Category Archives: math

How long could you make a suspension bridge?

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The above is one of the engineering questions that puzzled me as a student engineer at Brooklyn Technical High School and at Cooper Union in New York. The Brooklyn Bridge stood as a wonder of late 1800s engineering, and it had recently been eclipsed by the Verrazano bridge, a pure suspension bridge. At the time it was the longest and heaviest in the world. How long could a bridge be made, and why did Brooklyn bridge have those catenary cables, when the Verrazano didn’t? (Sometimes I’d imagine a Chinese engineer being asked the top question, and answering “Certainly, but How Long is my cousin.”)

I found the above problem unsolvable with the basic calculus at my disposal. because it was clear that both the angle of the main cable and its tension varied significantly along the length of the cable. Eventually I solved this problem using a big dose of geometry and vectors, as I’ll show.

Vector diagram of forces on the cable at the center-left of the bridge.

Vector diagram of forces on the cable at the center-left of the bridge.

Consider the above vector diagram (above) of forces on a section of the main cable near the center of the bridge. At the right, the center of the bridge, the cable is horizontal, and has a significant tension. Let’s call that T°. Away from the center of the bridge, there is a vertical cable supporting a fraction of  roadway. Lets call the force on this point w. It equals the weight of this section of cable and this section of roadway. Because of this weight, the main cable bends upward to the left and carries more tension than T°. The tangent (slope) of the upward curve will equal w/T°, and the new tension will be the vector sum along the new slope. From geometry, T= √(w2 +T°2).

Vector diagram of forces on the cable further from the center of the bridge.

Vector diagram of forces on the cable further from the center of the bridge.

As we continue from the center, there are more and more verticals, each supporting approximately the same weight, w. From geometry, if w weight is added at each vertical, the change in slope is always w/T° as shown. When you reach the towers, the weight of the bridge must equal 2T Sin Θ, where Θ is the angle of the bridge cable at the tower and T is the tension in the cable at the tower.

The limit to the weight of a bridge, and thus its length, is the maximum tension in the main cable, T, and the maximum angle, that at the towers. Θ. I assumed that the maximum bridge would be made of T1 bridge steel, the strongest material I could think of, with a tensile strength of 100,000 psi, and I imagined a maximum angle at the towers of 30°. Since there are two towers and sin 30° = 1/2, it becomes clear that, with this 30° angle cable, the tension at the tower must equal the total weight of the bridge. Interesting.

Now, to find the length of the bridge, note that the weight of the bridge is proportional to its length times the density and cross section of the metal. I imagined a bridge where the half of the weight was in the main cable, and the rest was in the roadway, cars and verticals. If the main cable is made of T1 “bridge steel”, the density of the cable is 0.2833 lb/in3, and the density of the bridge is twice this. If the bridge cable is at its yield strength, 100,000 psi, at the towers, it must be that each square inch of cable supports 50,000 pounds of cable and 50,000 lbs of cars, roadway and verticals. The maximum length (with no allowance for wind or a safety factor) is thus

L(max) = 100,000 psi / 2 x 0.2833 pounds/in3 = 176,500 inches = 14,700 feet = 2.79 miles.

This was more than three times the length of the Verrazano bridge, whose main span is ‎4,260 ft. I attributed the difference to safety factors, wind, price, etc. I then set out to calculate the height of the towers, and the only rational approach I could think of involved calculus. Fortunately, I could integrate for the curve now that I knew the slope changed linearly with distance from the center. That is for every length between verticals, the slope changes by the same amount, w/T°. This was to say that d2y/dx2 = w/T° and the curve this described was a parabola.

Rather than solving with heavy calculus, I noticed that the slope, dy/dx increases in proportion to x, and since the slope at the end, at L/2, was that of a 30° triangle, 1/√3, it was clear to me that

dy/dx = (x/(L/2))/√3

where x is the distance from the center of the bridge, and L is the length of the bridge, 14,700 ft. dy/dx = 2x/L√3.

We find that:
H = ∫dy = ∫ 2x/L√3 dx = L/4√3 = 2122 ft,

where H is the height of the towers. Calculated this way, the towers were quite tall, higher than that of any building then standing, but not impossibly high (the Dubai tower is higher). It was fairly clear that you didn’t want a tower much higher than this, though, suggesting that you didn’t want to go any higher than a 30° angle for the main cable.

I decided that suspension bridges had some advantages over other designs in that they avoid the problem of beam “buckling.’ Further, they readjust their shape somewhat to accommodate heavy point loads. Arch and truss bridges don’t do this, quite. Since the towers were quite a lot taller than any building then in existence, I came to I decide that this length, 2.79 miles, was about as long as you could make the main span of a bridge.

I later came to discover materials with a higher strength per weight (titanium, fiber glass, aramid, carbon fiber…) and came to think you could go longer, but the calculation is the same, and any practical bridge would be shorter, if only because of the need for a safety factor. I also came to recalculate the height of the towers without calculus, and got an answer that was shorter, for some versions, a hundred feet shorter, as shown here. In terms of wind, I note that you could make the bridge so heavy that you don’t have to worry about wind except for resonance effects. Those are the effects are significant, but were not my concern at the moment.

The Brooklyn Bridge showing its main cable suspension structure and its catenaries.

