quote; 1. Assumes low tech, not pre-charged piston type.
2. May depend on how and where they are installed
1. Mechanical styles are "shock absorbers" NOT air chambers
2. A filled air chamber is like a straw with your finger over the end of it. The liquid will NEVER drain out until air can get into it, and there is NEVER enough "pressure" to force air in, AND even if there were few systems are installed so that ALL the water can be drained.
As far as the location, good hydraulics specifies that the inlet to the shock absorber be in line with the direction of water flow.
Air chambers can be drained. Air chambers should not be compared to a drinking straw. Shock absorbers have o-rings that will eventually fail and could toxic like other old o-rings that melts on our fingers turning it black. But some are saying air chambers can't be recharged?
Yes, if the lines teeing off to the air chambers are full of water, the air chambers will never drain. If the lines are empty of water and air pressure is open to the atmosphere, it will drain. Law of physics tells us it must drain.
Using a drinking straw, the first thing to ask is what keeps the water from dripping down? The partial vacuum on the upper part of the tube can't be the full answer, because it doesn't explain why little droplets don't simply detach from the bottom side and fall without changing the size of the vacuum. The key physical ingredient of the explanation must be
surface tension, the extra free energy that's needed per area of water-air interface. Formation of a droplet increases the surface area, so unless the gravitational force is strong enough, it will take extra energy for a droplet to form, and the water will stay stuck in the tube.
How does that limit the tube radius, R? I'll do a crude calculation, not keeping track of small numerical factors, i.e. using some dimensional analysis. The extra energy required to form a drop goes as its area times the surface tension s, very roughly sR2 for the biggest drop that would form from a tube of radius R. The energy lost by lowering the water in that drop by a distance comparable to its size is roughly ρgR4, where ρ is the mass density and g is the gravitational acceleration. So for the bottom surface of the water not to shed drops we need roughly sR2>ρgR4 or R2< s/ρg.
The surface tension of water near room temperature is about s= 70 ergs/cm2, ρ=1 gm/cm3, g=980 cm/s2, so we end up with roughly R2< 0.07 cm2. That would correspond to a tube with diameter of about 5mm. Given that I haven't tried to solve the problem exactly, that answer could easily be off by a factor of 3. Is it anywhere near the measured cutoff between tubes that hold the water and those that don't?
Mike W.
p.s. A first test showed that ~3mm diameter gave stability and ~ 10 mm diameter was unstable, so this calculation seems to be pretty close.