Sunday, November 8, 2009

Oscillations of Shoes and Barges

One specific example of the practice of Natural Philosophy, or the scientific method, appeared in Eliza’s journal of her trip across the boundary between eastern France and the western parts of Germany. In the entry of 20 August 1688 (pages 830, 831 of Quicksilver), she described her situation and immediate objective. "For several days we have been working our languid way up the Marne." "This vessel is what they call a chaland, a long, narrow, cheaply made box with but a single square sail ..." "... there is nothing for a spy to look for, except, perhaps, certain military stocks." "... certain items, such as gunpowder, and especially lead, might be shipped up the river from arsenals in the vicinity of Paris." "So I peer at the chalands making their way upriver and wonder what is stored down in their holds. To outward appearances they are all carrying the same sort of cargo as the chaland of M. LeBrun, viz., salted fish, salt, wine, apples, and other goods ...."

In the entry of 25 August 1688 (pages 831, 832), she described her thoughts and experimental observations. "... was there any outward sign by which I could distinguish a chaland loaded as M. LeBrun’s is, and one that had a few tons of musket-balls in the bilge with empty barrels above ...?" "Even from a distance it is possible to observe the sideways rocking of one of these chalands by watching the top of its mast– ..." "I borrowed a pair of wooden shoes from M. LeBrun and set both of them afloat .... Into one of these I placed an iron bar, which rested directly upon the sole of the shoe. Into the other, I placed an equal weight of salt, ... Though the weights of the shoes’ cargoes were equal, the distributions of those weights were not, for the salt was evenly distributed through the whole volume of the shoe, whereas the iron bar was concentrated in its ‘bilge.’ When I set the two shoes to rocking, I could easily observe that the one laden with iron rocked with a slower, more ponderous motion, because all of its weight was far from the axis of the movement." "... I timed one hundred rockings of the chaland I was on, and then I began to make the same observation of the other chalands on the river. ... I noticed one or two that rocked very slowly. ..., the first one turned out to be laden with quarried stones."

The method appears to be excellent, with steps of modeling, theoretical explanation, and observation. Unfortunately, the real world doesn’t match any of the steps as described.

Rather weirdly, I happen to own a pair of wooden shoes. Mine are of Dutch design, rather than French, but perhaps that is irrelevant. Each shoe weighs 14 ounces, and 2 pounds of rock salt filled one, to a depth of about 1 1/2 inch and an interior width of about 3 inches.. The other held 2 pounds of steel, cut from flat stock. The stack of pieces was 5/8 inch deep and 1 1/2 inch wide. Those heavy loads left only about 1/4 inch of 'freeboard', when the shoes were floated, so that the angular amplitude had to be small. I didn't have a stopwatch, and the shoes tended to drift against the edge of the basin, thereby quickly damping the motion. However, the shoe loaded with steel rocked at more than twice the frequency of the one loaded with salt. This is exactly the opposite of Eliza's report!

The theoretical explanation as given by Stephenson is grossly oversimplified. It matches too closely the simple pendulum, in which a point mass supplies the inertia, and a restoring force is supplied by the combination of its weight and the tension in the supporting string. For a simple pendulum, a longer pendulum does indeed oscillate with a longer period.

Here the shoe or the barge is rotating, and must be considered as a special case of a 'physical pendulum'. In this situation, the inertial factor is the 'moment of inertia', which must be taken about the appropriate axis. The moment of inertia can also be expressed as mass x ('radius of gyration') squared. The radius of gyration is the root-mean-square distance of the mass from that appropriate axis. For a true physical pendulum, that axis passes through the fixed point of suspension. For a floating object, that axis passes through the center of mass, which effectively does not move as the object rotates. (Any acceleration of the center of mass would involve hydrodynamic forces, whereas this analysis is limited to hydrostatic forces.)

For a true physical pendulum, the restoring action is the torque about the suspension point, due to the weight of the body, acting at its center of mass (or 'center of gravity'). For a floating object, the torque is supplied by the pressure of the surrounding water. The effect is always in the form of an upward buoyant force, equal to the weight of the displaced water, acting at the centroid of the volume of the displaced water (the 'center of buoyancy'). Perhaps Stephenson considered that "the axis of the movement" passes through the center of buoyancy, but that is not a fixed point.

For an object floating at rest, the weight and the buoyant force must act along the same vertical line, or equivalently, the center of gravity and the center of buoyancy are both on that same vertical line. If the center of buoyancy is above the center of gravity, this equilibrium position is absolutely stable. If the center of gravity is above the center of buoyancy, the equilibrium must be investigated more carefully.

Boats, barges, and ships are typically bilaterally symmetric, and shoes are approximately so. When a barge is level (side to side), the center of buoyancy is on the vertical plane of symmetry. The load is typically adjusted so that the center of gravity is also on that plane of symmetry. Let us consider a virtual displacement by a rotation about the roll axis, which is the horizontal line in the plane of symmetry and through the center of gravity. In this displacement, the center of buoyancy typically shifts sideways, away from the plane of symmetry, because the displaced water has changed its shape. The line of action of the buoyant force intersects the plane of symmetry at the 'metacenter'. If the metacenter is above the center of gravity, the equilibrium is stable, at least for small rotations. If the center of gravity is above the metacenter, the equilibrium is unstable, and the barge or shoe will roll over.

For any physical pendulum, including a floating object, the period of oscillation about a position of stable equilibrium is always the same as that of a simple pendulum of length = (radius of gyration) squared / (length of arm associated with restoring torque). Here that length of arm is the distance between the metacenter and the center of mass. The concept which Stephenson overlooked is that a physical pendulum has two characteristic lengths, both of which depend upon the distribution of its mass. The simple pendulum has only one characteristic length, because its mass is not distributed at all, so that radius of gyration = length of arm.

