Calculating tire forces under combined braking and lateral acceleration for 1 wheel

Hi Guys,

putting the last touches on my bachelor thesis about the design of a FSAE suspension system
for the team that maybe will exist in the future of my former college.

I am going through the last calculations of tire forces experienced in the worst cases.

I assumed braking and cornering as the worst condition in normal circumstances. I excluded hitting the kerb, because my professor is just interested in
the forces under normal conditions.

I will use the formula of the inverse matrix. Ax=B as discussed in an other thread for the forces in the suspension members.

Let's tell you guys how I calculated my final forces for the right front wheel in a left hand turn.

In front view
x Axis: positive to the front of the car
y Axis: positive to the right side
z Axis: positive vertical upwards


The accelerations I am using for braking are 1,5G
and for the lateral acceleration is 1,5 G.

First I calculated the weight transfer between front and rear axle during braking. The car has a weight including driver of 321,4 kg.
With my CG heights and wheelbase, I come to a weight transfer of 90,38 kg.

That gives me a total weight at the front axle of 239,4 kg.
Assuming the car is symmetrical side to side that gives me an initial loading of the front right tires of
119,6 kg.

Next I add the lateral acceleration, which gives me a weight transfer from the left to the right wheel
of 101,4 kg.

That makes a total of 221 kg at the front right wheel. Making it 2168,01 N. At the left tire 18,2 kg. Almost tipping ;)

That`s for the Z component of the B matrix.

For the X component, I made following calculation.
The car is breaking at 1,5G.
321,4 kg * 1,5 G *9,81 m/s² = 4729 N.

Doing the force equilibrium in x direction and assuming a break distribution of 50:50
each wheel takes the same amount of force to equal the decelerating force.

I simplify the tire load vs. grip model quiet a bit here.

This gives me 1182,35 N at the front right wheel in x direction.

Now the Y direction.

The centrifugal force just for the front axle weight to simplify the calculation
at 1,5 G is 2189,6 N.

This centrifugal force has to be put into equilibrium by the tires.
Assuming the load cases for the outer tire with 221 kg = 2168 N and the inner
tire, 18,2 kg = 179 N. Now lets look at the load vs. slip angle vs. lateral force (unfortunatly I had to use a random one I found on the interwebs, so lateral force
with FS tires might be higher)

At an assumed slip angle of 5° the outer tire is capable of roughly 1850 N lateral force.
The inner tire now has to cope with 339 N.

Now I have force in the x, y, z direction and can apply the moments in respect to the cg
to complete my matrix.

X: 1182,35 N
Y: 1850 N
Z : 2168,01 N

Are these sensible numbers to calculate my stresses in the suspension from?

I suggest extending this calculation to all four wheels. Then ask yourself if the results make sense. Make sure you haven't "lost" any tire forces along the way--i.e., you still get your target 1.5g lateral and longitudinal. Also look at the ratio of tire force to normal load for each of your tires. What friction coefficient is required to get that result, and is it reasonable?

Two other questions:
1. What is your justification for choosing 1.5g?
2. It sounds like you chose a 50:50 fore:aft brake distribution. Why?

Hello Edward!

Assuming the tire friction coefficent is 0,9. That means if my tire has 2168 N normal force, it is capable of 1951,2 N longitudinal breaking force.
This would suggest that my number in the x direction is a little bit to small ;) . But I do combine the longitudinal acceleration with the lateral acceleration, so it
should be in the ballpark, according to the friction circle, that not all force in the longitudinal direction can be used, because of the combined condition.

For the use of 1,5 G negative acceleration:
Taking Beaver Racing in the Austrian Event in 2009.
Their Top-Speed was 94,7km/h.
They stopped after 23,35m
That gives me a negative acceleration of 1,5g. Looking at other competitors the negative acceleration ranged from 1,1 G up to 1,9 G.

We haven't built a car yet. So I need to make some calculations before and look at comparable numbers. Otherwise, it would
be easy to install an accelerometer in a car and test it ourselves.

To the break distribution. Yes, I have used the 50:50 split, although it won't reflect reality. It made the calculation simpler. The end distribution would
definitely be biased towards the front, because the front tires are capable of more negative acceleration due to the increased vertical load they experience under braking.

