Boussinesq and Newmark

April 25, 2024 at 1:31 PM by Dr. Drang

In a post this morning, John D. Cook listed several types of partial differential equation that are unusual in that they can be solved analytically. The last one in his list caught my eye because it’s one that I know pretty well from the theory of elasticity: Boussinesq’s equation.

Consider a linearly elastic body with a planar boundary that is infinite in extent below that boundary. Now apply a point load normal to the plane on that boundary:

Boussinesq problem from Malvern

Figure from Introduction to the Mechanics of a Continuous Medium by Lawrence Malvern

The solution for the vertical stress, $σ_{z}$ , at a depth $z$ and a horizontal distance $r$ from the point of application of load $P$ is

σ_{z} = - \frac{3 P}{2 π} \frac{z^{3}}{(r^{2} + z^{2})^{5 / 2}} = - \frac{3 P}{2 π} \frac{z^{3}}{R^{5}}

This solution was derived by Joseph Boussinesq in 1885. Geotechnical engineers later found this solution interesting because it gave them an opportunity to calculate subsurface pressures in soil due to the weight of a building founded on that soil. It isn’t perfect because soils aren’t linearly elastic, but it’s a reasonable approach to making estimates, which is what engineering is all about.

Now, the Boussinesq equation is for a point load, and buildings aren’t points—buildings distribute their weight over their foundation footprint. In theory, this isn’t a problem, because we can handle the distributed load by integration. In practice, it’s a pain in the ass, because the integration has to be done numerically for realistic building footprints and, furthermore, has to be done repeatedly for different depths and different locations under the building. A nonstarter in the pre-computer era.

In 1938, Nathan Newmark, then a young professor in the Department of Civil Engineeering at the University of Illinois (my alma mater), came up with a graphical solution to the integration problem. He devised an influence chart that looked like this:

Newmark influence chart

Figure from Engineering Experiment Station Bulletin No. 338

Here’s a more readable view of the scale and other information in the bottom right corner:

Influence chart scale information

To figure out the vertical pressure in the soil at a given depth, you take that depth to be the distance OQ shown above and draw an outline of the foundation plan to that scale on tracing paper. You then put the drawing on the influence chart and position it so that the horizontal position at which you want the soil pressure is at the center of the chart. For example, if our building plan is a fat ell shape, 50′×75′ with a 25′×25′ chunk taken out of the southwest corner, and we want to calculate the soil pressure under the inside corner at a depth of 25′, we take OQ to be 25′, draw an outline of the building to that scale, and place the drawing with the inside corner at the center of the concentric circles. Like this:

Newmark influence chart with foundation plan

Now we count the number of “squares” the building covers and multiply that by two values:

The average pressure of the building (its weight divided by the area of its footprint).
The chart’s influence value of 0.001, which is given in the lower right corner of the chart.

The result is the soil pressure 25′ below the inside corner. To get the pressure at other horizontal positions, we slide the drawing around and count the squares again. To get the pressures at another depth, we redraw the building footprint at a scale appropriate for that depth.¹

I’m not saying there isn’t a degree of tedium to this, but it’s a hell of a lot easier than doing multiple numerical integrations on paper with a slide rule and a desk calculator.

The soil pressure influence chart was hardly Newmark’s only contribution to engineering. He was a genius at coming up with alternate ways of seeing problems and devising solution techniques that could be used by everyday engineers working in design offices. Many of these techniques have been superseded by computer-based methods, but at least one—the beta method for structural dynamics problems—is still going strong.

Newmark had moved to emeritus status by the time I was a student; he died during my senior year. The department shut down for a day so the faculty could attend his funeral. Not only had he been their colleague, for many he had been their thesis advisor. This group included my advisor; academically, I am one of Newmark’s many grandchildren.

