Archive for the ‘mechanics’ Category

The Toyota pedal problem

February 3rd, 2010 at 9:59 pm

I’m not a big fan of the Chicago Tribune,1 but I have to say it did a pretty good job yesterday explaining the Toyota pedal recall that’s been all over the news lately. The article is only so-so, but the graphic that accompanies it, by Phil Geib and the improbably-named Max Rust, answered many of the questions I had.

[Click on the graphic to get a slightly larger version.]

Comparing the drawings to a photo of an actual pedal (on my 2007 Camry), we see that the parts in question are all inside a plastic housing near the pedal’s pivot.

I couldn’t shoot a photo with the same point of view as the drawing because its a little cramped down there, but I think you get a sense of where everything is. In addition to the gas pedal, you can see the brake pedal (lower left foreground) and part of the steering mechanism (the shaft with the yellow-orange stripe at the upper left corner).

The gas pedal is a “drive-by-wire” system. There is no mechanical connection between the pedal and the engine; the pedal sends electrical signals that describe its position to a controller which runs the fuel injectors. (In my photo you can see the wires coming up off the top of the pivot housing and leading into a black corrugated conduit.) This is very different from the way cars used to work, and to give the pedal the traditional “feel,” Toyota has incorporated two sets of plastic teeth that rub against one another and provide some of the resistance felt by the driver’s foot.

Additional resistance comes from the pedal’s return spring, which, for clarity, isn’t included in the drawing. The return spring is what, under normal circumstances, pushes the pedal back up when you take your foot off the gas.

The toothed portions of the blue and tan parts slide on one another, the blue part rotating with the pedal and the tan part remaining stationary. The friction between the teeth is what generates the resistive force. Apparently, moisture on these plastic parts can increase the friction between them, and if the friction gets large enough, the return spring can’t overcome it and the pedal won’t return to the up position when you remove your foot. You may have heard this called “sudden acceleration” in news reports, which gives the impression of the gas pedal moving down on its own. “Stuck throttle” or “stuck accelerator” would be a more accurate name for the problem.

[Aside: Friction is an interesting force because it always opposes motion. When you press down on the gas, the friction and the return spring are acting in concert, both pushing up to resist the downward motion. When you take your foot off, the return spring continues to push up, but the friction suddenly changes direction to push down, resisting the spring.]

Toyota’s page for the pedal recall mentions wear of the plastic as another factor in the sticking pedal problem, so it’s not clear whether the problem is caused by an increased coefficient of friction, surface damage from galling, or some combination of both. Regardless, Toyota’s solution is to change the relative positions of the plastic teeth and reduce the friction.

Dealers will be installing what Toyota is calling a “reinforcement bar” (but should really be called a “spacer” or “shim” because it isn’t changing the strength of the parts, it’s changing their position) behind the tan part. The Tribune drawing does not do a good job of explaining how that shim is going to reduce the engagement of the two sets of teeth, but it seems clear that that’s the intent.

One thing I still don’t understand, and something I’ve not heard Toyota address publicly: How will the new, reduced friction affect the feel of the gas pedal? Presumably, the pedal resistance I get now will be lessened after I take my car in for the recall fix. Will it feel too floppy? If not, why did Toyota have the higher resistance to begin with?

Frankly, I’m curious why a friction mechanism was chosen in the first place. Toyota obviously wanted to get a particular relationship between the resistive force and the pedal deflection, but I don’t see why that couldn’t be achieved with a spring package of some sort. Mechanical engineers have been designing spring mechanisms for a long time and can get all kinds of force-deflection relationships.2

Maybe I’m just out of date and friction mechanisms like this are common in brake systems nowadays. If you know what other manufacturers do, I’d like to hear about it in the comments.

Update 2/4/10
I’ve known for a few days that there are “good” pedals and “bad” pedals, but it wasn’t until this morning that I learned who their manufacturers are. The good pedals, not subject to the recall, are made by Denso, and the the bad pedals, which are going to have the shim fix described above, are made by CTS.

