Developing a Proper Suspension Model

[u/Josh]

Originally Posted by Shankenstein

If you want to trick it up, you can do:
- progressive spring rate = f(x-x_0)
- linear damping = k * (x-x_0)'
- digressive damping = f((x-x_0)')
- damper hysteresis = nasty equations)
- pressure-sensitive tire rate = k_sidewall + k_air * pressure (absolute)
- temperature-sensitive tire rate = k_sidewall + k_air * n*R/V * temperature (absolute)
- tire temperature estimation = T_atm + k * log (T_tread - T_air_in_tire)
- tire tread temperature = T_atm + k * (heat generated - heat dissipated)
... lots of ways to complicate the problem.

Of these, I think comparing different damping curves, along with testing the effects of changing unsprung weight are most interesting. Also, the frequency response would be interesting just to show why it is smoother when you drive faster over bumps. I'll plug this in to MATLAB when I get a minute free at school.

I have thought about trying to convert this to a 4 wheel model and I can't quite get my head around how y1, y2, etc should be constrained to represent an actual chassis. What I have come up with so far is that if you consider the chassis to be the x-z plane then the points (x1,y1,z1) and (x2,y2,z2) etc, would have to all lie on the same plane; Where xi and zi are the coordinates where the i'th wheel is. It's late, I hope this makes sense.

[Hancha Group]

Originally Posted by u/Josh

Of these, I think comparing different damping curves, along with testing the effects of changing unsprung weight are most interesting. Also, the frequency response would be interesting just to show why it is smoother when you drive faster over bumps. I'll plug this in to MATLAB when I get a minute free at school.

I have thought about trying to convert this to a 4 wheel model and I can't quite get my head around how y1, y2, etc should be constrained to represent an actual chassis. What I have come up with so far is that if you consider the chassis to be the x-z plane then the points (x1,y1,z1) and (x2,y2,z2) etc, would have to all lie on the same plane; Where xi and zi are the coordinates where the i'th wheel is. It's late, I hope this makes sense.

Wait. First, let's all get on the same reference frame. Are we all thinking X is the longitudinal direction of the car, Y is the lateral direction of the car, and Z the vertical?

Also, where are we going to place the origin? Center of the wheels on the ground plane?

[u/Josh]

Originally Posted by Hancha Group

Wait. First, let's all get on the same reference frame. Are we all thinking X is the longitudinal direction of the car, Y is the lateral direction of the car, and Z the vertical?

Also, where are we going to place the origin? Center of the wheels on the ground plane?

I just went with the notation of y being vertical (in the direction of bump travel) like Earl used to derive the transfer function. Then I guess X can be longitudinal and Z is lateral. The origin should probably be in the center of the car so for example the coordinates for the front passenger mass would be (I am guessing at the numbers here):

(50",y,30")

I know the X and Z coordinates would actually be a function of Y but for small angles I think this is negligible?

[EarlQHan]

Originally Posted by u/Josh

I just went with the notation of y being vertical (in the direction of bump travel) like Earl used to derive the transfer function. Then I guess X can be longitudinal and Z is lateral. The origin should probably be in the center of the car so for example the coordinates for the front passenger mass would be (I am guessing at the numbers here):

(50",y,30")

I know the X and Z coordinates would actually be a function of Y but for small angles I think this is negligible?

The Hancha user is me (or plucas) ;] I used the vars x, y, and u for the transfer function rather than x1, x2, etc. since it makes it easier to keep track of the variables. In terms of the car as a global coordinate system, I use the X, Y, Z variables as stated above.

If we can create a sophisticated enough model, the suspension would see movement in all three coordinates along their respective arcs.

[u/Josh]

I just plugged the quarter car transfer function you derived into MATLAB. The constants are mostly guesses, so let me know what needs to be changed. One thing that seems weird to me is that the step response has a final value of zero when I think it should be one.

edit - 'unsprung weight' should be 'unsprung mass' and 'tire spring weight' should be 'tire spring rate'

http://i.imgur.com/osZSMGs.png

[EarlQHan]

The impulse response is correct and you want the function to eventually get to zero. This is representing the wheel oscillation, which will oscillate when excited by a bump and eventually settle.

Side note: from a subjective standpoint, from what I've been told, typically around four oscillations for the settling time said to "feel good," rather than shooting for a critically damped model. A critically damped suspension will be back breaking and the vertical accelerations happen too quickly for a driver to really respond to them.

[u/Josh]

Originally Posted by EarlQHan

The impulse response is correct and you want the function to eventually get to zero. This is representing the wheel oscillation, which will oscillate when excited by a bump and eventually settle.

Side note: from a subjective standpoint, from what I've been told, typically around four oscillations for the settling time said to "feel good," rather than shooting for a critically damped model. A critically damped suspension will be back breaking and the vertical accelerations happen too quickly for a driver to really respond to them.

