The online racing simulator
Understeer problems
2
(50 posts, started )
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(franky500) DELETED by franky500 : *cough*
i'm not feeling picked on. but not removing the 2nd post as well could contribute to me feeling picked on. what i am feeling is that my post was brushed under the mat (so to speak) so that the conversation could continue. had you moderated correctly, the op would have been investigated a bit more to find out whether he was infact using pirated software. the conversation would still continued without him/her being able to offer further input. the fact that my post stood out should have been enough for you to act and moderate accordingly.
you have to remember that one mans rubbish is another mans treasure. so although you might have the opinion that there's lots of rubbish in the forum, others are under the opinion that it's a fun place to be. as a moderator, you should recognise this.
is one post in this thread enough for you to delete my post claiming i was banging on?
Quote from dadge :?

dude, just let it go. as a former moderator for a high-ranking forum, i know these guys can get you pretty miffed because of how they moderate (or lackings thereof). but the fact is, they can't be bothered to change their status quo, so it's better to just leave it alone.
Quote from bunder9999 :dude, just let it go. as a former moderator for a high-ranking forum, i know these guys can get you pretty miffed because of how they moderate (or lackings thereof). but the fact is, they can't be bothered to change their status quo, so it's better to just leave it alone.

eh? you seem to be misunderstanding my view. i made one post in this thread. it was deleted. i wanted to know why.
if we all "just let it go" then what's the point of us buying LFS when all we'd have to do is crack the game and still get support from the official forum? why did we pay £24 if cracking the game get you the same privileges?
As a current LFS user (and customer), i feel that i am entitled to post my opinion in this thread. again, it was a light hearted post i made (that was deleted). i don't think it was reported so i don't understand why it was deleted. maybe as an ex-mod, you could help clear that up for me?


(not directed at anybody in particular) a moderator that can't be bothered is lazy moderator and should not be moderating any further. if doing the right thing is too much hassle, then maybe it's time to pass on the mantle so someone who is willing to do the job correctly?
all we need is for the OP to post how he's able to use the FXO with demo status and the conversation will get right back on track i'm sure. and if i'm wrong i'll apologise.
Quote from dadge :eh?

you won't get any disagreements with me about the current state of moderation... but rather than continue this conversation in public, i sent you a pm.
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?
Hi pajkul.

would you be as kind and let the forum know how you are able to drive the FXO with a demo account? there seems to be a bit of a misunderstanding and this information would really help clear things up.

thanks for your time.
Quote from pajkul :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?

Correct, although the cornering stiffness of a tyre, and the maximum lateral force, do not necessarily increase at the same rate (usually the stiffness increases less, thus the required slip angle ends up increasing), and neither increase as quickly as load.
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(pajkul) DELETED by pajkul
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(pajkul) DELETED by pajkul
Some good info in this thread and reminds me of the challenges of setting up tbo class cars. Imagine if we had rain in LFS. That would make things really interesting.
@dadge: hey forum police - did it ever occur to you that not everybody knows that he can log into this forum with their s1/s2 account?
maybe he is just a user that uses another nick on the forums than ingame?

and do you think it is impossible to play at another persons computer?

please do not assume everybody is as you think he is...

and if you accuse someone of misbehaviour - then don´t spam treads yourself...

thank you, back to topic...

peace, mo
spam? last i checked i was posting my opinion. it's your issue if you think it's spam. if you can't respond to a post without resorting to calling people names like "forum police" then maybe you should go back to watching catroons until the grown ups have finished talking?
please tell me how it's spam? my posts are thought out and have not insulted anyone. unlike your little rant about forum police and spam. seriously dude, get a proper opinion, the one you currently have is making you look like a kid.

i don't mean to throw a spanner into your little theory but you forgot to include the point i was actually making, what if he had cracked the game? i've already done the job of the moderator by asking him if he has actually paid for the game. a moderator has already informed me that the ip that the OP is posting under does not match that of any LFS S2 user.
do you think i just blindly claimed that this guy was using cracked software? because you'd be wrong kid.

