Got to read the fine print, guys:

"If you don’t like VRC Pro you get your money back, no questions asked**"

That includes the whopping 5 Euros for a full month of play. I spent more than that on lunch today.

It's been a long six years, but we've finally got it ready and out to the public.

https://www.vrcworld.com

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.

Last edited by jtw62074, Mon, 13 Jun 2011 04:48.

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

Last edited by jtw62074, Sat, 11 Jun 2011 01:49. Reason : &quot;lb/ft&quot; should have been &quot;lb*ft&quot;

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.

Last edited by jtw62074, Sat, 11 Jun 2011 01:45. Reason : &quot;lb/ft&quot; should have been &quot;lb*ft&quot;

"Game nerd" were her exact words.

Yes, and I like horsey food. Good to see you, Rene

Last edited by jtw62074, Wed, 29 Dec 2010 09:32.

I suspect his meaning of "played with" meant something a bit more than that

I have an idea too: Let Scawen do whatever he wants. It's his creation. He knows better how to proceed with LFS development than any of us do

Just ran across something that reminded me of this thread. Regarding the shapes that the poster was coming up with in Excel, what I was trying to get at is the end result should look something like this:

http://white-smoke.wetpaint.com/page/Combined+forces

There shouldn't be any valleys and so on in the middle regions. This is probably old news by now, but wanted to post it anyway for others in the future.

Quote from RasmusL :

Mathematically, I still think the OP is actually talking about a CoG below the ground. Not really possible, but he's just imagining things

If that's the case then indeed the weight transfer would go the other way as others pointed out

Many RC cars do have the center of gravity below the middle of the tire and run the front roll center below the ground as well. Weight transfer still works the same way. It's strictly due to lateral acceleration, center of gravity height above the ground, and track width (if we're talking pure cornering). The roll center position just shifts some of the weight transfer directly to the suspension arms so it doesn't "show up in the springs and anti-rollbars," so to speak.

Does a car really roll around its roll center? This one is a little tricky to answer because in a sense it does and in a sense it doesn't.

If you were to draw the suspension geometry at 0 body roll and draw the roll center, it's tempting to solve for the geometry at 5 degrees body roll by first rotating the body some tiny amount around the roll center, say 0.01 degree, and then draw out the new roll centers. Then repeat the process by rotating the car around the new roll center, over and over until you hit 5 degrees of body roll. Even though at one instant in time the car does roll around the roll center, this particular method does not really show how things will move. Once you hit 5 degrees of body roll I'm pretty sure you'll have a different geometry than you will in reality.

What that approach misses is the jacking force effect that someone quoted earlier. You do probably have a rotation precisely around the roll center (roll axis), but you also get a vertical motion of the body in addition to it that changes the results you'd get from a purely kinematic approach. In other words, as the body is rolling the roll centers are not moving the same way a purely kinematic approach would predict. So from that perspective it makes sense to say the car doesn't really roll around the roll centers, even though if you look purely at the rotation separately from the vertical motion, you'll probably find that at any given instant the rotation is right around the roll center. In that case the answer is "yes."

For the non-math/physics head I would be happy saying that the motion does not really go around the roll center and answering "no" to the question. If all the car did was roll around the roll center the center of gravity would always get lower with roll (unless it was Bob's "roll center is higher than the CG and rolls into the corner like a motorcycle" car). The jacking can lift the center of gravity.

The key thought is "at this instant it is rolling around the roll center, but it is also moving vertically which we are ignoring."

Last edited by jtw62074, Wed, 24 Nov 2010 23:27.

Quote from DevilDare :

This. The only gun that needed changing was the M60. I mean that thing was crazy.

Dammit. I just started getting used to it.

Quote from Juls :

Hm yeah my exemple with spring dampers was too simple. I was thinking something like take a simple model like a LUGRE stiction/friction and I figured it would not look simple at all for the reader

I wanted to say something like that: IMO if you really model the tyre/road action/reaction like a stiction/friction mechanism, even a very simplified one, you can avoid many pitfalls.
- some stiction/friction models are robust at speed 0
- they can immediately take into account hysteresis
- they seem scalable and very convenient for object oriented programming

Of course you have lots of experience I clearly don't have. But as a software developer, my feeling is that implementing such model instead of the very usual slip curve model could be a win win situation. Possibly richer model at the end, probably less problems adjusting it, more intuitive. Don't see the problems precisely in advance...but usually I am not too bad guessing which general direction is less crowded with problems than the other

Juls, you're intuition on this one is spot on. I've been doing things in a very similar manner for quite some time now and it indeed works very well

For any others here who have heard of "problems with Pacejka" when it comes to car sims, you've now seen exactly what developers run into and the dilemmas that pop up. Historically most sim developers have been completely unaware of these very issues. So when folks say "yes, Pacejka's fine, but you need to be very careful about combined grip" you now have an idea of what they're talking about

Pacejka has published solutions to this, so again I encourage anyone who wants to use Pacejka's magic model (he's written tons of other models too, this is just the only one that sim folks hear about), make sure you use one of his full, combined solutions that solves these problems or at least minimizes them.

