Do we need suspension?

[hardingfv32]

DaveW

Yes, the position I was proposing has run its coarse. I can respect everyone belief that I am wrong but no one has really provided a reason I can appreciate at this point. I can keep researching for answers on my own.

I should state I did not start this thread.

I would say my original thought was: No suspension needed for a dead smooth track. I can now see the tires raise a complication to such a simple proposal. From the discussion I doubt anyone has learned anything. So we move on...

Brian

[GSpeedR]

DaveW

Yes, the position I was proposing has run its coarse. I can respect everyone belief that I am wrong but no one has really provided a reason I can appreciate at this point. I can keep researching for answers on my own.

I should state I did not start this thread.

I would say my original thought was: No suspension needed for a dead smooth track. I can now see the tires raise a complication to such a simple proposal. From the discussion I doubt anyone has learned anything. So we move on...

Brian

Your argument is completely based upon the following formula:

Lateral Accel [m/s^2] * Mass [kg] * CH_Height [m] / TrackWidth [m] = LoadTransfer [N]

This formula creates a relation between Moments (Newton's 2nd Law)about a reference point and it is only applicable for a suspended chassis. As mentioned above, it is only explicitly valid for steady-state motion and only a single degree of freedom (Roll -- Mx) is accounted for. Any true vehicle has at least "unsprung" degrees of freedom, even your no-suspension race-car assuming it has sensible tires. A more general treatment is:

Lateral Accel [m/s^2] * Mass [kg] * CH_Height [m] = RollMoment [Nm]

This roll moment is reacted by the suspension and then tires (both sides). It is only when it has been reacted by the tire contact patches (or the ground) that it can be called "Load Transfer". There is absolutely a delay between the lateral acceleration of the sprung mass and the appearance of load transfer (or a change in normal load at the contact patches). In fact, if you had an infinitely compliant suspension with an infinite amount of travel, there would never be a load transfer at the ground. The chassis would simply accelerate in roll until the vehicle stopped cornering (obviously ignore the details of this theoretical exercise). Load transfer requires that the sprung mass roll moment be REACTED by the ground. Load transmits though components, it does not arrive instantaneously on the other side.

You've repeatedly made the claim that this magical car has "no suspension". However, it appears to me that you actually imply a "rigid suspension", which is...a suspension...one that reacts the sprung mass roll moment instantaneously (if your tires are also perfectly rigid); this is the only situation where load transfer does not delay vehicle lateral acceleration. If you truly mean "no suspension" then the situation I described above will occur: the sprung mass will accelerate and there will be no load transfer.

In any case, load transfer is absolutely dependent upon time. You are using the wrong equation for your exercise.

[hardingfv32]

What is the nature of this delay? Is it only relevant to the sprung mass, not unsprung mass?

What measures control the delay, Lateral Accel , Mass , CH, etc.? How is this delay perceived by the driver?

Brian

[Jersey Tom]

What is the nature of this delay?

F=ma

[GSpeedR]

What is the nature of this delay? Is it only relevant to the sprung mass, not unsprung mass?

What measures control the delay, Lateral Accel , Mass , CH, etc.? How is this delay perceived by the driver?

Brian

The delay is motion as mentioned above, in this case roll acceleration. Suspension and tire compliances control the delay, and it is certainly felt by any driver complaining about sprung mass motion. Driver's likely perceive displacement changes visually comparing the chassis (or dashboard) to the track horizon/walls/etc; they can also generally sense inertial pitching/rolling accelerations, particularly above 2Hz.

[hardingfv32]

Is this delay only associated with the roll of the sprung mass? Very little roll, very little delay?

During corner exit does this same delay also slow the unwinding of the load transfer?

Brian

[Jersey Tom]

Change in force can only be realized through motion, be it spring displacement, shock velocity, whatever. There is inertia and therefore time transients in every suspension mode - ride, roll, pitch, and yaw.

So yes.. corner entry, exit, you name it.. there is a non-zero time delay associated with reacting loads through the force elements.

[ubrben]

Is this delay only associated with the roll of the sprung mass? Very little roll, very little delay?

During corner exit does this same delay also slow the unwinding of the load transfer?

Brian

Depends on what you mean by "little" I'm told men in general are poor estimators of size.. :-p It's not insignificant if that's what you're asking.

Ben

[hardingfv32]

So the force that creates load transfer in steady states is used to roll the chassis in transitions?

Is there a split between of the force levels, load transfer and roll movement, during transitions?

Or does the roll acceleration add to the load transfer?