Yeah, the impulse looks good, but the step reponse should have a final value of 1, not zero. I'm not sure where the error is coming from though.

Edit - just found the error. I made a mistake entering your transfer function. I'll upload new plots in a second.

[EarlQHan]

Is there a way to get the picture a bit clearer? I'm struggling to read the key XP

[u/Josh]

The only odd thing left is the natural frequency seems to be around 5 Hz, which I think is way to high to be true. I guess one (or more) of my assumed values is off by quite a bit.

[EarlQHan]

The rates are off. Shankenstein has already done the base estimates:

http://www.ft86club.com/forums/showp...38&postcount=1

[u/Josh]

Originally Posted by EarlQHan

The rates are off. Shankenstein has already done the base estimates:

http://www.ft86club.com/forums/showp...38&postcount=1

I think there may be a mistake in his post because I am getting the front rate to be 131 lb/in = 22,000 N/m. I'll plug that number in tomorrow and check the results.

And it seems my damping coefficient was way off as well. I don't remember where I got this plot from but it is probably useful to some of you. This is a front stock BRZ damper. I'll try 6700 N*s/m.

[Shankenstein]

To contribute:
m_1 (unsprung mass) = 37.6 kg
m_2 (sprung mass) = 280.4 kg (front) 244.5 kg (rear)
k_1 (stiffness of tire) = 350 N/mm (at 30 psi)
k_2 (stiffness of suspension) = motion ratio * spring rate = 2.11 N/mm (front) 2.8453 N/mm (rear)

We can construct a piece-wise interpolation that's easily mapped using a "spline" fit. How to Spline Like a Boss

note: Sign convention usually has compression as positive displacement/velocity and rebound as negative displacement/velocity.

damping coeffficient = Force/velocity

damping ratio = actual damping coefficient / critical damping coefficient

We want a damping ratio of nearly critical (1.0) for roll and pitch modes, but ride can be 0.5 - 0.8 for some driveability.

If anyone has not read these, please do! OptimumG Technical Papers

WRT ride frequencies:
spring rate = 4 * pi^2 * ride frequency^2 * sprung mass * motion ratio^2
2294.2 = 4 * pi^2 * f_r^2 * 280.4 * (1/0.92)^2
f_r = 0.42 Hz

That's a very soft ride. Sway bars stiffen it up in roll though. That's another calculation for another day.

[u/Josh]

Originally Posted by Shankenstein

To contribute:
m_1 (unsprung mass) = 37.6 kg
m_2 (sprung mass) = 280.4 kg (front) 244.5 kg (rear)
k_1 (stiffness of tire) = 350 N/mm (at 30 psi)
k_2 (stiffness of suspension) = motion ratio * spring rate = 2.11 N/mm (front) 2.8453 N/mm (rear)

We can construct a piece-wise interpolation that's easily mapped using a "spline" fit. How to Spline Like a Boss

note: Sign convention usually has compression as positive displacement/velocity and rebound as negative displacement/velocity.

damping coeffficient = Force/velocity

damping ratio = actual damping coefficient / critical damping coefficient

We want a damping ratio of nearly critical (1.0) for roll and pitch modes, but ride can be 0.5 - 0.8 for some driveability.

If anyone has not read these, please do! OptimumG Technical Papers

WRT ride frequencies:
spring rate = 4 * pi^2 * ride frequency^2 * sprung mass * motion ratio^2
2294.2 = 4 * pi^2 * f_r^2 * 280.4 * (1/0.92)^2
f_r = 0.42 Hz

That's a very soft ride. Sway bars stiffen it up in roll though. That's another calculation for another day.

I'm still getting 21.11 N/mm. Perhaps your are still in units of Kg/mm?

The good news is this works out to be 1.6 hz ride frequency. Anyone know why we see the highest gain in the bode plot at around 5 hz not 1.6 hz?

[Shankenstein]

Originally Posted by u/Josh

I'm still getting 21.11 N/mm. Perhaps your are still in units of Kg/mm?

The good news is this works out to be 1.6 hz ride frequency. Anyone know why we see the highest gain in the bode plot at around 5 hz not 1.6 hz?

Mistake was exactly what you said. Used an online calculator, and I shouldn't have trusted it.

The original post is updated.

spring rate = 4 * pi^2 * ride frequency^2 * sprung mass * motion ratio^2
22970 = 4 * pi^2 * f_r^2 * 280.4 * (1/0.92)^2
f_r = 1.325 Hz (front)

Notice that the motion ratio is inverted. If you leave it as 0.92, you will get 1.5 Hz as you got.

The rear comes out to be 3.8154 Hz. It is definitely stiffer in the rear, but it doesn't quite feel THAT stiff. Once we can confirm the rear motion ratios, we can state it with confidence.