# Inertia of the drivetrain components of a 4x4 vehicle

• Chimaera987
In summary, the inertia of the car's drivetrain is calculated as: clutch_Out_Inertia = clutch_In_Inertia * clucthLocking +clutch_Inertia; where clucthLocking is a 0-1 value where 0 is clutch is disengaged and 1 is locked? Or should I transfer all of the engine_Inertia once the clutch is engaged??
Chimaera987
TL;DR Summary
Hi, I’m making a vehicle simulator and I'm unsure how to calculate the inertia of the drivetrain.
Hi, I’m making a vehicle simulator and I’m not entirely sure if my inertia calculations are right or are completely wrong. What I currently have now is something like this:
1. Engine:
• engine_Out_Inertia = engine_Inertia; this should contain the inertia of the engine and all of it’s auxiliaries, the flywheel and the clutch parts connected to it
2. Clutch:
• clutch_In_Inertia = engine_Out_Inertia
when clutch disengaged:
• clutch_Out_Inertia = clutch_Inertia; the clutch plate rotates with the gearbox, I guess it also could be included in transmission_Inertia, then this would be 0
when clutch locked:
• clutch_Out_Inertia = clutch_In_Inertia + clutch_Inertia
when clutch is slipping:
• clutch_Out_Inertia = clutch_In_Inertia * clucthLocking +clutch_Inertia; where clucthLocking is a 0-1 value where 0 is clutch is disengaged and 1 is locked? Or should I transfer all of the engine_Inertia once the clutch is engaged??
3. Transmission:
• tranmsmission_In_Inertia = clucth_Out_Inertia
• transmission_Out_Inertia = (tranmsmission_In_Inertia + transmission_Inertia) * gearRatio^2
4. Transfercase:
• transfercase_In_Inertia = transmission_Out_Inertia
if 2WD, assuming the rear wheels are driven:
• transfercase_Out_Inertia_FrontAxle = transfercaseFront_Inertia ; front axle is disconnected, so only the inertia of the front driveshaft has to be taken into account?
• transfercase_Out_Inertia_ RearAxle = (transfercase_In_Inertia +transfercaseRear_Inertia) * gearRatio^2
this is where it gets really confusing:
if 4WD:
• transfercase_Out_Inertia_FrontAxle = (transfercase_In_Inertia +transfercaseFront_Inertia) * gearRatio^2
• transfercase_Out_Inertia_ RearAxle = (transfercase_In_Inertia +transfercaseRear_Inertia) * gearRatio^2
Should I divide the transfercase input inertia into two, given that the transfercase splits the torque 50/50?In the case of rear axle inertia should I also add the inertia of the front axle components, given that the whole drivetrain is locked together? Something like this?
• transfercase_Out_Inertia_RearAxle = (transfercase_In_Inertia + transfercaseRear_Inertia + FrontRight_Wheel_Inertia + FrontLeft_Wheel_Inertia) * gearRatio^2
If this is the case, then full time 4WD cars that have a center differential should also behave like this? or only when the center differential is locked?
5. Differential:
Like in the case of the tranfercase, do I split the input inertia based on the left/right torque distribution in the differential? Do I include the left wheel side inertia in the right wheel side inertia calculation?
• Rear_Differential_In_Inertia = transfercase_Out_Inertia_RearAxle
• Rear_Differential_Out_Inertia_Left = (Rear_Differential_In_Inertia +differential_inertia) * finalDrive^2
• Rear_Differential_Out_Inertia_Right = (Rear_Differential_In_Inertia +differential_inertia) * finalDrive^2
6. Axle:
• Rear_Axle_Out_Inertia_LeftWheel = (Rear_Differential_Out_Inertia_Left + halfshaftLeft_Inertia)
• Rear_Axle_Out_Inertia_RightWheel = (Rear_Differential_Out_Inertia_Right + halfshaftRight_Inertia)
Sometimes you have unequal length halfshafts, part of the reason why cars torque steer?
7. Wheel:
• Right_Wheel_Inertia = Rear_Axle_Out_Inertia_RightWheel +wheelAssembly_Inertia + brakeDisc_Inertia

Chimaera987 said:
TL;DR Summary: Hi, I’m making a vehicle simulator and I'm unsure how to calculate the inertia of the drivetrain.

clutch_Out_Inertia = clutch_In_Inertia * clucthLocking +clutch_Inertia; where clucthLocking is a 0-1 value where 0 is clutch is disengaged and 1 is locked? Or should I transfer all of the engine_Inertia once the clutch is engaged??
If the clutch is slipping then the engine inertia does not contribute at all to the effective inertia of the car.

With the clutch slipping, the force applied to the transmission is determined by clutch friction. Neither the engine's effective moment of inertia nor its applied power are an immediate factor in the car's motion.

With the clutch locked, the force applied to the transmission is determined by applied engine power and by the moment of inertia and angular acceleration of the engine. The engine's moment of inertia and its throttle setting are both immediate factors in the car's motion.

If the clutch is slipping, you need to be tracking the engine's rotation rate separately from the car's speed.

jbriggs444 said:
If the clutch is slipping then the engine inertia does not contribute at all to the effective inertia of the car.
Thank you for explaining this. I had a suspicion that multiplying engine inertia by how much the clutch is slipping is not right.

jbriggs444 said:
If the clutch is slipping, you need to be tracking the engine's rotation rate separately from the car's speed.
I do, I can rev the engine freely when the clutch is disengaged and then the output torque of the transmission is its internal friction torque only, which is based on its angular velocity.

