Less than 1G exerted on rider?!?

Question

Do you know of a ride in Six Flags that less than 1 G of force exerted on the rider? I'm doing a project and I neeeeed help!!!!

Answer 1

First, let's define what "less than 1 g" means. Gravity pulls us on Earth's surface with a 1 g = 9.81 m/sec^2 or 32.2 ft/sec^2 acceleration into the center of the planet. That's always there no matter what amusement ride we take.

But some rides create other forces (F) that counter the force of gravity (W) so that the net effect is to reduce that gravitational force. In math talk, we have f = W - F; where W is the force due to gravity (aka weight) and F is some other force counteracting your weight.

Some guy named Newton found that f = ma; which simply say that a force on a thing of mass m will yield an acceleration a. Well, this f is the same one in f = W - F = ma. W is a force that can give us an acceleration of g, one g. So, as a force, W = mg; where m is the same mass as before.

If the other, counteracting forces are F = mA; then these other forces want to accelerate (A) that mass in a direction opposite to the direction of g. In math talk we have f = ma = mg - mA = m(g - A); so that a = g - A by canceling out the m's on either side of the equation.

From a = g - A, we can see that when there are counteracting forces, like F, the effective acceleration on the mass m is less than g. In other words, as you ride your ride where there are counteracting forces, it feels like "less than 1 g" on your body (m).

So where do these "countacting forces" come from? When a moving body changes direction, a force called centripetal or centrifugal force is created by that change in direction. That's F. And where might you find a change in direction in an amusement ride...just about everywhere. That's usually the fun in amusement rides; so they make it a point to change direction a lot.

Now understand, a change in direction can make F add to your weight as well as subtract from it. It depends on which direction the ride changes to. For example, when a roller coaster pulls out of a hairy dive, the force at the bottom of the dive, as the car pulls out of that dive, will add, not subtract, from your weight (W). In which case a = g + A and the effective G's are greater than 1 g.

On the other hand, as the coaster car goes over the top of a high part of the roller coaster track, the force F subtracts from your weight as the car changes direction and goes into its dive. Thus we have, as shown earlier, a = g - A and there is less than 1 g on the rider.

So what other rides subtract forces from your weight? Often at amusement parks there is a ride where you go around in a car that dips suddenly while rotating. When the car dips, that's a change in direction. A counteracting force will occur and the effective acceleration acting on the rider will be lessened from 1 g.

(By the way, there are a lot of Six Flags theme parks throughout the U.S. Each one is different from the other. So, without being more specific, there is no way to be precise on what rides can cause the counteracting forces F.)

Answer 2

Any rollercoaster that gives you the sensation of flying out of your seat has less than 1G at that point. Which Six Flags are you talking about? If it's Magic Mountain near LA, I know Goliath does at the top of some of the hills. Viper has some hills too. X has G<1 all over the place. But basically, any good roller coaster should have G<1 at the top of the hill (not the very first hill but the ones after the ride started.)