Is lateral force positively connected to weight transfer 5

[TMcRally]

I feel I have a fundamental problem with my understanding of the physics of weight transfer V's lateral forces.

When weight transfers around the car and we control where it goes does this have an effect on the lateral loads that the tyres resist.

We say if we transfer more weight to the outside front on turn in we get less grip from that tyre, and we do, I've seen it and driven it.

Because the weight pushes the tyre down it has more grip albeit that it is a reduced proportional rate to the weight, if it has more grip it should be able to do more work. If the work has not increased then it should turn in better. Unless the lateral force from the mass not wanting to turn has increased the work load as well on that tyre.

Does the lateral force move around the car with the weight transfer - are they connected. It would seem in some cases it is and in others less so.

 

[hemipanter]

Just for a starter, the weight transfer is a product of G-force. So they are in direct relations. Wt=(G*Cgh*W)/Tw.
Wt=weight transfer.
Tw=track width.
Cgh=centre of gravity height.
G=sidway G load.
W= weight.
The weight transfer will be distributed between the front and rear axle in proportion to spring settings.
Goran Malmberg

 

[GregLocock]

If you load a wheel more heavily vertically then it will provide more lateral force for a given slip angle. Generally.

Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[TMcRally]

Thanks

Is it because the tyre with the most vertical load does provide more grip and that that in turn causes the lateral load to be transferred to it (the point of most resistance).

So is this as a consiquence of the weight transfer rather than being part of it.

 

[GregLocock]

I was carefully avoiding that part of the discussion!

I was describing how tires work, not how vehicles work.

At an axle level weight transfer tends to reduce the grip, that is the inside wheel loses more than the outer wheel gains, in lateral force, although the difference is typically small.



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[hemipanter]

"Is it because the tyre with the most vertical load does provide more grip and that that in turn causes the lateral load to be transferred to it (the point of most resistance).

So is this as a consiquence of the weight transfer rather than being part of it."
If the car set up is like what I call "Zero" which means an equally tire loading construction, all tires will get the same vertical and side load, either by a pulling wire at CGH or from driving at "steady state" cornering. When the car starts sliding sideways it will do so in a balanced manner.

To make one tire receive more loads we must alter the condition, like an uneven road or one axle having stiffer springs, things like that. Or else, a single tire will not start raising side load capacity. Stiffer springs on one axle make it take a larger part of the weight transfer thereby loading its outer wheel harder. This makes that tire to take care of a larger portion of the total weight transfer, and thereby side load. But, it will not be capable to handle the same amount of side load, as when lighter loaded, in relation to the vertical load. This means that the car will not be able to reach the same cornering power. The amount of side load in N of the axle is being higher, but the G-force capacity number is being lower.
Goran

 

[TMcRally]

I’m still not sure I’m with it. I see the side load and weight as 2 separate and unrelated things, and maybe this is where I’m wrong.

In the instance where we have, statically, 60% of the weight distributed on the front and 40% on the rear and we drive around a steady state corner or wired up as you say. At any point prior to any tyre sliding the front wheels are resisting 60% of the side load and the rear 40% (disregarding discrepancies in slip angles).

If we mix this up a bit and change the dynamic balance of the car by increasing the spring rate in the rear, then we drive around that same corner, at that the same speed, we will transfer more vertical weight on to the rear and as a result the tyre will be able to do more work. (disproportional to the weight applied but still it can do more work). If it can do more work then we should have more grip at the rear and less at the front because the work load is still 60F/40R ?

By changing the spring rate you haven’t moved the CoG much, so the amount of side loading laterally across the front is still the same. So in my mind, the need to resist the unchanged mass from continuing in a straight line up front hasn’t changed up front and by taking some of the ability off the front to resist that force and giving it to the rear tyres you should create understeer.

Why then does the opposite happen ?

The only explanations I can find to justify what is happening is if the side loads somehow follow the weight distribution around the car, but the mass hasn’t moved just its effect (weight). Is it because the slip angle on the rear with more weight now has a higher slip angle and runs the rear wide but on a square tyre slip angles must be very small or an I missing a fundamental of weight transfer and that with the weight also goes an increase in the side loading.

While I’m here annoying you, why is the polar axis between the front and rear wheels and not between the rear wheels, front wheel steer cars seem to pivot around a point between the rear wheels.

 

[hemipanter]

Hi again!
You are in no way annoying me, I like theese sort of problems. Better put this time, I hope I catch the way think. Ill be back shortley.
Regards
Goran Malmberg

 

[NormPeterson]

"So in my mind, the need to resist the unchanged mass from continuing in a straight line up front hasn't changed up front and by taking some of the ability off the front to resist that force and giving it to the rear tyres you should create understeer.

