Thread: Independent Suspension Design

Big day today, but it did not look like we achieved much. All the bits are now in place ready for field testing, i could not get any images of it complete as it was to dark.
A couple of images of the cut out of the hitch mount.

Cheers
David

Took same images this morning of the weekend progress.

Hitch in place.
Shock in finished position as well as the limiting traps.
Control testing this coming weekend

Cheers
David

I have a question for the members that are running air springs.
We have the same air pressure in both air springs but the trailer is high on one side, the load is even across the trailer.
Does anyone have an idea why this is happening?

We have just completed our off road(slow) 700kg load test, every thing went really well, tried really hard to roll the trailer pushing it over some really interesting terrain. I'm not saying we could not get it to roll over but i needed to know that the suspension was not aiding the trailer to roll over, considering the load is a water tank full with the load centre 1.4m of the ground.

Very happy with the weekend results.

Cheers
David

Right some images from the weekend.
Very happy with the departure angle.
I really love the hitch, never used this type before.

Cheers
David

Originally Posted by Witchdoctor

I have a question for the members that are running air springs.
We have the same air pressure in both air springs but the trailer is high on one side, the load is even across the trailer.
Does anyone have an idea why this is happening?

We have just completed our off road(slow) 700kg load test, every thing went really well, tried really hard to roll the trailer pushing it over some really interesting terrain. I'm not saying we could not get it to roll over but i needed to know that the suspension was not aiding the trailer to roll over, considering the load is a water tank full with the load centre 1.4m of the ground.

Very happy with the weekend results.

Cheers
David

Many people incorrectly assume pressure adjusts height of air springs.

Pressure x effective area = load. So the pressure changes to suit the load and vice versa.

To change the height you add more air to increase height or remove air to lower. For the same load, the pressure is constant as air is added or removed, assuming the change in air spring height is about the middle of the recommended operating range.

That said, the effective area does change over the operating height of the air spring, particularly associated with the piston diameter. Note all of my comments are for reversible sleeve type air springs like you are using.

Thanks John,

My understanding was if the air springs are equal & the load is also equal one would expect that with the same pressure in each spring the load would then sit level, very simplistic. I'm now thinking that the springs are not identical after all. I do not have a problem with putting more air in one spring to make the load look level.

We will just have to do more testing, i'm still very happy with were we are at so far.
This weekend i will be starting the electrical to get it ready for registering.

Thanks Martin for the Boost wheels for this project.

Cheers
David

Originally Posted by Bush65

Many people incorrectly assume pressure adjusts height of air springs.

Pressure x effective area = load. So the pressure changes to suit the load and vice versa.

To change the height you add more air to increase height or remove air to lower. For the same load, the pressure is constant as air is added or removed, assuming the change in air spring height is about the middle of the recommended operating range.

That said, the effective area does change over the operating height of the air spring, particularly associated with the piston diameter. Note all of my comments are for reversible sleeve type air springs like you are using.

Originally Posted by Witchdoctor

Thanks John,

My understanding was if the air springs are equal & the load is also equal one would expect that with the same pressure in each spring the load would then sit level, very simplistic. I'm now thinking that the springs are not identical after all. I do not have a problem with putting more air in one spring to make the load look level.

We will just have to do more testing, i'm still very happy with were we are at so far.
...

Cheers
David

Comparing air springs to coil springs for our suspensions is like comparing chalk to cheese.

Static and kinetic energy from the application of load is transformed into strain energy in the material of coil springs as they deflect (extension and compression) – recall that natural laws deem that energy cannot be created or destroyed, but only change form.

For air springs, static and kinetic energy from the application of load change the pressure, volume and temperature of the trapped air in compliance with gas laws.

The usual/useful data provided for coil springs are diameter, length (free and minimum compressed), and spring rate. From these values it is simple to evaluate load vs deflection, etc.

Because air springs are often used as vibration isolation mounts and actuators, as well as suspension springs, manufacturers provide a wide range of air spring data. Firestone refer to these three applications as Airmount, Airstroke, and Airride in their air spring data.

