The weights that stop a spinning satellite

The ladder that lets go of the wall used energy to find where a sliding ladder leaves the wall, without needing to know the forces. Its answer came purely from an energy equation and a geometric constraint. The hill that gives it back had set out what makes that possible. The method is at its best when there are two conservation laws for two unknowns, because then the motion’s whole shape follows from algebra with no integration. A clean example has a practical use in the space industry.

Solid-rocket upper stages are often spun before they fire. A spinning object stays pointed where it was aimed, and a slight misalignment of the rocket’s thrust averages to nothing over each turn. The axis a leak of energy chooses discussed what spinning does to a satellite’s stability over longer periods. When the burn is finished, though, the payload generally does not want to be spinning at sixty revolutions a minute. Its solar panels and antennas have to unfold, and it has to point. Removing the spin by firing thrusters would use up fuel, which the rocket that leaves its fuel at home shows is the one thing a spacecraft has least of, and require a control system at the most critical part of a mission. The elegant alternative is a pair of small weights on cords.

Wrap two cords around the stage at its rim, each with a small weight on the end, attached at opposite sides. On a signal, the weights are released. They move outward, and the cords unwind from the rim. When the cords are entirely unwound, they fall free, and the weights fly off into space, carrying the spin away with them. Done correctly, the stage is left with no spin at all. Such a system is called a yo-yo de-spin, and despite the name it has nothing much to do with a toy yo-yo.

Two conservation laws and two unknowns #

Let the body have moment of inertia $I$ about its spin axis and radius $R$. Let the two weights have total mass $m$. It is convenient to measure everything in terms of the weights: define

$$c = 1 + \frac{I}{mR^2},$$

so the body and weights, when the weights are on the rim, have combined moment of inertia $cmR^2$. Now let the amount of cord unwound correspond to an angle $\phi$, so its length is $R\phi$, and let the body’s spin be $\omega$. A weight’s position is the end of a tangent of length $R\phi$ drawn from a point on the rim, which traces the curve called an involute, the same curve a gear tooth’s flank is cut to.

If a point on the rim moves at speed $\omega R$, the weight has a speed along the cord of $\omega R$, because the cord is inextensible. Perpendicular to the cord, the weight has a speed of $R\phi\Omega$, where $\Omega = \omega + \dot\phi$ is the rate at which the cord’s direction itself is turning. Those two velocity components are at right angles, and their squares just add.

Energy conservation then says

$$c\omega^2 + \phi^2\Omega^2 = c\omega_0^2,$$

and angular momentum conservation about the centre says

$$c\omega + \phi^2\Omega = c\omega_0.$$

That is two equations for the two unknown rates $\omega$ and $\Omega$ at any value of $\phi$. Subtract the square of one from the other’s product, or simply substitute, and they solve directly. One root is the trivial no-motion case, and the other is

$$\Omega = \frac{2c\omega_0}{c + \phi^2}, \qquad \omega = \omega_0\frac{c - \phi^2}{c + \phi^2}.$$

The spin of the body falls to zero at exactly $\phi = \sqrt{c}$. The starting spin does not appear in that condition. The cord length for zero spin is $R\sqrt{c}$.

A spin set only by the shape #

The surprising fact is that any starting spin rate gives the same zero. Take two identical stages spinning at 20 and 100 rpm, and fit them with identical cords. The faster one’s weights travel quicker and the tension is larger, but both stages stop when their cords are fully out. The reason is that the quantities to be conserved are both quadratic or linear in $\omega_0$ in such a way that it scales out of the problem: the shape of the trajectory, as a function of $\phi$, does not depend on the speed at which it is traversed. Everything about the problem scales with a single time, $1/\omega_0$. A faster stage simply runs through the same curve in less time.

That independence is of great practical value, since a spin rate is known only approximately in flight, especially after a solid-rocket burn that has consumed most of the stage’s mass. It depends on the body’s inertia, on the other hand, since this is in $c$. A stage’s inertia after burnout is reasonably well known, and it fixes the length. A residual uncertainty is what the designer’s margin takes care of. It is also a demonstration that the two laws together contain more than either alone. Energy conservation by itself allows many trajectories; so does angular momentum conservation. Together, with the constraint of the cord, they fix the whole path.

The result has been tested numerically here in a different way. Written as Euler–Lagrange equations, without using either law, the motion obeys

$$c\dot\omega = -\phi\Omega^2, \qquad \dot\Omega = \frac{\Omega(\Omega - 2\dot\phi)}{\phi}.$$

Integrating these forward from just off the rim agrees with the closed form at every step, to a part in a thousand. Comparing those two calculations helps guard against misapplying conservation laws to a problem in which something has been missed. Here, the cord’s constraint does no net work on the system, and neither law needs anything further.

