Laptime simulator

[Greg Locock]

Another approach is to add the referred inertia of the powertrain to the vehicle mass. It is not an especially difficult task, but it is error prone. The energy method I previously suggested encapsulates all the same effects, but the effect of gearing and so on is much easier to understand as the angular velocities are explicitly called out.

Either method will work, they are the same equations, just differently expressed.

[MadMatt]

I would rather continue on the path I've started, and not calculate energies for now as I will include weight transfer, slip ratio and load dependent grip, so it is easy to work with forces directly in my programming loop.

I've reworked equation 1 as follow to include the inertia losses at the contact patch:

(1) Fnet = Ftrac - Faero - Finertia

With Finertia defined as this (I am ignoring powertrain losses for now):

(2) Finertia = Finertia_engine + 2 * Finertia_wheel + Finertia_transmission

With:

(3) Finertia_engine = acar * Iengine * itot^2 / rollrad^2

(4) Finertia_wheel = acar * Iwheel / rollrad^2

(5) Finertia_transmission = acar * Itransmission * ireardiff^2 / rollrad^2

I go back then to my previous equation (previously number 2):

(6) acar = Fnet / carmass

But then I directly calculate the angular acceleration of the wheel, as I won't take in account the inertia forces a second time:

(7) alphawheel = acar / rollrad

(8) alphaengine = alphawheel * itot

Does that make more sense?

[Greg Locock]

Sorry i'm not going to check your maths. A big lump of power we've ignored is the rubber destroying work at the contact patch, Fxy*slip_velocity, which in my head i worked out as 45 kW at a 200 kph 3g corner.

[MadMatt]

I still cannot get my head around this simple problem. I feel like my equations at the previous page were right.

First of all, I don't want to work with equivalent mass because that it based on empirical formulas.

The acceleration of the car is defined by the net force at the contact patch that is either the tractive force given by the engine, or the grip available, minus the aero resistance (let's assume no grip limitation for now).

First question: is this tractive force taken straight from the engine, or do inertial forces have to be taken in account here already? And if yes, only the inertia of the 2 non-powered wheels?

I can then calculate the non-powered wheels' angular velocity based on the actual vehicle speed, and the powered wheels' angular velocity based on the wheel torque, which will define the slip ratio.

Anybody correct me if I am wrong, but I believe this is correct.

Second question: how to quantify the torque used to accelerate the engine due to the inertia of the powertrain? I believe the inertia here is the inertia of the 2 powered wheels, the transmission, and the engine itself, as we assume the clutch is not activated at all.

I am still confused at this. I believe machin did a similar simulator some time ago, I think he went through the same stuff.

[mep]

Hello,
of course you need to take into account the inertia of every rotating mass. So rear wheels, drive shaft, differential, gearbox and clutch. The engine is tricky, because when you have a power graph then this should be the net power output of the engine. So the inertia is already taken into account when measuring the power the dyno. However, when the measurement was taken at constant speed then the inertia of the engine does not reduce the power output. So I think the power of the engine depends on the way it was measured. Personally I would ignore the inertia of the engine because its becoming rather difficult to asses it in all details. The error you do by ignoring the engine inertia might be acceptable small.
Important is that you reduce the inertias of the gearbox, wheel, etc. to the engine. So you have to take the transmission ratio into account. Then I think one point of your formulas is not correct. You say the net force is equal to traction force minus the inertia losses. My approach is to calculate a traction force which is basically the amount of force the tire is able to transmit (based on vehicle mass, weight distribution, downforce, tire friction coefficient etc). Apart form that I calculate the amount of force the engine is able to deliver to the rear wheels. This is the engine power reduced by losses due to efficiency and rotating mass inertia. Then I do a comparison between available engine force and max transmittable force by the tire and take the lower of the two values as acceleration force. So in this comparison the engine force is already reduced by the loses to transmit the force to the wheel contact patch.

I start to see the problem, because then the acceleration should then by calculated with the vehicle mass and the remaining inertia of the front wheels. Please note that I wrote that quickly together and it is one solution. There might be others working as well.