The useful air spring data for our purposes are those for static and dynamic performance. The info in this post is for reversible sleeve type air springs.

Static data is often given in chart or table form. The first pics below are static data for a Firestone air spring assembly using a 1T14C-1 bellows, which I have guessed, similar to what you are using. For the chart, the 'Y' axii are 'Volume' (of trapped air) on the left, and 'Force' on the right side. Spring 'Height' is read from the 'X' axis (bottom).

Note the family of 'Constant Pressure Curves' for a number of pressure values within the allowable operating range. These curves plot the 'Force' as the air spring is compressed from the maximum to minimum height while the air pressure in kept constant.

Force = pressure x effective area. The shape of each constant pressure curve is a function of how the 'Effective Area' changes between the maximum to minimum height. The sketch below illustrates effective area for air springs. Effective area of reversible sleeve air springs is smaller near maximum height when the piston is withdrawn from the bellows. Near minimum height the shape near the piston base increases the effective area, while the bellows travels over a parallel region of a piston the effective area is constant.

The range of recommended static ride height happens to be within the region where the the effective area remains reasonably constant. On the constant pressure curves it is the range of heights where the force remains reasonably constant. For these particular air springs with 1T14C-1 bellows, between approximately 180mm (7”) and 280mm (11”).

The shaded area on the left side of the chart, indicates the region of spring height to be avoided when the spring is under load (high pressure), otherwise bellows damage can/will occur. It is permissible however for the spring to extend into this area when the suspension unloads (droop or rebound).

It should be apparent now that to maintain a particular static ride height (within the recommended range), pressure rises when the load is increased, and lowers when the load is reduced.

The other curve (from top left to bottom right) is the 'Volume Curve' which is a measure the volume of air left inside the air spring as the height changes while the pressure is constant, 7 bar or 100 psi, in this case.

Now it should be apparent that to change the static ride height (within the recommended range), at constant load, the air pressure remains reasonably constant, but the amount (volume) of air in the spring needs to be increased to raise the height, and reduced to lower the height.
Note the above discussion on air pressure and volume is for static (stationary) conditions.

When the vehicle moves the springs compress and rebound to absorb bumps and motion can be great as the axle articulates while 4 wheel driving. The amount of air in the spring remains constant as it compresses (or extends). Pressure increases and the volume reduces as the spring is compressed, and vice versa. The amount of air is constant, but its density changes with pressure and volume variations. The air temperature also changes slightly, but with normal use it is not significant.

The last pic is a diagram of Dynamic Load vs Deflection for a 1T15M-11 reversible sleeve air spring – I don't have a dynamic load diagram for the 1T14C-1 bellows. The starting point for each curve is the static height in the mid point of the recommended range (18” in this example). Starting pressures of 20, 40, 60, 80 and 100 psig are plotted. Dynamic Load is read on the 'Y' axis and spring height on the 'X' axis.

Looking at the curve starting at 18” for 100 psig, we read the dynamic load on the left axis of about 7000 pounds force.

If we find the point on this same curve for a compressed height of 15” (bottom axis), we can read the corresponding dynamic load from the left axis at 9800 lb. If we were to look at the static data chart for this air spring, we could read the new pressure, 138 psig (for 15” and 9800 lb). If the the spring extends to 21”, we see that the dynamic load has reduced to 5200 lb and the pressure has reduced to 77 psig.

Similar dynamic data for air springs can be determined using the static data sheet and the following gas law for pressure and volume:
P1 V1 ^1.38 = P2 V2 ^1.38 (for normal vehicle operation).

A spreadsheet makes it reasonably simple to determine the dynamic data for the Design Load and Ride Height, given a static data sheet for the air spring we are using.

Create a table with 8 columns for; Height, Static Load, Volume, Effective Area, Absolute Pressure, Gauge Pressure, Dynamic Load, and G-force. The table should have rows for different heights over the range of interest.

