How to calculate the inertia of a gearbox?

[Sri_Vars]

Do I have to sum the inertia of all the gears and shafts? If so, how to do that?

 

[Baluncore]

Welcome to PF.
That depends on what type of gearbox it is.
The traditional automotive gearbox uses an input shaft that drives the lay (or counter) shaft. The output shaft is, in effect, a single unit. The ratio of the input shaft to the output shaft is gear dependent.

The moment of inertia of three subsystems must be measured or calculated.
1. The engine and flywheel should include the clutch pressure plate.
2. The gearbox input shaft and lay shaft should be grouped with the clutch friction disc.
3. The gearbox output shaft should be grouped with the drive shaft, differential, drive axles and all wheels.

 

[Sri_Vars]

Thank you Mr. Baluncore for your answer.
We are currently using the traditional gearbox only. But I am not sure how to calculate the inertia of the subsystems. It would be so helpful if you could brief it with formula.

 

[Baluncore]

You will need to dismantle the gearbox, then weigh and measure every component. Take a look at a gearbox parts diagram to see how many items are held on the main shaft.

The pitch diameter of a gear will be misleading due to the radius squared. You will have to approximate gear wheels as polygons, or find some other way.

A ball bearing race will probably have the inner race moving, the outer race static, and the balls spinning, somewhere in between.

https://en.wikipedia.org/wiki/Second_moment_of_area#Examples
https://en.wikipedia.org/wiki/List_of_moments_of_inertia

 

[Sri_Vars]

Thank you so much for your help!

 

[jack action]

Do I have to sum the inertia of all the gears and shafts? If so, how to do that?

No, because all gears and shafts don't necessarily have the same angular velocity. You can do a free body diagram for each rotating component which will give you a set of equations to solve (1 equation for 1 unknown).

This post shows how to do it with the equivalent kinetic energy method.

 

[jrmichler]

Do I have to sum the inertia of all the gears and shafts?

Yes. And you have to calculate the inertia of each component reflected to a common point in order to include the effects of gear ratios. Having done this calculation more than a few times, here's the procedure that I use. It's the same as the procedure in the post linked in Post #6 by @jack action above, just stated differently.

1) Calculate the inertia of each rotating assembly.
2) Calculate all gear ratios.
3) Define the starting point. It's normally the input shaft.
4) The input shaft is geared to the second shaft with a gear ratio. The inertia of the second shaft reflected to the input shaft is the inertia of the second shaft multiplied by the square of the gear ratio between the two shafts.
5) Proceed to the third shaft. The inertia of the third shaft reflected to the input shaft is the inertia of the third shaft multiplied by the square of the gear ratio between the input and third shafts.
6) The inertia of each shaft is the total inertia of the rotating components attached to that shaft.
7) When you have the inertia of all rotating components reflected to the input, then sum them. That's the inertia of the gearbox at the input shaft.

If you want the inertia at the output shaft, the procedure is similar, except that you start at the output shaft and work back to the input shaft.

 

[kreuchbag]

Yes. And you have to calculate the inertia of each component reflected to a common point in order to include the effects of gear ratios. Having done this calculation more than a few times, here's the procedure that I use. It's the same as the procedure in the post linked in Post #6 by @jack action above, just stated differently.

1) Calculate the inertia of each rotating assembly.
2) Calculate all gear ratios.
3) Define the starting point. It's normally the input shaft.
4) The input shaft is geared to the second shaft with a gear ratio. The inertia of the second shaft reflected to the input shaft is the inertia of the second shaft multiplied by the square of the gear ratio between the two shafts.
5) Proceed to the third shaft. The inertia of the third shaft reflected to the input shaft is the inertia of the third shaft multiplied by the square of the gear ratio between the input and third shafts.
6) The inertia of each shaft is the total inertia of the rotating components attached to that shaft.
7) When you have the inertia of all rotating components reflected to the input, then sum them. That's the inertia of the gearbox at the input shaft.

If you want the inertia at the output shaft, the procedure is similar, except that you start at the output shaft and work back to the input shaft.

Hello,

I'm questioning how to calculate the gear ratio for this type of problem. (for both the reflected inertia at the input and reflected inertia at the output examples) what’s the rationale for deciding which tooth count goes in the numerator and which in the denominator? How does the convention driven/driving relate to this situation? I appreciate any feedback you’ve got.

Thank you!

 

[jrmichler]

I am turning Shaft A. Shaft A is the input shaft. It is connected to Shaft B through gears. Inertia B is connected to Shaft B. If B is turning slower than A (speed reduction), then the inertia of B reflected to A is lower than the inertia of B by itself. If B is turning faster than A (speed increase), then the inertia of B reflected to A is larger than the inertia of B by itself. That's the basic principle, arrange the equations accordingly.

 

[Ranger Mike]

I assume you are trying to find the amount of energy it takes in a transmission to rotate the gears, shafts and bearings to quantify total energy required? The tricky part is to determine this requirement when all these parts are assembled correctly. One thing to add - Gear Oil.
Let me digress-
Total drive line parasitic drag is 15-20 percent but why?

1000 hp nets about 850 hp on the chassis dyno numbers.
Why is it that a lower hp car will still lose the same percentage? It does not require more hp to rotate axles and such on a higher hp car compared to the lower hp car.
There are variances such as weight, 2 wheel vs 4 wheel design, etc. Just for sake of the conversation lets say everything is equal on a 500 hp car and on a 1000 hp car. Why does one lose 150 compared to 75 on the other?


Parasitic losses come mainly from energy that's changed from torsional energy to heat energy via friction. The more power you put through a drive train, the more heat you lose inside. I can't tell you if it's a linear relationship, or exponential, but I suspect there's a relatively simple correlation between the two, which is why the results are so consistent across power levels, transmissions, fluids, etc.

More powerful cars require heavier duty drive lines so the torque of the engine doesn't break things. Heavier gears, axles, driveshafts, u-joints ,bearings, etc, by their very nature require more power to turn so this 15-20% loss stays consistent. Also, the faster a bearing turns, the more friction is created and more power is removed. We will not discuss gear ratios, differential gearing and the like.

If we have a passenger car with a stock 300 hp engine (at 6000 rpm on engine dyno) and the drive line designed for that car ( similar to a Mustang GT), put the car is put on a chassis dyno it will probably put down about 255 hp due to 15% drive train loss.

Now if you replace that engine with a 600 hp engine (at 6000 rpm on engine dyno) and keep the same clutch, transmission and rear end, , I would think that if it was put on the chassis dyno (assuming nothing blows up), the loss would be less than 15% (maybe 13%) because it is turning lighter duty drive line components. The downside of course is that it won't be a reliable car if driven hard.

One area no one looks at - You will lose 1 to 3 HP thru gear oil parasitic drag. The parasitic drag loss due to gear oil, typically referred to as "windage loss," can account for a small percentage of horsepower loss within a transmission, usually estimated between 1-3 horsepower depending on the gear design, oil viscosity, and operating speed; however, in high-performance applications, this loss can become more significant, especially at high RPMs.

It isa data driven fact special race oil, like Red line or Royal Purple engine oil ads 3 to 5% horsepower. Same with gear oil. I ran it in my Formula car and improved track time just by changing gear oil.



https://www.sciencedirect.com/science/article/abs/pii/S0301679X23003596

 

[Tom.G]

Don't forget the the energy needed to constantly deform the tires!

Cheers,
Tom