Last edited by jtw62074, Sat, 17 Apr 2010 02:12.

Quote from AndRand :

Don't tell me Excel calculates it wrong as 939,84

@Nikn: I think you should zoom it in to get the feel of one quater 'cos it is symmetrical

BTW: I perceived those graphs as geometrical and forgot about units. In fact it doesnt matter if I use %, degrees or rads but the numbers should be used in the proper places for chosen units if taken from other charts

Whoops!! I was wrong Not sure how I came up with 1371. Guess I better call it a day

And we have the error confirmed...

I picked a cell that was in the valley.

Column 17 (along horizontal axis) and row 25 (along vertical axis) (8 degree slip angle and 6 slip ratio).

Your top graph shows for FX = 792
Your middle graph shows for FY = 506

Vector addition should gives us (792*792+506*506)^0.5 = 1371 which is what we should see in your bottom graph. (This would change the curves dramatically but they'd still be wrong).

Your bottom graph shows 940 for this cell.

Quote :

this is because I used linear or exponential proportion instead of hyperbolic for changing of force curve along the opposite coordinate because I didnt bother for big values of SR and SA.

I haven't the slightest clue what you're talking about there.

Anyway, here's the main point: You said you were plotting the vector sum. You aren't. I have no idea what you're doing, but it's definitely not right

Deleted. Was having a brain fart

Last edited by jtw62074, Thu, 8 Apr 2010 20:50.

Quote from AndRand :

Fx - Slip Ratio from 0 to 90%, Slip Angle for 2, 5, 8 and 12 degrees on Figure 6.
Fy - Slip Angle from 0 to 30 degrees and SR 10, 20, 40 and 80% on Figure 7.

You say they dont match?

I was referring only to the bottom graph of figure 6, sorry. The top graph indeed is largely complete sliding.

I'm not sure I understand your question. My earlier post for figure 7 (top diagram) shows clearly no valleys or bumps. I'm not going to repeat the exercise for the top graph of figure 6. What I'm trying to show you is that your graphs are not showing what you think they are showing.

Last edited by jtw62074, Thu, 8 Apr 2010 20:53.

Quote :

top 3d chart is result (vector sum) of charts below.

Check out my attachment. I challenge you to find any condition plotted to the right of my red slip lines (high slip) where you get a value as low as 600 like you get in the back left area of your graphs. You can not possibly be really plotting the vector sum. Check your math.

Quote from AndRand :

top 3d chart is result (vector sum) of charts below.

No they aren't. How are you getting values at extreme slip less than 600 given those Fx and Fy graphs? At the highest slips neither of the 2D graphs goes below 800 by itself, let alone what happens if you add the corresponding 900 to 1100 of the other F to it vectorially.

Quote from AndRand :

Did you use Figure 7 or Figure 7 and Figure 6?

Figure 7. That one is the graph that is showing what is going on at extreme slips where your valleys and bumps are showing up.

Figure 6 is mostly below the limit of traction except in the areas where they hook back to the left as they go down. The rest of it is near the origin where your graphs start at 0 and climb extremely quickly and then go off and do goofy things.

Combination slip below the peaks and combination slip passed them are probably going to be two different things with the Magic Model.

Quote from AndRand :

Sorry to put it bluntly... but here it IS MEASURED (for both Fx and Fy). And it sure doesnt look like turning that slip ratio 0 curve 360°.

And your graphs don't reflect this data at all like you thought they did, per my last post. I agree with Android in that it looks like you aren't really understanding what your own graphs are saying and how they relate to the ones in the paper

Quote from AndroidXP :

@AndRand: I'm not sure if you're actually understanding the graphs you're posting here, since if you did you'd immediately notice they're nonsense.

To give you a rough idea of what it should look like, let's for a moment forget about the longitudinal (slip ratio) grip curve. Just take the lateral one, put a pin at the 0-slip-angle point, grab it by its tail and spin it around 360°, leaving a 3D-trail while doing so. The shape you're left with is one of a rather simplisticly simulated tyre that behaves exactly the same no matter which direction it is being pushed. Now granted, just thinking about the tyre shape and its basic workings a little makes it clear that a real tyre doesn't work like this, but if used in a simulation it would result in a fairly drivable car, already several orders of magnitude better than what your magic fairy dust canyon curve would produce.

