ME1: Due to conservation of angular momentum, when the Citadel closes, objects towards the ends of the Citadel will experience a roughly .13 g acceleration perpendicular to the usual artificial "gravity". That doesn't sound like much until you realize that your best china set will head for the walls at 1.27 m/s2. Smash city!
For the three of you who care about the maths, let L=mvr where L equals angular momentum, v equals the rotational velocity (perpendicular to the radius of rotation) and r equals the radius of rotation.
Since L is constant for an unaccelerated system, and the mass of our objects (the Citadel and my china set) also remains constant, we are left with
V1*R1=V2*R2
Where the left side represents the open Citadel velocity (V1) and radius (R1) and the right the closed. Pretty obvious so far, I assume.
The thing is that an object's radius depends upon the centre of mass, so in the open Citadel's case, this will mean averaging the narrow part with the wide part. The Presidium is the narrow part when open, and for ease, I
assume that's the uniform radius for the entire Citadel when closed, or R2. We'll call the wide radius Rw for now.
We then set Rw=K*R2 where K equals the ratio of Rw/R2 Again, pretty obvious.
To get the average radius of rotation of the open Citadel, we average the radii. This gives us
R1=(Rw+R2)/2=(KR2+R2)/2=(K+1)R2/2
Since
V1*R1=V2*R2
V1*(K+1)*R2/2=V2*R2
((K+1)*V1)/2=V2
or simply,
V2= (K*V1)/2 + (V1)/2
(extra parentheses for clarity)
We can see how this is true, because if the citadel open and closed were to be the same radius, K would equal 1. Plugging 1 in there for K, we see that there would be no change in V in that case.
Anyway, that's for the Citadel. Looking at the wiki, we derive a K of 1.78 (12.8km/7.2
km) Its rotational velocity at the end is 191.5 m/s when open, based on one rotation every 3.5 minutes. Plugging our numbers in, we get a V2of 266.2 m/s when closed.
So far, so good! But what about the china set in my flat on the end of an arm? Unlike the Citadel, its
centre of mass is solely located at the wide end, for a rotational radius of Rw, or K*R2. It will therefore have a different final rotational velocity, let's call it Vc
So for the cup,
V1*K*R2=Vc*R2
which gives us Vc=K*V1
Plugging in the numbers gives us a Vc of 266.2 m/s for a difference in velocity of 74.7 m/s
between the cup and the Citadel. The cup would want to slide in a spinward direction (again, nothing to do with the centripetal vector) You'll note that exactly halfway down the arm, delta V is 0, and at the Presidium end, delta V will be in an anti-spinward direction (the Citadel accelerates, but the table service does not).
AND YET NO ONE EVER MENTIONS THIS!!!!!111
Oh yes, ME2. Boo the Space Hamster was onboard when EDI opened the ship to space (except for the engine room). HOW DID HE SURVIVE????? In a recent thread about this, it became clear that the ONLY way Boo could survive would be Reaper Tech. Which brings up the obvious suspicion---
Modifié par SuperMedbh, 10 juin 2010 - 02:56 .