There’s also a pressure component due to the weight of the soil itself, but that’s just the unit weight of the soil multiplied by the depth. Add that to the value calculated via the influence chart to get the total vertical pressure in the soil. ↩

Doug McIlroy and Bing Copilot

April 23, 2024 at 10:03 AM by Dr. Drang

This morning, I read a short article describing certain deficiencies in Bing Copilot when it comes to doing math. The article interested me for two reasons:

First, I had run into ChatGPT-3’s trouble with arithmetic last year, and I was happy to see someone investigating other types of math error.
Second, the article was written by Doug McIlroy. Yes, that Doug McIlroy.

The article investigates three math problems: one having to do with time zones, one an elementary logic problem, and the third a simple calculus problem. At first, I thought the time zone problem wasn’t really a math problem, but I soon learned that it was. McIlroy asked Copilot for the minimum time difference between Oregon and Florida. Copilot knew that both states have two time zones, and it could regurgitate the definition of “minimum,” but it didn’t know how to put those two pieces of information together.

I’ll let you explore the logic problem on your own. Suffice it to say that you shouldn’t worry about beating Copilot in a game of rock-paper-scissors.

The calculus problem was the most complicated: given a general ellipse defined by the usual equation,

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

find the points on the ellipse where the magnitude of the slope is one.

McIlroy gave Copilot several opportunities to solve this problem. Although it always got the formula for the slope correct, it failed in different ways each time to apply that formula to the problem at hand.

The value of the article isn’t in simply pointing out that LLMs can be wrong. It’s in McIlroy’s detailed review of Copilot’s answers and where it went wrong in every step. By taking small problems and exploring the errors thoroughly, McIlroy does a better job of finding faults in large language models than any of the broad-brush criticisms I’ve read.

In the notes at the end of the article, McIlroy gives us its editing history. First written in December of last year; last edited just a few days ago. I should mention here that McIlroy will be celebrating his 92nd birthday tomorrow.

Typing special characters

April 21, 2024 at 12:01 AM by Dr. Drang

A couple of days ago, two tweets appeared in my Mastodon timeline¹ that got me thinking about how and why I type special characters on my Mac. The first was from Cabel Sasser,

… these are permanently hardwired into my brain: Option-2 for ™, Option-R for ®, Option-G for ©, Option-. for …, and Option-8 for •

and the second was a reply from David Friedman,

Great idea. Personally I use macOS text replacement: hhalf for ½, ccent for ¢, ttm for ™, ccopy for ©, aapple for , bbull for •, etc.

What struck me was that I use a mixture of these two techniques (OK, I use Typinator instead of the macOS’s text replacement system, but the idea is the same).

For example, like Cabel, I type the bullet (•) and ellipsis (…) characters directly; But like David, I use abbreviations (the Typinator term for text substitutions) for the cent (¢), trademark (™), and apple () symbols. Is there any logic to my system?

Certainly, some symbols that I use fairly frequently, like ⌘, ⌥, ⇧, and ⌃, have to be handled by abbreviations because they can’t be typed directly using the Option or Option-Shift keys.

Keyboard viewer with option key pressed

Keyboard viewer with option and shift keys pressed

But as you can see, the  and ¢ characters can be typed directly. So why don’t I? I guess it comes down to habit. Long, long ago—back in the 80s, before the Mac had text substitutions or abbreviation utilities—I learned how to type certain characters using Option and Option-Shift because I needed them and there was no other way to get them. So bullet (•), degree (°), ellipsis (…), the n- and m-dashes (– and —), and all the curly quotation marks (‘,’,“, and ”) became part of my muscle memory—muscle memory that was somehow retained even after I took an eight-year Mac hiatus in the late 90s and early 00s.

But I never needed ™, ©, or  enough to train my fingers how to type them. When I returned to the Mac and started using text expansion utilities,² these were among the special characters I created abbreviations for. Expansion made it unnecessary to learn new key combinations.