Tip, next time you talk to someone and you would like them to engage in a mature conversation, don't start it by calling the guy childish names.
Last question, are vertical load on a tyre and the current slip angle linear?
Everything Bob said is right. I'll try to explain this another way using some simple math that illustrates what happens with a car that's cornering at the limit.

The questions are basically these:

1) Why does a car tend towards understeer when you have most of the weight on the front tires?

2) Since those tires are more heavily loaded than the rears, shouldn't the force at the front be greater than at the rear and therefore tend to make the car more oversteer instead of the other way around?

Here we'll be looking only at the case where all four tires are producing maximum force (at the limit).

Forget about centrifugal/centripetal force and momentum. There's no need to look at these concepts for a simple vehicle dynamics problem like this one (or any that I can think off the top of my head). All we have are four lateral (sideways/cornering) tire forces trying to pull the car around the corner, plus an extra vertical force at each tire.

Those four lateral forces, in addition to pushing the car sideways into the turn, each create a torque around the center of gravity of the car. When a car is in a steady corner the front torques try to spin the car into the corner (let's call this "positive torque") while the rears do just the opposite and try to straighten up the car (let's call this "negative torque"). When you add the torques together in a steady corner they equal 0. This is when the car is rotating at a constant speed and tracking a perfect circle as it corners.

Let's break out some very simple math. We'll take an example car that has 50/50 front/rear weight distribution. The car weighs 2000 lb. The front tires support 1000 lb of weight between the pair. Same for the rear. (That's our definition of weight distribution, so we can be sure that's right).

weight is 2000lb
1000 lb weight on front tires
1000 lb weight on rear tires

Let's assume the wheelbase (distance from front to rear tires) is 8 feet. The distance from the center of gravity to the front axle is 1/2 this distance. Same with the rear:

distance from center of gravity to front axle = 4 feet
distance from center of gravity to rear axle = 4 feet

Let's assume the tires are all the same and have a friction or grip coefficient of 1, meaning if you push downward on the tires with 1000 lb of force they can produce 1000 lb of cornering force to get you driving around the circle.

Cornering force = vertical load X friction coefficient

So our front tires (combined) produce:

cornering force at front = 1000 * 1 = 1000
cornering force at rear = 1000 * 1 = 1000

Now we can find the yaw torques.

Yaw torque = lateral force X distance from axle to center of gravity. In this case:

yaw torque front = 1000 * 4 = 4000 lb*ft
yaw torque rear = 1000 * 4 = -4000 lb*ft

The total yaw torque = 4000 - 4000 = 0.

Ok, so now we have a car driving around a circle at 1g. What happens if we move the center of gravity forward? If we repeat these same calculations for a front/rear weight distribution of 62.5%/37.5%, here's what we get:

Distance from cg to front axle = 3 feet
Distance from cg to rear axle = 5 feet

1250 lb weight on front
750 lb weight on rear

Assuming tires have friction coefficient of 1

lateral force at front = 1250 * 1 = 1250
lateral force at rear = 750 * 1 = 750

yaw torque front = 1250 * 3 = 3750 lb*ft
yaw torque rear = 750 * 5 = -3750 lb*ft

total yaw torque = 3750 - 3750 = 0

The yaw torque is still 0 and the total tire force is still 1250 + 750 = 2000. So this car is behaving exactly the same way at the limit as the car with the 50/50 weight distribution did. What's going on??

This brings us to the answer to question 2:

First off, the front tire forces are larger just as you'd expect: 1250 versus 1000. However, the distance from the front axle to the cg is a lot lower, so the torque is actually lower than it was before. The rear has less force now (750 versus 1000), but the distance from the rear axle to the cg is higher. The end result is you still have 0 total torque on the car so it doesn't understeer or oversteer at the limit any more than the 50/50 car did.

So what's missing here? Why does a car understeer when you move the weight forward when our analysis here seems to show that it wouldn't make any difference?

