Maybe for you.

No force-feedback improvements? I think the force feedback effects are DIRECTLY connected with the tyre physics, as everything the driver feels through the steering wheel results from the tyre forces.

LFS devs assume that no matter how long it will take to finish the new updates, we will love the new LFS so much that we'll forget about that waiting and forgive them immediately.
They are probably right, but I'd rather be kept informed about what's going on and how much waiting is left there.

We will get Big Nothing for Christmas
No report, no test patch... nothing as always.

Name: P.Kulijewicz
LFSF name: pajkul
Team: [SIW]
Car: FZR
Number: 73

I'm not sure if this bug has been adressed already.

So the thing is when in the hotlap mode a replay is played in slow motion, after pressing shift+R the restart takes too much time, the process is slowed down too. That's annoying.

Offtop: Where can I find the description of the forces displayed after pressing F button?

Last edited by pajkul, Wed 15 Jun 2011, 15:52.

So even if there is the skidmark made by the front left tyre, it doesn't mean the tyre doesn't have grip?

What I meant is when car is turning, the tyres have certain slip angle. If I was taking a turn of the same radius at the same speed, but with a 2 times higher load on the tyres (assuming that there's still enough lateral force, or friction for tyres that they can have grip) would the slip angles increase also 2 times? Assuming that there's no weight transfer from left to right or vice versa to make things more simple.

With regard to your last words, there's such a thing called trail braking.

http://soliton.ae.gatech.edu/p ... tsiotra/Papers/ecc07a.pdf

As it was described here, the techniqe basically consists in moving the weight to the front tyres and thus increasing their grip through the balanced use of brakes and throttle.

As the vehicle decelerates, the weight of the vehicle
transfers from the rear to the front axle and thus, the front
tires generate higher friction than the rear ones.

So it actually generates more oversteer. Or maybe I misunderstood you and what you said is more like braking and accelerating while cornering is basically totally different from the pure weight distribution of a static car. Then yes, I want to discuss that too.

Oh, and probably the most importantly: would you be so kind and explain why the front left tyre is losing grip at TURN 1 during the hotlap, not the front right? The natural thing would be that there's more load on the front left tyre, so it has MORE grip than the front right. Probably this is somehow connected to what you've described before - the higher the vertical load, the lower the friction coefficient. But I can't be sure.
Replay:
http://www.lfsworld.net/get_spr.php?file=72052

P.S. And what is the relation between the steering input I apply and slip angle? Linear, non-linear? I guess up to the point linear, then when at the limit of adhesion, the slip angle is increasing much faster than the steering input.
P.S.2: I've heard many times that this is impossible to maintain a constant rotational speed of a understeery car at constant speed. So this is probably correct for high speeds and when on the limit of adhesion.
P.S.3: So the slip angle decreasing you've described when cornering is caused by the self-aligning torque, right? The lateral force is trying to reduce the slip angle. I forgot about that.

Last edited by pajkul, Sun 12 Jun 2011, 08:37.

Wow, thanks. But my last question remains unanswered: does current slip angle increases linearly with the vertical load on the tyre?

And: why is it that the rear yaw torque decreases to the value of the front tyres?

And no. 2: Is the max. lateral force the maximum friction the contact patch can generate?

Quote from jtw62074 :

63.5/37.5 car:



yaw torque front = 1125* 3 = 3375 lb/ft

yaw torque rear = 825 * 5 = -4125 lb/ft



The last car is unbalanced. In reality what would happen is the rear slip angles would decrease so that the rear yaw torque was -3375 which would balance out the front yaw torque. How much lateral force would be required to get -3375 lb/ft torque at the rear?



yaw torque rear = lateral force * 5 = -3375lb/ft



lateral force at rear to stabilize the car = -3375 / 5 = 675 lb



What's happening here with a real car driving in a circle is that the rear tire force drops down to 675 lb through a reduction in the rear slip angle. It can still make 825 lb if you flick the car hard into the turn momentarily, but once the car settles into the turn the rear will produce 675 lb instead of the maximum 825 lb.