Why then does the opposite happen ?"

Because understeer is a relative displacement effect (front minus rear) rather than a force effect directly? Don't confuse the ability to resist a lateral force with the slip angle required to get there.

I don't think that a bicycle force-only model looking only at the outside tires is sufficient.


Norm

 

[hemipanter]

Let’s see what this gives.
Vehicle 1.
If we think about a vehicle with equal roll rate all around,
having a sprung weight of 800, (zero unsprung weight just for this example), it will statically load each tire by 200. The vehicle is then subject to 200 in roll weight transfer.
Then the inner pair of wheels will now have 100 and the outer pair 300 each of load.

Vehicle 2.
Let us take the roll situation to the extreme.
I should now create a vehicle having NO roll resistance for one axle (A), like a roll central bearing at ground level, and the other axle (B) is given 100% roll stiffness. When this vehicle is subject to the same side load the A axle will not transfer any loads, but it will still carry the static measurement of 200 each wheel. Axle B is handling ALL the weight transfer alone which means that its inner wheel will be at zero load and the outer is having 200+200=400.

Vehicle 1 will be having the same sideway grip for both axles as the tire contact patch and load for both axles are the same.

Vehicle 2 is having the wheels of the "in roll open axle" loaded to 200 each since roll Tq is of no influence. This means that all the weight transfer will take place at the infinite in roll stiff axle, leaving for a zero loaded inner wheel and a by 400 loaded outer wheel.

The axle with the infinite stiff sway bar, show half the rubber area for the same side load as compared to the open axle.

Goran

 

[cibachrome]

Just to stick to the original post, go to a department store (Ikea ?), grab 4 bathroom scales and a 4 legged kitchen table. Put each table leg on a scale. Have your friends hold the table in position over the scales. Then push laterally at the leg/scale interface. Have 4 other trustworthy friends note the scale weights. Given the circumstances, you have just answered your own question. Anyone who uses a K&C machine knows that this method serves to calculated the instant centers of the "roll Axis" at any trim, regardless of whether the suspesnion is symmetric/square or asymmetric (as in having a Panhard/Track bar). And it is far superior to methods which assume the "roll axis" is in the center of the vehicle. Generally it is, but in many cases, is not.

 

[TMcRally]

Thanks everyone, I'm still struggling a bit because it is a large step from where my mind was, but I think I have it now, I just have to justify it to myself before I can make it second nature.

Heres where my thinking has slipped a gear. Goran, in your example I assumed that when the weight transferred it transferred equally from front and back and transferred that weight proportionally to the leg with most resistance. In your example I thought we wound up with axel (A) having 200 and 100 and axel (B) with 400 and 100. My understanding was the weight would jack diagonally across the car as there was no resistance at (A) to apply the transfer, so it loaded (B).

In my process above (B) had more grip and no change in the load it needed to resist so it had more ability to resist that force.

If I’m still off track can you let me know.

Thanks again
Dave

 

[hemipanter]

1
Goran, in your example I assumed that when the weight transferred it transferred equally from front and back and transferred that weight proportionally to the leg with most resistance.
2
In your example I thought we wound up with axel (A) having 200 and 100 and axel (B) with 400 and 100. My understanding was the weight would jack diagonally across the car as there was no resistance at (A) to apply the transfer, so it loaded (B).

1
The sprung chassis is getting a roll motion from cgh and side load G-force. The chassis is stiff as a safe (on our superb race cars) so the force of roll is equal along the chassis.
In my example moment of roll was 200 that as you suggest, must be distributed according to roll resistance, which was zero one axle and 100% the other.
2
In your example there is no weight transfer in wedge present for the two inner pair of wheels.
Think of my car as a tree wheeler, where the front ground level roll bearing represent the front wheel.

Cibachrome:s table is the type of experiment I have been performing myself, but on a down scaled model using one spring and an electronic height scale each corner.
The four "wheels" could be positioned at any track width and stiffness, and the "chassis-table" could be freely loaded in roll or vertically.
Using this model we can see that if the track width is less on one axle, the other axle with wider track with will take more of the roll resistance, showing a less loaded inner wheel and heavier loaded outer wheel.

