GSpeedR

Superpilun: You don't need a longitudinal force to have longitudinal weight transfer...you only need a moment about the y axis (just like how the couple about the x-axis produces lateral weight transfer). The point where people are dissagreeing on is whether the suspension will cause both just a lateral couple or both a lateral and longitudinal couple.

I believe that the above agrees with descartesfool's last point, but looking at the situation a different way (diagonal load transfer from RF to LR being a combination of a lateral and longitudinal couple).

Scott: I think the problem comes down to a changing model. If you apply enough load to lift both tires, then you are left with a bike, which has different eq's of motion (and thus reacts differently to lateral force) compared to a 3 wheeler or a 4 wheeler.

Chris - who is going to live in a cardboard box if he keeps posting here during work.

Jack Black

At this point, I've applied the engineering/physics principals with which I am familiar and come to the conclusion that load does not transfer longitudinally under a steady state cornering condition except possibly by centrifugal effects on an inclined mass centroid.

I would love to be proven wrong, but so far I haven't seen anything that I could consider proof. If load does in fact transfer longitudinally, please explain the mechanism by which it is transferred or refer us to something that explains it.

superpilun

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">Superpilun: You don't need a longitudinal force to have longitudinal weight transfer...you only need a moment about the y axis (just like how the couple about the x-axis produces lateral weight transfer). The point where people are dissagreeing on is whether the suspension will cause both just a lateral couple or both a lateral and longitudinal couple.
</TD></TR></TABLE>

Where did I imply that? I thought my post pretty clearly showed that either a longitudinal force or a moment about the pitch axis could cause longitudinal load transfer. Yes the suspension can have jacking or antisquat or whatever it has but regardless of suspension, you cannot violate physics. Front to back weight ratio has to remain the same unless you have either a longitudinal force or a moment about the pitch axis. So I'd argue that a lateral force cannot turn into a longitudinal couple by means of suspension effects.


<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">
Scott: I think the problem comes down to a changing model. If you apply enough load to lift both tires, then you are left with a bike, which has different eq's of motion (and thus reacts differently to lateral force) compared to a 3 wheeler or a 4 wheeler. </TD></TR></TABLE>

Once you cross the threshold of lifting both tires, the car immediately rolls on to it's side - there's no more load transfer available to balance the roll moment. But Scott's still right - at the precise point of two-wheelin' fun the front-back weight ratio is still the same as static.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">
Chris - who is going to live in a cardboard box if he keeps posting here during work.</TD></TR></TABLE>
Where do you work?

descartesfool

Amongst other things, I like equations, so here goes. Words can be so distorted by philosophers!

We have 3 tires, each with 2 forces, one horizontal and one vertical.

Viewing the car from the rear more or less, the forces are name as follows

LF tire: F1 vertical up, FA horizontal left
RF tire: F2 vertical up, FB horizontal left
LR tire: F3 vertical up, FC horizontal left

The vertical up forces are balanced by the downward force due to gravity, W.
The horizontal forces are balanced by the lateral acceleration force to the right, mA

We balance the horizontal and vertical forces to zero such that we have a steady state. Thus:

F1 + F2 + F3 = W (1)
FA + FB + FC = mA (2)

Now we must balance the moments about the centre of gravity point G, also for steady state. Track is T, half-track is T/2, c.g. height is H. This gives:

F1*T/2 - F2*T/2 + F3*T/2 - (FA +FB +FC)H = 0 (3)

Substituting from above (2) into (3), we get:

T/2*(F1 - F2 + F3) = mAH (4)

or

(F1 - F2 + F3) = 2mAH/T (5)

from (1) we have F2 = W - (F1 + F3) (6)

(6) into (5) gives:

F1 - (W - (F1 + F3)) + F3 = 2mAH/T

which simplifies to:

(F1 + F3) = W/2 + mAH/T (7)

Substituting (7) into (1) gives:

F2 = W/2 - mAH/T (8)

where mAH/T is the lateral weight transfer, independent of suspension type, roll centres or anything else inside the black box. Equations (7) and (8) are very similar to the ones you get for just two wheels, except we have the LF and LR vertical forces F1 and F3 on the left side and only the RF force F2 on the right side. Now how do we know what the split is between F1 and F3, since we have F2. We need another equation. We have to look at the car from the side to get one. The cg is a distance "a" from the front and "b" from the rear, with the wheelbase l=a+b.