Now to discuss catenaries, the diagonal wires that support many modern bridges and that, on the Brooklyn bridge, provide  support at the ends of the spans only. Since the catenaries support some weight of the Brooklyn bridge, they decrease the need for very thick cables and very high towers. The benefit goes down as the catenary angle goes to the horizontal, though as the lower the angle the longer the catenary, and the lower the fraction of the force goes into lift. I suspect this is why Roebling used catenaries only near the Brooklyn bridge towers, for angles no more than about 45°. I was very proud of all this when I thought it through and explained it to a friend. It still gives me joy to explain it here.

Robert Buxbaum, May 16, 2019.  I’ve wondered about adding vibration dampers to very long bridges to decrease resonance problems. It seems like a good idea. Though I have never gone so far as to do calculations along these lines, I note that several of the world’s tallest buildings were made of concrete, not steel, because concrete provides natural vibration damping.

Statistics for psychologists, sociologists, and political scientists

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In terms of mathematical structure, psychologists, sociologists, and poly-sci folks all do the same experiment, over and over, and all use the same simple statistical calculation, the ANOVA, to determine its significance. I thought I’d explain that experiment and the calculation below, walking you through an actual paper (one I find interesting) in psychology / poly-sci. The results are true at the 95% level (that’s the same as saying p >0.05) — a significant achievement in poly-sci, but that doesn’t mean the experiment means what the researchers think. I’ll then suggest another statistic measure, r-squared, that deserves to be used along with ANOVA.

The standard psychological or poly-sci research experiments involves taking a group of people (often students) and giving them a questionnaire or test to measure their feelings about something — the war in Iraq, their fear of flying, their degree of racism, etc. This is scored on some scale to get an average. Another, near-identical group of subjects is now brought in and given a prompt: shown a movie, or a picture, or asked to visualize something, and then given the same questionnaire or test as the first group. The prompt is shown to have changed to average score, up or down, an ANOVA (analysis of variation) is used to show if this change is one the researcher can have confidence in. If the confidence exceeds 95% the researcher goes on to discuss the significance, and submits the study for publication. I’ll now walk you through the analysis the old fashioned way: the way it would have been done in the days of hand calculators and slide-rules so you understand it. Even when done this way, it only takes 20 minutes or so: far less time than the experiment.

I’ll call the “off the street score” for the ith subject, Xi°. It would be nice if papers would publish these, but usually they do not. Instead, researchers publish the survey and the average score, something I’ll call X°-bar, or X°. they also publish a standard deviation, calculated from the above, something I’ll call, SD°. In older papers, it’s called sigma, σ. Sigma and SD are the same thing. Now, moving to the group that’s been given the prompt, I’ll call the score for the ith subject, Xi*. Similar to the above, the average for this prompted group is X*, or X°-bar, and the standard deviation SD*.

I have assumed that there is only one prompt, identified by an asterix, *, one particular movie, picture, or challenge. For some studies there will be different concentrations of the prompt (show half the movie, for example), and some researchers throw in completely different prompts. The more prompts, the more likely you get false positives with an ANOVA, and the more likely you are to need to go to r-squared. Warning: very few researchers do this, intentionally (and crookedly) or by complete obliviousness to the math. Either way, if you have a study with ten prompt variations, and you are testing to 95% confidence your result is meaningless. Random variation will give you this result 50% of the time. A crooked researcher used ANOVA and 20 prompt variations “to prove to 95% confidence” that genetic modified food caused cancer; I’ll assume (trust) you won’t fall into that mistake, and that you won’t use the ANOVA knowledge I provide to get notoriety and easy publication of total, un-reproducible nonsense. If you have more than one or two prompts, you’ve got to add r-squared (and it’s probably a good idea with one or two). I’d discuss r-squared at the end.

I’ll now show how you calculate X°-bar the old-fashioned way, as would be done with a hand calculator. I do this, not because I think social-scientists can’t calculate an average, nor because I don’t trust the ANOVA function on your laptop or calculator, but because this is a good way to familiarize yourself with the notation:

X°-bar = X° = 1/n° ∑ Xi°.

Here, n° is the total number of subjects who take the test but who have not seen the prompt. Typically, for professional studies, there are 30 to 50 of these. ∑ means sum, and Xi° is the score of the ith subject, as I’d mentioned. Thus, ∑ Xi° indicates the sum of all the scores in this group, and 1/n° is the average, X°-bar. Convince yourself that this is, indeed the formula. The same formula is used for X*-bar. For a hand calculation, you’d write numbers 1 to n° on the left column of some paper, and each Xi° value next to its number, leaving room for more work to follow. This used to be done in a note-book, nowadays a spreadsheet will make that easier. Write the value of X°-bar on a separate line on the bottom.

T-table

T-table

In virtually all cases you’ll find that X°-bar is different from X*-bar, but there will be a lot of variation among the scores in both groups. The ANOVA (analysis of variation) is a simple way to determine whether the difference is significant enough to mean anything. Statistics books make this calculation seem far too complicated — they go into too much math-theory, or consider too many types of ANOVA tests, most of which make no sense in psychology or poly-sci but were developed for ball-bearings and cement. The only ANOVA approach used involves the T-table shown and the 95% confidence (column this is the same as two-tailed p<0.05 column). Though 99% is nice, it isn’t necessary. Other significances are on the chart, but they’re not really useful for publication. If you do this on a calculator, the table is buried in there someplace. The confidence level is written across the bottom line of the cart. 95% here is seen to be the same as a two-tailed P value of 0.05 = 5% seen on the third from the top line of the chart. For about 60 subjects (two groups of 30, say) and 95% certainty, T= 2.000. This is a very useful number to carry about in your head. It allows you to eyeball your results.

