Let's say I have perfected a boomerang throwing machine - in Earth surface conditions the boomerang is always thrown the same way, and always returns to the throwing arm. And let's say I've sponsored the building of a kilometre radius 1 atmosphere gym for the ISS. I then mount the machine on the wall of this room and let rip. What sort of trajectory would the boomerang follow? It seems equally possible that since the lift would not be counteracting any gravity it would follow some sort of upward (relative to the original horizontal plane) spiral or alternatively that it would start rotating on all three axes.

To be clear, by "weightless" I meant in microgravity, not having zero mass. I'm not asking about magical materials here.

  • $\begingroup$ Zero mass plus nonzero volume means it'll float to the top of the atmosphere and probably drift off into spaaaaaaaaaaace . BTW does your massless material have any surface energy (which leads to friction)? Lift depends on that. $\endgroup$ – Carl Witthoft Sep 24 '15 at 11:33

Apparently, a sufficiently sturdy boomerang does return to its thrower in microgravity, as shown in the NASA video "International Toys in Space: Boomerang" in 2013:



The boomerang orbit is a result of differential lift (rotation plus velocity) that provides torque and this precession of the angle of rotation.

Normally a boomerang is thrown with its axis of rotation at an angle to the horizontal - this provides a force to counter gravity (the component of lift is partly vertical, and partly inwards). Throwing the boomerang in the same way, it would execute a helix upwards. Better have a high ceiling on your gym.

Now the rate of spin vs precession depends on the rate of rotation - if you throw with more spin or less spin it will change the trajectory.

I wrote a couple of earlier answers with some of the math behind the motion of boomerangs:



  • $\begingroup$ I especially liked your second link, where you show how to do a miniature boomerang. This could be tried in the international space station. $\endgroup$ – Mike Dunlavey Sep 25 '15 at 19:06
  • $\begingroup$ @MikeDunlavey - when I looked at my old answer, the same idea occurred to me. But it's quite hard to make it work in a very confined space - you need something the size of a physics classroom. Did you try it? $\endgroup$ – Floris Sep 25 '15 at 19:08
  • $\begingroup$ Maybe when my grandsons goad me into entertaining them :) I'm leaning more toward the helix idea because I would expect the axis of rotation to precess in a circle, just like a top slowing down. $\endgroup$ – Mike Dunlavey Sep 25 '15 at 19:50

First of all, what would a model airplane do?

Normal aircraft are designed with a center of mass forward of their aerodynamic center, which keeps them pointed forward.

They also have decalage meaning that they are designed to curve up. This is done either by having a tail that pushes down, or canards that push up, or a single wing that is curved up at the back, or a combination of these. The strength of this tendency to curve up depends, like lift, on velocity squared. Since the center of gravity is forward, pulling the nose down, the plane tends to seek a particular speed at which the forces are in balance. Most planes have a "trim wheel", which alters the curving-up tendency, which alters the plane's stable speed. More nose-up trim means slower speed.

Now, what would the airplane do if thrown in micro-gravity? I suspect its curving-up tendency would move it in a loop, bringing it back to where it started, no matter how slow or fast it is thrown. If all of the decalage were trimmed out, it would go in a straight line, which is just a loop of infinite radius.

The boomerang is also a lifting body that depends on gravity. Plus it has its own gyroscopic precession to tilt its lift direction, making it turn back to the thrower. (More lift on the forward-moving side, causing its rotation axis to "tilt back".) In a micro-gravity environment, tilting its lift direction would not work with gravity to make it fly in a horizontal curve. My guess, and it's only a guess, is that it would travel in a helical path away from the thrower.


Settle in and get a drink. This will take a while.

First, how does a boomerang work? It's all in the wrist. That is, rotation and torques.

Let's ignore the fact that, generally speaking, a boomerang does not follow a horizontal circle. That's for later. Furthermore, in order to make things clearer, let's pretend that a boomerang is a helicopter. In addition to the rotating rotor disk, it has a body with the intuitive orientations. And let's say that throwing a boomerang (right handed, flat side of the boomerang to the right) amounts to launching the helicopter nose first but rotated 90 degrees CCW so that the rotor disk is vertical.