Solving for moments about G in side view we get:

F3*b = (F1 +F2)*a (9)

or F3 = (F1 +F2)*a/b (10)

Substituting (10) into (7) gives:

F1 + (F1 +F2)*a/b = W/2 + mAH/T (11)

or F1(1 + a/b) + F2(a/b) = W/2 + mAH/T (12)

Substituting (8) into (12) gives:

F1(1 + a/b) + (W/2 - mAH/T )(a/b) = W/2 + mAH/T (13)

Re-arranging and simplifying we get for F1, the LF tire vertical force.

F1 = {(b-a)/(a+b)}*W/2 + mAH/T (14)

So F1 only depends on the lateral weight transfer and the location of the c.g. as per (14) and F2, the RF tire vertical force only depends the lateral weight transfer as per (8), it must be that these forces do not depend on the LR tire's vertical force F3.

Since (F1 + F3) = W/2 + mAH/T from (7), we get

F3 = W/2 + mAH/T - F1 (15)

Substiuting (14) into (15) we get

F3 = W/2 + mAH/T - [{(b-a)/(a+b)}*W/2 + mAH/T ] (16)

Simplifying we get

F3 = W/2*[1-(b-a)/(a+b)]

Or F3 = W/2*[(a+b-b+a)/(a+b) =W*a/(a+b) (17)

Voila, F3, the LR tire's vertical force is a constant. In our example, the front % weight was 60%, so a/(a+b) = 0.40, and F3 = 0.40W. Since W = 2700 lbs,

F3 = 0.40*2700 = 1080 lbs, the lift off weight, and it stays constant.

So in summary

(F1 + F3) = W/2 + mAH/T (7)

substituting (17) into (7)

F1 + Wa/(a+b) = W/2 + mAH/T

Thus if we are on 3 wheels (after the rear load transfer has reached the rear inside tire's static corner weight), we have

F1 = {(b-a)/(a+b)}W/2+ mAH/T (18)

F2 = W/2 - mAH/T (8)

F3 = {a/(a+b)}W (17)

Adding F1 + F2 + F3 gives W, as it should.

(As I have not seen this analysis in any book, someone please check it)

The rear tire becomes a spectator. Its job is done. Now we can all rest in peace and start worrying about slip angles and roll centres. Philosophers can explain why!

But what about if the car isn't pointed straight, and it has some yaw angle? Then the force mA is not perpendicular to the car's axis. Thus the force mA can be resolved into a component along the front to back axis of the car and one perpendicular to that axis. The front to back force component will cause some front to back weight transfer. And the outside rear spring's rest is over. A LapSim simulation of my ITR around Mont Tremblant shows a lot of inside rear off the ground with zero vertical load on it, and the outside rear tire's load keeps changing slightly, even with no longitudinal acceleration. I suspect yaw change is the cause.

Edit: fixed the 3 left front tires (damn cut and paste)


Modified by descartesfool at 9:30 AM 10/4/2005

Jack Black

That pretty well corresponds with my analysis. If you could just do me one favor and fix this part (three left front tires, hmmm... ):

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by descartesfool &raquo;</TD></TR><TR><TD CLASS="quote">
LF tire: F1 vertical up, FA horizontal left
LF tire: F2 vertical up, FB horizontal left
LF tire: F3 vertical up, FC horizontal left
</TD></TR></TABLE>

Regarding lapsim and the varying rear vertical force, there are several different things that could cause such variation: longitudinal acceleration, the yaw-related effect you mention, and the rotational effect on an inclined mass centroid that was discussed earlier. Any or all of these could be occuring while cornering.

It also occurs to me that lapsim probably uses an iterative numerical model. Depending on how big the changes you mention are, they could simply be artifacts of the model (due to round-off errors, grid coarseness, etc.).

GSpeedR

descartesfool, your math looks ok. I was dissapointed to find myself with a similar solution after putting static equations through (which I admittedly hadn't done until last night for the wheel lift case); I got worried when I realized that if the front tire lifts after the rear tire, you would have a non-static weight distribution about the outside wheels, which I don't agree with. I wasn't get a y-axis couple without considering dynamic effects unless I really pull something out of my ***, so I will finally agree with everyone else and say that they are outlawed by the steady-state clause (not "pure-roll" though).