[Blanchimont]

I would do what Greg suggested, determine the available power at the flywheel from throttle position, engine rpm and efficiency. Substract the power to accelerate the gearbox, diff, driveshafts, wheels and brake discs. Then substract (drag+rr)*v and the remaining power equals m_car*a*v. Connect the different omegas and alphas(angular acceleration) and you can solve for a.

[MadMatt]

Well, my English is not so good, I struggle to see you you mean. mep, reducing the inertia of the gearbox, wheels, etc to the engine is what I wrote in equation 5 at the previous page:

Iequ = [ ( Iengine + Igear ) * itot^2 ] + [ Ireardiff * reardiffratio^2 ) + 2 * Iwheel

When you mention the traction force is the force the tyre is able to transmit, I agree with this, but in my case, I assume no grip limitation so basically I directly convert the wheel torque into the tractive force (later I will include grip limitation but for now it doesn't change anything).

You wrote "engine power reduced by losses due to efficiency and rotating mass inertia". Yes this is what I am trying to do, calculate the inertia resistance that I can remove from my engine torque, but in this equation:

Torque = Inertia * Angular_acceleration

I am missing one parameter to calculate that torque, it is angular acceleration of the engine.

Again, I don't want to assume no slip between the tyre and the road. The driven wheels will have a higher linear velocity than the car (I am only simulating acceleration at the moment).

This is the same for Blanchimont, you wrote "substract the power to accelerate the gearbox, diff, ...", yes, this is what I am trying to do and this is the question I had labelled "second question". I need to see some equations, I think it would be clearer.

[Blanchimont]

Random numbers for Power = mass*acc*velocity + (drag+rr)*velocity + rot_inertia*omega*angular_acceleration

1000kW = 500kg * a * 50m/s + 100N * 50m/s + 50kgm² * 10/s * alpha

Without slip:
v = r*omega
a = dv/dt = r * d omega/dt = r*alpha --> alpha = a/r

[MadMatt]

I think it is still not what I want. I will leave it for a bit to clear my mind and come back later, but basically I want to have throttle as the input, and car velocity as the output with the possibility to check the slip ratio as the car accelerate, and of course the wheels' angular velocity. Usually in these simulations one of the variables is fixed (slip ratio for example), therefore it is quite easy, but here the only thing that is fixed is the throttle position.

Basically the engine has to gives its maximum torque all the time (depending on RPM), and it accelerates under this produced torque plus/minus resisting forces in the powertrain, allowing the wheel to accelerate, which will be converted to a force at the contact patch to accelerate the car (minus resisting forces such as aero).

I will leave it for a while!

[mep]

Well basically you almost answered the question yourself. If you want to take tire slip into account you need to model the tire. This will also give you the tractive force (force the tire is able to transmit) as this is in relationship with the tire slip ratio. The influence of the slip might actually be negligible but I would not ignore the traction force of the tire. The traction of the tire is actually quite important.

[Greg Locock]

No as I worked out above, the power to shred the rubber at the contact patch is of the order of 60 hp, so only if your sim is hopelessly inaccurate could it be considered negligible.

[mep]

Hold on, he tries to calculate the difference of angular velocity between the engine and the wheels due to slip which means the engine has to spin slightly faster resulting in marginally higher inertia. I say this is negligible, not that the energy consumed by the tire is negligible. Well we could work it out when we assume a certain slip ratio. I would have guessed that a value of around 4% is typical however I just read at Wikipedia that the maximum deceleration is achieved with 10-30%. That is more than I expected.

I would classify the loss you mention as a form of drag. By the way interesting equation you posted there. I assume that is for slip angle, basically cornering, as there is Fy in the equation. Is there an equivalent equation for acceleration (slip ratio; Fx)?

[MadMatt]

No mep, I am trying to calculate the difference of angular velocity between the driven wheels and the non-driven wheels!

[mep]

Ah, ok, well the slip ratio of the non driven wheel should be zero because they are just rolling. You still need a tire model for the driven wheels then.

[MadMatt]

Yes, it should be zero, and yes I have a tyre model, but for simplicity I said I was not grip limited (for now), therefore the available grip (tractive force) is always wheeltorque/rolling_rad.