Two odd things happen to the ship, and both of them are precessions. Calling the right side of the rotor disk the advancing side and the left the retreating side, it's clear that overall the advancing side has greater velocity wrt still air than the retreating side, due to the aircraft velocity. This means that the advancing side produces more lift than the retreating side, and the result is a torque that tries to rotate the ship CCW (right side up, left side down). But here's the thing: this is a gyroscope, and it won't move that way. Instead, the system will precess, and the result will be that the nose will try to pitch up (in the aircraft's body coordinates). This will cause the boomerang to circle to the left, and that's why it goes in a circle.

But another odd thing (precession) occurs, and the cause of that is not settled. The common explanation is the rotors in the front half of the disk (leading) encounter undisturbed air, while the rotors in the back half (trailing) encounter disturbed air in the wakes of the rotors in the leading half. This produces a net torque which tries to pitch the nose up. Another explanation is that the leading elements of the rotor disk interact differently with the translational air flow differently than the trailing edges, without taking wake disturbances into account, and the result once again is a differential lift which produces a torque which tries to pitch the nose up. Both arguments seem reasonable, so I suspect both are correct, but it's not at all clear which effect dominates. However, either way the torque is produced, the result is the same. Just as the boomerang cannot respond directly to the advancing/retreating torque, neither can it respond to this torque. The result is a second precession, this one rolling the system CW (again in body coordinates). In this case, this works out beautifully for real boomerangs: as drag slows the arms, the system rolls right, and the total lift becomes more vertical, so the boomerang does not crash into the ground. For a perfect throw, the boomerang comes around slowly rolling until it is horizontal just as it returns to the throwing point. Furthermore, the "nose" is pitched up slightly, braking it. As the forward speed drops, so do the induced torques and associated attitude angular rates. If the torques are proportional to velocity, the boomerang maintains a perfect circle. At some point the forward speed drops to zero, but the arms are rotating fast enough for the thing to hover, then gently start to fall as the arms slow. If you do it right, it's quite remarkable and easy to catch.

So, with that out of the way, what happens in microgravity. Well, that depends. First, let's ignore drag losses and assume both rotational rate and translational velocity remain the same. Then, depending on rate and velocity and a whole bunch of other stuff (air density, airfoil performance, mass, dimensions, etc) the 'rang will precess at two rates, P (pitch) and R (roll). For a perfect circle on the ground, the first precession rate P is exactly 4 times the second, R. In other words, when the boomerang has made a perfect 360 circle, it has also rotated 90 degrees to the horizontal, so P = 4R.

From this, it's clear that a circle will take T seconds, where $$T = \frac{2\pi}{\sqrt{{P^2}+{R^2}}}=\frac{2\pi}{\sqrt{5R}}seconds$$ and the "vertical" displacement D of the start and top points will be $$D =\frac{V\sqrt{5R}}{8\pi}$$ At the end of the first loop, the plane of the boomerang will be rotated 90 degrees, so the total path will form a closed loop with 4 segments of helix, with the start/stop points of each segment forming a square. In general for differing ratios of P and R I believe you'll get a series of helices whose central axis itself forms a circle, with the boomerang path lying within the surface of a toroid.

Ah, but that's in an ideal world. What about frictional losses? That gets tricky, and is beyond what I'm willing to spend the time on. Presumably P and R do not respond identically to changes in V, for instance. The important factor is the relationship between rotational drag (how fast the boomerang stops spinning) and translational drag (how fast the boomerang slows down). Because the boomerang is always "pitching up" the overall disk is constantly at a positive angle of attack and will suffer considerable losses. Obviously, the end point is a stationary boomerang, but the end game depends on which slows more quickly, translation or rotation. If translation dies first, as in the "perfect throw" the boomerang moves perpendicular to its plane of rotation due to lift from the arms. If the rotation dies first, the boomerang is no longer gyroscopically stabilized and will tumble in a chaotic fashion. In general, you'd expect the transition from helix to endgame to be very hard to manage, since it's very unlikely the various rates, velocities, torques and drags are simply and linearly related.

  • $\begingroup$ *headscratch* Sounds like it needs a simulation run. $\endgroup$ – l0b0 Sep 24 '15 at 22:04

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