It turns out that my pedal is a Denso. The Denso name is molded into the pivot housing, something you can’t see in my photo. Also, the housings of the two designs look very different; the CTS doesn’t have that circular area with radial spines sticking out of the side. You can see photos of the two designs at this site. Note: I am not endorsing anything said on that site, as I have not read through it in detail. I’m just linking to it as a source of photos showing the difference between the two types of pedal. It also has some nice photos taken with the housings opened.

Update 2/24/10
I’d like to say something about this week’s Congressional hearings into Toyota’s problems, especially since much of the testimony covered other explanations—mainly electronic—for sudden acceleration. But I don’t have a copy of the primary engineering report referenced in the testimony, so I don’t feel comfortable commenting yet. I was hoping the House subcommittee would post a PDF of the report, but it hasn’t so far. If you know where I can download a copy, leave a URL in the comments or send me an email directly.

  1. What’s wrong with the Trib? To start: its editorial position, its choice of columnists on the op-ed page, its placement of John Kass in the old Royko spot, and its many years of owning the Cubs. ↩

  2. If your experience with springs is limited to what you learned in physics class, you may think that all springs are linear. Ut tensio sic vis, and all that. But if you expand your concept of “spring” beyond helical coils of uniform wire many more things are possible. ↩

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Post #1166 in mechanics, media | 5 Comments

Meatballs and arches

November 28th, 2009 at 11:59 pm

My sons and I saw Cloudy with a Chance of Meatballs last night at a local second-run theater and enjoyed it thoroughly. Like all kids’ movies, it had Lessons For Us All (I think that’s a legal requirement), but it managed to keep them pretty well confined. And any movie able to use both James Caan, Mr. T, and Al Roker effectively is worthy of support. But I did have one problem with the film.

The story is about Flint Lockwood, a failed inventor, and the people who live with him on a little island in the Atlantic. The island is a dreary place where everyone lives on a diet of sardines. After years of screwups, Flint finally comes up with a device that works: a machine that alters the genetic components of water to create any kind of food. The machine gets shot up into the sky, where it gathers water from the clouds and rains food down onto the island. This starts out wonderful, but then turns dangerous.

You might raise an eyebrow at the notion of water having genes, but that’s an issue for a biochemist. I’m a civil engineer, and what horrified me was the brief view we get of the dam on the island.

The dam is built to hold back the excess food. Much more food falls from the sky than can be eaten, so Flint makes a machine that scoops up the leftovers and flings them across the island into a sort of garbage reservoir retained by a dam. The dam seems to be modeled on Hoover Dam, with a nice horizontal arch.

(Image adapted from snakefisch at Wikipedia.)

Arches are used in dams for the same reason they’re used in bridges and buildings: their geometry allows you to span large distances without generating tension in the structure. This is important when the structure is made of masonry or concrete, materials reasonably strong in compression but notoriously weak in tension. To work, an arch has to be curved into the load; upward in bridges and buildings to resist downward gravity loads, and into the water to resist sideways pressure in dams. You see how Hoover Dam is curved into Lake Mead.

And that’s what horrified me about the dam in Meatballs: it’s curved the wrong way!1 Being curved away from the load puts the dam entirely in tension. No wonder it—spoiler alert!—bursts and floods the town with leftovers during the climax. It should never have lasted as long as it did.

Normally I’d keep nerdy thoughts like these to myself, but releasing your inner nerd is one of Meatballs’ Lessons For Us All. I eagerly await a meteorologist’s comments on that clockwise-spinning spaghetti tornado.

Tags:

  1. I’ve scoured the Internets but can’t find an image of the Meatballs dam. Trust me, it’s curved the wrong way. ↩

Post #1120 in entertainment, mechanics | No Comments

A little background on the Bay Bridge failure

October 29th, 2009 at 6:17 pm

You probably heard about the structural failure on the San Francisco/Oakland Bay Bridge earlier this week. When I went looking for information on the nature of the failure, the news articles I found were disappointing. It’s not that the reporters did a poor job, its just that their focus—which was, quite rightly, on how the failure would affect their local readers and viewers—didn’t match mine. I wanted to know exactly what failed, and although that information was in the stories, I had to tease it out of the photos and certain tidbits in the text. This is what I found.

