I really want another answer to this question than just my own.

As many people probably know the late Prof Laithwaite in the UK (a brilliant electrical engineer) made the error of wandering outside his field of expertise and claiming Newton's Laws needed amending to deal with rotating objects.

He presents a whole series of experiments on gyros, that as far as I can see, perfectly follow Newton's Laws. There is just one video that totally baffles me. I can only think there is either intentional or unintentional deception by Alex Jones, who is demonstrating the device; see the video Patent DE2341245A1 — Vortrieb durch Präzession.

I believe what is going on is that Jones (either intentionally or otherwise) is applying a horizontal force during the initial motion. The gyro will, of course, be applying a torque, but it will be a pure torque with no net force.

Question: I know this sounds dumb, but have I got that right, or are there any alternative explanations? Is it the opinion of others seeing this video that the explanation is simply deception (as I believe it is)?

I admit to feeling a little embarrassed asking this, but assuming deception, it is fairly well disguised.

  • 3
    $\begingroup$ There is always a hand on the contraption. A demosnstration should be hands off. $\endgroup$ – anna v Dec 22 '19 at 19:03
  • 1
    $\begingroup$ I built what I thought was a gyroscopic thrust generator just like Laithwaite's in 1971; soon afterwards he went public with his gyroscopic antigravity theories and I corresponded briefly with him. He was as much in the dark about what was actually going on as I was, and I dropped the idea after figuring out that my device was bogus- but he did not. $\endgroup$ – niels nielsen Dec 22 '19 at 20:35
  • 3
    $\begingroup$ There is absolutely nothing violating Newton's laws here. The car is being moved forward by exactly the same force that moves forward every car: friction between the tires and the ground. And like every other car, if you made the ground frictionless, the car wouldn't go. $\endgroup$ – knzhou Dec 22 '19 at 21:14
  • 4
    $\begingroup$ The only difference versus a typical car is the reason there is a friction force. In a typical car it's because the engine is running, which tries to make the wheels turn, and this causes a friction force in response. In this car it's because the gyroscope is moving, which tries to make the wheels turn, and this causes a friction force in response. $\endgroup$ – knzhou Dec 22 '19 at 21:15
  • 1
    $\begingroup$ @knzhou I think you've hit the nail on the head with respect to the gyroscope causing a friction force in the wheels. That's what I was aiming for in referencing the bike wheel demo, that perhaps with a different set up, the transfer from the gyroscope would send the car forward, rather than spinning a person on a wheel in this case. Thanks for being able to condense it a bit better than I could. $\endgroup$ – Guthrie Douglas Prentice Dec 23 '19 at 6:09

Interesting. However narrator interpretation is totally wrong. Car moves because of reaction force which tries to compensate net force produced by gyroscope + arm system movement. Car movement stops when gyroscope+arm movement kinetic energy is exhausted.

To make things more simple, you can take into account a similar system - a ball wobbling on swings, which are established onto platform with wheels. Schematics : enter image description here

Tension + weight generates net force $F_A$ on ball, which according to third Newton law exerts same magnitude but opposite direction force $F_B$ to it's support : $$ \vec{F}_A = -\vec{F}_B $$ Because support is coupled with platform which has wheels on it - it starts to move in $F_B$ direction. It's simple as that. Of course gyroscope+arm system is more complex that this setup, but I am sure that basic principle is the same.

Conclusion - No any magic here and all Newton laws works as expected.


Just for completenes as @knzhou 's comments answer the question.

First and second laws refer to inertial frames:

1st: In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.

Italics mine.

An inertial frame of reference is one in which the motion of a particle not subject to forces is in a straight line at constant speed.

Hence, with respect to an inertial frame, an object or body accelerates only when a physical force is applied, and (following Newton's first law of motion), in the absence of a net force, a body at rest will remain at rest and a body in motion will continue to move uniformly—that is, in a straight line and at constant speed. Newtonian inertial frames transform among each other according to the Galilean group of symmetries.

Note that we are talking about laws. Laws are the axioms, distilled from data and observations, that physics theories use to pick up the mathematical model that is appropriate for the physical system under study.

The system described as a whole is not inertial, as motors have accelerating forces. In the initial inertial frame of the table the motion is defying no laws, because a force is exerted from the rotation of the motor.

All motors produce force , because of rotation, centripetal and centrifugal, whether electric or not, and depending on the set up can induce directional motion because once a rotation starts there is no longer an inertial system for Newton's laws to hold, and in the original inertial system a force appears, which according to the first law, can change the state of the body . As khnzu states in his comemnt, the rotational force creates a frictional force in the original inertial frame of the table.

Think of an electrical motor drone, it flies in the direction chosen by the design, energy supplied by the batteries electromotive force rotating the motor, the displacement of air (first law) producing the motion forward. ( same as with cars and friction of the wheels on the road).

So as knzhou says in the comments to the question, if there were no friction to the wheels of the device, the system would not move.

