Inertial torque


By Kevin Ash

Pictures: Yamaha Press, Cyborg, Caterham

Yamaha’s 2009 R1’s innovative cross-plane crankshaft has given us a new term: inertial torque (and plenty of crackpot theories to explain it). I’ve explained here (cross-plane crank) where this comes from and why the R1 doesn’t have it, but now I’ve taken it a step further and calculated just how much inertial torque there is in a conventional four. The answer is quite a shock: it’s massive! If you were expecting it to be around 3 or 4lb.ft (0.5kgm, 5.5Nm), like some sort of barely detectable background noise, try adding a couple of noughts and you’ll be closer...

inertial_torque_3Yamaha cross-plane crankshaft (click on image for full size)A brief résumé of the inertial torque phenomenon will help here: in a conventional inline four, all four pistons are stationary at the same instant, two at the top of their bores, two at the bottom, before they all accelerate to maximum speed a quarter of a crank turn later, gaining large amounts of kinetic energy which comes from the spinning crankshaft and slows it down. Another quarter turn and all four pistons have stopped again, returning that energy to the crank which in turn speeds up again. This rapid ebb and flow of energy generates positive and negative torque pulses entirely due to the pistons’ inertia, and how it’s changing – hence the name. You can see it happening here: Yamaha YZF-R1 cross-plane crank explanation and read about it in detail here: Ash on Bikes: the cross-plane crankshaft

In the cross-plane crank the crankpins are staggered at 90 degree intervals, meaning as any one piston is slowing down another is speeding up and the overall energy flow is almost zero, as the inertia one piston is gaining is cancelled out by the inertia another is losing. This results in steady, smooth crankshaft rotation which is better for throttle control and tyre grip.

While this is news to much of the motorcycle world (and some of the bike press are still struggling rather badly with the concept), the inertial torque effect has been known since early last century. High performance V-eights over the years as used by Lotus, Ferrari, TVR and in the Cosworth DFV F1 engine all use flat-plane cranks partly to eliminate inertial torque, although their light weight is another advantage compared with a V-eight’s cross-plane crank. And yes, this is the right way around, a cross-plane four has zero inertial torque but you need a flat-plane crank in a 90-degree V-eight to achieve the same.

crankshaftSuzuki flat-plane crankshaft (click on image for full size)
I’ve used a GSX-R1000 engine as an example, revving at 12,000rpm, and to make the calculations much easier, I’ve assumed the piston movement is symmetrical so it’s at maximum speed half way down the bore. In fact this is only true with infinitely long conrods but real ones make the maths too complex, and the result isn’t very different anyway.

I’m also assuming a piston mass of 0.2kg (I’m going metric for this as it makes life much easier...) although as the conrod and gudgeon pin move with the piston their weight should be included too, so 0.2kg is rather low and the real inertial torque figure will be higher.

The aim is to calculate the acceleration of the piston as it moves from stationary at TDC (top dead centre) to its maximum speed, and from that we can calculate the force exerted on it by the crankshaft through the conrod. Knowing this force and the distance of the crankpin from the crankshaft’s centre (half of the engine’s stroke) allows us to calculate the torque.

* Step 1 – Find the piston’s maximum speed:

In our virtual engine this occurs halfway down the bore where piston speed is the same as the crankpin speed. At 12,000rpm the crank rotates 200 times per second. The diameter of the circle the crankpin (big end) travels is Ï€ x the stroke, 57.3mm on the GSX-R. That’s a peak speed (v) of Ï€ x 0.0573 x 200 = 36m/s (metres per second) That’s 0-80mph in 28.65mm!

* Step 2 – Find how much time (t) the piston takes to travel from TDC to max speed:

At 200 revs per second, one rev takes 1/200 seconds = 0.005 seconds. Max speed is reached in 1/4 turn of the crank, so t = 0.005/4 = 0.00125 seconds.

* Step 3 – Calculate the piston’s acceleration (a), using a = v/t:

a = 36/0.00125

= 28,800 m/s/s (That’s huge. Earth’s gravity, 1g, is 9.8 m/s/s, so a piston experiences up to 3000g... this is why they need to be strong!)

* Step 4 – Calculate the force of the conrod pulling and pushing on the piston to accelerate it, using force = mass of the piston (m) x acceleration, F = ma:

F = 0.2 x 28,800

= 5760N (N is Newtons, and 9.8 Newtons is the same as 1kg, so a piston experiences more than five and half tonnes force as it accelerates at this rpm).

* Step 5 – The torque (T) generated by this force on the crankshaft, using torque = force x distance from rotation centre (half the stroke, 0.02865m):

T = 5760 x 0.02865

= 165Nm.

