mikeD10:
1lb = 4.45N, so 210.lb. = 935N; 935N/9.81 = 95.3kg.
10ft = 3.04m
So lets assume an elastic collision, and your body is in contact with the ground for .2 seconds (you hit the ground, bounce back up, or bounce the carabiner).
PE = mgh
KE = (.5)mv^2
and if mgh = .5mv^2, then v = (2gh)^(.5). So v = 7.72 m/s.
And assuming something like .2 seconds contact with the ground (carabiner), you say 7.72/.2 = A (deceleration) = 38.62 m/s^2.
F=ma = (95.3)(38.62) = 3680N.
So falling something 10 ft. would be the equivalent of multiplying your weight by four times or so. If you fell from that distance, you would need a carabiner rated at least 827 lbs.
Beware, though, because my .2 seconds is a completely arbitrary number. The strength of your carabiner entirely depends on the tensile strength of the rope that you are planning on using. Assume you have a rope that has entirely NO give (unnatural), then you would have to make a dead stop, and then you would need a carabiner that is infinitely strong. If you were using a nice enough bungee, such that it might counteract gravity at a slow enough rate (say, .7 seconds) then you could concievably have a carabiner that is rated at exactly your weight and get away with it.
Anyhow, I carry an aluminum carabiner, which is just strong enough to hold my keys.
Take care guys.