Silly question about air pressure in a shock/fork

[Minion]

 

 

And what the h3ll dose a cat have to do with it??

 

Schr?dinger's cat. The act of observing an event effects the event being observed.

 

http://www.imagehost.co.za/image-BC48_4A7BC7A5.jpg

Edman2009-08-11 11:56:29

[Capricorn]

 

oh nfm. honestly somethings are always overinflated.

Capricorn2009-08-11 12:25:34

[Goodbadugly]

The pressure you measure is the difference between the inside and outside pressure. At sea level the atmospheric pressure is about 1 Bar (101.3 kPa). One bar is 100kPa to be exact. In Gauteng the atmospheric pressure is around 0.7 Bar. So if you pump your tire 2 Bar at sea level, it will be around 2.3 Bar in Johannesburg. The amount of air molecules inside the tire stays the same. The pressure from outside is less. So the measured pressure goes up.

The shock pressure is MUCH higher. The pressure will go up by less than a bar. Nobody will notice that. 

[jugheaddave]

awsome, this is what I love about forums, you have a question, and it gets answered, thanks guys!!!

 

 

[Johan Bornman]

The pressure you measure is the difference between the inside and outside pressure. At sea level the atmospheric pressure is about 1 Bar (101.3 kPa). One bar is 100kPa to be exact. In Gauteng the atmospheric pressure is around 0.7 Bar. So if you pump your tire 2 Bar at sea level' date=' it will be around 2.3 Bar in Johannesburg. The amount of air molecules inside the tire stays the same. The pressure from outside is less. So the measured pressure goes up.

The shock pressure is MUCH higher. The pressure will go up by less than a bar. Nobody will notice that. 

[/quote']

 

That's where the cat comes in. Don't let the observed pressure fool you. It is like the aeroplane going up. Taking off from Cape Town International, it had a cabin pressure of 1 Bar and it maintained the pressure the same all the way to 10 000 meter cruising level. Sitting inside, we all still feel 1 Bar, but if you were to measure the inside pressure with your shock pump that's hanging outside the plane, it would tell you that the pressure is much more.

 

Same story with the shock. It is a rigid structure so the pressure remains the same and the shock still works the same, no matter what the observed pressure would be. 

 

You said "nobody will notice that", which is true. But the reason nobody would notice it is not because of a small difference in observed pressure but because of a zero difference in the shock's performance on the moon or in Cape Town. A shock doesn't care about ambient pressure, only about the pressure inside.

 

 

[Johan Bornman]

awsome' date=' this is what I love about forums, you have a question, and it gets answered, thanks guys!!!

[/quote']

 

Don't chicken out so quickly. We haven't skinned this cat quite yet.

 

 

[droo]

For once I'm going to have to disagree with you... and here's why:

 

 

 

In a shock (or tyre or weather balloon, whatever) the absolute pressure will remain the same no matter what the pressure outside is. The gauge pressure, which is the difference between the inside and outside pressures, will vary.

 

 

 

Where I disagree with your argument is here: the external pressure will act on the outside of the system, causing it to compress (yes, very slightly in this case, but I'm being pedantic.) So if the external pressure is decreased, it will result in a decrease in the external force on the system which would have to be compensated for in order for it to work as it did at the initial pressure.

 

 

 

So if the pressure inside is 7 bar, and the pressure outside is 1 bar, the difference will be 6 bar, and the resultant force on the piston will be 6a, where a is the piston area. If the outside pressure drops to 0.7 bar, the gauge pressure would be 6.3 bar, and the resultant force would increase to 6.3a.

 

 

 

Which means that to keep it working the same as it did at sea level, the gauge pressure is the critical one, as it determines the resultant force on the system.

[Johan Bornman]

For once I'm going to have to disagree with you... and here's why:

In a shock (or tyre or weather balloon' date=' whatever) the absolute pressure will remain the same no matter what the pressure outside is. The gauge pressure, which is the difference between the inside and outside pressures, will vary.

Where I disagree with your argument is here: the external pressure will act on the outside of the system, causing it to compress (yes, very slightly in this case, but I'm being pedantic.) So if the external pressure is decreased, it will result in a decrease in the external force on the system which would have to be compensated for in order for it to work as it did at the initial pressure.

So if the pressure inside is 7 bar, and the pressure outside is 1 bar, the difference will be 6 bar, and the resultant force on the piston will be 6a, where a is the piston area. If the outside pressure drops to 0.7 bar, the gauge pressure would be 6.3 bar, and the resultant force would increase to 6.3a.

