Re: What would be the diameter of revolving spacestation to maintain a 0.4g

Mike, you ask a very straightforward question.  I don't know your 
background, so I may be writing this a little below your level of 
education.  Please bear with me.  You want a space-station which is 
revolving once per minute to produce a centripetal acceleration of 0.4*g 
(0.4*9.8 m/s/s or 3.92 m/s/s)

Well, fortunately, there is a simple relationship between these quantities:

a = (v^2)/r     NOTE that "v^2" means v*v

where "v" is the speed of the particle (a person in this case, who is 
standing at a distance "r" from the center of rotation). "a" is the 
acceleration which the person feels.

Acceleration due to turning (or rotation) is often referred to as 
centripetal acceleration.  You can read about it at 
 http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html and, in case you've 
heard about centrifugal force, visit this page
http://franklin.icsd.k12.ny.us/...
a little more in-depth treatment of this whole circular-motion topic is at
 http://home.cord.edu/dept/physics/p128/lecture98_14.html
Finally, you may be interested to know that a similar question was asked 
here
 http://www.newton.dep.anl.gov/newton/askasci/1995/astron/AST240.HTM


In your question, we can determine the speed "v" by multipling the 
rotational rate "w" (w = 1 RPM) by the distance "r". Now, we have

a = ((w*r)^2)/r = (w^2)*(r^2)/r = (w^2)*r

we may solve for r with a little algebra....

r = a/(w^2)

We're going to need all our quantities in consistent units, so 1 RPM = 1 
revolution per 60 seconds or 1 RPM = (2*3.14159 radians)/(60 seconds)    
You'll notice that I used the value of Pi.  Why?  A full revolution is 360 
degrees but I'm not going to use degrees because the *natural* unit for 
angles is radians.  Radians are the natural unit for angles because a wheel 
of radius 1-meter that makes 3 revolutions as it rolls away travels a 
distance of
(3 revolutions)*(2*Pi radians per revolution)*( 1 meter ) = 
roughly 18.85 meters

Now we are ready to insert values into the formula
w = 1 RPM = (2*3.14159 radians)/(60 seconds) 
  = roughly 0.10472 radians per second
a = 0.4*9.8 meters per second per second 
  = 3.92  meters per second per second 
r = a/(w^2) = 3.92/(0.10472^2) meters 
  = 3.92/0.010966 = roughly 357.46 meters

This tells us that as long as the person's center (actually, center-of-
mass) is about 357.46 meters from the space station's center of rotation, 
then that person will experience about 0.4 times Earth's gravity.  
Obviously, you're going to want the floor a little further than 357.46 
meters from the center and the outer wall of the station will have some 
thickness.  Your space station will be need to be a little more than 714.92 
meters (2*357.46) in diameter.

I'm accustomed to working in metric units.  If you prefer feet to meters, 
know that 1 meter is about 3.28 feet.  714.92 meters is about 2,345.54 or 
about 4 tenths of a U.S. statute mile.

One last note: I don't know your background, but in exploring the web I 
found a dissertation on "The Architecture Of Artificial-Gravity 
Environments For Long-Duration Space Habitation" at 
 www.artificial-gravity.com/Dissertation/ You might
find it interesting.

Troy http://surf.to/tdg/