In order to use this T value, you will have to calculate the standard deviation, SD for both groups and the standard variation between them, SV. Typically, the SDs will be similar, but large, and the SV will be much smaller. First lets calculate SD° by hand. To do this, you first calculate its square, SD°2; once you have that, you’ll take the square-root. Take each of the X°i scores, each of the scores of the first group, and calculate the difference between each score and the average, X°-bar. Square each number and divide by (n°-1). These numbers go into their own column, each in line with its own Xi. The sum of this column will be SD°2. Put in mathematical terms, for the original group (the ones that didn’t see the movie),

SD°2 = 1/(n°-1) ∑ (Xi°- X°)2

SD° = √SD°2.

Similarly for the group that saw the movie, SD*2 = 1/(n*-1) ∑ (Xi*- X*)2

SD* = √SD*2.

As before, n° and n* are the number of subjects in each of the two groups. Usually you’ll aim for these to be the same, but often they’ll be different. Some students will end up only seeing half the move, some will see it twice, even if you don’t plan it that way; these students’ scares can not be used with the above, but be sure to write them down; save them. They might have tremendous value later on.

Write down the standard deviations, SD for each group calculated above, and check that the SDs are similar, differing by less than a factor of 2. If so, you can take a weighted average and call it SD-bar, and move on with your work. There are formulas for this average, and in some cases you’ll need an F-table to help choose the formula, but for my purposes, I’ll assume that the SDs are similar enough that any weighted average will do. If they are not, it’s likely a sign that something very significant is going on, and you may want to re-think your study.

Once you calculate SD-bar, the weighted average of the SD’s above, you can move on to calculate the standard variation, the SV between the two groups. This is the average difference that you’d expect to see if there were no real differences. That is, if there were no movie, no prompt, no nothing, just random chance of who showed up for the test. SV is calculated as:

SV = SD-bar √(1/n° + 1/n*).

Now, go to your T-table and look up the T value for two tailed tests at 95% certainty and N = n° + n*. You probably learned that you should be using degrees of freedom where, in this case, df = N-2, but for normal group sizes used, the T value will be nearly the same. As an example, I’ll assume that N is 80, two groups of 40 subjects the degrees of freedom is N-2, or 78. I you look at the T-table for 95% confidence, you’ll notice that the T value for 80 df is 1.99. You can use this. The value for  62 subjects would be 2.000, and the true value for 80 is 1.991; the least of your problems is the difference between 1.991 and 1.990; it’s unlikely your test is ideal, or your data is normally distributed. Such things cause far more problems for your results. If you want to see how to deal with these, go here.

Assuming random variation, and 80 subjects tested, we can say that, so long as X°-bar differs from X*-bar by at least 1.99 times the SV calculated above, you’ve demonstrated a difference with enough confidence that you can go for a publication. In math terms, you can publish if and only if: |X°-X*| ≥ 1.99 SV where the vertical lines represent absolute value. This is all the statistics you need. Do the above, and you’re good to publish. The reviewers will look at your average score values, and your value for SV. If the difference between the two averages is more than 2 times the SV, most people will accept that you’ve found something.

If you want any of this to sink in, you should now do a worked problem with actual numbers, in this case two groups, 11 and 10 students. It’s not difficult, but you should try at least with these real numbers. When you are done, go here. I will grind through to the answer. I’ll also introduce r-squared.

The worked problem: Assume you have two groups of people tested for racism, or political views, or some allergic reaction. One group was given nothing more than the test, the other group is given some prompt: an advertisement, a drug, a lecture… We want to know if we had a significant effect at 95% confidence. Here are the test scores for both groups assuming a scale from 0 to 3.

Control group: 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3.  These are the Xi° s; there are 11 of them

Prompted group: 0, 1, 1, 1, 2, 2, 2, 2, 3, 3.  These are the Xi* s; there are 10 of them.

On a semi-humorous side: Here is the relationship between marriage and a PhD.

Robert Buxbaum, March 18, 2019. I also have an explanation of loaded dice and flipped coins; statistics for high school students.

A probability paradox

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Here is a classic math paradox for your amusement, and perhaps your edification: (edification is a fancy word for: beware, I’m trying to learn you something).

You are on a TV game show where you will be asked to choose between two, identical-looking envelopes. All you know about the envelopes is that one of them has twice as much money as the other. The envelopes are shuffled, and you pick one. You peak in and see that your envelope contains $400, and you feel pretty good. But then you are given a choice: you can switch your envelope with the other one; the one you didn’t take. You reason that the other envelope either has $800 or $200 with equal probability. That is, a switch will either net you a $400 gain, or loose you $200. Since $400 is bigger than $200, you switch. Did that decision make sense. It seems that, at this game, every contestant should switch envelopes. Hmm.

The solution follows: The problem with this analysis is an error common in children and politicians — the confusion between your lack of knowledge of a thing, and actual variability in the system. In this case, the contestant is confusing his (or her) lack of knowledge of whether he/she has the big envelope or the smaller, with the fixed fact that the total between the two envelopes has already been set. It is some known total, in this case it is either $600 or $1200. Lets call this unknown sum y. There is a 50% chance that you now are holding 2/3 y and a 50% chance you are holding only 1/3y. therefore, the value of your current envelope is 1/3 y + 1/6y = 1/2 y. Similarly, the other envelope has a value 1/2y; there is no advantage is switching once it is accepted that the total, y had already been set before you got to choose an envelope.