Last edited by pajkul, Fri 10 Jun 2011, 10:04.

Last question, are vertical load on a tyre and the current slip angle linear?

Last edited by pajkul, Thu 9 Jun 2011, 21:55.

Quote from Bob Smith :

Essentially, yes. The net result is that the front tyres need to slip more to create the same (in this case, lateral) force as the rears, which is the definition of understeer.

So if we increase the load on the front tyres X times, the cornering stiffness (and thus the maximum friction the tyres can generate) is increased Y times when (Y<X), assuming that there's no weight transfer from left to right or vice versa during the turn?

Quote from roadrash17 :

Momentum isn't a force, it's a conserved quantity.

http://dictionary.reference.com/browse/centrifugal+force

Look at the second reference. It states it's a fictitious force. i.e. it's something we think is there but actually isn't. I have a college level physics book that states what I am telling you. Centrifugal force is something someone came up with to describe the feeling of being pushed outwards when going in a circle. The actual reason you feel like you're being pushed outwards is explained in Newtons First law of Motion: Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it. That external force is the force of friction between the tires and the road.



There is a limit of how fast you can go around a corner with a car. you will never be able to fully eliminate understeer, but you can do your best to increase the top cornering speed of a car. I suggest playing with Camber, tire pressure, and anti-roll.




The thing people call centrifugal force (actually Momentum, a conserved quantity) acts on all bodies in motion.



The suspension is helping the car go around the corner by taking a portion of the kinetic energy (also a conserved quantity) that the car has and dissipating it by compressing the shocks on the outside side of the car as it corners. The acceleration vector does point towards the center of the circle and the fictitious force is, in fact, centrifugal force.

I've done my best to explain this concept to you...it's something that took weeks of schooling to fully comprehend to a point where I could use it and explain it to people. I hope I helped.

I know this is a fictitious force, as I've mentioned before. It is actually the momentum in circular motions. So we have basically the same concepts, correct me if I'm wrong.

Last edited by pajkul, Sun 5 Jun 2011, 19:36.

"The higher load on the front tyres will make the front stiffness relatively less (cornering stiffness increases non-proprortionally with load), thus creating understeer on turn in."

Meaning this? I can't really understand this sentence.:

Anyway, I'm pretty much sure that the momentum (momentum is a fictitious force used in non-inertial reference frames) you wrote about is the centrifugal force. I read about it in some car magazines and web articles written by allegedly smart people. So we have two different opinions, can anyone go into facts?

Quote from roadrash17 :

The force that causes the tires to slip isn't a force, but it's called momentum. Momentum describes a moving objects desire to resist a change in direction. When you drive, the car and everything inside the car has momentum. When you turn, the car WANTs to continue in it's straight path but the force of friction between the tires and the road, if high enough, causes the car to turn, but because of its desire to resist change in direction, the car leans and you feel like there is a force pushing you towards the outside of the corner when there isn't. That feeling is what people like to think of as centrifugal force. Now, if momentum is too great, then the tires break the force of traction and you slide. Better?

Yes, thanks. Can you tell me why is it so hard to eliminate understeer? The front is just slightly heavier than the rear of the car. Even if I set the anti roll to the extreme values, it doesn't help that much.
Just to add one thing: when the car is finally taking the turn, the only grip it has is through its tyres, so the centrifugal force exerts on the car just as on a person sitting inside the car. But the suspension is dimnishing the real effect it has on the car's stablitiy. Am I right? In described case, there is the acceleration vector pointed towards the center of the circle being the path the car drives. So the fictitious force is the centrifugal force.

Quote from Ingolf :

No, it doesn't.
But there is a centripital force.

http://en.wikipedia.org/wiki/Centrifugal_force

It does. In non-inertial reference frames, e.g. from the car perspective.

Quote from roadrash17 :

Mmk. To start off:
-there is no such thing as centrifugal force.