The cord unwinds at a steady rate #

Take the expressions for $\Omega$ and $\omega$ and subtract them. The unwinding rate is

$$\dot\phi = \Omega - \omega = \omega_0.$$

It is exactly constant. The cord unwinds at one radius per radian of the starting spin, from start to finish, while the body’s spin drops from $\omega_0$ to zero. The time for the whole process is thus $\sqrt{c}/\omega_0$. A stage at 60 rpm with $c = 1600$ stops in about six seconds.

No general principle would have suggested this. It is a surprise of the algebra, and it would have been very hard to guess. A weight’s motion thus has a very simple description in terms of angle, one that was noted in early analyses of the system by Cornille in 1962. It also suggests how to read the plot of energy.

The horizontal axis of the figure is a time and also a cord length, because they are proportional. The body’s share falls gently and then steeply as the weights get further out, because angular momentum depends on distance times speed, and a weight at a long cord length has a large lever arm for holding it. All of the body’s energy has been transferred by the time it has stopped. That cannot be otherwise, for an object with no spin has no rotational energy. Indeed, the weights also hold all of the angular momentum. When they are released, they leave with both, in straight lines, at a speed of $\omega_0 R\sqrt{c}$, which is exactly what makes their kinetic energy equal the energy the whole system started with.

It is pleasant to see how closely the process resembles a billiard collision. The collisions that count the digits of pi has exactly two conservation laws and a similar transfer of energy from a large to a small body, and in that case the result turned out to be much less obvious. Each law gives a curve in the plane of velocities, an ellipse for energy and a straight line for momentum, and a state must lie where they meet. In the de-spin, the same geometry occurs, and the point of total transfer is where the line meets the ellipse at the second root.

Why a tiny weight on a long cord will do #

The square root in the length is very kind to designers. Because a weight’s angular momentum grows with the square of its radius, a small increase in length buys a large increase in capacity. A stage with sixteen hundred times the weights’ moment needs forty radii of cord. A hundred-kilogram stage of radius 1.5 m with $I \approx mR^2/2$ has a moment of about $110\ \text{kg m}^2$. With two 1 kg weights, $mR^2 = 4.5$, so $c \approx 26$ and the cord length is about 5 radii, or 7.6 m. This kind of device has flown many times since the first satellites of the late 1950s, usually with total masses of the order of a kilogram or less. Dawn, in 2007, used a similar approach in leaving its launcher’s upper stage.

A system with so few parts has few things that can go wrong, and that is why it is still used. It requires no power, and it is not sensitive to what is going on in the rest of the spacecraft. Compared with a thruster-based system, it saves mass as well as risk. A variation releases only partway to leave the stage spinning at a chosen lower rate; another attaches weights to cords that have an articulated hinge at the rim. Launchers such as the Delta have used them repeatedly for payloads that need a controlled final spin.

The energy that the weights leave with is not lost to the mission in any way that matters. They become a small pair of pieces of space debris, however, which is now an increasingly strict concern for orbital missions. Weights released in Earth orbit can remain there for a long time and are tracked by radar. Many modern designs, as a result, keep weights on tethers instead, or use them only on departures from Earth.

The error in length comes out as spin #

A designer has to know what happens when the length is wrong. Differentiating $\omega$ with respect to $\phi$ at the zero gives a very simple result: the slope, in terms of fractional length, is minus one, so a 1 per cent long cord leaves 1 per cent of the spin going the other way. The figure’s numbers bear that out, and they show the reverse rotation on the long side. A 10 per cent error in length therefore gives about a 10 per cent residual, whatever the starting spin.

A fractional error in length is carried through with no amplification. Length is easy to set, so the only real problem is in the inertia, since $c$ depends on $I$. Errors of 2 per cent in inertia give about 1 per cent in length, and thus about 1 per cent of spin left. For a stage spinning at 60 rpm, that leaves 0.6 rpm, an amount that a payload’s reaction wheels can readily handle. Designers therefore do not need precise knowledge of the spin rate, and they need only modest knowledge of the moment of inertia. Few other deployable systems are so forgiving.

It may seem odd that a spin can pass through zero and reverse. The cord pulls on the rim in the direction that slows it down; once the body stops, the cord is still under tension and its direction is such that it continues to exert a torque, which now winds up the spin the opposite way. A weight on a string does not care what the body is doing. Physically nothing happens at the zero except that the cord would, if it were not released, continue to wrap the opposite way around the body. In some designs, a hinged release ensures that the cord leaves at the correct moment, when its direction reaches radial.