There is one special character I’ve recently learned to type directly: Δ. I use this character to delimit equations here on the blog. I wanted something that was easy to type (⌥J) and had a mathy feel to it. So when I want to show something like

{&Integral;}_{0}^{a} \sin x d x

I type

ΔΔ \int_0^a \sin x \, d x ΔΔ

For inline equations I use a single Δ.

I have a feeling many longtime Mac users are like me: some special characters are typed directly, some are done through expansion, and the rest—never used before and never expected to be used again—come through the Character Viewer.

Character viewer

Since Twitter is now X, I feel “tweet” has fallen into the public domain and can be applied to any shortform social media system. ↩
I’ve used TypeIt4Me, TextExpander (originally as Textpander), and now Typinator. ↩

Gravitational potential energy

April 16, 2024 at 5:28 PM by Dr. Drang

One of the things I planned to do upon retirement was study topics that I had missed, or nearly missed, in school. Some of these would be things that were barely touched on in class; some would be material that was in the textbook but not covered in class; and some would be topics that were adjacent to the classes I took. When I was just out of school, I thought I’d be able to start following up on these topics pretty much right away. And if you allow for an elastic definition of “pretty much right away,” I’m on schedule.

My first topic is orbital mechanics, and the textbook I’ve chosen is Victor Szebehely’s Adventures in Celestial Mechanics, published in 1989 by the University of Texas Press.

Adventures in Celestial Mechanics

As textbooks go, it’s fairly thin at 175 pages, and the math is pretty elementary—a good book to start with, I think. I’m seven chapters in and taking a little break to review what I’ve learned so far. I’m treating this as if I were in class and Szebehely was using the text as his lecture notes. In other words, I’m not just reading the book; I’m going through all the derivations and solving all the problems. If I wanted a shallow understanding, I’d just watch a few Neil deGrasse Tyson videos on YouTube and call it a day.¹

One of the things I found odd about Szebehely’s approach is that he never derives—or even presents—the expression for gravitational potential energy. I decided to do it myself and then write it out here, so I could find it easily.

In outline, the derivation works this way:

We start with Newton’s Law of Universal Gravitation to get the force between two bodies.
We then determine the work done by that force as the distance between the bodies changes.
Finally, we use the relationship between work and energy to get an expression for the gravitational potential energy.

The force of gravity between two bodies is

F = \frac{G m_{1} m_{2}}{r^{2}}

where $m_{1}$ and $m_{2}$ are the masses of the two bodies, $r$ is the distance between their mass centers, and $G$ is the universal gravitational constant,

G = 6.674 \times 10^{- 11} \frac{m^{3}}{kg s^{2}}

This force acts on each body in the direction of the other body, so the work done by gravity as the distance between the bodies changes from $r_{1}$ to $r_{2}$ is²

W = {&Integral;}_{r_{1}}^{r_{2}} - F d r = {&Integral;}_{r_{1}}^{r_{2}} - \frac{G m_{1} m_{2}}{r^{2}} d r

The minus sign is there because the force acts in the opposite direction of the distance increment, $d r$ .

Carrying out the integration, we get

W = \frac{G m_{1} m_{2}}{r_{2}} - \frac{G m_{1} m_{2}}{r_{1}}

The work and energy principle says

W = - Δ U

where $Δ U$ is the change in potential energy, i.e.,

Δ U = U (r_{2}) - U (r_{1})

The negative sign in the work-energy equation comes from what potential energy is all about: the potential to do work. If a force does positive work, its potential to do further work has decreased.

Putting this together, we get

Δ U = U (r_{2}) - U (r_{1}) = - W = - (\frac{G m_{1} m_{2}}{r_{2}} - \frac{G m_{1} m_{2}}{r_{1}})

Sorting out the negative signs, we see that

U (r_{2}) = - \frac{G m_{1} m_{2}}{r_{2}}

and

U (r_{1}) = - \frac{G m_{1} m_{2}}{r_{1}}

so our expression for gravitational potential energy at any distance, $r$ , is

U = - \frac{G m_{1} m_{2}}{r}

If you’re worried about the negative sign in this expression, don’t be. Recall from your first course in physics that it’s the change in potential energy that matters, not the value at any given position. The key thing to note about this expression is that $U$ increases as $r$ increases, which is how it should be.