Brian

[Jersey Tom]

Load transfer and roll motion are very much hand in hand.

Here's the breakdown:

1. Car driving in straight line.

2. Let's say we have an instantaneous input of lateral acceleration, some step input. At this moment, sprung mass is still at roll = 0.

3. The lateral acceleration means there is a rolling moment, which means we now have roll acceleration.

4. For there to be steady state load transfer, the springs must change displacement. No other way for the system to be in equilibrium. For the springs to change displacement, we need roll motion.

5. After some non-zero amount of time, that roll acceleration leads to roll velocity, and roll velocity leads to roll displacement.

6. As the sprung mass continues to move, the springs, bars etc start to provide roll resistance which counteracts the roll moment.

7. The steady state solution is at the point where roll resistance equals roll moment.

If it's easier for you to visualize, it's all really equivalent to a spring-mass-damper system. Suppose I had a chair on a spring. If I sit on it, it's going to go down.. but it doesn't happen instantly. Takes time for the spring to compress enough until it balances out the additional static weight I've added to the system.

Out of curiosity Brian, what is your background? Still in some level of school? Out working? Science and engineering background, or just curious about these things?

[hardingfv32]

Thanks for the response. It was very helpful.

I'm just a club racer trying to improve my understanding of the car.

Brian

[bl79]

I will try to add my 2 cents to this thread, even with my limited understanding.

Firstly, as has been stated, load transfer is not something that occurs instantly. There is most certainly a period of time whereby the car is in a transient state and the complete load transfer has not been realized. I believe this is mentioned in Tune to Win when Smith mentions why there are amazing DTM drivers (among other series) who simply can't get it done in a "proper" racecar. The load transfer takes some finite time and that time is longer in a road going car. In general the shorter that transient time is, the more difficult a car is to drive at its limit (going back to the F1 vs DTM example). A car which takes a very long time to transfer its load will surely be seen as sluggish and lazy upon turn in because it wont "take a seat" for a while. On the other end of the spectrum, a car that transfers load in a very short amount of time is very difficult to control. Accordingly, a car with "no suspension" would transfer load with no time lapse and be undriveable. I know some karts have adjustable chassis for rigidity. Obviously in the real world, have a suspension is critical.

Then there is the fact that the load transfer equation is for steady state cornering. The racecar, when driven properly, is almost never in a steady state condition. A properly driven racecar should always be in a transient condition "on the brakes, WOT, or squeezing the throttle down" as it is generally said.

Lastly, my understanding of tires dictates that slip angles MUST be produced in order to generate any acceleration in any direction. If you have an infinitely stiff tire, I'm not convinced you could develop slip angles, ergo no grip. Also the tiny contact patch is another good argument. Although it seems a wooden wheel generating cornering force goes against my idea of slip angles being necessary. Anyone care to shed some light on this?


Also, loving the technical discussion taking place. Cheers.

[munks]

Lastly, my understanding of tires dictates that slip angles MUST be produced in order to generate any acceleration in any direction.

Narrow definition, if I'm being picky. A tire with camber, plysteer, conicity, or even a slip *ratio* might produce acceleration without slip *angle*. Not to mention, "any direction" could include vertical accelerations, too.

If you have an infinitely stiff tire, I'm not convinced you could develop slip angles, ergo no grip. Also the tiny contact patch is another good argument. Although it seems a wooden wheel generating cornering force goes against my idea of slip angles being necessary. Anyone care to shed some light on this?

Rubber tires don't work like wooden (or I should say theoretically perfectly stiff) wheels. Rubber deforms in the contact patch to match the slip angle/ratio and doesn't even need to slide to produce a force (although realistically, there's always a little bit sliding, even when free rolling in a straight line). In other words, it can react a force simply with static friction.

Stiff wheels, on the other hand, will react with *only* sliding friction. There really can't be static friction without deformation. When you turn your wooden wheel, it's now sliding sideways against the ground, just like a wooden block would. The only difference is that your wheel can compensate for longitudinal sliding by spinning. Also, the infinitely small contact patch of an infinitely stiff wheel on an infinitely stiff flat surface is immaterial if you used the Amontons-Coulomb Law (frictional force is proportional to load only, not area).

[tomazy]

Dont know if this was allready posted here, if it was, i appologise

http://www.youtube.com/watch?v=-LFmWd2J ... re=related

[Tim.Wright]

For what its worth, a wooden wheel will generate force and slip but the deformation will be on the micron scale so the slip angle will be barley noticable.

Tim