The failure occurred in the eastern portion of the bridge, between Oakland and Yerba Buena Island. This section is in the process of being upgraded; on Labor Day weekend, workers found a crack in a steel member of the bridge, and temporary reinforcement was added to that member.

The eastern section has a truss structure, and some of its members are eyebars, which are used to carry tension loads. It was one of these eyebars that was (and apparently still is) cracked.

Eyebars are flat plates cut into a sort of dogbone shape with a hole at each end.

Two or more of these plates are arranged parallel to each other to make a complete eyebar assembly (the word eyebar can refer to either a single plate or an assembly). In the Bay Bridge, the eyebar assemblies consist of four plates, as you can see in this photo taken from the side (the photos I’ll be using here are from this San Francisco Chronicle set).

This photo might be a bit confusing because it shows two eyebar assemblies, one in the foreground and one in the background. Ignore the background assembly and note that foreground assembly has four eyebar plates: two closely spaced plates on the left side and two closely spaced plates on the right.

Eyebars are connected to each other and to other parts of the structure via large steel cylinders, called pins, that are driven through the holes in the ends. The reason the eyebars are widened at the ends is to make up for the material lost by the holes.

(Take another look at the photo above. See the dark triangle in the innermost right eyebar plate under the pin? I can’t decide whether that’s the crack that led to the repair or just an oddly-shaped discoloration on the surface of the steel. If it’s the crack, it’s a nasty one—running all the way from the outer edge to the hole.)

Update 10/31/09
The fourth photo on this page confirms that what we’re seeing in the photo above is the crack that led to the repair, not just a rust stain. Scary.

If you’re thinking that an eyebar is like a giant, highly elongated link in a bicycle chain, you’re thinking right.

(Image from bike-parts-direct.co.uk.)

The reinforcement consisted of a set of long steel rods and a pair of anchorages, called saddles. The saddles are attached to the structure just beyond the ends of the cracked eyebar, and the rods are anchored in the saddles.

The rods were meant to reduce the load carried by the eyebar itself. Installing reinforcement like this in an existing structure can be tricky. If you don’t tighten the rods enough, they won’t reduce the load on the eyebar and will be ineffective. If you overtighten the rods, you can cause a premature failure in them or in a saddle or in the structure to which the saddle is attached. Overtightening can also distort the original structure.

According to the Chronicle article, two rods (out of what looks like four) and part of a saddle broke and fell onto the roadway. Don’t assume this means that the rods were overtightened; failures can come from lots of sources, and the current thinking is that vibration-induced fatigue—or possibly abrasion; the article mentions rubbing—led to the failure of the rods. Fatigue is failure due to repeated loading and unloading of a part. The classic example (mentioned by the Caltrans chief engineer in his explanation) is that of a paper clip being bent back and forth until it breaks.

If fatigue is the cause of the failure, the stresses in the rods must have been very high for them to have failed in less than two months. Hard to imagine a repair with such high stresses was allowed.

Post #1081 in mechanics, news | No Comments

An application of Castigliano’s Second Theorem with Octave

October 24th, 2009 at 5:53 pm

This week I did an analysis of a set of leaf springs and found myself using a favorite technique that I seldom get to apply anymore: Castigliano’s Second Theorem. There is some real, though not especially difficult, math ahead, so if you come here for the little Mac-based scripts I typically write you may want to skip this one. If you do want to follow along, I suggest you visit this page and download one of the TeX font packages there. This page, and the other pages in which I use Davide Cervone’s wonderful jsMath system, will load faster and print better if you have those fonts on your computer.