Well, that didn’t take too long. Let’s expand on this solution. You might also remember from physics class that the potential energy of a body of mass $m$ near the surface of the Earth is

U = m g h

where $h$ is the elevation of the body and $g$ is the acceleration due to gravity. How does this expression compare to what we just derived?

First, let’s rewrite our expression for $U$ to specialize it for a body near the surface of the Earth:

U = - \frac{G m_{E} m}{r}

where $m_{E}$ is the mass of the Earth.

Because the body is close to the surface of the Earth, we’ll expand our expression for $U$ in a Taylor series about $r_{E}$ , the radius of the Earth. That will be

U (r) = U (r_{E}) + U^{'} (r_{E}) (r - r_{E}) + \frac{1}{2!} U^{''} (r_{E}) (r - r_{E})^{2} + \dots

The further terms in the expansion will have higher derivatives of $U$ , higher powers of $r - r_{E}$ , and be divided by higher factorials.

Since

U^{'} (r) = \frac{G m_{E} m}{r^{2}}

U^{''} (r) = - \frac{2 G m_{E} m}{r^{3}}

U^{'''} (r) = \frac{6 G m_{E} m}{r^{4}}

and so on, the expansion works out to

U (r) = - \frac{G m_{E} m}{r_{E}} + \frac{G m_{E} m}{r_{E}^{2}} (r - r_{E}) - \frac{G m_{E} m}{r_{E}^{3}} (r - r_{E})^{2} + \dots

where the further terms will be multiplied by higher powers of $r - r_{E}$ and divided by higher powers of $r_{E}$ . Note that the factorials get canceled by the coefficients of the derivatives. Also, we get alternating signs because of the alternating signs of the derivatives.

At this point, we recognize that $r - r_{E}$ is the distance above the Earth’s surface, which is $h$ . And because we care only about the change in potential energy with changes in $h$ , we can drop the constant initial term. So if we collect the common terms, we get

U = \frac{G m_{E} m}{r_{E}^{2}} h (1 - \frac{h}{r_{E}} + {(\frac{h}{r_{E}})}^{2} - \dots)

Since $h &ll; r_{E}$ , the term in parentheses is approximately one, and

U = \frac{G m_{E} m}{r_{E}^{2}} h = m (\frac{G m_{E}}{r_{E}^{2}}) h

Is the term in the parentheses equal to $g$ ? According to NASA

m_{E} = 5.9722 \times 10^{24} kg

and

r_{E} = 6.371 \times 10^{6} m

Plugging in those values and the value of $G$ we saw earlier, we get

\frac{G m_{E}}{r_{E}^{2}} = 9.82 m / s^{2}

which is pretty close to the textbook value for the acceleration due to gravity, $g = 9.81 m / s^{2}$ .

But shouldn’t it be closer? We have enough digits in our values for $G$ , $m_{E}$ , and $r_{E}$ that we shouldn’t be off in the third digit of $g$ . What are we missing? Mainly that the Earth is spinning. The centrifugal effect on the surface of the Earth is small, but it’s the primary reason for the difference between $G m_{E} / r_{E}^{2}$ and $g$ .

There’s a place for Neil deGrasse Tyson videos, but that’s not where I want to go. ↩
Strictly speaking, the integrand of the work expression is the dot product of the force vector and the distance increment vector, but for gravity those two vectors are parallel, so we can get by with scalar terms. ↩

And now it’s all this

I just said what I said and it was wrong
Or was taken wrong

Boussinesq and Newmark

Doug McIlroy and Bing Copilot

Typing special characters

Gravitational potential energy

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And now it’s all this

I just said what I said and it was wrong Or was taken wrong

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Gravitational potential energy

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I just said what I said and it was wrong
Or was taken wrong