Re: Why does a Spin Fire Ring fly in a straight (horizontal) trajectory?

Greetings Jim & John:

I have never heard of the Koosh Vortex Tornado toy gun; however, a search of the web came up with pictures and a description of the toy gun and the projectile along with a comparison of it with several other toy guns located at the following web site: http://nerf.cent.org/gunsammo.html

This looks like an interesting device and along with your description the nerf center indicates that a cross wind will cause the projectile to drift.

The device makes use of the conservation of angular momentum produced by a rotating flywheel like mass. We use the effect to stabilize and maneuver our spacecraft in orbit with devices called momentum wheels.

For example, suppose that a bicycle wheel on an axle is sitting on top of two posts, one at each end of the axle, so that the axle is parallel to the surface of the earth. You then spin up the wheel to a high rotational velocity. The resulting angular momentum force vector L(a) will then be in the direction of the axle and it’s magnitude will be proportional to the number of revolutions per second (rps) that the wheel makes. The vertical force of gravity vector (mg = weight) on the wheel is equally supported by the two posts.

Now if we remove one of the posts the bicycle wheel will continue to spin and the axle will remain parallel to the earth with one end of the axle floating unsupported in the air (like a toy gyroscope). This occurs because the force of angular momentum is much larger than the force of gravity that is acting at a right angle on the wheel.

The moment that the rotating rubber ring leaves the Koosh gun the situation is similar to the floating bicycle wheel. The propulsive force delivered by the gun is delivered along the angular momentum vector causing the rotating ring to leave the gun along the vector L(a).

In the bicycle wheel example the wheel slows down from bearing friction and air resistance and the angular momentum vector L(a) decreases and approaches the magnitude of the gravity force vector (mg). As this happens the free end of the axle begins to oscillate in a complex wobbly circular motion called precession. Then fairly quickly the gravity force will overcome the momentum vector and the wheel will fall to the ground.

However, unlike the bicycle wheel, the spinning rubber ring leaves the gun along the momentum vector and aerodynamic forces begin to act on the rubber ring. The lift force produced by the circular curved surfaces is equal in all directions in a plane normal (at right angles) to the momentum vector. These forces are much stronger than gravity and act like a virtual axle with very little rotating friction (there are no bearings), holding the ring precisely around the guiding momentum vector. Because of its aerodynamic design, the thin ring encounters very little force from air resistance (drag) , and its forward velocity along the momentum vector enables the ring to quickly travel long distances in a straight line (like a gyroscope). However, as the ring rotation slows down, the gravity vector will eventually cause the ring to start precession which then quickly destroys the symmetry of the aerodynamic forces, the virtual axle is destroyed and the device will drop to the ground like a normal falling body.

The Koosh gun is a very clever device and I’ll have to look for one in the toy stores.

Best regards, Your Mad Scientist
Adrian Popa