The answer is the load sensitivity of the tire. For the examples above we assumed the grip/friction coefficient was 1 and didn't change at all when the vertical load on the tires changed. If real tires actually worked that way then it wouldn't make much difference what you did with the weight distribution. However, real tires have a friction coefficient that changes with load. All tires are different in terms of how much changes actually are, but they all have one thing in common: As you increase the load the friction coefficient gets smaller. So we might have a friction coefficient of 1 at 1000 lb vertical load, but it might drop to 0.9 with 1250 load, and increase to 1.1 with 750 load.

Let's try this same calculation on the 63.5/37.5 car, only now instead of assuming the friction coefficients are always 1, we will use these values instead:

750 lb load -> 1.1
1000 lb load -> 1
1250 lb load -> 0.9

This stuff is the same as before:

Distance from cg to front axle = 3 feet
Distance from cg to rear axle = 5 feet

1250 lb weight on front
750 lb weight on rear

Here's where things change. Instead of using 1 we use 0.9 and 1.1:

lateral force at front = 1250 * 0.9 = 1125
lateral force at rear = 750 * 1.1 = 825

yaw torque front = 1125* 3 = 3375 lb*ft
yaw torque rear = 825 * 5 = -4125 lb*ft

total yaw torque = 3375 - 4125 = -750

Check out the total yaw torque! It's not 0 anymore. The rear tires, even though they create less cornering force than they did on the 50/50 car, now create more yaw torque. At the front the opposite is happening. Let's take a look at those yaw torque numbers for both cars with the tire load sensitivity:

50/50 car:

yaw torque front = 1000 * 4 = 4000 lb*ft
yaw torque rear = 1000 * 4 = -4000 lb*ft

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 = -3375 lb*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.

So the rear pair of tires are running at only 675 / 825 = 82% of their potential. Since the total of front and rear tire forces is now lower, the lateral acceleration is less (you have to go slower through the corner) and the car is now more understeer since the rear slip angles decrease to the point where they make only 675 lb of force. The front tires are at their limit but there's still plenty more left in the rears.

Again, you can forget all about centripetal/centrifugal force and whether or not they even exist because they aren't necessary for this type of analysis anyway. Momentum is just mass X velocity so also has nothing to do with this.
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.

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(pajkul) DELETED by pajkul
Quote from pajkul :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?

Slip angle and vertical load: These are two separate things so I'm not exactly sure what you're really asking. You can have a vertical load of 100 and a slip angle of 0,1,2,3, etc, without changing the load. They're independent from each other. If you pose the question differently maybe I can shed some light on whatever it is you're wondering about. I think I'm just not understanding what you're really asking.

Rear yaw torque: In the example you're referring to where

yaw torque front = 1125* 3 = 3375 lb*ft
yaw torque rear = 825 * 5 = -4125 lb*ft

I'm not sure if you understand what yaw torque (also called "yaw moment") is. Let's imagine this is a right hand turn. What this positive 3375 on the front means is that the front of the car is being pulled around like a spinning top towards the right (clockwise when viewed from above). The negative value (-4125) on the rear means the rear tires are trying to twist it the other direction (back to the left). In this example the rear is pulling harder to the left than the front is pulling to the right. If you add the two yaw torques together you get:

3375 - 4125 = -750

A negative number means the rear tires are "winning the fight" so to speak and the end result is to twist the car back to the left. (In a sense. More accurately what is happening is the rotation speed of the car as it's going around the circle decreases).

As it's doing so the rear slip angles are getting smaller, which is making the lateral forces at the rear smaller, which is in turn reducing the yaw moment at the rear. The -750 goes to -650, -550, etc., until the rear slip angles reach a point where the lateral forces give a yaw moment at the rear that equals the front. The sum of the yaw moments is now 0 and the car's rotation speed becomes constant. The car is now tracking a bigger circle than it was before. The rear tires are not producing full capacity and their slip angles are smaller. That's increased understeer, basically, and is why a front heavy car tends to be an understeer car especially if it has the same tires at all four corners.

(A rear heavy car is just the opposite and tends to be an oversteer car for the same reason, which is why you'll usually see comparatively great big tires on the back of mid or rear engine cars. They need more lateral force at the rear to balance the car.)

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

Yes.

