Inertia Losses in Piston Engines

Does anyone know how to calculate the inertia losses associated with reciprocating piston engines? I know it involves reciprocating mass, piston speed, and bore & stroke ratios so it’s pretty complex, and little has been written about it…

But as I understand it, as RPM and piston speed increases, inertia (not pumping) losses start to increase dramatically absorbing more and more of the engines power…Also, there are those who say a headless engine could be spun up to very high RPM with very little power input required…It would be “free running” and absorb very little power.

What does the Car Talk Brain Trust think?

Here’s a discussion on that little matter. Enjoy!

Mechanical energy is constantly being traded between the flywheel (cranks have flywheel value too) and a piston and a portion of the rod. Energy needed to force a stopped piston at TDC or BDC into motion will be supplied by the flywheel. The piston will be at maximum speed at approx. halfway though the stroke and then will slow down as it approaches BDC or TDC. The slowing piston then gives up the energy invested in it by the flywheel, back to the flywheel.

The rod acts as flywheel mechanical energy on the big end and as piston mechanical energy at the small end.

If you had the means to precisely measure flywheel speed, you would see that it is not constant though one rotation as it gives and receives mechanical energy from the piston and upper portion of the rod.

As engine speed increases, frictional and pumping losses increase.

As the web site and other posts have said, the inertia is being traded between pistons/rods. A different way of thinking about it, if there was a lot of energy being consumed, it has to show up somewhere (heat, for example). That only come from friction, pumping losses, that kind of thing.

Thanks WhaWho, It finally snapped into focus for me…The energy required to accelerate the piston (supplied by the crankshaft) is mostly returned to the crankshaft as the piston is forced to decelerate…First the connecting rod is being compressed, then it is “stretched” as the piston is decelerated…It’s amazing pistons last as long as they do considering the stresses they must absorb…

So now, to explain the strong resistance to high speed rotation that is obtained when downshifting on a steep grade becomes more difficult…The old bromide “braking on compression” falls apart since at high vacuum, there is little to compress…Although the exhaust overlap timing might allow considerable air into the engine even though the throttle is closed…

I have seen this in slow motion on an oil well pump jack. Oil wells can be up to 25,000 ft deep although the ones in Luling are mostly about 5000 ft deep. The pump rod that goes down the pipe to the pump can weigh over 10,000 pounds so there is a lot of reciprocating mass in that system.
No matter how carefully you adjust the counterbalance on the crank, there are two spots near the end of the strokes where the pump jack overdrives the electric motor and the motor acts as a brake to hold the machine back instead of driving it. This is obvious by watching the little disk that goes around on the electric meter. That little disk spins forward really fast for a while and then comes to a stop and goes backwards for a little bit and then stops and goes forwards real fast and then backs up a little again over and over and over.

Two stroke engines almost freewheel compared to four stroke engines when the throttle is closed. A lot of early motocross racers discovered that opening their engine’s compression release would give two stroke engines four stroke like engine braking and some enterprising manufacturers even came up with a throttle that opened the compression release when turned past idle.

Part of it is that you are now trying to accelerate all the moving mass of the engine (crankshaft, flywheel, pistons, …) from the speed it was at to a now much higher speed due to the gear change.

“The old bromide ‘braking on compression’ falls apart…”

It falls apart because it’s not really the right term for what’s going on. A better term is “pumping loss”. It takes work to either compress air to a pressure higher than atmospheric or to pull a vacuum below atmospheric.

Sounds like a “poor mans Jake Brake”