Now that I've told you that, lets get down to business:

Understeer is caused by something called momentum. When you enter a corner with too much speed, the momentum of your car is "breaking" friction and causing the car to slide straight in the corner. The key to proper cornering is limited body roll. The key to limiting body roll is a stiff chassis. Body roll causes weight shifts which disrupt the line that the car was taking. However, from my understanding, a slight amount of weight transfer onto the front tires does help because it increases the force of friction. Hence why if you understeer, you can recover, or at least attempt to, by pressing the gas and brake at the same time. That transfers enough weight to the drive tires so they can "pull" you out of the understeer. And it really does work, I've tried it. lol

Anywho, feel free to roast me. I used my understanding of physics (I'm in school to be a mechanical engineer) to attempt to answer his question.

Why is it that there's no centrifugal force? In general sense of physics, this force exists. And I'm convinced this is one of the forces the car has to deal with when turning. From the car's perspective, this force is trying to beak the friction between the tyres and the road. If the loses grip, it means that there was some higher force pointed against the friction (how do I call this force)? Probably fictitious force, in this case - centrifugal force.

That's what I was afraid of, but no, this is not topic related.

What's more: when I drive RWD with very heavy rear and lock the tyre by braking in the corner, when I turn right, the one blocked is the front right. Even if the brake balance is set to 50%. This means that the load on rear tyres is higher making them more difficult to lock. No matter how many times higher is the centrifugal force generated on the rear of the car, the maximum friction should rise the same number of times, as it depends on the gravity force. I don't get it.

Last edited by pajkul, Sat 4 Jun 2011, 18:23.

Hello everybody,
I was trying to understand the very physical sense of most of the FWD cars being prone to understeer, but some things are getting too far.

http://www.suv-rollovers.com/a ... tions/understeer-01292010

According to this website:

A front-heavy vehicle with low rear roll stiffness (from soft springing and/or undersized or nonexistent rear anti-roll bars) will have a tendency to terminal understeer: its front tires, being more heavily loaded even in the static condition, will reach the limits of their adhesion before the rear tires, and thus will develop larger slip angles. Front-wheel drive cars are also prone to understeer because not only are they usually front-heavy, transmitting power through the front wheels also reduces their grip available for cornering.

To make things less complicated, let's talk about the car not accelerating or braking at all, just entering the corner. I've thought the higher the load on the front tires, the higher the grip. As it turns out, somebody forgot to mention (in my opinion) the important suspension settings factor.

In FXO, for example, the weight distribution is as follows: 57 front, 43 rear. When the car is turning right, the weight is going to the left more to the front of the car than, as the centrifugal force is higher, but at the same time the maximum friction the front can generate is higher than on the rear. We all know that when this max. friction is e.g. two times higher when the centrifugal force is two times higher. To prevent mentioned weight transfer, I set minimum anti roll on the rear and the maximum in the front. 57F and 43R on the paper doesn't seem to make a very big difference, however the car is still very understeery, as the first wheels losing grip are the front ones. Can anyone get me through this?

My bet is as follows:

The physics update won't be released before October 2011.

Why hasn't it been released yet? The answer is quite simple. Releasing a version changing everything in the game physics means erasing all the results on lfsworld.net. Any current WR will be gone for ever, and there's no way they can fix some bugs in the car physics after releasing the update. The car behaviour has to be perfectly predictable, there's no space for some inappropriate results of equations embedded in the physics model.

Thanks, now I understand the problem.

The fictitioue force is turned to the opposite direction to the movement of the wheel. If the fictitious force is turned to the right, then the wheel rotates to the left, thus counter clockwise. I have no idea what is not clear in my explaination.

I'd say the wheel rotates counter clockwise. The tyre surface on the left is substantially thinner than on the right side, and considering the fact there's also fictitious force involved caused by the accelerating movement of the wheel that makes the tyre thicker on the right side, the answer is B.
When wheel accelerates, every point of the wheel wants to stay where it was before acceleration, that's why there's more rubber on the right.