From: shoham@ll.mit.edu (Daniel Shoham)
Newsgroups: sci.physics,sci.math
Subject: Re: Elevator mayhem
Date: 17 Sep 92 02:29:18 GMT

burton@cfm.brown.edu (Joshua W. Burton) writes:
> So:  you're on the 34th floor, an even
>hundred meters up, when you hear the ~TWANG~ of four cables breaking at once,
>followed by a rusty *SCREECH* as the bolts holding two of the safety brakes
>give way from long neglect.  The remaining two brakes are both on the same
>side, and exert only enough force on the tracks to reduce your acceleration
>rate slightly and make a lot of noise.  At 8 m/s^2 (five-sixths of a gee or
>so), you have exactly five seconds to think this one through.

Actually, you shouldn't do that badly. Air resistance is on your side.

As this is a SCIENTIFIC newsgroup, lets consider some numbers. I will assume
that the elevator has a 1000kg mass and a 10m^2 bottom (choosing plausible,
round numbers). The density of air is about 1 kg/m^3. I don't know what the
drag coefficient is for elevators, so I will assume it is 1 (drag coef. is a
unitless number that is dependent {mostly} on an object's shape. Objects designed
to be aerodynamic - like missiles - have a DC well below 1, objects designed to be
aero-brakers - like parachutes - have a DC well above 1. Randomly shaped objects
have a DC around 1).

The force exerted by air against a moving object is:

F(air resistance) = DC * AD * A * V^2   (upward)

where:
DC = Drag Coef    = 1
AD = Air Density  = 1 kg/m^3
A  = Elevator area= 10 m^2
V  = Elevator velocity

The force exerted by gravity against you is given by:

F(gravity) = M * g (downward)

where:
M  = Mass = 1000 kg
gt = gravitational acceleration after taking brakes into account = 8 m/s^2
     (actually real brakes would increase resistance with velocity, but as this
      is part of the problem's defenition I will leave it unmolested).

Terminal velocity is reached when the downward gravitational force is equal to the
upward resistance force:

DC * AD * A * V^2 = M * gt

V = sqrt( M * gt / DC * Ad * A ) = 28 m/s = 63 MPH   (rounded off)

Well, a GOOD spring at the bottom should allow you to take this hit (air bags
seem to do the job with less "crunch-space") but lets see: If the spring is a
modest x=2m long exerting uniform resistance (again, spring increase resistance
with compression, but this is a quick order-of-magnitude calculation), you will
spend t=2x/V=0.14s suffering a=V/t=20g which is not fun and may break some bones
but not likely to burgerize you.


If this is your LUCKY day, this is a one-elevator-shaft. The air below you is
compressed and hence AD becomes very large. It is unlikely the air would have
anywhere to escape as fast as it is being compressed. Well, lets get to it:

F = - M * gt + DC * AD(y) * A * V^2

and AD(y) = AD(y0) * y0/y
where y is the elevator's height in the shaft, and y0 is the elevaator's original
height. AD(y0) is normal air density (= 1 kg/m^3). Using prime's for derivatives,
we can write:

y'' = (-gt) + (DC * A * AD(y0) * y0 / M) * y'^2/y

Where constant expression have been enclosed in (). Well, if someone knows an
analytic solution, why not post it. As I don't, i did a quick run on my PV-WAVE
and found that the elevator would come to a rest after about 8 seconds just 
around the ground floor.

Riders who are also subscribers of this newsgroup need not take any action, exept,
perhaps to extract near-death confessions/promises from an unsuspecting spouse.
Upon exiting, one would head to the nearest law office to bring a 7-figure suit
against some party or other for psychological damage, etc.

As we said, it is your lucky day. :-)

Dan Shoham
shoham@ll.mit.edu