And here, unfortunately is the lesson:The same issue applies in reverse when it comes to government taxation. If you assume that the total amount of goods produced by the economy is always fixed to some amount, then there is no fundamental problem with high taxes. You can print money, or redistribute it to anyone you think is worthy — more worthy than the person who has it now – and you won’t affect the usable wealth of the society. Some will gain others will lose, and likely you’ll find you have more friends than before. On the other hand, if you assume that government redistribution will affect the total: that there is some relationship between reward and the amount produced, then to the extent that you diminish the relation between work and income, or savings and wealth, you diminish the total output and wealth of your society. While some balance is needed, a redistribution that aims at identical outcomes will result in total poverty.

This is a variant of the “two-envelopes problem,” originally posed in 1912 by German, Jewish mathematician, Edmund Landau. It is described, with related problems, by Prakash Gorroochurn, Classic Problems of Probability. Wiley, 314pp. ISBN: 978-1-118-06325-5. Wikipedia article: Two Envelopes Problem.

Robert Buxbaum, February 27, 2019

Calculating π as a fraction

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Pi is a wonderful number, π = 3.14159265…. It’s very useful, ratio of the circumference of a circle to its diameter, or the ratio of area of a circle to the square of its radius, but it is irrational: one can show that it can not be described as an exact fraction. When I was in middle school, I thought to calculate Pi by approximations of the circumference or area, but found that, as soon as I got past some simple techniques, I was left with massive sums involving lots of square-roots. Even with a computer, I found this slow, annoying, and aesthetically unpleasing: I was calculating one irrational number from the sum of many other irrational numbers.

At some point, I moved to try solving via the following fractional sum (Gregory and Leibniz).

π/4 = 1/1 -1/3 +1/5 -1/7 …

This was an appealing approach, but I found the series converges amazingly slowly. I tried to make it converge faster by combining terms, but that just made the terms more complex; it didn’t speed convergence. Next to try was Euler’s formula:

π2/6 = 1/1 + 1/4 + 1/9 + ….

This series converges barely faster than the Gregory/Leibniz series, and now I’ve got a square-root to deal with. And that brings us to my latest attempt, one I’m pretty happy with discovering (I’m probably not the first). I start with the Taylor series for sin x. If x is measured in radians: 180° = π radians; 30° = π/6 radians. With the angle x measured in radians, can show that

sin x = x – x3/6 + x5/120 – x7/5040 + …

Notice that the series is fractional and that the denominators get large fast. That suggests that the series will converge fast (2 to 3 terms?). To speed things up further, I chose to solve the above for sin 30° = 1/2 = sin π/6. Truncating the series to the first term gives us the following approximation for pi.

1/2 = sin (π/6) ≈ π/6.

Rearrange this and you find π ≈ 6/2 = 3.

That’s not bad for a first order solution. The Gregory/ Leibniz series would have gotten me π ≈ 4, and the Euler series π ≈ √6 = 2.45…: I’m ahead of the game already. Now, lets truncate to the second term.

1/2 ≈ π/6 – (π/6)3/6.

In theory, I could solve this via the cubic equation formula, but that would leave me with two square roots, something I’d like to avoid. Instead, and here’s my innovation, I’ll substitute 3 + ∂ for π . I’ll then use the binomial theorem to claim that (π)3 ≈ 27 + 27∂ = 27(1+∂). Put this into the equation above and we find:

1/2 = (3+∂)/6 – 27(1+∂)/1296

Rearranging and solving for ∂, I find that

27/216 = ∂ (1- 27/216) = ∂ (189/216)

∂ = 27/189 = 1/7 = .1428…

If π ≈ 3 + ∂, I’ve just calculated π ≈ 22/7. This is not bad for an approximation based on just the second term in the series.

Where to go from here? One thought was to revisit the second term, and now say that π = 22/7 + ∂, but it seemed wrong to ignore the third term. Instead, I’ll include the 3rd term, and say that π/6 = 11/21 + ∂. Extending the derivative approximations I used above, (π/6)3 ≈ (11/21)3 + 3∂(11/21)2, etc., I find:

1/2 ≈ (11/21 + ∂) -(11/21)3/6 – 3∂(11/21)2/6 + (11/21)5/120 + 5∂(11/21)4/120.

For a while I tried to solve this for ∂ as fraction using long-hand algebra, but I kept making mistakes. Thus, I’ve chosen to use two faster options: decimals or wolfram alpha. Using decimals is simpler, I find: 11/21 ≈ .523810, (11/21)2 =  .274376; (11/21)3 = .143721; (11/21)4 = .075282, and (11/21)5 = .039434.

Put these numbers into the original equation and I find:

1/2 – .52381 +.143721/6 -.039434/120 = ∂ (1-.274376/2 + .075282/24),

∂ = -.000185/.86595 ≈ -.000214. Based on this,

π ≈ 6 (11/21  -.000214) = 3,141573… Not half bad.

Alternately, using Wolfram alpha to reduce the fractions,

1/2 – 11/21+ 113/6•213 -115/(120•215) = ∂ (24(21)4/24(21)4 – 12•112212/24•214+ (11)4/24•214)

∂ = -90491/424394565 ≈ -.000213618. This is a more exact solution, but it gives a result that’s no more accurate since it is based on a 3 -term approximation of the infinite series.

