Parts Outer Body

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Kuryakyn BAT LASHES OUTER FAIRING ACC'S body motorsports parts 419133 41-9133
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1988-2000 CHEVY/GMC C/K PICKUP (OLD BODY TYPE) REAR BUMPER OUTER BRACE SET NEW
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HOT BODIES D8S  67416 SUSPENSION PIN SILVER (FRONT/OUTER) - GENUINE NEW PART!
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NEW Motor Saver Outer Saver Pre-Filter for HPI Hot Bodies D8/T, part no. OS1005
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Parts Outer Body
How could you lose 10 pounds in one minute on a rotating space station?

The space station itself is a torus shape, and is designed so that one gee of gravity will be along the outer edge, and zero gees will be in the middle. Seems like it'd be fairly simple, but the answer's been eluding me. >.>
Oh, and you can't amputate any body parts. ;)
The space station itself is a torus shape, and is designed so that one gee of gravity will be along the outer edge, and zero gees will be in the middle. Seems like it'd be fairly simple, but the answer's been eluding me. >.>
Oh, and you can't amputate any body parts. ;)

EDIT: I know you'd move closer to the center of the torus, but is there any way to calculate how much closer? Like, a percentage of the ratio or something along those lines?

you are correct that you move towards the centre.
To calculate how far you need to look at Centripetal force - (Centrifugal force is the apparent affect of Centripetal force)

centripetal acceleration can be written in terms of the angular velocity (w)
C = w^2.r
angular velocity w=velocity/radius
C = v^2/r
Force = mass * acceleration
Centripetal Force Cf = (mass * velocity squared )/ radius
ie. Cf =(m.v^2)/r or Cf = mass * angularvelocity^2 * r
as you can see there are 2 key terms - the radius and the velocity . both are needed to work out the ratio change

so if at the edge of the torus at radius r1 you experience a force Cf1 and at an inner radius r2 you experience a force Cf2

then Cf1-Cf2 = m.w^2 *( r1 - r2)

to convert the 10 pounds to Force, convert to Newtons
10lbs is 44.48 Newtons
if the person is 70kg or 154.4lbs
(there are 2.205 pounds per kilogram)

44.48 = 70* w^2 * (r1 - r2)

lets decide on an angular velocity of 0.1 radian per second
this is 5.7 degrees per second or 0.96 revolutions per minute

as you specified 1 gee we can calculate r1
1 gee is 9.8 m/s - so 1kg weighs 9.8 netwons

r1 = C/w^2 = 980 m

note that the velocity v = w * r = 98 m/s = approx 352 kmh

r1 - r2 = 44.48/(70 * 0.1^2)
r1 - r2 = 64

ie you move in 64 meters to change your weight by 10lbs in this example. so if you want to lose the weight in 1 minute you have to move at just over 1 meter per second towards the interior of the space station.

there is a good reference on this discussing the max rotational speeds tolerated by humans. http://www.spacefuture.com/archive/artificial_gravity_and_the_architecture_of_orbital_habitats.shtml



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