The interesting thing about this idea might be the loads on all the other surfaces which suddenly become exposed to an overpressure of 0.1 bar.

The shaft wall has a vastly greater surface area than the cabin floor does. Even the doors on every floor are going to see a couple of unexpected tonnes of outward force - hopefully they'll blow out before the building structure fails (looking on the bright side, at least you won't have to excavate through a mound of rubble to recover the bodies of the hapless lift occupants).

Maybe Hiram Otis had it right all along.

A.

We saw that "collapsing column of compressed air = sideways force at bottom of building" when both WTC towers compressed vertically. The lower floor windows blew out before the upper floors reached their level <- caused by the very-fast-rising internal air pressure of the building as the top fell down.

If the buildings had fallen sideways (tilted over as a single piece) then you would not see the air pressure rise inside the building -> There would be no change in volume.

If you have an elevator system serving 30 floors, you will have multiple, side by side, elevator shafts, with no intervening walls, therefore do not expect much of an effect from of an air column. Even a single shaft elevator will not produce much of an air column in case of free fall( and that will happen only if the braking system does not work)as there is too much clearance between the shaft walls and the elevator. Regardless of the type of elevator, there is a braking system in the event the hoisting cable brakes.

Hi chicopee
Is it an option to reduce such clearances (between shaft walls and elevator) in order to create damping in case of free fall ? Will that work ?

"If you want to acquire a knowledge or skill, read a book and practice the skill".

You could dig deeper into the maths.

You've already identified one of the simultaneous equations - the one that says that for constant descent rate, shaft pressure equals car mass times floor area.

The others are that at the same constant descent rate, the volume of pressurised air that leaks past the seal per second must be equal to the volume swept by the descending car per second and that the volume that leaks past a given seal clearance per second is a function of the shaft pressure.

Solving those will allow you to calculate the terminal velocity that the lift car will tend towards given a chance. (The Particular Integral of what is going to be a messy differential equation solution).

The interesting bit is the transient response - Imagine that your 1m^2 1000kg cabin is at the top of a perfectly sealed shaft. The perfectly sealed shaft implies that the terminal velocity should be zero - but the lift is actually going to exceed this. It will accelerate for the first 10% of the shaft length (this being how far it has to go in order to compress the air in the shaft to 0.1 bar) and then overshoot that point while it decelerates to the theoretical terminal velocity. The same, but messier, is going to happen for an imperfectly sealed shaft.

My guess is that the insurance companies aren't going to be interested in an emergency braking system whose terminal velocity depends in a complex way on where you are when you need to use it.

A.

My guess is that a perfectly sealed shaft, might also to a certain extent prevent the lift from moving (normal operation) as it acts as damper.

So what about having tight wall clearances, pressure proof walls (don't know what is the cost of this, might be huge...), and together with this install a valve that is set in open position during normal operation to continuously balance the pressures. An accelerometer can trigger the valve to close in emergency (free fall) making the two pressure systems (air columns top and bottom of the lift) isolated.

I will dig deeper into the maths ...:)

"If you want to acquire a knowledge or skill, read a book and practice the skill".

Better to wait to the last split second and jump up as high as possible.

you must get smarter than the software you're using.