First, let’s talk about leaf springs. Leaf springs are long, flat steel strips bundled together to provide part of the suspension system of heavy vehicles like trucks. They sit between the axle and the frame and smooth out the ride by flexing. You can think of them as allowing the body of the vehicle to bounce above the axle (roll your mouse over the images to see the springs flex),

or you can think of them as allowing the axle to bounce up toward the frame as the vehicle goes over bumps in the road.

In fact, the usual behavior is a combination of the two, but since we will be investigating the spring stiffness—a static property—the analysis is the same no matter which frame of reference you choose.

In some leaf spring packages, the individual leaves are curved to nestle against one another; in others, like the kind I dealt with this week, the leaves are curved to meet only at the center and at the tips, as in the simplified schematic drawings above.

To determine the stiffness of a spring package like that in the drawing, each leaf can be treated as a simply-supported beam with a central point load, and the stiffnesses of the package is the sum of the stiffnesses of the individual leaves. Because the initial curvatures of the leaves are relatively small (the ratio of leaf thickness to initial radius of curvature is much less than one), we can treat them as straight beams with no significant error.

The deflection under the load of a simply-supported beam of uniform thickness and width loaded at the center is

\Delta = \frac{F L^3}{48EI}

where E is the modulus of elasticity of the steel (29,000,000 psi in US Customary units), and I is the area moment of inertia of the cross section,

I = \frac{b t^3}{12}

for a rectangle of width b and thickness t. The equivalent spring stiffness, k, is the ratio of load to deflection, so

k = \frac{F}{\Delta} = \frac{48EI}{L^3} = \frac{4Ebt^3}{L^3}

Repeat this calculation for each leaf, add them up, and you get the equivalent spring stiffness for the package. Very simple if you have leaves of uniform thickness.

Unfortunately, the leaves I was analyzing were tapered, about twice as thick in the center as out at the tips. It’s because there is no simple formula for a tapered beam that I went to Castigliano’s Second Theorem.

Carlo Castigliano was a 19th-century Italian mathematician/physicist/engineer who studied at the Polytechnic of Turin. His dissertation presented the two theorems that bear his name. The First Theorem is interesting but doesn’t have a lot of practical application. Castigliano’s Second, though, is very helpful in solving many interesting real-world problems in structural and mechanical engineering. Including, as I said at the top, the problem of the tapered beam.

Castigliano’s Second Theorem states that the derivative of the complementary strain energy of a body with respect to a point load acting on that body is equal to the deflection, in the direction of the load, of the point on the body to which the load is applied. In mathematical terms, this is

\frac{\partial U^*}{\partial F} = \Delta

A trivial demonstration of this can be found in the behavior of a simple linear spring. The complementary strain energy in the spring, U^* is the area above the usual force-deflection diagram, written in terms of the applied force and the spring stiffness. Recall that the regular strain energy, U, is the area below the force-deflection diagram, and is written in terms of the deflection and the spring stiffness.

According to Castigliano’s Second, the deflection of the spring is

\Delta = \frac{\partial U^*}{\partial F} = \frac{\partial}{\partial F}\left(\frac{F^2}{2k}\right) = \frac{F}{k}

which is exactly what we expect. The genius of Castigliano was his generalization of this trivial case to any type of structure.

Update 10/25/09
I meant to add here that the strain energy and the complementary energy are equal for a linear elastic spring, but would not be equal for a nonlinear spring.

To apply Castigliano’s Second to our tapered beam problem, we’re going to first take advantage of the symmetry of the beam and analyze just the left half. The right half will behave as a mirror image of the left, so once we’ve analyzed the left half, we have the solution for the right half.