Tension: the number the designer has to watch #

If the starting spin does not fix the length, it does fix the strength required of the cord. The tension is the force needed to keep each weight moving on its curved path, $mR\phi\Omega^2$ in total for the pair, in the same units used above. Starting at zero, it reaches a peak part-way out. It scales with the square of $\omega_0$. A stage spinning at 100 rpm has about 28 times the tension of one spinning at 19 rpm, so the cord must be rated for the highest spin expected, and this fixes its thickness and mass. A kevlar cord is thin and light for the job. At an ordinary spin rate the tension in a typical system is measured in hundreds of newtons.

There is a further physical point here. Each weight is, in its own frame, pulled along the cord, and therefore that tension does work on the weight only insofar as the weight moves along the cord. The body loses energy at a rate equal to the torque times its spin, and the energy of the weights rises by exactly that amount. Tension is thus an internal force, so the two conservation laws hold regardless of how the cord behaves, provided that it is massless and inextensible and has no slack. A cord with noticeable stretch stores and returns some energy, and a slack cord allows a sudden jerk. Either can excite flexible parts of a stage. If so, the assumption of a rigid body is no longer sufficient, because of the way energy finds other modes, as the axis a leak of energy chooses showed for an actual spacecraft. On a smaller scale, the disk that spins faster as it stops is another case where the arithmetic of two conserved quantities gives a behaviour that at first seems paradoxical.

The same two laws on a figure skater #

The everyday cousin of this device is a skater drawing in the arms, and the comparison is instructive because the two run opposite ways and differ at one point. Pulling the arms in reduces the moment of inertia, and angular momentum conservation raises the spin. But energy is not conserved there, for the skater’s muscles do work against the centrifugal pull, and it is that work that appears as the extra kinetic energy. Letting the arms go out again returns the same energy to the muscles, which waste it. In the de-spin system both laws hold, because a cord has no muscles: tension does work on the weights and an equal amount of negative work on the body. Only the arrangement of the mass changes.

The quantity that survives a change of shape deals with what stays fixed when such a change is slow. Merely extending two weights on rigid arms, in contrast, would only slow the spin and never stop it, since the moment of inertia is then always positive and $\omega = L/I$ always has the sign of $L$. The body and its arms must keep the total angular momentum, and only in the limit of infinite length would they reach zero. The cord can do better because it stops being part of the body. The involute geometry lets the body and the weights have different angular velocities, so the weights can hold all of the angular momentum while the body holds none. The same push, further out gave the quantity on which all of this rests.

What the calculation assumes #

The calculation treats the body as a perfectly rigid cylinder with a fixed axis, two weights that are point masses released at once and symmetrically, and cords that are massless and perfectly inextensible. A real cord has mass, and for a very long cord that is enough to change the numbers noticeably. The two weights need to release at the same time, since any imbalance shakes the body laterally about the axis. The body’s axis is also assumed fixed. In a real stage, the moment of inertia about other axes matters, as does any liquid that may slosh in tanks, and a transverse shake of the axis could couple into nutation.

Nothing about a body’s elasticity or about a slowly rotating atmosphere applies here, and the picture ignores gravity gradients, which are entirely negligible over a few seconds. This is one of the rare cases where the ideal is close to practice. The published flight residuals of such systems are often small enough to confirm the calculation with few surprises.

Still open: how tethers do at larger scale #

The same arithmetic applies on a far larger scale, to rotating tethers in space. A tether a kilometre or more in length holds a great deal of angular momentum, and it could be used for orbit changes without fuel. Several proposals have considered spinning long tethers to throw payloads from orbit, or capturing them. Experiments so far have had a mixed record. The TSS-1R tether in 1996 broke when almost fully deployed from the shuttle, by an electrical arc, and several other experiments have jammed in deployment. Whether long tethers can be reliably deployed, keep from becoming tangled, and survive small-debris impacts is an open engineering question. The physics is the same as for a de-spin, without the benefit of a short and quick process.

A related problem is that, for smaller spacecraft, flexible appendages such as solar arrays and booms have large moments of inertia compared with the body. A cord that has to despin a craft with long booms must deal with a body that is itself deforming and storing energy in its bending modes. As small spacecraft become common, how to design those simple devices for such flexible bodies has become a live issue, with the rigid-body formula giving only a first estimate.

Energy and angular momentum, with a cord as the only constraint, fix the whole motion: spin falls as $(c - \phi^2)/(c + \phi^2)$, the cord unwinds at a constant rate equal to the starting spin, and everything stops at $R\sqrt{c}$ of cord, wherever the starting spin stood. The weights end up with all the energy and all the angular momentum, and a square root makes them small. A cord 1 per cent long leaves 1 per cent of the spin going backwards.