# Archive for March 22nd, 2013

### Showing My Math

So here’s the deal. I’m working on a new project and I want to make sure at least some of the details have actual technical grounding. I’m okay with a little handwavium, it’s probably unavoidable, but I want to at least have some grounding in reality. Problem is, I’m not all that great at actual technical grounding, as the last physics class I took was non-AP physics in high school where I barely got a B. So I might occasionally make these posts, I might make just this one, in an attempt to crowdsource some of my equations. The questions I have are whether I’ve got the right equations, and whether I’m using them correctly, and to also play around with LaTeX a little. But mostly the first two things.

So here’s today’s problem: Given a cylindrical space ship with an internal radius of 6km, how fast must it be rotating to create a centripetal acceleration equivalent to earth gravity for someone standing on the inside surface? I didn’t know any of these equations, but found them at this rather helpful forum post. First, we must find the speed at that 6km point that would produce an acceleration of 9.8m/sÂ²: $a=\frac{v^2}{r}$ $a=9.8 \frac{m}{s^2}$ $r=6000 m$ $9.8=\frac{v^2}{6000}$ $9.8*6000=v^2$ $58800=v^2$ $v=242.49 \frac{m}{s}$

That number sounds awfully damn fast, but consider the speed of rotation of the earth at sea level on the equator is roughly 465 m/s. Next step, at least what I’m assured is the next step, is converting this into radians/second: $W=\frac{v}{r}$ $v=242.49$ $r=6000$ $W=\frac{242.49}{6000}=0.0404$

Finally this can be converted to revolutions per minute. The conversion formula I found is: $1 \frac{rad}{s} = \frac{60}{2pi} = \frac{30}{pi} rpm$ $0.0404 \frac{rad}{s} = \frac{30*0.0404}{pi}=\frac{1.212}{pi}=0.386 rpm$

Therefore the ship is rotating at a rate slightly faster than once every three minutes. What I didn’t expect is that, since the rate of rotation is a constant, centripetal acceleration increases linearly from the axis of rotation. I’m so accustomed to formulas for gravity having squares all over the place, but this isn’t, strictly speaking, gravity. It’s an acceleration equal to gravity. So at half the distance from the axis of rotation, we can work backwards with W as a constant… $W=\frac{v}{r}$ $0.0404=\frac{v}{3000}$ $v=0.0404*3000=121.2\frac{m}{s}$ $a=\frac{v^2}{r}$ $a=\frac{121.2^2}{3000} = \frac{14689.44}{3000} = 4.9 \frac{m}{s^2}$

Which is equivalent to half gravity.

My next trick will be to find a formula that describes the rate of descent for a body falling through linearly increasing gravity. That’s less likely to come up in-story, but more for my own curiosity.

Edit: Some further poking around (which, I’m ashamed to say, has mostly been at Wikipedia so far) suggests that 2rpm is about the maximum rotation that most humans can adjust to with no ill effects, so my rotation of nearly 1/6 that rate is shockingly safe in and of itself. So that’s good to know. Now if only it didn’t have a “citation needed” tag.