Let us say we have a front and rear tw of 1600mm, a cornering load of 125-250kg front and 125-250kg rear axle.
There is 125kg weight transfer each axle.
We will now narrow the rear tw to 800mm. This will give a front axle load of 50-450kg and rear axle load of 150-350.
The froth axle will now transfer 200 kg and the rear only 100 kg.
There is now a greater load transfer 300 vs 250kg, due to less average tw.
Goran

 

[alangbaker]

TMcRally,

I think the factor you're missing from your understanding is that the coefficient of friction of rubber is not the same under all loads. The amount of friction a tire can generate doesn't go up precisely in proportion to increased vertical load; it goes up slightly less quickly with increasing loads.

Hence two tires carrying the load of their end of the car equally will produce more friction than when one tire is carrying more load and the other less.

So when you use spring rates or anti-roll bars to make one end of the car resist more of the rolling forces, you create a more uneven distribution of loads at one end and a more even distribution of load at the other.

So if you stiffen the front, you decrease the overall side load at the front and increase it at the rear, leading to more understeer (or less oversteer). And obviously, vice versa if you stiffen the rear.

Does that help?

 

[TMcRally]

1
Goran, in your example I assumed that when the weight transferred it transferred equally from front and back and transferred that weight proportionally to the leg with most resistance.

1
The sprung chassis is getting a roll motion from cgh and side load G-force. The chassis is stiff as a safe (on our superb race cars) so the force of roll is equal along the chassis.
In my example moment of roll was 200 that as you suggest, must be distributed according to roll resistance, which was zero one axle and 100% the other.

Thanks for you patience.

In this exercise we would have a greater total weight on the axel with resistance - 500 at one end and 300 at the other. While I understand what alanbaker is saying about the proportionally diminishing grip as weight is added, surely the axel with 500lbs will have more grip then the one with 300 regardless of the distribution from inside to outside ? I understand that there will be less total grip available for the car but to me it seems the end with more weight should hang on better than the lighter end.

 

[alangbaker]

" While I understand what alanbaker is saying about the proportionally diminishing grip as weight is added, surely the axel with 500lbs will have more grip then the one with 300 regardless of the distribution from inside to outside ?"

Ah, but!

You must remember: the axle with more weight only has more weight on it because the centre of gravity is near to it and also the centre of *mass*. So all else being equal, the larger lateral force it can produce is in direct proportion to the greater *mass* it must accelerate and thus the steering is neutral.

Now throw in the effects of differing roll stiffness at each end. If both axles share equally in the roll stiffness, then the overall grip is diminished equally as the weight is transferred and everything remains neutral. Stiffen one end more than the other and it will suffer greater diminishment of grip than the other end and that will unbalance the handling from neutral to oversteer (if it is the rear that is stiffened) or understeer (if it is the front).

I hope that clarifies things.

 

[TMcRally]

In the example we had even weight distribution; 200lbs on each wheel at rest and no roll resistance in the front and 100% resistance in the rear with 200lbs of total weight transfer in turn.

a). So in the turn we have 400 outside rear 100 inside rear, 200 outside front 100 inside front

OR did we have

b). 400 outside rear 0 inside rear, 200 outside front 200 inside front - thinking about the 3 wheeled car I guess this is right.

So I am thinking that the mass does not move so its resistance to change of direction stays with the front and the same goes for the back. It is not subject to moving around the car like the vertical force does when it moves to the wheel with most resistance. So if we have a). I’m right and the whole world is wrong - I might win a prize and if we have b). it will make more sense.

But in a race car as rigid as a safe (what is it by the way) I can’t understand how the weight transfer off the inside front doesn’t travel to the outside rear if it can’t put it down on the outside front. Why doesn’t it move and all the weight come from the inside rear ?

 

[alangbaker]

There is no reason that all the roll resistance cannot come from one end of the vehicle... .

...at least in theory.

In theory, a car without any sort of rear suspension at all would resist all roll at the rear, while the fronts thus had to resist none of it (because in order to change the forces on a sprung suspension, the suspension must be extended or compressed, and that wouldn't happen if the rear end had no suspension at all.

So in that idealized situation (which couldn't really exist, because even a nominally "rigid" body would deflect somewhat and if nothing else, so would the tires), the weight distribution would go to fronts 200lbs each, outer rear 400, inner rear 200.

And in that situation, the inner rear would be developing 0 lateral force while the outer rear would *not* be developing twice the force it had developed at 200lbs load.

So the rear end -- relatively speaking -- loses lateral force compared to the front and you get oversteer if you were neutral prior to that, more oversteer if you were already oversteering, or less understeer if you were understeering previously.

Even if the overall weight balance were 40% front, 60% rear, it would change nothing.