Now, that's only one curve which sucks, so lets make that a combination of two curves. But instead of spending ages tinkering with it and somehow ending up with the forces cancelling each other out, just morph the curve from the lateral to the longitudinal one as you do your 360° spin (so I guess you make a linear interpolation between both curves modulated by the current rotation angle or something). That's probably not quite how a real tyre works either, but by doing this really simple procedure you'd have already far surpassed everything you've brought up till now.

Sorry for putting this a bit bluntly, I just can't help myself sometimes :o

Right on the money.

Quote from AndRand :

Well, I took data from Figure 7 for lateral force and for Figure 6 for longitudinal and then combined them vectorially (which result is shown on my 3d graphs)

Obviously not because your data is showing something quite different from mine. One of us made an error in our math. Quite sure it wasn't me

You posted earlier a link to figure 7 here, but I don't want to go find it now so here:

http://hal.archives-ouvertes.f ... 00/05/14/75/PDF/vsd05.pdf

I think you'd made a comment that your graphs are following what's happening in the top diagram. In my own tire models I frequently plot exactly what you have in the top diagram. However, I also plot the resultant force on that same graph. What you see is a line that looks very much like a pure force curve. It goes up very quickly to the point where you have total slippage in the patch and then perhaps decreases slowly.

There is never some weird combination of slip out in this area that results in a valley or saddle shape like all your graphs show.

I just whipped up a quick little proggy to take the data in the top graph of figure 7 and add it all together vectorially. Can't be bothered right now with making a 3D app to display it like you're doing, but perhaps seeing the numerical data laid out in the same manner as yours is might suffice to show that there are no such bumps or valleys along the diagonals as you're getting. Part of your argument was that your graphs reflect this particular diagram. I assure you they most certainly do not

In the top graph of figure 7 you're looking at combined slip very close to and well passed the friction limit. In other words, it's all outside in that "falling down towards a plateau" region. What I want to show here is that there is NOT a valley or a region where the combined forces suddenly drop way off at some weird combination of slip angle and slip ratio. The drop fans out in all directions in a similar manner as you go farther from the origin.

In my attachment the forces are laid out in the same manner as in your graph. The bottom right (2700) is the spike where the combined force is highest. This is slip ratio 0.1 and slip angle 0. This is pure longitudinal force. And it is the highest point on the graph along with the 2703 at slip ratio 0.2 and slip angle 0. Again, I want to point out that the highest point on the graph is right along the pure slip line. It is not off somewhere appearing as a hump in the combined area. If it were, nobody would have bothered coming up with all this friction ellipse theory stuff since it would obviously be wrong. We'd need a friction cross/bubble thing theory instead like I showed earlier.

Moving to the left the slip angles are increasing 5 degrees at a time up to 30 degrees slip angle. Moving upwards the slip ratios are increasing from 0.1 to 0.2, 0.4, and 0.8. (This is what you get when you're drifting with the wheels on fire )

For others following along, what I'm doing is taking a point on the dotted line at say, slip angle 5 degrees, for Sx = 0.1 and adding it to the point on the solid line at the same slip angle vectorially. (Fx * Fx + Fy * Fy). So this is the total force the tire is producing at all of these combinations of slip angles and slip ratios. They correspond to the height of various spots in AndRand's combined force graphs. This says nothing at all about the actual direction of the force which is another topic entirely that we probably will not get to given how the thread is developing, unfortunately.

You'll see the same trend as in that green graph I posted earlier. Near the origin (bottom right) you have a spike, then as you move to the left and "up" the values quickly drop to a sort of plateau where they slowly continue dropping. However, along the diagonal there is no trend at all toward having any sudden upturn in force again (especially above what you get in pure slip, my goodness...) nor is there a valley along any diagonal where the forces are far lower than what you get in surrounding areas. Note that the "back row" (top row) all have pretty much the same value. It doesn't suddenly collapse to 50% or something at the top left corner like it does in some of your graphs.

Keep in mind that I am loosely eyeballing all this data reading it off the graph, so there is some noise in there. I've been writing tire models for a good 10 years now. I think my first attempt at it was in 1993 or so with nothing but Fred Puhn's "How to Make Your Car Handle" on a 386SX-16Mhz. It doesn't mean I know everything, but I'm not a noob to this stuff and do know what I'm talking about on this one

Last edited by jtw62074, Thu, 8 Apr 2010 20:05.

Forgive the goofy artwork. If those combined force graphs looked like that (saddle shape) you'd wind up with not a friction ellipse theory, but something more like the shape in my attachment. Here, along some combination of slip angle/slip ratio you couldn't make as much force as you could along pure slip or any other combination of slip angle/ratio. Sorry to say, but this is complete nonsense