Which means that to keep it working the same as it did at sea level, the gauge pressure is the critical one, as it determines the resultant force on the system.[/quote']

 

I think you have a klap coming your way. Not from me...but possibly from a chick with a dictionary.  But before someone accuses me of aggression, I'm referring to your signature. It's quite a good one.

 

Back to the problem at hand.

 

The original question Jughead posed was:

 

"Does outside air pressure (ie sea level vs 2000m above sea level) affect shock/fork air pressure?"

 

I'm taking the liberty to put words in his mouth by adding this rider:  "and therefore do I have to compensate for it when I travel to places of different altitude?"

 

I know this is not his words but I think it is a fair assumption as to what he really wanted to know. Dave...chip in here.

 

I will assume that the fork is rigid on all sides in respect of compression from atmospheric air. When air pressure outside the fork increases or decreases, the pressure inside is not affected since no extra air is added nor is the size of the chamber altered.

 

Next, lets look at the strucutre of the fork. It does not represent a simple syringe in your hand but is more complex. It is a piston inside a piston or in some cases, a double-champer positive/negative piston inside a single piston.

 

The single piston I refer to is the stanchion moving inside the slider. The stanchion can be pushed inside the slider. The air pressure inside does not change since the volume of the slider is compensated for by the movement of the inside piston upwards in the stanchion. In other words, the pressure inside the slider remains constant in operation. I'm assuming that the slider seal is airtight.

 

The  piston that affects ride height is inside the stanchion. If the piston is moved upwards (the fork is compressed), it compresses a column of air above the piston and against the crown of the fork.  If it is a dual-air fork, the piston's movement wrt to the pressure in the slider remains the same as the single-air example and I'll argue that it makes no difference to the answer to Dave's question if it is a single or dual air piston.

 

OK, lets go for a ride and see what happens.

 

We're at sea level and the fork is pumped to 6 bar. This is with no-one sitting on the bike. The barometer reads 1 bar.

 

We ride up a mountain to a point where the barometer reads 0.5 bar.

 

Question: Does Dave's bike now have a different sag (compared to sea level levels) with Dave on or not?

 

 

 

 

 

 

 

[Capricorn]

droo: when you speak of resultant pressure, are you saying that the ambient pressure is acting on the same piston surface area as the internal pressure? unless they are acting in the same plane, there wont be a change of the force on generated by the internal pressure acting on the stanchion cross section.

[droo]

@ Johan: I've got a klap coming MY way? You're the one assuming a woman's going to need a dictionary to understand my sig

 

 

 

Back to the point: I assumed a simple system. For negative springs and the like, I'd have to break out the old engineering textbooks, although I suspect (as, it seems, do you) that it wouldn't make a difference.

 

 

 

@ Capricorn: Yup, that's exactly what I'm saying. By Newton's 3rd law (balancing forces) and Pascal's law (fluid pressure exerting a uniform force) any force acting along any axis other than the axis of the piston will be balanced by an opposing force and can therefore be ignored. (If this wasn't the case, the fork would randomly scoot from side to side.)

 

 

 

It doesn't matter what shape the outside of the piston is, the effective CSA on which the force along the axis of the piston acts will effectively be the same as the internal surface area of the piston (said another way, look at the piston along its axis from both sides, and the CSA you see is the same)

 

 

 

@ anyone who's still reading: In this case much of this is irrelevant though, because it's all constant. The only thing that changing is the atmospheric pressure. And by my previous argument, that means that Dave's bike will sag less at altitude due to the force being exerted by the atmospheric pressure on the outside of the piston being less, so the resultant of the two (atmospheric and internal) would be (in this example) [0.5 bar] x [CSA of piston] greater, requiring an equivalently greater force (N3 again) to compress it.

 

 

 

EDIT: Johan, just saw your plane example. If there was no difference in resultant force between sea level and altitude, fuselages wouldn't be built as pressure vessels.droo2009-08-13 03:25:21

[Rocket-Boy]

Wow, Dangerous Dave, my advice to you. Take a shock pump with you and if you need to adjust it when you get there then do so.... otherwise just ride it like it is 

[droo]

Wow' date=' Dangerous Dave, my advice to you. Take a shock pump with you and if you need to adjust it when you get there then do so.... otherwise just ride it like it is?

 

[/quote']