help me understand the rotating ring artificial gravity thing

[TuxDave]

http://pics.bbzzdd.com/users/xevadex/RotatingRing.GIF

So without any equations you can intuitively see that jumping vertically lands you ahead of where you started. The reasons are twofold.

1) The path you travel in space is shorter than a stable object on the surface (tree in this case) because you don't move along the arc.

2) The magnitude of your velocity in the direction of your path is greater because you not only have the surface velocity but also your contributed velocity.

But to double check, we know that as the radius of the ring become arbitrarily large, that we shouldn't see these effects and that's true because for #1, at very large R's and decreasing values of theta, the distance 'D' gets closer to the arclength and for #2, your contributed velocity becomes negligable to the surface velocity.

So in this case, you will ALWAYS land ahead of where you started (assuming you're facing the direction of rotation) assuming a finite sized ring. Now if it's significant enough to make you sick, that's another question but at least it gives you a picture of how big the ring needs to be in order to reduce seeing things that you wouldn't see on earth. A 32m radius ring is pretty damn large but landing 6cm ahead after jumping straight up is still noticable.

 

[KIAman]

Wow, goes to show how much I pay attention when I read?!?!?

The diameter is 1.8km so the circumference is actually 5.652km!!!

In that respect, you'd have to run 207.24mph (not 66mph) to counter the acceleration.

Also, I am assuming TuxDave's calculation using using METERS for the ring's radius so in that respect, a 32m radius is not large at all.

Can you repeat the formula for this specific example (radius = 900m)? Looks like it will be less than 1cm movement which will most likely not be noticeable.

 

[PolymerTim]

Originally posted by: TuxDave
http://pics.bbzzdd.com/users/xevadex/RotatingRing.GIF

So without any equations you can intuitively see that jumping vertically lands you ahead of where you started.

Great job TuxDave! Somehow in my head, I had competing factors that I though would cancel, but your diagram helped clear things up. I'm curious to see your equations if you can post them. I tried to figure it out, but I am way out of touch with geometry and ended up having to do some numerical calculations along the way. But I could at least see that the jumper and his point of origin do not reach the same point at the same time.

So yeah, 1.8km! I hadn't actually read the wiki article before now. I guess that removes a lot of problems like this and the coriolis effect. Pretty cool stuff.

-Tim

 

[plastick]

the simplest answer is that everything has relative values.

 

[hellokeith]

Originally posted by: KIAman
Wow, goes to show how much I pay attention when I read?!?!?

The diameter is 1.8km so the circumference is actually 5.652km!!!

In that respect, you'd have to run 207.24mph (not 66mph) to counter the acceleration.

Also, I am assuming TuxDave's calculation using using METERS for the ring's radius so in that respect, a 32m radius is not large at all.

Can you repeat the formula for this specific example (radius = 900m)? Looks like it will be less than 1cm movement which will most likely not be noticeable.

You are correct. The larger the diameter, the "flatter" ring will look from the perspective of the inhabitant. Still, you could put some crossed lines on the "ground" and see the change when you came "down".

Now I'm curious about basics. What kind of strange toilet would you need? Kitchen sink? And could you effectively grow crops, would you be able to water them in a normal fashion?

"Stop jumping in the beds, kids.. or you'll end up outside."

 

[PolymerTim]

Actually, I think most of those things would be about the same. The water in your toilet doesn't know the gravity is artificial any more than when you swing a bucket of water over your head. I think most of the differences would just be due to the enclosed atmosphere, having a ceiling, etc.

As for the effect of jumping, I think at 1.8 km diameter, the distance you travel jumping straight up would be smaller than your ability to jump straight up. I don't think you could really measure it by just jumping.

The only thing I find funny that someone else mentioned is that you'd have to get used to the stars gonig by pretty fast since the 1.8k diameter ring is spinning at 1 rpm. Funny enough, I was only at 2.6 rpm when looking at the 128 m diameter ring and then realized that the rpm needed for 1 g is proportional to the square root of the ring radius. Then I asked myself, how big would the ring need to be to get to one revolution pre day. The answer is 1.8 million km (radius, not diameter)!

-Tim

 

[TuxDave]

Originally posted by: KIAman
Wow, goes to show how much I pay attention when I read?!?!?

The diameter is 1.8km so the circumference is actually 5.652km!!!

In that respect, you'd have to run 207.24mph (not 66mph) to counter the acceleration.

Also, I am assuming TuxDave's calculation using using METERS for the ring's radius so in that respect, a 32m radius is not large at all.

Can you repeat the formula for this specific example (radius = 900m)? Looks like it will be less than 1cm movement which will most likely not be noticeable.

Yup it's in meters. I dunno, I thought a 32m radius is pretty realistic. It makes the ring longer than the international space station. Building one that's 1.8km in diameter will be quite a feat.

 

[Nathelion]

hey they did it in Halo, so why not...