I was wondering how the air suspension managed to both lower the ride height and spring stiffness by altering the air in the airbag. Wouldnt adding air increase the ride height and stiffen the ride, since the pressure rises? Exactly the opposite of what you want high and hard. How do they do yhe opposite just with air? Am I missing something?

 

You have a number of errors there.

Imagine a cylinder filled with air, with a weight on top of it. The weight exerts a downward force. Pressure inside the cylinder exerts an upward force. If the upward force is less than the downward force, the cylinder compresses, the pressure increases, and the upward force increases. Similarly, if the upward force is higher, the cylinder expands, the pressure decreases, and the force decreases.

In either case, the cylinder is only at equilibrium if the pressure is exactly balanced against the weight. When the Panamera adds or removes air, the end pressure is always precisely the same. The only difference is the total volume of air used to create that pressure.

When the car hits a bump, the cylinder compresses, and the force it exerts increases. If the volume of air is small, and the travel distance short, this happens much faster than if the volume of air is large and the travel distance long. Thus the "hardness" of the ride isn't about the initial force, it's about how fast the suspension reacts to a bump.

Steel springs, by the way, behave the same way. Which is why cars with lowered suspensions and shorter springs have a much harder ride.

 

So what you are saying is, no matter what it does, only the height will vary and the spring rate will be the same. However http://www.porsche.com/usa/models/panamera/panamera/chassis/adaptive-air-suspension/ here it says the spring rate gets higher as well as the the ride height getting lower.





Plus here, https://accuair.com/intro

In the q and a section it says that as the ride height increases, the spring rate will go LOWER which is exatly the opposite of what we want

 

No, I didn't say that. I said the resting pressure remains the same at all ride heights.

I also said that for smaller volumes, compressing the cylinder increases the pressure and hence the upward force faster. Which is another way of saying that the spring rate increases. Which, as you read in the FAQ, is what happens.

A lower spring rate is exactly what we want for a softer ride. That you think it's the opposite implies that you don't understand what spring rate means.

Spring rate is the amount of force required to compress a spring a given distance. It's always given as force/distance, i.e. pounds / inch. A spring rate of 125 lbs / inch means that if you compress the spring 1", it exerts 125 lbs of force. If you compress it 2", it exerts 250 lbs of force. The more you compress a spring, the more force it generates.

A bump deflects your front wheel upward. The initial upward velocity depends on things like the size of the bump and your speed, but not your springs. Your springs exert force as they compress, slowing the wheel and eventually pushing it back.

The lower the spring rate, the longer it takes for the wheel to stop and return, the less acceleration you feel, and the softer the ride.

 

Thanks, that helped me.

 

@iammulva, go seinfeld!

 

yeah i mixed up the terms but the thing i an asking remains the same. If we increase the ride height by putting air in the suspension it will make it stiffer. And we end up with STIFFER AND HIGHER car. If we decrease the ride height by taking air out of the suspension, we will have SOFTER AND LOWER car which is the exact opposite of what we want. I just mixed up the terms in my other post since english is not my mother tongue. The problem remains though. And it says that it indeed is the opposite of what we want in faq section i posted the link to.

 

No, that's not what happens. I've explained the physics behind why this isn't so. The FAQ doesn't say what you think it does, because you don't understand the terms.

Look, if you can't follow the physics, try the empirical method. The cars exist. They have air suspensions. They're softer when the ride is higher, and harder when it's lower. You could go test drive one today and see that for yourself.

Are you trying to prove they don't exist? Or are you trying to understand how air suspensions work?

I'm certainly willing to take other approaches to explaining it, but would it help? Because I'm not sure where you're losing the path. It almost seems like you've decided on your conclusion, and you're rejecting explanations because they don't fit that conclusion.

I just want to know if you're genuinely interested in the answers.