The left half acts like a cantilevered beam, as shown in the diagram below. The complementary strain energy of a beam in bending is governed by this equation:

U^* = \int_0^L \frac{M^2}{2EI} dx

where M is the bending moment in the beam due to the applied force. Both M and I are functions of x,

M = Fx
I = \frac{b (t_0 + \alpha x)^3}{12}

where \alpha = (t_1 - t_0)/L is the rate at which the thickness increases as we move from the tip to the wall. Thus,

U^* = \int_0^L \frac{F^2 x^2}{2 E \frac{b (t_0 + \alpha x)^3}{12}} = \frac{6 F^2}{E b} \int_0^L \frac{x^2}{(t_0 + \alpha x)^3} dx

Putting this into the Castigliano equation, we get

\Delta = \frac{\partial U^*}{\partial F} = \frac{12F}{Eb} \int_0^L \frac{x^2}{(t_0 + \alpha x)^3} dx

Back when I was a student, we didn’t have all these fancy personal computers, so we’d have to do that integral analytically, probably through integration by parts. Now I can’t be bothered. I just fire up Octave, enter the values, and let it do the integration numerically. Here’s my Octave session for a particular set of tapered beam dimensions (I used Octave’s diary command to save the session as a text file):

octave-3.2.3:2> global E=29e6 b=3 t0=.5 t1=1 L=25;
octave-3.2.3:3> function retval = A(x)
> global t0 t1 L;
> retval = x^2/(t0 + (t1-t0)/L*x)^3;
> endfunction
octave-3.2.3:4> [int,icode,nfun,err] = quad("A",0,L)
int =  8518.4
icode = 0
nfun =  21
err =  9.4573e-11
octave-3.2.3:5> k = E*b/12/int
k =  851.10

So the stiffness of one-half of a leaf is 851 lb/in, which makes the stiffness of the full leaf 1702 lb/in. Compare this to

\frac{4\cdot29,000,000\cdot3\cdot1^3}{50^3} = 2785\: \textrm{lb/in}

for a 1″ flat leaf, and

\frac{4\cdot29,000,000\cdot3\cdot0.5^3}{50^3} = 348\: \textrm{lb/in}

for a 0.5″ flat leaf. How close would we have come if we’d used the flat-leaf formula with the average thickness? We’d get

\frac{4\cdot29,000,000\cdot3\cdot0.75^3}{50^3} = 1174.5\: \textrm{lb/in}

which underestimates the stiffness by more than 30%. Another way to think of this is that by tapering the beam from 1″ to ½″ we get 30% more stiffness for the same amount of steel as a flat beam of ¾″.

For the purposes of this post, I made the taper linear, which allowed me to use Octave’s quad function for integration. In fact, the leaf thickness tends to vary in a more complicated way, so the more general way to solve the problem is to

  1. Express the thickness as a vector of values uniformly-spaced along the length of the spring.
  2. Use Octave’s many matrix/vector operators to generate a vector of the integrand.
  3. Use the trapz function to perform the integration.

It takes a bit longer to do it this way because there are more thickness values to enter, but conceptually it’s the same problem.

The tapered leaf spring problem is tailor-made for Castigliano’s Second Theorem.

When you run into a problem with these characteristics, it’s good to have Castigliano in your toolbox.

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Post #1073 in math, mechanics, programming | No Comments

Bike light failure

June 21st, 2009 at 7:00 pm

Last year I bought a Blackburn Voyager 4.0 light for my bicycle. While I’ve always liked the amount of light it put out, the way it mounts to my handlebar has been a source of concern and frustration. Today the body of the light broke off the mounting hardware, due in large part to the design of the light. I won’t be replacing it with another Blackburn.

Here’s the Voyager 4.0 as depicted on Blackburn’s web site.

The light comes in two large pieces. The body of the light is a tapered cylinder that weighs 8-9 ounces. On the bottom of the body is a slot that connects it to the mounting piece.

(Don’t worry about the fracture just yet; we’ll talk about it later on.)

The mounting piece has a clamp that grips the handlebar and a pair of flexible arms that squeeze inward as they slide into the slot of the body. The arms spring out and a barb on each arm engages with a detent inside the slot when the two pieces are slid together fully.