For the following example, assume 9” is our design static ride height, so make 3 rows for 11”, 9” and 7” spring length. Assume also that our design spring load is 2000 lb.

In the middle row, for static ride height, the data for each column are found or calculated as follows:
1.) Static Load, This value is obtained on the closest constant pressure curve, at the required ride height. This static load may be a little higher or lower than our actual design load. Use the 60 psi curve, or the 60 psi column in the Force Table, and read static load 1890 lb

2.) Volume, This value is taken from the 'Volume at 100 psig ' in the Force Table at the ride height, or from the 100 psig volume curve. For this example we see the volume is 280 cubic inches at 9” height

3.) Effective Area, Calculated using the pressure value of the constant pressure curve and the static load found in the step 1.). Effective Area = Static Load / Pressure, i.e. EA = 1890 lb / 60 psi = 31.5 square inches.

4.) Gauge Pressure, The Design Load divided by the effective area from the previous step 3.).
P1 = 2000 lb / 31.5 in^2 = 63.5 psi

5.) Absolute Pressure, This is gauge pressure from step 4.) plus atmospheric pressure (14.7 psi).
P1a = 63.5 + 14.7 = 78.2 psia

6.) Dynamic Load at point 1, equals the Design Load, i.e. 2000 lb

7.) G, This is the ratio of dynamic load / design load , i.e. 2000/2000 = 1.0 for this row.

Now find the data for the extended height (11") row as follows:
8.) Static Load, This value is obtained using the same constant pressure curve (or column in the Force Table) as before (step 1.), and the new extended spring height. Static load = 1770 lb at 60 psi and 11” height.

9.) Volume, This value is taken from the 100 psig Force Table at the new extended spring height, or from the 100 psig volume curve. Volume = 347 cubic inches at 11” height

10.) Effective Area, Calculated using the pressure value of the constant pressure curve and the static load found for the new extended spring height at step 8.). Effective Area = Static Load / Pressure = 1770 / 60 = 29.5 square inches.

11.) Absolute Pressure, This is calculated using the gas law given above;
P2 = P1 x (V1 / V2) ^1.38 where:
P1 is the absolute pressure found at step 5.)
V1 is the volume found at step 2.)
V2 is the volume found at step 9.)
P2a = 78.2 x (280 / 347)^1.38 = 58.2 psia

12.) Gauge Pressure, The absolute pressure from step 11.) minus atmospheric pressure (14.7 psi).
P = 58.2 – 14.7 = 43.5 psi

13.) Dynamic Load = Effective Area x pressure using: effective area at step 10.), and pressure at step 12.)
Dynamic Load = 29.5 x 43.5 = 1282 lb

14.) G, This is the ratio of dynamic load divided by design load, using dynamic load at step 13.)
G = 1282 / 2000 = 0.64

Repeat steps 8.) to 14.) for the compressed height (7") row, using the new compressed height instead of extended height.

After this our Dynamic Load table should look like:
11, 1770, 347, 29.5, 58.2, 43.5, 1282, 0.64
9, 1890, 280, 31.5, 78.2, 63.5, 2000, 1.0
7, 1920, 211, 32.0, 115.6, 100.8, 3227, 1.61

Now plot the dynamic load vs deflection curve through the 3 points representing dynamic load and spring height.
1282 lb at 11”, 2000 lb at 9”, and 3227 lb at 7”.

Note in this example the curve is approximate, but we could improve it by determining values over more heights at smaller deflection increments.

We can also calculate the spring rate (load / deflection) from our curve (spring rate = the slope of the tangent to the dynamic load vs deflection curve) and natural frequency if required.

Natural frequency is a useful value that should be used more often when choosing vehicle suspensios. For good performance over a range of general and off road conditions use values of natural frequency around 1.35 Hz for front springs and 1.688 Hz at the rear – motion/ride will be bad if the front is not less than the rear.

Thanks John,
Really appreciate the air spring info.

Made a start on the wiring on the trailer.

Cheers
David