That's the force here:

lateral force at front = 1250 * 0.9 = 1125
lateral force at rear = 750 * 1.1 = 825

That's the vertical load (weight on the tire) multiplied by the friction coefficient of the tire. In this example the load sensitivity was included so the friction coefficient is 0.9 for the heavily loaded front tires (1250 lb), and 1.1 for the lightly loaded rear tires (750 lb). It is the maximum force the tires can make in any direction, whether it's cornering, braking, accelerating, or some combination of these (friction circle/ellipse theory).

I called it "cornering force" instead of "lateral force" in another example. It's the same thing as lateral force. Just different terminology. I should have stuck to one term or the other to avoid confusion

To add something regarding FWD cars: In the example calculations above, one thing that happens when when you hit the throttle while cornering is the lateral force at the front gets smaller. And any time you reduce the lateral force you also reduce the yaw moment (basically that's the case although there are a couple things left out here for simplicity). The point here is that moving the weight distribution to the front has this same basic increased understeer effect regardless of whether it's a FWD or RWD (or AWD). When we start looking at cars that are accelerating/braking while cornering then we can see some major differences. I'd be happy to go into that stuff if you'd like. This type of discussion is fun. Haven't had one of these in awhile
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.
Drive properly and the car won't understeer
Quote from wolfshark :Drive properly and the car won't understeer

That is totally wrong. Understeering happens because of the setup most of the times. So with a tight setup, he won't be able to turn a car at the same speed thus understeering it.
Pajkul, take a look at this:

http://features.evolutionm.net/imageview.php?image=1538

This shows lateral force as a function of slip angle for three different loads (weights on the tire). If this tire was loaded to 900 lb and you wanted to get maximum lateral force out of it you would make sure your slip angle is at about 6 degrees. If it's the front tire we're talking about, you control that slip angle directly with the steering wheel. If you then doubled the load to 1800 lb you would need to increase the slip angle a little bit to reach maximum force. Generally with tires there is some small increase in the slip angle where the peak force occurs, but it does not double when you double the load. See how the line of peaks moves to a slightly higher slip angle as the load increases? I think that might be what you're looking for here..

If you were to double the weight of a car the slip angle at peak lateral force would increase a little bit, generally, as you can see in this graph. It depends on the tire though. With some tires the line of peaks is slanted quite a bit and with other tires there's hardly any difference until you get to really high loads beyond what the tire was designed to operate at.

Trailbraking: Yes, I think you've got the idea. That generally would tend to create some oversteer unless you've got a lot of forward brake bias that ends up reducing the front lateral force to the point where you end up in understeer. We should probably keep the discussion to maximum, limit tire performance where we're driving the tires at the peaks of the lateral force curve. When you're below that limit all bets are off and lots of different things can happen.

Let's look at the difference between getting more weight on the front wheels through a static weight distribution change and then getting that same weight transfer through trailbraking the car a little bit. The effects are rather opposite and the reason is because the length of the torque arms changes. For instance, here's our 50/50 static weight distribution car again:

weight is 2000lb
1000 lb weight on front tires
1000 lb weight on rear tires

wheelbase is 8 feet which gives:

distance from center of gravity to front axle = 4 feet
distance from center of gravity to rear axle = 4 feet

Cornering force = vertical load X friction coefficient

Using tire load sensitive friction coefficients like before (0.9, 1, and 1.1):

So our front tires (combined) produce:

cornering force at front = 1000 * 1 = 1000
cornering force at rear = 1000 * 1 = 1000

Yaw torque = lateral force X distance from axle to center of gravity. In this case:

yaw torque front = 1000 * 4 = 4000 lb*ft
yaw torque rear = 1000 * 4 = -4000 lb*ft

The total yaw torque = 4000 - 4000 = 0.

