Drivetrain loss.. How much do you really lose?

[oldcorollas]

for every diff and every gearbox, there will be a 3D map of speed and load versus losses.
it would take a fair bit of effort to map this out for every component

[BrianRA23]

Since the percentage is just a "nice to know" value then any amount of effort to determine it is probably too much.

However if the rwkw can be increased (or better economy given) just by selecting the right components then some basic tests might be worth while.

BTW drivetrain loss must all be due to friction since when any mass is moving at a constant speed the only thing slowing it down is friction. Stronger items tend to have bigger bearings and increased contact area.

[sillycar chick]

The reason you "lose" power through a drive train is also partly because it is a reduction system. All those gears inside your gearbox are there to reduce the rpm that the motor puts out to a useable rpm. (ie it would not be very practical to have your back wheels spinning at 3200rpm, you just wouldnt get traction).

This reduction system ALONE will change the power figure that you get at the end of the line because it changes the rpm and the torque through the gears. Then to further reduce your power figure at the end of the line, you get friction and heat losses through bearings, lube and rubber wheels that others have mentioned earlier.

It really would be a pain in the ass to calculate out the power loss manually, there would be so many formulas. If you ignore the friction losses etc., you can figure out the loss through gear ratio calcs. Every gearbox will have a reduction rating eg. 50:1 overall reduction. Some workshop manuals may tell you this, or more likely the reduction between each gear. Just depends on the manufacturer I guess. And you may get torque figures instead of power figures.

It may be handy to know that:

TORQUE (in-lbs) = (HP x 63025) / RPM
(where 1in-lbs = 0.1129848 Nm)

Therefore, power, torque and rpm are inter-related, meaning all factors are going to change through a reduction system.

I forget exactly how to calculate out the reduction between gears, I will try find it in my notes. From memory, if you know the # teeth, gear diameter and the driver gear rpm between two gears you can work out the reduction.

As for applying this to work out the power loss between the flywheel and the back wheels, im not sure how accurate it would be. I have thought of doing it before just for fun, but havent had any reason to do it yet. These days, dynos can tell you approx what to expect at the wheels in terms of power anyway! Good Luck with your quest for knowledge

[BrianRA23]

Assuming no friction:

If you half thee speed of something you double to torque. If you run it through a gearset giving you a ratio of 1:50 your torque should be 50 x more. Set up a piddly little electric motor through a 50:1 reduction and you will not stop it without loosing some skin because of the enormous torque it has just gained. Power hasn't been lost (except for th burning skin)

In each case the power is the same as per your formula. Energy can't be lost in a enviroment without friction (noise making excluded) so you should be able to change the speed back to what it was before without power (energy since kW) being lost.

Unfortunately we have friction so power is lost.

A gearbox with straight cut gears will allow more power through it, but since they are so noisy we have angle cut groves in the gears to increase the contact area and reduce movement.

[YelloRolla]

I like newtons third law when this question is raised - for every action (force applied) there is an equal and opposite reaction.

That pretty much tells us that as you try to accelerate the gearsets harder, the harder they are going to be to turn. Whether that resistance is through heat, inertia or Mary Poppins.

I think that 20% is a more likely drivetrain loss for a rear wheel drive vehicle (higher for auto). I think that where this percentage gets lost or skewed is when people are trying to work out the engine power from the rr wheel (or front) power figure.

IE: 20% drivetrain loss: 100 flywheel kilowatts x 0.8 = 80rwkw (or a loss of 20kw). 20kw is actually 25% of 80 - so to calculate back to flywheel kilowatts 25% becomes the figure. 25% is not the drivetrain loss though.

Whenever reversing an equation like this I like to just divide by the original multiplication IE: 80 (rwkw) divided by 0.8 = 100. This maintains the original value of 20% rather than trying to use another number (like 25%).

Now I believe that "wild drivetrain loss equations" come from someone then saying 25% is the figure. (100kw with 25% loss = 75 wheel kw) to which the reverse is 33.3%. Now you have people saying the loss is 35% - which is of course bollocks.....