We find that π/6 ≈ .523596, or, in fractional form, that π ≈ 444422848 / 141464855 = 3.14158.

Either approach seems OK in terms of accuracy: I can’t imagine needing more (I’m just an engineer). I like that I’ve got a fraction, but find the fraction quite ugly, as fractions go. It’s too big. Working with decimals gets me the same accuracy with less work — I avoided needing square roots, and avoided having to resort to Wolfram.

As an experiment, I’ll see if I get a nicer fraction if I drop the last term (11)4/24•214: it is a small correction to a small number, ∂. The equation is now:

1/2 – 11/21+ 113/6•213 -115/(120•215) = ∂ (24(21)4/24(21)4 – 12(11221)2/24•214).

I’ll multiply both sides by 24•214 and then by (5•21) to find that:

12•214 – 24•11•213+ 4•21•113 -115/(5•21) = ∂ (24(21)4 – 12•112212),

60•215 – 120•11•214+ 20•21^2•113 -115 = ∂ (120(21)5 – 60•112213).

Solving for π, I now get, 221406169/70476210 = 3.1415731

It’s still an ugly fraction, about as accurate as before. As with the digital version, I got to 5-decimal accuracy without having to deal with square roots, but I still had to go to Wolfram. If I were to go further, I’d start with the pi value above in digital form, π = 3.141573 + ∂; I’d add the 7th power term, and I’d stick to decimals for the solution. I imagine I’d add 4-5 more decimals that way.

Robert Buxbaum, April 2, 2018

Beyond oil lies … more oil + price volatility

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One of many best selling books by Kenneth Deffeyes

One of many best-selling books by Kenneth Deffeyes

While I was at Princeton, one of the most popular courses was geology 101 taught by Dr. Kenneth S. Deffeyes. It was a sort of “Rocks for Jocks,” but had an unusual bite since Dr. Deffeyes focussed particularly on the geology of oil. Deffeyes had an impressive understanding of oil and oil production, and one outcome of this impressive understanding was his certainty that US oil production had peaked in 1970, and that world oil was about to run out too. The prediction that US oil production had peaked was not original to him. It was called Hubbert’s peak after King Hubbert who correctly predicted (rationalized?) the date, but published it only in 1971. What Deffeyes added to Hubbard’s analysis was a simplified mathematical justification and a new prediction: that world oil production would peak in the 1980s, or 2000, and then run out fast. By 2005, the peak date was fixed to November 24, of the same year: Thanksgiving day 2005 ± 3 weeks.

As with any prediction of global doom, I was skeptical, but generally trusted the experts, and virtually every experts was on board to predict gloom in the near future. A British group, The Institute for Peak Oil picked 2007 for the oil to run out, and the several movies expanded the theme, e.g. Mad Max. I was convinced enough to direct my PhD research to nuclear fusion engineering. Fusion being presented as the essential salvation for our civilization to survive beyond 2050 years or so. I’m happy to report that the dire prediction of his mathematics did not come to pass, at least not yet. To quote Yogi Berra, “In theory, theory is just like reality.” Still I think it’s worthwhile to review the mathematical thinking for what went wrong, and see if some value might be retained from the rubble.

proof of peak oilDeffeyes’s Maltheisan proof went like this: take a year-by year history of the rate of production, P and divide this by the amount of oil known to be recoverable in that year, Q. Plot this P/Q data against Q, and you find the data follows a reasonably straight line: P/Q = b-mQ. This occurs between 1962 and 1983, or between 1983 and 2005. Fro whichever straight line you pick, m and b are positive. Once you find values for m and b that you trust, you can rearrange the equation to read,

P = -mQ2+ bQ

You the calculate the peak of production from this as the point where dP/dQ = 0. With a little calculus you’ll see this occurs at Q = b/2m, or at P/Q = b/2. This is the half-way point on the P/Q vs Q line. If you extrapolate the line to zero production, P=0, you predict a total possible oil production, QT = b/m. According to this model this is always double the total Q discovered by the peak. In 1983, QT was calculated to be 1 trillion barrels. By May of 2005, again predicted to be a peak year, QT had grown to two trillion barrels.

I suppose Deffayes might have suspected there was a mistake somewhere in the calculation from the way that QT had doubled, but he did not. See him lecture here in May 2005; he predicts war, famine, and pestilence, with no real chance of salvation. It’s a depressing conclusion, confidently presented by someone enamored of his own theories. In retrospect, I’d say he did not realize that he was over-enamored of his own theory, and blind to the possibility that the P/Q vs Q line might curve upward, have a positive second derivative.

Aside from his theory of peak oil, Deffayes also had a theory of oil price, one that was not all that popular. It’s not presented in the YouTube video, nor in his popular books, but it’s one that I still find valuable, and plausibly true. Deffeyes claimed the wildly varying prices of the time were the result of an inherent quay imbalance between a varying supply and an inelastic demand. If this was the cause, we’d expect the price jumps of oil up and down will match the way the wait-line at a barber shop gets longer and shorter. Assume supply varies because discoveries came in random packets, while demand rises steadily, and it all makes sense. After each new discovery, price is seen to fall. It then rises slowly till the next discovery. Price is seen as a symptom of supply unpredictability rather than a useful corrective to supply needs. This view is the opposite of Adam Smith, but I think he’s not wrong, at least in the short term with a necessary commodity like oil.