We have 1000 lb cornering force at both front and rear. This is the car in a steady state turn. As we compare it with the front heavy car (63.5%/37.5% front/rear static weight distribution) we found that the car without any load sensitivity would behave the same way as the 50/50 car did even though there was a lot more lateral force at the front than at the rear. This is because the torque arm lengths:

50/50 car:
front = 4
rear = 4

changed to:
front = 3
rear = 5

With load sensitivity added in we ended up with an understeering car where the yaw torque was negative if both tires were limiting. From a previous post:

Quote :
lateral force at front = 1250 * 0.9 = 1125
lateral force at rear = 750 * 1.1 = 825

yaw torque front = 1125* 3 = 3375 lb*ft
yaw torque rear = 825 * 5 = -4125 lb*ft

total yaw torque = 3375 - 4125 = -750

See the 3 and the 5? That's a huge part of the equation. The way it works out is the torque arm length changes balance out the fact that you have more lateral force. The lateral force is only half the picture (one of two terms in the yaw torque equation). The other half is the distance from the cg to the axle.

Ok, let's make a new example that compares this front heavy car to a 50/50 car that is trail braking just hard enough so that the tire/axle loads are exactly the same as the front heavy car was without trailbraking. In other words our 50/50 car is braking a little bit. Just enough to add 250 lb of weight to the front axle and remove 250 lb of weight from the rear:

This part is exactly the same as our front heavy car:

lateral force at front = 1250 * 0.9 = 1125
lateral force at rear = 750 * 1.1 = 825

But this next part is different. The distance from the axles to the cg is 4 for both front and rear instead of being split 3 and 5:

yaw torque front = 1125 * 4 = 4500 lb*ft
yaw torque rear = 825 * 4 = -3300 lb*ft

total yaw torque = 4500 - 3300 = +1200 lb*ft

Here we have a positive yaw torque, which means the car is oversteering now. The front end is rotationally accelerating the car into a spin, faster and faster as it goes around. This is where you countersteer and try to reduce the front slip angle, which reduces the front lateral force, which reduces the front yaw torque, just enough so that the car doesn't snap around.

Quote :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.

I can't view any replays at the moment so I'll just assume that you're understeering on corner entry somewhere.

Now we can introduce a bit more stuff that might help explain. Let's do an example like the other ones, only this time we'll look at all four tires separately. Let's use something like front heavy car as an example since that seems to be what you're most concerned with. Here's a car that's almost the same, but it's 63%/37% instead of 63.5%/37.5%.

1260 lb on front tire pair
740 lb on rear tire pair

Gives us static tire loads (with the car resting) like this:

630 630 <--front tires
370 370 <--rear tires

I'll spare showing most of the calculations since I'm using a program I wrote to display the wheel loads quickly. With this particular configuration (17 inch cg height, 108 inch wheelbase, 65 inch front/rear track width), if we corner the car at 1G these are the wheel loads we'll see:

892 368
632 108

If we assume our friction coefficients follow our 1.1,1.0,0.9 at 750,1000,1250 load, we are getting 0.0004 mu change ("mu" is short hand for friction coefficient) per lb load change. That's the definition of the load sensitivity. It's 0.0004 mu/lb for our tire. (In reality it's usually a little non-linear, but this works for the example).

Here are the friction coefficients that each tire would see at its individual load:

1.0432 1.2528
1.1472 1.3568

What I'm doing here is assuming that 50% of the lateral load transfer occurs across the front axle. I.e., the springs and rollbars and suspension systems and so forth are identical on the front and rear of the car and we have infinite chassis stiffness. 892-368 = 524 lb weight transfer. 632-108 = 524 lb weight transfer too, so we are getting the same weight transfer across both ends of the car.

We're making a right hand turn here and you can see that the outside (left) front tire is the most heavily loaded at 892 lb, just as would be expected for a front heavy car. We're not trailbraking or speeding up at all here. This is a constant speed and radius turn.

If we do something to stiffen up the front of the car so much that 90% of the weight transfer occurs across the front axle (done with harder springs, stiffer anti-rollbar, and/or higher roll center, or just the opposite on the rear), we get these wheel loads:

1101 159
422 318

And the friction coefficients:
0.9596 1.3364
1.2312 1.2728

Because of the tire load sensitivity the friction coefficients are different at all four tires. The outside front tire DOES have more grip in terms of force, but the friction coefficient is lower.