I presume you can't shed some light on how inertia is calculated inside a differential? Thinking about it now, with a locked differential you would have a solid rod connecting the wheels, basically a single unit, like in a slot car pictured below. I guess in this case the effective left AND right wheel inertia would be:

effective left/right wheel inertia = input inertia * final drive ratio^2 + left wheel inertia + right wheel inertia + left halfshaft inertia + right halfshaft inertia

As far as I understand the transfer case should be the same, when engaging 4WD or locking the center differential, however what happens when the differential is not locked together? In an open differential for example one wheel can be stationary while the other one rotates freely. Do you need to include the inertia of the stationary wheel in this case?

Chimaera987 said:
In an open differential for example one wheel can be stationary while the other one rotates freely. Do you need to include the inertia of the stationary wheel in this case?
Are either of the tires slipping on the road? If neither is slipping then you still have non-slipping contact with the road and drive train inertia is passed through.

If one is slipping then the inertia manifesting through the non-slipping wheel will be roughly equal to the left wheel inertia plus the right wheel inertia.

Welcome @Chimaera987 !

I would eliminate all those complicated calculations.
The most significative inertia is the mass of the chassis+fluids+passengers, as well as the rotational inertia of the wheels and tires.

Parts that rotate but have small diameters (like gears and shafts) don't make a huge difference, since their moment of inertia is relatively small.

That inertia (or resistance) manifests itself only when those things are increasing or decreasing their velocities (linear and rotational).

Those values can be easily calculated based on documented performance of real vehicles.

Regarding "rotational resistance of the engine", you must consider how much its breathing is restricted by the accelerator.
How quickly an engine can be rev'ed up while in neutral, is a good indicator of the inertia of its internal parts and flywheel and clutch at wide-open-throttle.

jbriggs444
Doing a Free Body Diagram would be really helpful here.

jbriggs444
jbriggs444 said:
Are either of the tires slipping on the road? If neither is slipping then you still have non-slipping contact with the road and drive train inertia is passed through.

If one is slipping then the inertia manifesting through the non-slipping wheel will be roughly equal to the left wheel inertia plus the right wheel inertia.
In an open diff the whole thing can be “locked” together like here:

and if it’s “locked”, again acts like a single unit, meaning you need both left and right wheel inertias to calculate one side.

but at the same time one wheel could be essentially still, while the other spinning as fast as the engine lets it like here:

This is what confuses me, that even though in the second example the two axles are connected by the spider gears, it’s acting like it drives one wheel, so to include the stationary sides inertia doesn’t really makes sense to me here.

Then what if you take a limited slip diff where the differential limits the speed difference, essentially slightly locking it. Then what? You add a slight part of the other wheels inertia to the equation? That sounds even weirder...

Chimaera987 said:
Then what if you take a limited slip diff where the differential limits the speed difference, essentially slightly locking it. Then what? You add a slight part of the other wheels inertia to the equation? That sounds even weirder...
It means that modelling the car with a single parameter named "inertia" is inadequate. You will need a richer model.

Lnewqban said:
I would eliminate all those complicated calculations.
Well, I want to use this in games and given that there will be an upgrade system, where you can change your prop shaft for example from steel to carbon or changing your brake discs to lighter or heavier ones it makes sense to have at least some of these. It doesn't really hurt my fingers to write brakeDiscInertia * 0.9f or the put some values in an Excel sheet and get a rough estimate.

jbriggs444 said:
It means that modelling the car with a single parameter named "inertia" is inadequate. You will need a richer model.
What do you mean by richer model. Can you point me in a general direction at least were to look?
Also I don't quite understand what do you mean by "a single parameter named "inertia"". When you sum up all the drivetrain inertias at the wheel it is a single parameter.

Chimaera987 said:
What do you mean by richer model. Can you point me in a general direction at least were to look?
Also I don't quite understand what do you mean by "a single parameter named "inertia"". When you sum up all the drivetrain inertias at the wheel it is a single parameter.
The behavior of a real car is not summarized so simply.

Chimaera987
Chimaera987 said:
Well, I want to use this in games and given that there will be an upgrade system, where you can change your prop shaft for example from steel to carbon or changing your brake discs to lighter or heavier ones it makes sense to have at least some of these. It doesn't really hurt my fingers to write brakeDiscInertia * 0.9f or the put some values in an Excel sheet and get a rough estimate.
In that case, as the geometry (radius and lengths) of parts will not change (at least appreciably), any increased performance due to reduced moment of inertia and linear inertia (lighter materials) will be a simple linear ratio of carbon fiber/steel masses.

I would use available data for equivalent vehicles with steel parts as base of those ratios.

Chimaera987
jbriggs444 said:
The behavior of a real car is not summarized so simply.
Perhaps, but I don't think I need super intricate formulas to know if I should treat the left and right wheels as one piece, separate or anything in between when it comes to inertia in a differential. Probably I will try some testing and see what feels better, compare it real footage and with other sims. Thank you for your help.

jack action said:
I have it bookmarked already. Great site.

Lnewqban said:
In that case, as the geometry (radius and lengths) of parts will not change (at least appreciably), any increased performance due to reduced moment of inertia and linear inertia (lighter materials) will be a simple linear ratio of carbon fiber/steel masses.

I would use available data for equivalent vehicles with steel parts as base of those ratios.
Yeah, I already have some data for all of this that I can extrapolate from. Not that this is super important, but if I already have most of the building blocks to implement it, why not. Little details add up.

Lnewqban

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