Academics accepted the peak oil prediction, I suspect, in part because it supported a Marxian remedy. If oil was running out and the market was broken, then our only recourse was government management of energy production and use. By the late 70s, Jimmy Carter told us to turn our thermostats to 65. This went with price controls, gas rationing, and a 55 mph speed limit, and a strong message of population management – birth control. We were running out of energy, we were told because we had too many people and they (we) were using too much. America’s grown days were behind us, and only the best and the brightest could be trusted to manage our decline into the abyss. I half believed these scary predictions, in part because everyone did, and in part because they made my research at Princeton particularly important. The Science fiction of the day told tales of bold energy leaders, and I was ready to step up and lead, or so I thought.

By 2009 Dr. Deffayes was being regarded as chicken little as world oil production continued to expand.

By 2009 Dr. Deffayes was being regarded as chicken little as world oil production continued to expand.

I’m happy to report that none of the dire predictions of the 70’s to 90s came to pass. Some of my colleagues became world leaders, the rest because stock brokers with their own private planes and SUVs. As of my writing in 2018, world oil production has been rising, and even King Hubbert’s original prediction of US production has been overturned. Deffayes’s reputation suffered for a few years, then politicians moved on to other dire dangers that require world-class management. Among the major dangers of today, school shootings, Ebola, and Al Gore’s claim that the ice caps will melt by 2014, flooding New York. Sooner or later, one of these predictions will come true, but the lesson I take is that it’s hard to predict change accurately.

Just when you thought US oil had beed depleted for good, production began rising. It's now higher than the 1970 peak.

Just when you thought US oil was depleted, production began rising. We now produce more than in 1970.

Much of the new oil production you’ll see on the chart above comes from tar-sands, oil the Deffeyes  considered unrecoverable, even while it was being recovered. We also  discovered new ways to extract leftover oil, and got better at using nuclear electricity and natural gas. In the long run, I expect nuclear electricity and hydrogen will replace oil. Trees have a value, as does solar. As for nuclear fusion, it has not turned out practical. See my analysis of why.

Robert Buxbaum, March 15, 2018. Happy Ides of March, a most republican holiday.

math jokes and cartoons

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uo3wgcxeParallel lines have so much in common.

It’s a shame they never get to meet.

he-who-is-without-mathematics

bad-vector-math-jokemath-for-dummies

sometimes education is the removal of false notions.

sometimes education is the removal of false notions.

pi therapy

pi therapy

Robert E. Buxbaum, January 4, 2017. Aside from the beauty of math itself, I’ve previously noted that, if your child is interested in science, the best route for development is math. I’ve also noted that Einstein did not fail at math, and that calculus is taught wrong, and probably is.

The game is rigged and you can always win.

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A few months ago, I wrote a rather depressing essay based on Nobel Laureate, Kenneth Arrow’s work, and the paradox of de Condorcet. It is mathematically shown that you can not make a fair election, even if you wanted to, and no one in power wants to. The game is rigged.

To make up for that insight, I’d like to show from the work of John Forbes Nash (A Beautiful Mind) that you, personally, can win, basically all the time, if you can get someone, anyone to coöperate by trade. Let’s begin with an example in Nash’s first major paper, “The Bargaining Problem,” the one Nash is working on in the movie— read the whole paper here.  Consider two people, each with a few durable good items. Person A has a bat, a ball, a book, a whip, and a box. Person B has a pen, a toy, a knife, and a hat. Since each item is worth a different amount (has a different utility) to the owner and to the other person, there are almost always sets of trades that benefit both. In our world, where there are many people and everyone has many durable items, it is inconceivable that there are not many trades a person can make to benefit him/her while benefiting the trade partner.

Figure 3, from Nash’s, “The bargaining problem.” U1 and U2 are the utilities of the items to the two people, and O is the current state. You can improve by barter so long as your current state is not on the boundary. The parallel lines are places one could reach if money trades as well.

Good trades are even more likely when money is involved or non-durables. A person may trade his or her time for money, that is work, and any half-normal person will have enough skill to be of some value to someone. If one trades some money for durables, particularly tools, one can become rich (slowly). If one trades this work for items to make them happy (food, entertainment) they can become happier. There are just two key skills: knowing what something is worth to you, and being willing to trade. It’s not that easy for most folks to figure out what their old sofa means to them, but it’s gotten easier with garage sales and eBay.

Let us now move to the problem of elections, e.g. in this year 2016. There are few people who find the person of their dreams running for president this year. The system has fundamental flaws, and has delivered two thoroughly disliked individuals. But you can vote for a generally positive result by splitting your ticket. American society generally elects a mix of Democrats and Republicans. This mix either delivers the outcome we want, or we vote out some of the bums. Americans are generally happy with the result.

A Stamp act stamp. The British used these to tax every transaction, making it impossible for the ordinary person to benefit by small trade.

A Stamp act stamp,. Used to tax every transaction, the British made it impossible for ordinary people to benefit by small trades.

The mix does not have to involve different people, it can involve different periods of time. One can elect a Democrat president this year, and an Republican four years later. Or take the problem of time management for college students. If a student had to make a one time choice, they’d discover that you can’t have good grades, good friends, and sleep. Instead, most college students figure out you can have everything if you do one or two of these now, and switch when you get bored. And this may be the most important thing they learn.