Let's calculate the tire forces now for both cars. This is the vertical force times the friction coefficient.

For the 50% weight transfer across front axle case:
Lateral forces are:
930 461 --> Total force across front pair of tires is 930+461 = 1391
725 146 --> Total force across rear pair of tires is 725+146 = 871

There's more force across the front axle than the rear, just as would be expected from the front heavy car. Does it mean the car spins? No, it doesn't, because this is determined by the yaw torque, not the lateral force directly. This was shown in an earlier post so I'll skip the yaw torque calculation.

Now for the stiff front end with 90% weight transfer across the front axle case:

Lateral forces are:
1056 212 --> Total force across front pair of tires is 1056+212 = 1268
519 404 --> Total force across rear pair of tires is 519+404 = 923

Look what happened when we went to 90% front weight transfer. The lateral force of the outside front tire DID increase, but the front pair of tires, when added together, actually lost some force. The front end lost some grip overall, even though the outside front tire gained some. The total dropped from 1391 lb down to 1268 lb. At the same time the rear did just the opposite. It increased from 871 lb to 923 lb. So in the second case we can see that even though we have loaded up the outside front tire and it IS producing more force than it was before, the net effect over all four tires is an understeer effect.

The outside front tire gained lateral force, but not as much as it could have because the friction coefficient dropped off a bit. At the same time the inside front lost some lateral force and gained some friction coefficient, but the net result across that axle is that the pair of tires, when added together, lost some grip (produced lower forces). This is all due to the load sensitivity of the tires.

Moving along to your other questions:

Quote :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.

The front slip angles are controlled pretty much directly with the steering. All the slip angle is is the angle between where the tire is pointing and where it's moving. The rear slip angles more or less adjust themselves to whatever yaw torque keeps the car balanced, at least in steady state cornering. Steering angle at the wheel controls front steer angle pretty linearly in most cars, although not perfectly generally speaking. However, you can safely imagine it as being linear.

Front slip angle is also affected by the yaw of the car too of course. Imagine putting the car into a 90 degree slide and then straightening up the steering wheel. All four tires are at 90 degrees slip angle. You can think of the front slip angle as being your steering wheel angle times some constant number, plus the rear slip angles basically. That's not exactly right, but close enough.

Quote :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.

The rotational speed is just how many degrees per second the car is rotating around like a spinning top. If it's understeering you can drive a car in a constant speed and size circle very easily. More so than an oversteer car because you don't have to keep trying to fight the car.

Quote :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.

No. The yaw torque because of the difference in forces front and rear is causing the car to straighten up just as shown in the earlier examples. Self-aligning torque causes the torque you feel on the steering wheel and is a separate thing.
Quote from pajkul :
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

I can't really see what you describe.
In which way you detect the grip loss? From corner entry to mid corner, both front tires are slipping above optimum slip angle.. so they are both "loosing grip". (pict 3)
(press F to get forces view)
I guess you saw at the skidmarks the outside tires make. But that doesn't mean the inside tires are not slipping.

In case you are referring to the grip loss that happens earlier while braking, I think it’s quite obvious that curb rumble strips don't let the left suspention settle, making tire load oscillate from normal to almost 0. (pict 1 , 2)


PS
Todd, I admire your patience in explaining the same thing again and again... Probably 98% of your posts here are about tires and vehicle dynamics in general.
Attached images
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2.jpg
3.jpg
So even if there is the skidmark made by the front left tyre, it doesn't mean the tyre doesn't have grip?
As long as it is forced against the ground, It always has grip... in the above case it just got past its optimum slip angle.
What I was really saying though is that also the inside tires where slipping excessively. Just the vertical load was so low that couldn’t justify a skidmark. Hence what you saw in game.

The answer on your questions is given by jtw62074 though. So if there is something more to discuss here, you should refer to the last jtw's post.
here is some HARDCORE pyhsics going on
Quote from jtw62074 :<Post after post of maths>

Wow that's awesome. Thanks for the insight - as it turns out it's all really just quite simple physics Big grin
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Understeer problems
(50 posts, started )
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