There are two problems with this design. One problem became evident on the very first ride I took after installing the light. There’s no popping sound as the arms spring outward, nor is there any significant clicking feel when you hit the detent. There is, however, a decent amount of friction between the slot and the arms; the resistance to further sliding can lead you to believe you’ve got full engagement when you haven’t. That’s what happened to me when I installed the light and went on my first ride with it. After a few decent bumps, the body slid off the mounting hardware and fell to the pavement.

Fortunately, the light still worked, and from that point on I tried to ensure that the arms had sprung back out every time I reinstalled the light. Despite my care, it did pop off one other time, earlier this year.

The second problem with the design is what caused today’s failure. The plastic along the edges of the slot is simply too thin and weak to support the mass of the body. I had my bike parked on grass next to a sidewalk as I relubed the chain. The ground around here is soft from all the rain we’ve had this spring, so the kickstand sunk and the bike fell over to the left onto the sidewalk. The plastic along one edge of the slot broke and the body of the light popped off.

Here’s a more detailed view of the broken area.

The whitened area is common in plastic fractures. It’s known as “stress whitening”1 and is due to subsurface micro-cracking at high strains.

You might be wondering if the fracture occurred when the light body hit the sidewalk. If that were the case, my complaints about the design would be unfair to Blackburn, as that would be a pretty significant impact. I can assure you that’s not how it happened. Here’s a front view of the two parts of the light, oriented as they would have been as the body broke away from the mounting hardware.

For the fracture to be where it is, on the upper edge of the slot, the light body must have been rotating as shown in the picture—clockwise as viewed from the front. Had the body of the light hit the sidewalk, the rotation would have been in the opposite direction and the edge of the other side of the slot would have fractured. (The geometry of the handlebar and the light’s position on it also make it very unlikely that the light would strike the sidewalk while still mounted.)

So the force that cracked the edge of the slot was the inertial force of the light body itself. The bike and the mounting piece suddenly stopped when the bike hit the sidewalk, but the light body wanted to keep going. The plastic at the edge of the slot wasn’t strong enough to support the mass above it during this fairly common occurrence.

Time to start looking for another bike light. This time I’ll pay as much attention to the structural design as I do to the light output.

Tags:

  1. Engineers are not known for clever nomenclature or imaginative flights of fancy. ↩

Post #823 in biking, mechanics | No Comments

Swingin’

April 12th, 2009 at 9:18 pm

A couple of days ago I saw this poster of time traveler hints via a tweet by @gnomedad. It’s cute, but

Anyway, there was one thing on the poster I found interesting.

The exact speed of light in a vacuum is 299,792,458 meters per second. Good to know. A meter is defined in terms of light, but if you can’t measure it accurately, the length of a pendulum that takes one second to swing from end to end will do the trick.

Let’s ignore the odd notion that you’re going to be able to use the speed of light to lay out a line one meter long (no problem, just invent a clock accurate to 10^{-9} seconds), and focus on that thing about the pendulum. Why would a one-meter pendulum take exactly one second to swing from side to side? I’ve analyzed the behavior of pendulums as both a student and a teacher, and I don’t ever remember running across that fact. But it’s fairly easy to derive from first principles, and since I’m not teaching anymore, I’ll inflict the derivation on you.

Here’s a simple pendulum consisting of a small ball of mass m hanging from a pivot point by a string of length L.

The ball is so small (compared to the length of the string) that we don’t have to account for its radius, and the string is so light (compared to the ball) that we don’t have to account for its mass. We’ll call the angle of the string from the vertical \theta and construct the equation of motion through Euler’s second law, a rotational analogue of Newton’s second law2:

\sum M_O = I_O\:\ddot{\theta}

\sum M_O is the sum of all the moments about the pivot point, O; I_O is the mass moment of inertia of the pendulum about the pivot point; and \ddot{\theta} is the angular acceleration of the pendulum, the second time derivative of the position angle \theta. Since we are taking the entire mass of the pendulum to be m concentrated at a distance L from the pivot,