Tyson's post about the distributed weight on the right triangle was what got me philosophizing...it made a lot of sense. I think the reason why that doesn't make sense is that the rear load has already transferred to the outside tire (and a triangle under acceleration does not have to distribute weight equally). I think this is equivalent to the rear mass concentration located above the outside rear tire...which means you don't have the moment arm from LR to CG to cause a resisting couple (there isn't one). So the load on the rear tire is constant. Since it doesn't have a moment arm to resist roll then it won't and is reliant on it's buddies up front to do it...on the rigid sprung mass.

Does this mean that the load transfer phenomenon (mass under acceleration) is equivalent to a migrating mass centroid axis? Hmmm...maybe I should take a break...

GSpeedR

Summary post. We had many discussions, might as well summarize them.

1.) Wheel lift under steady-state cornering does not cause longitudinal load transfer.

2.) However you can get long load transfer under lateral accel if you allow dynamic effects:

- Large yaw rates and small turn radii cause a moment about the y-axis on an inclined mass centroid axis. I still think this is steady-state, but others dissagree.

- A skewed roll axis (which the vehicle rolls about) means the couple about the x-axis from the lateral force is transferred into a couple about the x and y-axis which is reacted by long and lateral load transfer. Not steady-state since we need to account for rolling motion of vehicle.

- Gyroscopic precession of the rotational inertia of the sprung mass precessed by the vehicle's yaw rate causes a couple about the y-axis. Not steady-state, and a small effect due to low roll inertia and small yaw rate. I've never heard of this described for a vehicle except by me to prove a point, but it is a known problem in intermediate dynamics. Might be significant when you spin, but then you have bigger problems.

3.) Inclined roll axis could cause long load transfer, but it is something else that I created based on my own thinking. It's a stretch to say the least.

Chris - philosophizer

RR98ITR

Did you Dis me by calling me a philosopher?

Scott, who, as a caveman in good standing, resents the imputation and would roll his wheel right over you but it's up on jackstands right now...

RR98ITR

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">...Scott: I think the problem comes down to a changing model. If you apply enough load to lift both tires, then you are left with a bike, which has different eq's of motion (and thus reacts differently to lateral force) compared to a 3 wheeler or a 4 wheeler.

Chris - who is going to live in a cardboard box if he keeps posting here during work.</TD></TR></TABLE>

If I accept that, don't you have to accept that they produce the same values at the boundaries? And if so then what causes the reversal of long load transfer under the 3 wheel function as we approach the 2 wheel function?

Scott, who is too long past real computational mathematical fluency...philosophy is the residue of my education...

GSpeedR

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by RR98ITR &raquo;</TD></TR><TR><TD CLASS="quote">If I accept that, don't you have to accept that they produce the same values at the boundaries? And if so then what causes the reversal of long load transfer under the 3 wheel function as we approach the 2 wheel function?
</TD></TR></TABLE>

Yep. I realized that I was missing something when I had all this load transferring to the rear tire, and then I would lift the other inside and have the car basically doing a wheelie on one tire....all from a lateral force. Cars doing wheelies during steady-state cornering sure would be fun to watch though.

I am currently in the middle of my advanced education and now I forgot how to do simple problems. My span of knowledge never grows, it justs shifts.

descartesfool

But the curious thing with the math for our 3 tired car vs our 4 tired car is that once the inside rear lifts, the load transfer no longer cares what either the front or rear roll resistances are. The total LLT only ever depends upon the car's basic dimensions, but with four tires on the ground, the roll resistance distribution has a profound effect on what is happening, because I can get to lift off the inside rear with 540 lbs of load transfer there in my example, but I could have only 200 lbs of transfer at the front, or 650 lbs, and this ratio depends on the front/rear roll resistance ratio, which is a factor I need to add to the equations to get the four wheel loads before lift-off. However when I lift one tire off the ground, all roll resistance ratios are off, as we no longer care since we are just getting load transfer between a pair of wheels (the front ones), as if the front springs are just along for the ride, since they don't figure in the wheel load equations.

Of course the front springs will determine the roll angle there after lift-off, but they have been rendered dumb because they will just adjust to make the tire loads on the front tires match the weight transfer there as determined by the factor mAH/T. Before lift-off, springs seem much more important.

Seems a curious finding to have roll ratio and even front roll resistance disappear from effectiveness. Perhaps we need to start to design our suspensions better to control what they do with one tire off the ground to maximize performance. A whole new chapter begins.

