Centrifugal Force

Game Engine Game Engine Support and Discussion
#1

I’m trying to simulate Installation 04 from the Halo Universe, and I know only three things that can help me in this simulation:

1: The diameter oh Installation 04 is 10.000 kilometers

2: The gravity of installation 04 is that of earth or very near to it… 9.82 m/s^2

3: The gravity of installation 04 is achieved by Centrifugal Force
Now by using this information I should be able to figure out the rotational speed of the ring… But I can’t!
I simply don’t know which formula to use. I searched wikipedia but I couldn’t figure it out, so if any of you know how to do this I would be very thankfull.

Oh and please correct me if some of the info is wrong :slight_smile:
(I’ll add more info as I go)

#2

I found this on wiki

#3

Thanks alot mate, I will look into it :slight_smile:

#4

“One down, fifty billion to go!”

Well now that I’ve got the formula, it’s time to mess around with XD

As far as I know I need to isolate rpm since it’s the only factor I don’t have. But I don’t know what to move nor where…

So here are original formulas.
Where:
g = Decimal fraction of Earth gravity
R = Radius from center of rotation in meters
π =3.14159
rpm = revolutions per minute

I hope this helps you, help me :slight_smile:

#5

Thats easy…
you have to figure out how fast it has to turn in order to acheive a “g” of 9.82 on the outside of the ring.

that being said. using the first equation on the site that larryboy gave you. and plug a 1 into the rpm spot and figure the question…
then put a 2
then a 3
ect 4
ect 5
untyil you reach a 9.82 as the answer… then you have your rpm…
PEACE
Nickadimos

#6

It’s far too hard to do it that way! But thanks anyway…
note: 10.000.000 in the formula should be 5.000.000.

#7

geezz simple math
http://upload.wikimedia.org/math/4/2/5/4252162fb5973d9f3fa12696d8f16d3f.png
–> omega = SQRT(a/r) roughly SQRT(10[m/s^2] / 5 e^6 [m] ) …
SQRT(2) 10e-3 [1/s]
0.001414 … [1/s]
5 [1/hour]
BM

#8

Say again? XD

#9

5 rounds per hour.
Not sure if the omega ( ω ) above should even be divided by 2PI so you"ll end up with once around in an hour.

Ah! yes 1 rad ~ 60 degree /second

so it is finally a little less than once per hour

guess why the earth, which is almost the same size only does one in 24 hours