 

I actually think you re the one who cant keep up with the physics, since what i said makes perfect sense and you didnt explain why it didnt. Plus i also dont think you know how to read. I am now going to quote from the faq that you think it say what you think. Lets see if you know hot read properly. Q: Can I adjust the Spring Rate of my air springs along with the height?
A: Yes, but the Spring Rate will be directly connected with spring height. As you raise the air spring’s height, you will also be increasing the air pressure inside, thus increasing its Spring Rate. Unfortunately, this is opposite from what most of us would want for performance applications, (low and firm for handling, high and soft for mobility). Here is it is. See what it does? And i would love to test it if you let me drive tour panamera actually

 

Oooook. You don't understand, and instead of trying to understand, you're trying to turn this into an internet fight.

I'm not really interested in trying to prove anything to you. I was willing to help you learn, but I don't have to "win." The cars exist, they behave as I describe. I happen to own a Panamera that has air suspension, so I know first hand.

To be honest, I didn't bother reading that AccuAir FAQ before you quoted the precise wording. It's wrong. Very wrong. It makes a lot of the same errors you did. Makes me wonder about whoever is running that business. Maybe it's just the web-page author who has no practical experience with air suspensions?

It says increasing the height increases pressure. It doesn't. It increases the mass of air inside the cylinder, the pressure remains constant. Spring rate is unrelated to static pressure.

I'll try one more time, then I'm done. I started with proportions because so many people are math-averse, but now I'll haul out the mathematics. This is the http://en.wikipedia.org/wiki/Ideal_gas_law:

PV = nRT.

where P = Pressure, V = Volume, n = mass of the gas in moles, R is a constant (8.314 Joules / mole-degree Kelvin), and T is the temperature (in Kelvin).

Solving for P:

P = nRT / V.

The force the spring exerts is equal to the pressure (P) multiplied by the cross section area of the the cylinder. Increases the diameter of the cylinder, and you get more force for the same pressure.

This is not the spring rate. The spring rate is the change in force when we compress the cylinder.

Suppose our cylinder currently exerts 1000 lbs of force, is 6.2" in diameter with a cross-sectional area of 30 square inches, and is currently 12" in length. The internal pressure is 1000lbs / 30 square inches = 33 lbs / square inch.

What happens if we compress the cylinder? If the mass of air (n) and the temperature (T) remain constant, the "nRT" part of the Ideal Gas Law won't change, so P*V will remain constant. In this case P * V will always be 33 lbs / sq inch * 360 cubic inches = 11,880 inch-lbs.

1" of compression reduces the volume to 11 * 30 = 330 cubic inches, P = 11,880 inch-lbs / 330 cubic inches = 36 lbs / sq inch. 36 lbs / sq inch * 30 square inches = 1080 pounds.

Since we compressed the spring 1" and increased the force by 80 lbs, the spring rate for the first inch is 80 lbs / inch.

If we drain out half the mass (n) of air, so PV = 5,940 inch-lbs, the cylinder contracts to 6". The initial force is still 1000 lbs, and initial pressure is still 33 lbs / square inch.

Decreasing the cylinder length by 1" decreases the volume to 5 * 30 = 150 cubic inches. P = 5,940 / 150 = 39.6 lbs / square inch, and the force is 39.6 lbs / square inch * 30 square inches = 1,188 lbs.

The spring rate for the first inch of the shorter cylinder has increased from 80 lbs / inch to 188 lbs / inch. The shorter the cylinder is at equilibrium, the higher the spring rate.

This happens because 1" is a much bigger percentage of 6" than 12".

EDIT: OK, the forum is doing weird things to my Wikipedia link to the Ideal Gas Law. I didn't intend to insert a graphic there, just a link for further reading. I hope it's still clear.

 

Thanks for explaining it this way. Now I see your point, and it makes sense.

 

Good, I'm glad I was finally helpful. While I'm naturally inclined to just go straight to the mathematics, I've found that fairly often that gets in the way. Some people freeze up at the sight of equations and automatically dismiss them as "gobbledygook."

 

Today I had opportunity to do deep suspension check and only one bad thing was found - both sides cracked 'ADDITIONAL SPRING REAR AXL' as named in catalog. What was a surprise that such a simple thing doesn't have aftermarket replacement .... autoparts shops even were not able to see part number 970.333.10505 (004951000375) !!!