MechE00

This post is just a placeholder post.. I don't have time to formulate a reply right now and probably won't until next Tuesday or Wednesday at the earliest (in case anyone cares)...

RR98ITR

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by MechE00 &raquo;</TD></TR><TR><TD CLASS="quote">This post is just a placeholder post.. I don't have time to formulate a reply right now and probably won't until next Tuesday or Wednesday at the earliest (in case anyone cares)...</TD></TR></TABLE>

This post is just to acknowledge your post and to convey the message that I care.

BUT - if you are gonna wave your hands and make vague references to Euler, Fermat, and Beelzebub then that's different.

Scott, who see's life as long gauntlet of Vampires...and this one, like most, just Will Not Die!

descartesfool

So how do we tune the car to be good under 3 wheel operation. Drive it like a tricycle? I will have to did deep in my old bags of tricks for those skills!

If springs don't matter except for camber control since load transfer is determined by nothing but the g's, what does matter? You can shift load front or rear with the gas pedal (or brake) and affect the load on the tires, but what is really happening under that temporary steady state. Car always has some yaw and thus there is always front to back load transfer as you increase g's? If you first consider no yaw, then after the wheel lifts you start to get more g's with a push on the loud pedal to a new plateau of steady state, how did you get there. Say we went from wheel lift at 1.00 g and then got to 1.05 g's. Slip angle has to change on the rear tire to increase grip since vertical load didn't change, but that means yaw angle changed since you can't steer the rear wheels. And then you got more load transfer on the front wheels which is a grip reducing event at a constant slip angle for the front tire pair, so slip angle has to change there too. Rear end comes out more (more yaw due to more rear slip) and front end has more slip made possible by more steering. Could be having oversteer and understeer all at the same time. Who cares as long as I know how to tune it on 3 wheels, which I don't, yet! Then again 4 wheels has to have more grip than 3, so maybe we just need to shift the c.g. backwards and stop behaving like dogs.

RR98ITR

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by descartesfool &raquo;</TD></TR><TR><TD CLASS="quote">....Then again 4 wheels has to have more grip than 3....</TD></TR></TABLE>

A slayers work is never done.

Yes...4 must have more grip than 3...but where is it and do you need it there?

Scott, who realizes Claude isn't really a Vampire...he's just been bitten too many times...

descartesfool

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by RR98ITR &raquo;</TD></TR><TR><TD CLASS="quote">
Scott, who realizes Claude isn't really a Vampire...he's just been bitten too many times...</TD></TR></TABLE>

But....I just came out of the crypt.

Retire now to your tents and to your dreams. To-morrow we enter the town of my birth. I want to be ready.

RR98ITR

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by descartesfool &raquo;</TD></TR><TR><TD CLASS="quote">

But....I just came out of the crypt.

Retire now to your tents and to your dreams. To-morrow we enter the town of my birth. I want to be ready.</TD></TR></TABLE>

You scary guy - not right in head.

Scott, who knows what he's talking about...anybody?

GSpeedR

More rear slip angle with more g's means more yaw which means more rear weight transfer... I wonder if you could pull enough lateral g's to put the inside rear tire back down on the ground. I guess that depends on how fast the fronts transfer load and loose grip. Understeer and oversteer at the same time...does that count as neutral handling?

Shift CG backwards? Your front tires will object.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by RR98ITR &raquo;</TD></TR><TR><TD CLASS="quote">Scott, who knows what he's talking about...anybody?</TD></TR></TABLE>

Britney Spears lyrics?


Modified by GSpeedR at 10:22 AM 10/7/2005

RR98ITR

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">...Britney Spears lyrics?</TD></TR></TABLE>

If so I wouldn't know...probably just one of those monkey & typewriter things.

Scott, who thinks that if the outside rear starts sliding out from under the kaa then surely the inside rear would drop back down - it's just simple fysiks.

Black R

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by descartesfool &raquo;</TD></TR><TR><TD CLASS="quote">

But....I just came out of the crypt.

Retire now to your tents and to your dreams. To-morrow we enter the town of my birth. I want to be ready.</TD></TR></TABLE>

[edit]

n/m, it's the doors:

http://home.c2i.net/psletsjo/lizard.htm

[/edit]

GSpeedR

Just linking to the Ortiz answer thread for those who want it.

Ortiz Thread!