  • $\begingroup$ Surely it's still an inertial frame unless were describing the system from the frame of reference of the motor? The lab frame is not rotating so shouldn't be inertial regardless of the motor is doing? $\endgroup$ – bemjanim Jan 30 at 7:09
  • $\begingroup$ @bemjanim The motor creates forces in the original inertial frame, so the thing accelerates, , a force is exerted. in the original inertial frame, by the motor rotation, so the first law is not relevant. I will correct. $\endgroup$ – anna v Jan 30 at 7:23
  • 1
    $\begingroup$ @knzhou said it much better. Nobody thinks of associating an inertial frame to a car. Why would you want to associate one to this device in the first place? $\endgroup$ – lcv Jan 30 at 11:33
  • $\begingroup$ Ah ok I get what you mean now, thanks for clarifying $\endgroup$ – bemjanim Jan 30 at 13:20

I don't think that the demonstrator is giving the contraption a shove.

Suggesting that the demonstrator gives a shove is not necessary.

Let me call the direction in which the four wheels are pointing 'forward' and 'backward'. The contraption has a bottom plate (to which the four wheels are attached) and an attached vertical plate. Then there is a movable vertical plate, the connection allows the movable plate to swing from side to side. Finally, a movable arm is connected to the movable vertical plate.

I'm spelling out all of this because I need names. From here on I will refer to 'the movable plate', and 'the movable arm'.

I will refer to the freedom of motion of the the movable arm as motion with respect to the movable plate. That is, even if the movable plate is swung sideways to 90 degrees, I will still refer to the motion of the movable arm as up/down.

Let me first discuss what the demonstrator would have to set up if the gyro wheel is not spinning.
Then he can simply lift the movable arm, and when he releases the movable arm the contraption as a whole will start to move, simply because the weight at the end of the movable arm is swinging down.

Now to the actual demonstration, with a spinning gyro wheel.

In the starting position the demonstrator turns the movable plate sideways, but the movable arm isn't moved with respect to the movable plate.
The demonstrator releases the contraption.
The gyro wheel is in a high position so it starts to fall down. If the contraption as whole would not be on wheels but on an air table then the contraption would not be in a position to exchange momentum with the outside world. On an air table the contraption would mainly shift from side to side, with its center of mass remaining in the same position.
On the wheels: since the contraption as a whole won't move sideways the swing of the movable plate is relatively larger.
As the movable plate swings down the axis of the spinning gyro wheel is forced to change orientation. That forced change of orientation causes the movable arm to swing up (gyroscopic precession). As the movable arm swings down again the contraption as a whole is dragged forward by the weight of the down-moving arm.

By the looks of it:
I do see that the downswing of the movable plate overshoots the lowest point. But I think by that time the movable arm is back to resting against the movable plate. From that point on the only direction the movable arm can swing is away from the movable plate, or back to resting against the movable plate.

In the demonstration you see that for each run the system is freshly prepared.
The movable plate is lifted sideways, then the contraption is released.

The case is really not different from what you would see if the gyro wheel is not spinning. It's just that if the gyro wheel is not spinning you go straight to lifting the movable arm.

When the gyro wheel is spinning you add the intermediate step of lifting the movable plate sideways, and then the gyroscopic precession lifts the movable arm.

A closer look at the case of the contraption suspended on an air table:
Even if the contraption is suspended on an air table it would still exchange momentum with the outside world when the sideways lifted movable plate is released. When the movable plate is released, and it starts to swing, the air suspension needs to exert a net torque to keep the bottom plate of the contraption in the horizontal orientation. So that is still exchange of momentum with the outside world.

  • $\begingroup$ Thanks for your respective replies. I a green with you both that there must be an exchange of momentum with the outside world. In my view this is either through friction in the wheels and or the demonstrator is applying a force. I am still nclined. to think the latter. He may not.be pushing it. He may simply not be letting allowing any rearward motion initially when the frame of the device should move backward to counter the gyro arms forward motion (such that the centre of mass is stationar). With a lack of understanding of the physics the demonstrator may not realise his.sabotage. $\endgroup$ – Rory Cornish Dec 26 '19 at 15:46
  • $\begingroup$ @RoryCornish I concur. When the movable arm swings up you expect the contraption as a whole to move rearward a little, but it doesn't. It's only after the movable arm starts sagging that the demonstrator releases. About the movable arm: as it sags it really moves in a diagonal line with respect to the bottom plate. About the suspension: in each run the contraption is close to tipping over, you see two of wheels lifting off. The wheels prevent sideways sliding, but only barely so; I think I see one run where the contraption does slide towards the edge, and the demonstrator catches it. $\endgroup$ – Cleonis Dec 26 '19 at 17:08
  • $\begingroup$ Thanks for the answers. Sorry I have not accepted any. I think is poor form to accept answers that seem wrong, but I was trying not to be rude. Regarding the one discussing inertial frames, the laboratory frame is an inertial frame, so Newtons laws ARE expected to be satisfied, thus I could not "accept". Regarding the "reaction" suggestion, the only forces acting on the device are friction, gravity and possible force due the demonstrators hand. Aside from mg, the action of the gyroscope on the device will be almost pure Torque,there is no other net force to react against, thus I couldnt accept $\endgroup$ – Rory Cornish Feb 4 at 16:39

Not the answer you're looking for? Browse other questions tagged or ask your own question.