When I just looked at the front axle, I "lost" the tire forces of the back axle. Although the car is braked at it's maximum capability (assumed capability) the rear axle is still able to take some lateral and longitudinal forces.
In respect to that, I assume that my resulting forces for the front tire are smaller than I calculated them.

If you split the horizonal loads (Fx and Fy) so that they match the same distribution (front/rear and left/right) as the calculated vertical loads (including the static weight and aero forces) your result in X and Y will be pretty close to what happens in reality. For braking, the rear tyres are deliberately used well within their limits in order to retain some directional stability, so the brake distribution would be a bit more to the front than the forward load transfer would suggest.

I'm not surprised your car is nearly at tipping point, your load case of 1.5G simultaneously lat + long is a bit over the top. Sure its better to overestimate the forces but in my opinion its better to add these over-estimations as safety factors or uncertainty factors at the end (togther with some justification) instead of starting with an unrealistic loadcase to start with. There are a lot of (valid) simplifactions in your model, so to account for these you will need an uncertainty factor larger than 1 if you want to use the resulting link loads as design inputs.

I think a better method is to do a single 1.5G lat calc, a single 1.5G long calc then a combined 1.06G Lat/Long (which gives a magnitude of 1.5G) and see which load case gives the worst case link loads.

Also I don't understand how you reached 1.5G if you assume a µ of 0.9 at the tyres.

You are reasoning through the components pretty well it seems, nice work.

Quote:

---

x Axis: positive to the front of the car
y Axis: positive to the right side
z Axis: positive vertical upwards

---

Just from trouble I have had in the past, I would be careful about using a coordinate system that is not right-handed, especially if you're solving things using matrices (particularly coordinate rotations like you will probably use for the links).

I am also confused about 0.9g mu with 1.5 g acceleration.

Hi Tim!

I didn't use the coefficient of 0,9 during the initial calculations for the first post. After Edward asked what friction coefficient would be needed to get
to my initial results, I looked for an appropriate friction coefficient and found a 0,9 coefficient for slick racing tires in a paper, just to compare my calculations with a realistic expectable tire force under my normal load configuration.
As I compared it with my findings I saw that my initial numbers were a bit to high.

Combined loading at 1,5 and 1,5 Gs would probably just be possible with downforce, but than I have nothing to compare with. F1 brakes at 4,5-5,5 Gs and corners at ~ 3,5 to 5 Gs
when the braking phase has ended.

I will use a combined 1,06 G and calculate for this case. Thanks for the suggestion Tim.

Using the combined acceleration of 1,06 G gives following forces:

X: 835,4 N
Y: 1090,1 N
Z: 1610,5 N

These should be more realistic.

Hi MP4/4,

Slightly off topic, but out of curiosity. You told us that your college doesn't have an FS team, yet your signature states that you're at the University of Stuttgart.
Did you, or will you transfer to Stuttgart? Or are you guys starting a second (actually third) team?

Hi Thijs,
I am not yet in Stuttgart, although already have my dorm room and will start there in 2 weeks on the 7. of April for the summer semester.
I was referencing to my former college, a small 3000 people college (was an engineering/technical college back in the GDR) in the center of Germany where I have to defend my Bachelor Thesis in a week
to get my Bachelor Of Engineering.

I absolutely can't tell you anything, because as my signature states, I just start in the SS 2014 in Stuttgart and this starts in 2 weeks.
Better ask active members if you have any questions.

MP4/4,

Thanks for the thoughtful answers to my questions. I'm glad you're thinking about the problem: You have some justification for your values and you understand why I was questioning your 50:50 brake distribution. I don't think you have a good estimate of the worst case yet, but you're moving in the right direction. It's a question of finding the possible highest load and determining what lateral and longitudinal forces the tire can generate from that load.

Have you thought about evaluating three "worst cases": Maximum pure braking, maximum pure cornering and perhaps the case you're considering (or similar combined loading which maximizes the load in a critical member)? Each case will load the suspension differently.

I think the next step is to talk with some of your (almost) team members at Stuttgart. They will be able to share data from their car and provide you with their own description of what each wheel experiences. I think you'll have a good discussion. Also, Stuttgart is a TTC member, so you can look at some tire data. Of course, the usual caveat applies--these tires show more grip when tested at Calspan than on real surfaces, but the Stuttgart team can discuss how they handle that, too. This will give you some realistic, single-tire mu values for your estimates.

