Here''s an interesting look at the elevator problem. From my perspective the back-up breaking systems tend to work very well and thus if you're actually in free-fall I'd just totally disregard the breaking from the cables and what not since it has to be a pretty decent catastrophe to keep the emergency breaks from kicking in. If we do that, then all that's left is air resistance.

Basic equation:

d = v*t + (a*t^2)/2

assuming initial velocity is negligible:

t = sqrt (2*d)/sqrt(a)

Now let's say you'll be falling at the full 9.8m/s/s but the elevator will be falling slower due to air resistance. Let's say the acceleration difference is 'ar' so the elevator is falling at an acceleration of (9.8-ar)m/s/s. Now let's discover what ar has to be so that you have hit bottom of the elevator by the time the elevator hits the bottom of the shaft, calling the height of your center of gravity 'cg' and the height your up in the elevator 'h'.

sqrt (2*h)/sqrt (9.8-ar) >= sqrt (2*(cg+h))/sqrt(9.8)

This turns into:

ar >= 9.8-(9.8*h)/(cg+h)

Let's say you're 15 meters up and you center of gravity is 0.75 meters high. That would result in a necessary air resistance of 0.47m/s/s Of course a major problem there is the assumed constant, de-acceleration from the air resistance which would not be the case. Quick back of the envelope calculation shows that the speed half-way through the descent is about 9m/s. If you trust the math and base assumption in the above link then the air resistance at a constant velocity of 9m/s is 0.8m/s/s. So in this example you would be OK. Note that the total time of drop here is about 1.8 seconds, and it would take the majority of that time for you to go from your standing position to ground: a rather odd sensation one imagines. Of course if you hit the ground and the elevator hits its ground at the same time you've done just made it worse by removing the effect of the cushioning the slightly slower drop the air resistance gave you. Hopefully being on your back compensates!

Some calculations to find the break even height of drop if your cg was one meter above the elevator floor:

h+1 = 9.8/2*t^2

v_avg = 9.8/2*t

ar = 1*1*10*v_avg^2/1000

ar = 9.8-9.8*h/(h+1)

That give you a break-even at about 13.1 meters. I'd want time to spare and not hit at coincident times so I'd say 15 meters but the minimal effect of air resistance on end speed probably isn't that big a deal at this point. As a reference, 13-15 meters translates into 43 to 49 feet, or at typical 12 foot stories, that's around four floors.

Basic equation:

d = v*t + (a*t^2)/2

assuming initial velocity is negligible:

t = sqrt (2*d)/sqrt(a)

Now let's say you'll be falling at the full 9.8m/s/s but the elevator will be falling slower due to air resistance. Let's say the acceleration difference is 'ar' so the elevator is falling at an acceleration of (9.8-ar)m/s/s. Now let's discover what ar has to be so that you have hit bottom of the elevator by the time the elevator hits the bottom of the shaft, calling the height of your center of gravity 'cg' and the height your up in the elevator 'h'.

sqrt (2*h)/sqrt (9.8-ar) >= sqrt (2*(cg+h))/sqrt(9.8)

This turns into:

ar >= 9.8-(9.8*h)/(cg+h)

Let's say you're 15 meters up and you center of gravity is 0.75 meters high. That would result in a necessary air resistance of 0.47m/s/s Of course a major problem there is the assumed constant, de-acceleration from the air resistance which would not be the case. Quick back of the envelope calculation shows that the speed half-way through the descent is about 9m/s. If you trust the math and base assumption in the above link then the air resistance at a constant velocity of 9m/s is 0.8m/s/s. So in this example you would be OK. Note that the total time of drop here is about 1.8 seconds, and it would take the majority of that time for you to go from your standing position to ground: a rather odd sensation one imagines. Of course if you hit the ground and the elevator hits its ground at the same time you've done just made it worse by removing the effect of the cushioning the slightly slower drop the air resistance gave you. Hopefully being on your back compensates!

Some calculations to find the break even height of drop if your cg was one meter above the elevator floor:

h+1 = 9.8/2*t^2

v_avg = 9.8/2*t

ar = 1*1*10*v_avg^2/1000

ar = 9.8-9.8*h/(h+1)

That give you a break-even at about 13.1 meters. I'd want time to spare and not hit at coincident times so I'd say 15 meters but the minimal effect of air resistance on end speed probably isn't that big a deal at this point. As a reference, 13-15 meters translates into 43 to 49 feet, or at typical 12 foot stories, that's around four floors.

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