerted for half an hour. So let’s say you have a one-kW microwave oven; if you run it for five minutes, you’ll use one-twelfth of a kilowatt-hour.  A typical house uses about 900 kWh per month (cite).
    2. Your bathtub probably holds about 40 gallons of water during a typical bath.  There are 7.48 gallons of water in a cubic foot of water.  A keg of beer is a little over two cubic feet of beer.  A gallon of water weighs about eight pounds near its boiling point, or about 8.3 pounds at room temperature.
  2. Since we all understand kilowatt-hours and gallons, the goal is to relate kilowatt-hours to gallons of water under force of gravity.  So, step one, let’s break down the kilowatt-hour into its component units until we find something that makes sense in terms of water and gravity.
    1. A Watt is a unit of power (i.e., how hard you’re pushing), and a Watt-hour is a unit of energy (how hard you’re pushing for how long).  The Joule is also a unit of energy, equal to Watt-second.  There are 3,600 seconds in an hour, so there are 3,600 Joules in a Watt-hour.  Then, because there are 1,000 Watt-hours in a kilowatt-hour, there are 3,600,000 Joules in a kilowatt-hour.  (Keeping all this relatable: a Calorie is 4.187 Joules, and is the heat energy required to heat one gram of water one degree Celsius – so a kilowatt-hour would be enough energy to heat about 3 gallons of water from room temperature (20 C) to boiling (100 C) – do bear in mind, however, that the “calories” that appear on your food’s nutrition labels are actually kilocalories – each one is 1,000 of the SI unit, the Calorie… therefore in order to burn off a 250-calorie Krispy Kreme doughnut, you’ll need to do about 0.30 kilowatt-hours of work).
    2. A Joule is a Newton-meter, which is starting to get us somewhere because Newtons are a measure of force, and the action of gravity on water is a force.  So the only piece of the puzzle still missing is: how many Newtons of force does gravity exert on a gallon of water?
    3. A Newton is a kilogram-meter per second per second.  Gravity pulls on objects at Earth’s surface with a force equal to their mass times 9.8 meters per second per second.  That means a kilogram of water weighs 9.8 Newtons… and because a kilogram weighs 2.2 pounds, that means a gallon of water at room temperature weighs about 37 Newtons.
    4. So then in order to get to Joules, we have to factor in meters.  If we have a gallon of water with two meters of drop, that’s 74 Joules, which is… only about 0.00002 of a kilowatt-hour.  So in order to get any meaningful power out of hydroelectric generation, it takes either a lot of water or a lot of height.
  3. Solving the equation we have set up, it turns out that:
    Wh = 0.01028 * gallons of H2O * meters of drop


    … or, since there are 7.48 gallons in a cubic foot of water (and it’s useful to know cubic feet since the flow rates of most bodies of water are measured in cubic feet per second):
    Wh = 0.07688 * feet3 of H2O * meters of drop


    … or, to calculate wattage expressed in terms of flow rate (gallons or cubic feet per minute):
    W = 0.62 * (gallons of H2O / minute) * meters of drop


    W = 4.6 * (feet3 of H2O / minute) * meters of drop

Now let’s play with those equations some…

First off, let’s say you have a license to dam the Yadkin River for purposes of hydroelectric power generation. Let’s say you dam the Yadkin, giving you the Yadkin’s entire flow rate of about 2,000 cubic feet per second to work with. And let’s say you build your dam to create a 60 foot elevation drop (about 19 meters). That means every second, you can generate about three kWh… which means that every hour, you can generate about eleven megawatt-hours. However, if you really wanted to maximize your profits, you’d probably build in about 40 MW worth of generating capacity, so that you could let the water level rise during non-peak times of day, and then generate like crazy during peak usage hours when energy is selling for a high price. How much money could you make? Well, if you could sell power to the grid for $0.05/kWh, you could make $4.8 million annually.

As another scenario, let’s do a basic analysis to see if it would be worthwhile to install a home hydro system. Say you’ve measured your flow rate and found that a stream on your property has a flow rate of 60 gallons per minute, and it undergoes a 20′ change in elevation on your property. Equation says this can produce somewhere around 67 Watts. Because this is a home setup, figure some losses to inefficiency, so maybe it will actually produce about 57 Watt-hours per hour. That’s about 41 kWh per month… which comes to about four bucks a month if you pay $0.10/kWh for energy. So you probably have better things to do than try to produce energy off that stream.

How much water or drop do you need to make it worthwhile? Well, I’d be pretty happy to have a system that could both erase my power bill and supply me with energy in the event of a major blackout. For such a system, I’d want to generate about 1000 kWh per month. There are about 720 hours in a month, so the system needs to generate about 1.4 kW. Applying an 85% efficiency factor, that means I need about 1.65 kW worth of water and drop. So, at a drop of 6 meters, I’d need a flow rate of 444 gallons per minute. Or, at a drop of 12 meters, I’d need a flow rate of 222 gallons per minute. Either way, that’s a pretty good-sized waterfall.