I must admit I'm a little concerned that the "finishing touches" of your thesis involve asking whether or not your initial assumptions are reasonable. Hopefully this relates to a final calculation at the end of the thesis, not a running example throughout!

Edward,

thanks for your valuable input. It is great to have this forum to get your questions answered.
It has been a great resource for the thesis for a first year team.
Somebody suggested in a former thread, when I was still determining what to write about in the thesis, to
write a sort of "handbook" for a team at the old university to follow along and to get good inputs. I followed that route
with a concentration on the suspension.

It is indeed the final chapter I had to finish. All the others have been finished before!

For the case of pure braking:

If we take 1,5 G and assume a mu of 0,9 in the longitudinal direction (which I recon is a little bit on the low side for a FS tire) the front axle
would experience a vertical loading of 239,4 kg = 2348,5 N. With a coefficent of 0,9
that would give me a braking force of 2113,6 N on the front axle. Making it 1056,7 N at one wheel.

As for pure cornering at 1,5 G:
Taking Wl=W/2+(W*Ay*h)/t

Wl= 148,84 kg/2 + (148,8 kg * 1,5 G * 0,3) = 141,38 kg = 1386,94 N vertical load. Higher than
just braking!
In the lateral direction we have to look at the centrifugal force. The centrifugal force acts on
the CG towards the outside and both tires have to counteract that force.
The outer tire takes more load because of the weight transfer. The higher the CG, the higher the weight transfer.
1,5 G * 148,8 kg * 9,81 m/s² = 2189,59 N

Lateral force is not linearly increasing with increased wheel load. Fy= Fz*mu is not entirely
correct for tires, but I will use it here:
Right tire:
1291,29 N*0,9 = 1162,16 N lateral force

Left tire:
168,43 N*0,9 = 151,587 N lateral force

combining these two give 1313,747 N lateral force that the axle can produce. Countering the 2189,59 N under the 1,5G lat. acc
the tire with that mu won't even go there and sliding before.
As I have said the mu in reality is probably higher than that.
I've seen diagrams with 2,47 mu lateral friction coefficient at 1000 N Fz.
So this would become more a question of not tipping the car :)

So concluding the findings out of this single investigation, my value I have now found for Fy are now higher
than in the last post. The vertical force is higher in the combined case.

What I really need, as you have said Edward, are some realistic numbers for the coefficient of friction for the tire.
But soon this will clear up :)

MP4/4

Think the car is just a big mass on one tire.

F = M*A (whether it is Fy - M * Ay or Fx = M * Ax) and at the same time Fy - Fz * Muy or Fx = Fz Mux.

Unless you have aerodynamic downforce (which you did not stated) how can you have 1.5 G with a coefficient of friction of 0.9?

If you think 4 tires and if you want to make sure your calculations are correct you will have to make a "sanity" check to be sure that the SUM of all Fy on the 4 tires = M * Ay and that the SUM of all Fx on the 4 tires = m * AX

FYI 0.8 to 1.1 is the coefficient of friction of a good passenger car tire on dry, clean asphalt. 1.4 to 1.8 is the coefficient of friction of a warm slick tire on dry, clean asphalt

MP4/4,

Going back to my original reply, if you assume 0.9 for mu on your most heavily loaded tire, what mu will be required from your other tires to achieve 1.5g braking and cornering? I think you'll get some unrealistically large mu values to counter your unrealistically small 0.9. For example, your original straightline braking calculations require the rear tire mu to be about 3.0. Having done that calculation you can ask yourself if that makes sense (the answer is "no". Not if you're assuming mu = 0.9 for the front tires--they're not *that* load sensitive!)

And as Claude said, if all the tires were at mu=0.9 you couldn't do 1.5g.

Claude's comments are perhaps a clearer version of my encouragement to make sure you aren't "losing" any load and to think about all four tires as a system, even though you are doing detail analysis only one. In the end this is the level I want you to get to. Once you see what's happening at all four tires you'll have a much easier time deciding whether or not you have a reasonable condition for your analysis.

FSAE tires are at the grippy end of the spectrum. Doubling your mu assumption would not be a bad place to start. Talk with your Stuttgart teammates, review their data, poke at the TTC data and then you'll make an informed decision about your test case.