But then it's a circular, and not parabolic arc, isn't it?

And if you were somehow able to measure your "arc velocity" you'd measure it as a constant, wouldn't you?

Nope. Jumping in a halo appears to you to seem like jumping on Earth. You jump in the air and seem to come back down. In reality you jumped up, then the halo hit you on your feet, while you're still moving up!

Remember, there's no gravity, and thus no parabolas. Your trajectory (from an observer standing next to you in the halo) is straight up, then straight down, just like on Earth. But an observer standing near you, but floating outside the halo, will see something completely different. Your airborne trajectory will be a perfectly straight line! It sounds freaky, but it works, thanks to the fact that centrifugal force is fictitious. But you sure do feel it!

And yes, your arc velocity is constant (aka tangent velocity).

I want a halo to play with.

I want one to play with too, but I beg to differ with your math.

I just did some back of the envelop calculations. To keep this simple, let's assume a halo 500 meters in radius spinning so that it's tangential velocity is 70 mps, this gives the 9.8 acceleration we're looking for on the surface. Now let's take a pole 9.8m long and drop a ball from the top of it. Looks like there's a 2/100ths of a second difference in how long this takes to fall in the halo vs RE/FE. Here's the math:

**First RE: **At 9.8 meters high, and d = 1/2 * a * t

^{2} we can solve for t and get sqrt(2) = 1.414 seconds

**Second HE:****part 1: how fast is it "falling"**At 9.8 meters high, the radial velocity is a little slower than at the surface. The surface is 70m/s while at this height it's 68.628 m/s (Justification: 70 mps at 500 meters radius gives rotational speed of .14 radians per second w = v/r, now w is constant, so when r is 490.2 v is then 68.628)

**part 2: how far does it "fall"**When we "drop" the ball it starts moving along the chord from 9.8m high. How long is this chord? It's pretty easy to calculate from Pythagorean theorem, as the full radius is the hypotenuse, the semi-chord length (what we're interested in) i

s one leg and the (radius - the pole) is the other leg. Anyway, do the math and this length is 98.51m.

**Part 3: how long does the "fall" take**98.51m / 68.628 m/s gives 1.435 seconds 2/100ths of a second longer than in FE/RE

**Part 4: Where does it land?**Fortunately there are 2 different equations for this. But each equation is easier if you think about the angle that the halo turns through in both cases instead of linear distances. Let's start with the easier one.

The ball took 1.435 seconds to fall and we've already established that the halo rotates at 0.14 radians/second so in 1.435 seconds the pole that the ball was dropped from will have rotated by 0.20096 radians. On the ground of the halo this is 100.48 meters. That is to say that after 1.435 seconds the halo rotates 100.48 meters.

The other equation is where the chord intersects with the halo. That's an inverse trig function from the triangle in part 2. Here it's the inverse cosine of 490.2/500 which is .1983 radians. On the ground of the halo this is 99.16 meters.

**Conclusion**:

At this scale, the HE ball drops .02 seconds slower than RE/FE and "mysteriously" lands about 1.3 meters

*behind *the pole from which it was dropped.

EDIT: I just did the calculation at an earth sized halo, approximately 5,000 kmeters in radius and the answers are much closer to each other, though at this scale the ball still falls behind the pole that it was dropped from by a little more than a centimeter.