I_O = m L^2

The reaction force at the pivot doesn’t generate a moment about the pivot because its lever arm is zero. Therefore, the only force that contributes to \sum M_O is the weight of the pendulum, mg, where g is the acceleration due to gravity. Its lever arm is L \sin \theta, so

-m g L \sin \theta = m L^2 \ddot{\theta}

where the negative sign indicates that the moment is in the opposite direction of \theta. Canceling the common terms and rearranging, we get

\ddot{\theta} + \frac{g}{L} \sin \theta = 0

This is a rather nasty differential equation because of the sine term. If we assume the swing angle stays relatively small throughout the motion, we can make the further simplification, \sin \theta \approx \theta. How good is this approximation? It’s easy to make up a little table and check.

θ (deg) θ (rad) sin(θ) error
0.00 0.0000 0.0000
1.00 0.0175 0.0175 0.01%
2.00 0.0349 0.0349 0.02%
3.00 0.0524 0.0523 0.05%
4.00 0.0698 0.0698 0.08%
5.00 0.0873 0.0872 0.13%
6.00 0.1047 0.1045 0.18%
7.00 0.1222 0.1219 0.25%
8.00 0.1396 0.1392 0.33%
9.00 0.1571 0.1564 0.41%
10.00 0.1745 0.1736 0.51%
11.00 0.1920 0.1908 0.62%
12.00 0.2094 0.2079 0.73%
13.00 0.2269 0.2250 0.86%
14.00 0.2443 0.2419 1.00%
15.00 0.2618 0.2588 1.15%

So the error is less than 1% if the pendulum doesn’t swing more than 14° from vertical and less than 0.5% if it swings no more than 10° from vertical. Not bad.

The small angle approximation leads to

\ddot{\theta} + \frac{g}{L} \theta = 0

This is a second order linear differential equation with constant coefficients, which is, as these things go, quite easy to solve. The solution will have the form

\theta = A \cos\omega t + B \sin \omega t

where \omega = \sqrt{g/L} is the the circular frequency of the oscillation, and A and B depend on the initial conditions, the angle and angular velocity of the pendulum at time zero. For a pendulum pulled to the side and released from rest, the solution takes the form

\theta = A \cos \omega t

with A being the amplitude of the oscillation, the largest value of the swing angle. The periodic motion of the pendulum—left to right to left to right, and so on—is reflected in the periodic nature of the cosine function.

In our solution, the pendulum never slows down or stops because we haven’t modeled air resistance or friction at the pivot.

The period of the pendulum—one back-and-forth cycle—is

\frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{L}{g}}

So the time it takes to swing from one side to the other is half that:

\pi \sqrt{\frac{L}{g}} = \sqrt{\frac{\pi^2 L}{g}}

In SI units, g = 9.81\:\mathrm{m/s^2}. For a pendulum one meter long, the time to swing from one side to the other is

\sqrt{\frac{\pi^2 \cdot 1 \mathrm{m}}{9.81\:\mathrm{m/s^2}}} = 1.003\:\mathrm{s}

I think it’s fair to say that we can ignore the 3 millisecond difference and declare that a 1-meter pendulum has a half-period of 1 second.

This nice result comes about because the gravitational acceleration on the surface of the Earth in m/s^2 is just about equal to \pi^2. In fact, there was a time when the meter was going to be defined in terms of this pendulum length, but that definition lost out to the meridonal definition.

Pi is a universal constant because it’s based on mathematics alone, but g isn’t; so if your time machine also transports you to another planet, don’t expect this trick to work.

  1. Yes, I know it’s just for fun, but much of the fun of science fiction lies in being as realistic as possible after accepting one or two unrealistic premises. ↩

  2. As always, the equations in this post are formatted by jsMath, a wonderful JavaScript library for LaTeX rendering written by Davide Cervone. I prefer jsMath to ASCIIMathML because it works in all browsers, not just those with MathML support. You can fiddle with the jsMath settings by clicking the little jsMath button that should be in the lower right corner of your window. To get the most out of any browser-rendered math—jsMath or ASCIIMath—you should have a set of TeX fonts installed on your computer.