In order to examine the effect of the density (or specific gravity) of the load upon the motion of a barge, we need to model the system. The description as "a long, narrow, cheaply made box" serves to inspire a model. A "box" should have a uniform rectangular cross-section, with an outside width and an overall depth from bottom of hull planks to top of deck (if any). That it is "long", suggests that we can ignore the effects of the ends as an initial approximation. That it is "cheaply made", means that the load should be uniformly distributed over the entire area of the bottom and of the deck. That way the load is supported almost directly by the pressure of the water against the hull planks, and there is no need for beam strength to support concentrated loads. Thus we can describe the entire barge by a single cross section, which can also be considered to represent a unit length ('one meter') of the barge.

We can now evaluate the mechanical properties of such a barge, under various loads. Every item shown in the cross section, whether part of the barge or of the load, can be represented by a rectangle, and can be treated in the same manner. (The bottom, sides, and deck would actually be longitudinal planks over spaced transverse ribs, but the mass of the ribs can be 'averaged' over the unit length.) The mass of a unit length of the item is its density times its area, and its individual center of mass is at the center of its rectangle. The mass and the center of mass of the entire system can be found by summing the contributions.

The moment of inertia of every rectangular item, about its individual center of mass, is given by 1/12 of its mass x the square of its diagonal (or the sum of the squares of its edge lengths). Its moment of inertia about any other point, such as the overall center of mass, is the sum of that central moment and its mass x the square of the distance between the desired point and the center of the item. The moment of inertia of the entire system about the overall center of mass, is the sum of the individual contributions. A further significance of "cheaply made", is that the mass and moment of inertia are dominated by the load, rather than by the wooden structure of the barge.

The calculation of the metacenter is particularly simple for this rectangular cross section. When the barge is level, the displaced water for a unit length appears as a rectangle, the width of the hull x the 'draft'. The draft is the vertical distance from the bottom of the hull to the waterline, and it depends upon the load in the barge at the time. The center of buoyancy is thus at half of the draft above the bottom of the hull. When the barge has a virtual displacement such that the deck has a particular (small) slope relative to the waterline, the displaced water for a unit length appears as a trapezoid, having the same area as the original rectangle. The centroid of the trapezoid is displaced horizontally from the plane of symmetry, by 1/12 x slope x width squared / draft. Thus the height of the metacenter above the hull bottom is 1/2 draft + 1/12 width squared / draft.

Let us then consider identical barges, each loaded with cargo of uniform but different density. The total load will have the same mass in each case, so that the draft will be the same in each case. Thus the position of the metacenter will also be the same in each case. The width of the load will fill the space between the inner faces of the walls. The height of the load will be inversely proportional to the density of the load. The center of mass of the system will be close to the centroid of the load (the "cheaply built" condition). Similarly, the central moment of inertia of the system will be dominated by the central moment of inertia of the load, which has the factor (1/12) [(inner width) squared + (load height) squared].

The overall radius of gyration obviously decreases as the load height decreases, or as the load density increases. At the same time, the length of the torque arm, from the overall center of mass to the metacenter, increases as the load density increases. Thus the load of greatest available density corresponds to a simple pendulum of the shortest possible length, or shortest period. This is exactly opposite to Stephenson’s assertion.

Some other conditions of equilibrium should be mentioned. An empty barge of reasonable width would have the smallest possible draft and the largest possible height of metacenter. It would be quite stable, and would not require any ballast before being loaded with useful cargo.

There is a smallest critical load density which allows the system to be stable, and to oscillate. When the load has density just greater than that critical value, the overall center of mass is below the metacenter, but close to it. The equivalent simple pendulum is very long, and the period of oscillation is correspondingly large. This is the condition which produces "ponderous motion", and not high density as Stephenson asserted.

When the load has any density less than that critical value, the overall center of mass is above the metacenter, and the barge would roll over. If the load has exactly the critical density, the system would be in neutral equilibrium. It ought to 'hang' at any angle of displacement, but it would be in danger of capsizing if anything on board were moved.

The exact value of this critical load density depends on the dimensions of the barge, but it ought to be roughly half the density of the wood used in the barge. The barge would appear dangerously overloaded, with cargo piled on the deck to about the height of the hull itself. Thus the most dangerous cargo for a barge is enclosed air! It could take the form of uncompressed fiber (wool or tow), or of partly empty barrels or crates. In the modern era, it could be empty shipping pallets, which actually represent a dangerous load for a flatbed semitrailer.

Stephenson should have known better than to ignore these effects of load distribution when he was writing Quicksilver, because on page 302 appears: "Once loaded, the carronnades are being run out to the gunwales–hugely increasing the ship’s moment of inertia, accounting for the change in the roll period–" Unfortunately, he used perhaps the most inappropriate specialized form of naval artillery as his example. The carronade was not invented until about 1769 (see Wikipedia), so that it could not have been available on Minerva in November 1713. One of its special features, beyond those described on that same page, was its very low mass. (It was typically used on upper decks of ships, without raising the center of mass too high.) The ship’s long guns would have made a greater increase in the moment of inertia by being run out, but probably not by enough to be called "hugely".

It is likely that Stephenson knows even better now, because in his latest book Anathem, the narrator (Fraa Erasmas) remarks on page 683: "Then I went back to work estimating the inertia tensor of the Geometers’ ship." What I have done above amounts to considering a single component of that tensor, for a chaland. None of the other components are needed here, when considering only rotation about the roll axis. However, Erasmas did have to allow for simultaneous rotations about all three axes. Essentially, Stephenson knows many of the right words, but he seems not to have the full significance of "moment of inertia" at his fingertips.

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