This is my solution to Israel’s classic identity dilemma. David Ben-Gurion famously noted that Israel had the following three choices: they could be a nation of Jews living in the land of Israel, but not democratic. They could be a democratic nation in the land of Israel, but not Jewish; or they could be Jewish and democratic, but not (for the most part) in Israel. This sounds horrible until you realize that Israel can elect politicians to deliver different pairs of the options, and can have different cities that cater to thee options too. Because Jerusalem does not have to look like Tel Aviv, Israel can achieve a balance that’s better than any pure solution.

Robert E. Buxbaum, July 17-22, 2016. Balance is all, and pure solutions are a doom. I’m running for water commissioner.

Weir dams to slow the flow and save our lakes

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As part of explaining why I want to add weir dams to the Red Run drain, and some other of our Oakland county drains, I posed the following math/ engineering problem: if a weir dam is used to double the depth of water in a drain, show that this increases the residence time by a factor of 2.8 and reduces the flow speed by 1/2.8. Here is my solution.

A series of weir dams on Blackman Stream, Maine. Mine would be about as tall, but somewhat further apart.

A series of weir dams on Blackman Stream, Maine. Mine would be about as tall, but wider and further apart. The dams provide oxygenation and hold back sludge.

Let’s assume the shape of the bottom of the drain is a parabola, e.g. y = x2 , and that the dams are spaced far enough apart that their volume is small compared to the volume of water. We now use integral calculus to calculate how the volume of water per mile, V is affected by water height:  V =2XY- ∫ y dx = 2XY- 2/3 X3 =  4/3 Y√Y. Here, capital Y is the height of water in the drain, and capital X is the horizontal distance of the water edge from the drain centerline. For a parabolic-bottomed drain, if you double the height Y, you increase the volume of water per mile by 2√2. That’s 2.83, or about 2.8 once you assume some volume to the dams.

To find how this affects residence time and velocity, note that the dam does not affect the volumetric flow rate, Q (gallons per hour). If we measure V in gallons per mile of drain, we find that the residence time per mile of drain (hours) is V/Q and that the speed (miles per hour) is Q/V. Increasing V by 2.8 increases the residence time by 2.8 and decreases the speed to 1/2.8 of its former value.

Why is this important? Decreasing the flow speed by even a little decreases the soil erosion by a lot. The hydrodynamic lift pressure on rocks or soil is proportional to flow speed-squared. Also, the more residence time and the more oxygen in the water, the more bio-remediation takes place in the drain. The dams slow the flow and promote oxygenation by the splashing over the weirs. Cells, bugs and fish do the rest; e.g. -HCOH- + O2 –> CO2 + H2O. Without oxygen, the fish die of suffocation, and this is a problem we’re already seeing in Lake St. Clair. Adding a dam saves the fish and turns the run into a living waterway instead of a smelly sewer. Of course, more is needed to take care of really major flood-rains. If all we provide is a weir, the water will rise far over the top, and the run will erode no better (or worse) than it did before. To reduce the speed during those major flood events, I would like to add a low bicycle path and some flood-zone picnic areas: just what you’d see on Michigan State’s campus, by the river.

Dr. Robert E. Buxbaum, May 12, 2016. I’d also like to daylight some rivers, and separate our storm and toilet sewage, but those are longer-term projects. Elect me water commissioner.

if everyone agrees, something is wrong

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I thought I’d try to semi-derive, and explain a remarkable mathematical paper that was published last month in The Proceedings of the Royal Society A (see full paper here). The paper demonstrates that too much agreement about a thing is counter-indicative of the thing being true. Unless an observation is blindingly obvious, near 100% agreement suggests there is a hidden flaw or conspiracy, perhaps unknown to the observers. This paper has broad application, but I thought the presentation was too confusing for most people to make use of, even those with a background in mathematics, science, or engineering. And the popular versions press versions didn’t even try to be useful. So here’s my shot:

Figure 2 from the original paper. For a method that is 80% accurate, you get your maximum reliability at the third to fifth witness. Beyond that, more agreement suggest a flaw in the people or procedure.

Figure 2 from the original paper. For a method that is 80% accurate, you get your maximum reliability at 3-5 witnesses. More agreement suggests a flaw in the people or procedure.

I will discuss only on specific application, the second one mentioned in the paper, crime (read the paper for others). Lets say there’s been a crime with several witnesses. The police line up a half-dozen, equal (?) suspects, and show them to the first witness. Lets say the first witness points to one of the suspects, the police will not arrest on this because they know that people correctly identify suspects only about 40% of the time, and incorrectly identify perhaps 10% (the say they don’t know or can’t remember the remaining 50% of time). The original paper includes the actual factions here; they’re similar. Since the witness pointed to someone, you already know he/she isn’t among the 50% who don’t know. But you don’t know if this witness is among the 40% who identify right or the 10% who identify wrong. Our confidence that this is the criminal is thus .4/(.4 +.1) = .8, or 80%.

Now you bring in the second witness. If this person identifies the same suspect, your confidence increases; to roughly (.4)2/(.42+.12) = .941,  or 94.1%. This is enough to make an arrest, but let’s say you have ten more witnesses, and all identify this same person. You might first think that this must be the guy with a confidence of (.4)10/(.410+.110) = 99.99999%, but then you wonder how unlikely it is to find ten people who identify correctly when, as we mentioned, each person has only a 40% chance. The chance of all ten witnesses identifying a suspect right is small: (.4)10 = .000104 or 0.01%. This fraction is smaller than the likelihood of having a crooked cop or a screw up the line-up (only one suspect had the right jacket, say). If crooked cops and systemic errors show up 1% of the time, and point to the correct fellow only 15% of these, we find that the chance of being right if ten out of ten agree is (0.0015 +(.4)10)/( .01+ .410+.110) = .16%. Total agreement on guilt suggests the fellow is innocent!