    If you’re reading this in an RSS reader, the equations won’t render and you’ll just see the raw LaTeX. I’m looking into fixing that. ↩

Post #724 in math, mechanics | No Comments

Mechanics solution manual

January 22nd, 2009 at 10:45 pm

A few years ago I had a second blog, the goal of which was to provide a solution to every problem in J.P. Den Hartog’s Mechanics textbook. I stopped updating the blog after about six months, for reasons I’ll discuss in a bit, and eliminated it entirely a year ago when I switched web hosts and blogging platform. Over the weekend, I resurrected it from the backup and turned it into a good old static website with a home page at http://www.leancrew.com/mechanics/.

Mechanics is an undergraduate textbook in statics and dynamics, pitched primarily at sophomores and juniors in mechanical and civil engineering. It was a fairly popular book in its day, which was 60 years ago, but I don’t think it’s being used as a classroom text any more. It does, however, have a wonderful set of problems, and because it’s a Dover reprint, it’s quite inexpensive. If I’m more diligent this time around, the finished web site will be a good resource for engineering students—they’ll be able to see how an expert solves these types of problems without being given direct answers to problems in the texts they’re using in class.

Expert? Yes. The stuff I tend to do on this blog (with one or two exceptions) is hobby stuff, not what I do professionally. I got my Ph.D. in civil engineering from the top-ranked civil department in the US and later taught mechanical engineering—including statics and dynamic—at a fairly well known university just north of Chicago. One of the things I miss most from my teaching days is working out the solutions to problems like those in Mechanics.

So why did the blog peter out in the first place? I ran into two roadblocks.

The first roadblock was pedagogical. When I hit Chapter 4, which starts off with truss problems, I had trouble deciding how to approach the problems. There are two methods of solving truss problems by hand: the method of joints and the method of sections. Den Hartog, like most authors, assigns some problems to be done by one method and some to be done by the other. My first inclination was to simply do the problems using the method he assigned, but I immediately had doubts. Some problems are best done by joints, some by sections, and some by a mixture of the two methods. Wouldn’t it be better to choose the more efficient technique and let the readers in on why I made the choice? As I held up posting until making this decision, I realized that another roadblock awaited.

The second roadblock was practical. Solving a truss problem usually means making several drawings of different parts of the truss. This goes pretty quickly when you’re making the drawings by hand, but I didn’t want scans of hand-drawn figures. I wanted crisp, clean, computer-generated figures, like I’d drawn for the earlier problems. I’d drawn the earlier figures in Illustrator and exported them as PNGs; I was happy with their quality, but not so happy with the speed at which I produced them. The prospect of needing four or more figures per problem instead of just one or two stopped me cold.

Now I think I’ve got a way around both roadblocks. I’ve decided to do the problems the way I think is best, regardless of how Den Hartog assigned them. I’m pretty sure I’ll be repeating this decision when I get to rigid body kinematics later in the book; Den Hartog doesn’t use vector notation because it wasn’t the style back then, but I definitely will.

As for the drawing roadblock, I now have a copy of OmniGraffle, and ever since I bought it I’ve been thinking that it’s ability to link elements together is perfect for drawing trusses quickly and accurately. I’ll soon learn if I was right.

Switching to topics more common on this blog: The soutions are written in a customized Markdown format that includes LaTeX-style equations via jsMath. The HTML pages are generated by a few Python scripts and a couple of Makefiles. The overall approach was adapted from my no-sever personal wiki (described here, here, and here), and I’ll probably write a couple of posts on how the Mechanics pages are made after I’ve been in production for a while and have worked out the kinks. I’m using Git to keep both the solutions and the page-generating scripts under version control; one of the scripts creates an RSS feed from the commit messages for the solutions.

As it stands, there are currently 80 solved problems from Chapters 1–3. I’ll start adding to it this weekend.

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