Measuring your flow rate

To measure your flow rate, grab a five-gallon plastic bucket, divert your stream’s water into it, and see how long it takes to fill. You could also measure the cross-sectional area of your stream and measure the water’s speed and then run those numbers, but that’s probably more work than the plastic bucket. You’re going to have to build some sort of diverter to get all the water into the hydro system anyway. Remember that flow rates can vary a lot with weather and time of year, so it may be advisable to take a few measurements.


Common Questions

Q: Do I need a drop?
A: Yes. Now, I know what your next question is: what about water wheels on old mills? True, those work. The thing is, to get any kind of meaningful electricity generation out of one, it would have to be huge.

Q: Can I keep my waterfall and also use the water and drop to generate electricity?
A: Not really, no. You have to channel the water from the higher elevation to your turbine, which sits at a lower elevation. It can’t just drop; it has to be channeled so that the water at the turbine is under pressure. Of course, you could just channel part of the water from the waterfall, and let the rest cascade over. That works.

References:

What’s in a Pottle?

a short guide to the American units of liquid measure

Ever since I was a little kid learning the units of liquid measure, I’ve been a bit bothered by a couple of gaps in the progression.  I mean, everyone knows a pint is two cups and a quart is two pints, but then out of nowhere a gallon is four quarts.  I think some good portion of the population probably remembers the song “I Love You a Bushel and a Peck” and knows that a peck is two gallons and a bushel is four pecks (and a barrel is four bushels – but I don’t know how a heap is defined).  Then if you know beer, you probably know that a keg is two bushels, and a barrel is two kegs.  Also, on the smaller end of things, an ounce is two tablespoons but then a cup is eight ounces, and eight is the third power of two, so it seems like there are a couple of missing units there too.  So why the breaks in this otherwise-binary progression from tablespoon to barrel?

Well, the other day I got around to investigating this old pet peeve, and I found out that my elementary school curriculum was sadly lacking.  In fact there is a word for two quarts (or a half gallon), and that word is pottle.  So now I no longer go to the store for a half-gallon of milk; instead I procure a pottle of milk.

Likewise there is no naked jump from peck to bushel; in between lies the kenning, with a volume of four gallons or two pecks.  In fact, there’s an extended binary progression extending below cups and above barrels!  Here’s what I’ve pieced together from various sources:

a minim is basically about a drop (except that a minim is a defined quantity, whereas a drop varies in volume);
a fluid drachm is 60 minims;
a tablespoon is 4 fluid drachms;
(a tablespoon is 3 teaspoons;)
a fluid ounce is 2 tablespoons or 8 fluid drachms; (but there are 16 drams in an ounce-force [which is a bit confusing since dram sounds like drachm]);
(a jigger is 3 tablespoons [so there’s a sort of trinary progression from teaspoons to tablespoons to jiggers];)
a double-shot (or tot) is 2 fluid ounces;
a gill is 2 tots; (pronounced “jill”)
(a wineglass or lowball is 3 double-shots;)
a cup is 2 gills;
(a highball is 3 gills;)
a pint is 2 cups;
a quart is 2 pints;
a pottle is 2 quarts;
a gallon is 2 pottles;
a peck is 2 gallons;
a kenning is 2 pecks;
a firkin or a bushel is 2 kennings;
a rundlet or a keg or a kilderkin is 2 firkins or bushels;
a barrel is 2 rundlets or kegs or kilderkins;
(a hogshead is somewhere around 1.5 to 2 barrels;)
a tierce is 2 barrels; a puncheon is also approximately 2 barrels;
(a butt is typically about 3 barrels;)
a vat is 2 tierce; a pipe is approximately 2 puncheons or hogsheads;
a tun is 2 pipes (and, interestingly, also weighs about a ton);
and a cauldron or chaldron is 4 vats or about 2 tuns.

… and here’s a nice table of conversions:

minim drachm tablespoon fl. ounce jigger tot or double-shot gill wineglass cup highball pint quart pottle gallon peck kenning bushel keg barrel tierce vat or pipe tun cauldron
minim 1 1/60 1/240 1/480 1/720 1/960 1/1,920 1/2,880 1/3,840 1/5,760 1/7,680 1/15,360 1/30,720 1/61,440 1/122,880 1/245,760 1/491,520 1/983,040 1/1,966,080 1/3,932,160 1/7,864,320 1/15,728,640 1/31,457,280
drachm 60 1 1/4 1/8 1/12 1/16 1/32 1/48 1/64 1/96 1/128 1/256 1/512 1/1,024 1/2,048 1/4,096 1/8,192 1/16,384 1/32,768 1/65,536 1/131,072 1/262,144 1/524,288
tablespoon 240 4 1 1/2 1/3 1/4 1/8 1/12 1/16 1/24 1/32 1/64 1/128 1/256 1/512 1/1,024 1/2,048 1/4,096 1/8,192 1/16,384 1/32,768 1/65,536 1/131,072
fl. ounce 480 8 2 1 2/3 1/2 1/4 1/6 1/8 1/12 1/16 1/32 1/64 1/128 1/256 1/512 1/1,024 1/2,048 1/4,096 1/8,192 1/16,384 1/32,768 1/65,536
jigger 720 12 3 3/2 1 3/4 3/8 1/4 3/16 1/8 3/32 3/64 3/128 3/256 3/512 3/1,024 3/2,048 3/4,096 3/8,192 3/16,384 3/32,768 3/65,536 3/131,072
double-shot 960 16 4 2 4/3 1 1/2 1/3 1/4 1/6 1/8 1/16 1/32 1/64 1/128 1/256 1/512 1/1,024 1/2,048 1/4,096 1/8,192 1/16,384 1/32,768
gill 1,920 32 8 4 8/3 2 1 2/3 1/2 1/3 1/4 1/8 1/16 1/32 1/64 1/128 1/256 1/512 1/1,024 1/2,048 1/4,096 1/8,192 1/16,384
wineglass 2,880 48 12 6 4 3 3/2 1 3/4 1/2 3/8 3/16 3/32 3/64 3/128 3/256 3/512 3/1,024 3/2,048 3/4,096 3/8,192 3/16,384 3/32,768
cup 3,840 64 16 8 16/3 4 2 4/3 1 2/3 1/2 1/4 1/8 1/16 1/32 1/64 1/128 1/256 1/512 1/1,024 1/2,048 1/4,096 1/8,192
highball 5,760 96 24 12 8 6 3 2 3/2 1 3/4 3/8 3/16 3/32 3/64 3/128 3/256 3/512 3/1,024 3/2,048 3/4,096 3/8,192 3/16,384
pint 7,680 128 32 16 32/3 8 4 8/3 2 4/3 1 1/2 1/4 1/8 1/16 1/32 1/64 1/128 1/256 1/512 1/1,024 1/2,048 1/4,096
quart 15,360 256 64 32 64/3 16 8 16/3 4 8/3 2 1 1/2 1/4 1/8 1/16 1/32 1/64 1/128 1/256 1/512 1/1,024 1/2,048
pottle 30,720 512 128 64 128/3 32 16 32/3 8 16/3 4 2 1 1/2 1/4 1/8 1/16 1/32 1/64 1/128 1/256 1/512 1/1,024
gallon 61,440 1,024 256 128 256/3 64 32 64/3 16 32/3 8 4 2 1 1/2 1/4 1/8 1/16 1/32 1/64 1/128 1/256 1/512
peck 122,880 2,048 512 256 512/3 128 64 128/3 32 64/3 16 8 4 2 1 1/2 1/4 1/8 1/16 1/32 1/64 1/128 1/256
kenning 245,760 4,096 1,024 512 1,024/3 256 128 256/3 64 128/3 32 16 8 4 2 1 1/2 1/4 1/8 1/16 1/32 1/64 1/128
bushel 491,520 8,192 2,048 1,024 2,048/3 512 256 512/3 128 256/3 64 32 16 8 4 2 1 1/2 1/4 1/8 1/16 1/32 1/64
keg 983,040 16,384 4,096 2,048 4,096/3 1,024 512 1,024/3 256 512/3 128 64 32 16 8 4 2 1 1/2 1/4 1/8 1/16 1/32
barrel 1,966,080 32,768 8,192 4,096 8,192/3 2,048 1,024 2,048/3 512 1,024/3 256 128 64 32 16 8 4 2 1 1/2 1/4 1/8 1/16
tierce 3,932,160 65,536 16,384 8,192 16,384/3 4,096 2,048 4,096/3 1,024 2,048/3 512 256 128 64 32 16 8 4 2 1 1/2 1/4 1/8
vat or pipe 7,864,320 131,072 32,768 16,384 32,768/3 8,192 4,096 8,192/3 2,048 4,096/3 1,024 512 256 128 64 32 16 8 4 2 1 1/2 1/4
tun 15,728,640 262,144 65,536 32,768 65,536/3 16,384 8,192 16,384/3 4,096 8,192/3 2,048 1,024 512 256 128 64 32 16 8 4 2 1 1/2
cauldron 31,457,280 524,288 131,072 65,536 131,072/3 32,768 16,384 32,768/3 8,192 16,384/3 4,096 2,048 1,024 512 256 128 64 32 16 8 4 2 1

There’s a lot of very interesting history behind these measures. For example, as a beer brewer, I know that a keg is actually 15.5 gallons and a barrel is 31 gallons. This probably has something to do with beverage merchants historically shortchanging their customers – or possibly when the measures were set out legally, it was deemed desirable that anything over 15.5 gallons should satisfy a contract to deliver a “keg,” thereby allowing for such things as evaporation and minor spillage.

If you’re a fan of Alton Brown’s cooking show Good Eats (as I am), then you’ve heard the rhyme, “a pint’s a pound the world around.” I would add, “and a ton’s a tun, or most of one.” Of course, these rhymes only apply to water, upon which Earth’s gravity acts with approximately one ounce-force per ounce-volume… depending on how hot the water is among other factors. The rhyme also, apparently, does not apply in the UK, where an “Imperial pint” is 20 ounces, which is a pound-and-a-quarter, because an Imperial gallon is defined as 10 pounds.