The graph above, the second in the paper, presents a generalization of the math I just presented: n identical tests of 80% accuracy and three different likelihoods of systemic failure. If this systemic failure rate is 1% and the chance of the error pointing right or wrong is 50/50, the chance of being right is P = (.005+ .4n)/(.01 +.4n+.1n), and is the red curve in the graph above. The authors find you get your maximum reliability when there are two to four agreeing witness.

Confidence of guilt as related to the number of judges that agree and your confidence in the integrity of the judges.

Confidence of guilt as related to the number of judges that agree and the integrity of the judges.

The Royal Society article went on to a approve of a feature of Jewish capital-punishment law. In Jewish law, capital cases are tried by 23 judges. To convict a super majority (13) must find guilty, but if all 23 judges agree on guilt the court pronounces innocent (see chart, or an anecdote about Justice Antonin Scalia). My suspicion, by the way, is that more than 1% of judges and police are crooked or inept, and that the same applies to scientific analysis of mental diseases like diagnosing ADHD or autism, and predictions about stocks or climate change. (Do 98% of scientists really agree independently?). Perhaps there are so many people in US prisons, because of excessive agreement and inaccurate witnesses, e.g Ruben Carter. I suspect the agreement on climate experts is a similar sham.

Robert Buxbaum, March 11, 2016. Here are some thoughts on how to do science right. Here is some climate data: can you spot a clear pattern of man-made change?

Marie de Condorcet and the tragedy of the GOP

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This is not Maire de Condorcet, it's his wife Sophie. Marie (less attractive) was executed by Robespierre for being a Republican.

Marie Jean is a man’s name. This is not he, but his wife, Sophie de Condorcet. Marie Jean was executed for being a Republican in Revolutionary France.

During the French Revolution, Marie Jean de Condorcet proposed a paradox with significant consequence for all elective democracies: It was far from clear, de Condorcet noted, that an election would choose the desired individual — the people’s choice — once three or more people could run. I’m sorry to say, this has played out often over the last century, usually to the detriment of the GOP, the US Republican party presidential choices.

The classic example of Condorcet’s paradox occurred in 1914. Two Republican candidates, William H. Taft and Theodore Roosevelt, faced off against a less-popular Democrat, Woodrow Wilson. Despite the electorate preferring either Republican to Wilson, the two Republicans split the GOP vote, and Wilson became president. It’s a tragedy, not because Wilson was a bad president, he wasn’t, but because the result was against the will of the people and entirely predictable given who was running (see my essay on tragedy and comedy).

The paradox appeared next fifty years later, in 1964. President, Democrat Lyndon B. Johnson (LBJ) was highly unpopular. The war in Vietnam was going poorly and our cities were in turmoil. Polls showed that Americans preferred any of several moderate Republicans over LBJ: Henry Cabot Lodge, Jr., George Romney, and Nelson Rockefeller. But no moderate could beat the others, and the GOP nominated its hard liner, Barry Goldwater. Barry was handily defeated by LBJ.

Then, in 1976; as before the incumbent, Gerald Ford, was disliked. Polls showed that Americans preferred Democrat Jimmy Carter over Ford, but preferred Ronald Regan over either. But Ford beat Reagan in the Republican primary, and the November election was as predictable as it was undesirable.

Voters prefer Bush to Clinton, and Clinton to Trump, but Republicans prefer Trump to Bush.

Voters prefer Bush to Clinton, and Clinton to Trump, but Republicans prefer Trump to Bush.

And now, in 2015, the GOP has Donald Trump as its leading candidate. Polls show that Trump would lose to Democrat Hillary Clinton in a 2 person election, but that America would elect any of several Republicans over Trump or Clinton. As before,  unless someone blinks, the GOP will pick Trump as their champion, and Trump will lose to Clinton in November.

At this point you might suppose that Condorcet’s paradox is only a problem when there are primaries. Sorry to say, this is not so. The problem shows up in all versions of elections, and in all versions of decision-making. Kenneth Arrow demonstrated that these unwelcome, undemocratic outcomes are unavoidable as long as there are more than two choices and you can’t pick “all of the above.” It’s one of the first great applications of high-level math to economics, and Arrow got the Nobel prize for it in 1972. A mathematical truth: elective democracy can never be structured to deliver the will of the people.

This problem also shows up in business situations, e.g. when a board of directors must choose a new location and there are 3 or more options, or when a board must choose to fund a few research projects out of many. As with presidential elections, the outcome always depends on the structure of the choice. It seems to me that some voting systems must be better than others — more immune to these problems, but I don’t know which is best, nor which are better than which. A thought I’ve had (that might be wrong) is that reelections and term limits help remove de Condorcet’s paradox by opening up the possibility of choosing “all of the above” over time. As a result, many applications of de Condorcet’s are wrong, I suspect. Terms and term-limits create a sort of rotating presidency, and that, within limits, seems to be a good thing.

Robert Buxbaum, September 20, 2015. I’ve analyzed the Iran deal, marriage vs a PhD, and (most importantly) mustaches in politics; Taft was the last of the mustached presidents. Roosevelt, the second to last.