So us British and Americans can understand, that’s a massive 121lb.ft of torque (16.8kgm), more than the engine makes in the usual way! And this is only from one piston, and there are four each doing it... In other words, spinning at 12,000rpm, the inertial torque in a GSX-R1000 engine due to the pistons peaks at 484lb.ft. As Prodrive designer Damian Harty said when I ran the numbers past him: “Engines are surprisingly violent things!” Even he raised an eyebrow at the scale of inertial torque though, so he calculated it for himself using an entirely different method dependent on what’s called the Simple Harmonic Motion of the piston/crankshaft system: the numbers were much the same.

inertial_torque_4Springs set into clutch basket of this KTM 990 damp out the effects of the inertial torque pulses (click on image for full size)It’s important to emphasis that this torque flips from positive to negative, speeding up and slowing down the crankshaft, twice every crank revolution, so the net torque output is zero. But it does have to be taken into account by engine designers: massive torque pulses like this fed directly into the transmission could easily cause damage or at least shorten its life. This is why clutches have springs built into their backplates to absorb this rotational ‘vibration’ and protect the gearbox. inertial_torque_1Familiar car dry clutch plate with inertial torque damper springs (click on image for full size)These springs are familiar set into the centre of many cars’ dry clutch plates – inertial torque affects all inline fours with conventional flat plane cranks, car, bike or anything else. This also underlines how inertial torque is not some new discovery but a very long established factor in engine and transmission design.

We don’t feel it directly because of the direction change every quarter turn of the crank. But it’s there in a normal four and because the driveline has to include some kind of ‘springiness’ to deal with it, the feel of the throttle response is not as direct and tactile as it could be.

Joined: 03/01/2010

How does the cross plane crank result in zero inertial torque?

The inertial torque experienced by the crank for one piston occurs twice in 180 degrees of crank rotation at both TDC and BDC.

At TDC the inertia of the piston from its' upward movement in the cylinder is trying to pull the crank pin upwards, out of the cylinder head, and at BDC the ineria of the piston from its' downward movement in the cylinder is trying push the crank pin downwards, through the sump.

So on a traditional I4 layout with a flat plane crank all 4 pistons exert an inertial force upon the crankshaft simultaneously, all of equaly magnitude (assuming variables such piston weight and crank throw are all equal) in 2 directions, opposed by 180 degrees.

On the cross plane crankshaft the crank pin layout is, when the crank is veiwed end on, No 1 piston at 0 degress, No 2 piston at 90 degrees, No3 at 270 degrees and N 4 at 180 degrees.

So when No 1 piston is at TDC... No 4 piston will be at BDC...No 2 will have accelerated to (or tending toward) maximum velocity after TDC on the way down the cylinder and No 3 will be decelerating (or tending away)from maximum velocity before TDC.

So you've always got two pistons that are stationary simultaneously, and exerting inertial torque on the crankshaft, because of the massive percentage change in their velocity from the instant before they were stationary to when the stop moving.

Which ISN'T cancelled out by the inertia of the other two pistons because the other two pistons have a statiscally insignificant percentage change in their velocity as they tend toward or away from the their maximum velocity ~ halfway up/down the cylinder, so exerting little or no inertia.

At best you've got inertial torque at the crankshaft which is aproximately half that of the tradtional flat plane crank, but it's not zero.

This is further negated by the fact that maximum piston velocity does not occur exactly halfway up/down the bore. So one piston will be closer to its' maximum velocity than the other, whether rising or falling.

kevash's picture
Joined: 05/10/2008

Have you also looked at the cross-plane crank feature on here? Click on this link which explains it in detail, the feature on this page is about the magnitude of inertial torque, which is surprisingly large.

At TDC a piston is stationary and about to accelerate down the bore, and in a free-spinning system leaving aside the combustion-generated torque, the crankshaft is pulling it down through the conrod. This accelerates the piston and slows down the crank.

In an inline four, all four pistons are stationary at the same moment, two at TDC, two at BDC, so the cranks shaft at that moment is accelerating all four and itself slowing down. This continues for 90 degrees of crank rotation, at which point all four pistons are now slowing down again, and in doing so are pushing against the crank, returning their kinetic energy to it and speeding it up.

So in half a crank rotation in an inline four, where two pistons are moving from TDC to BDC and the other two are going from BDC to TDC, the crank slows then speeds up again.

In the cross-plane engine, when one piston is at TDC, another is at BDC, and these two are just starting to accelerate, and are therefore gaining kinetic energy which comes from the crank. At the same time, the other two pistons are at 90 degree intervals have just reached maximum speed and are now just starting to decelerate, returning their kinetic to the crank. This counters the energy being taken from it by the first two pistons, and the crank speed remains constant.

Okay, the pistons' movement isn't quite symmetrical (you could only achieve this with infinite conrod length) so there are still small fluctuations in the energy's ebb and flow and the crank speed does change a little, but it's far less than in a flat-plane crank where all the pistons stop, accelerate, decelerate and stop again simultaneously. It's a lot less than the half which you mention and more like 100 times less.

I think you've been mixing up crank balance with crank inertia. You're right that two pistons in a conventional four are pulling up while the other two are pulling down, but this is a balance issue, not a kinetic energy and inertia issue. At this point there is no energy transfer between pistons and crank because they're stationary, even though they're pulling with a large force. That force anyway is pulling through the crank's axis of rotation so it can have no effect on the crank's rotation speed.

I hope that helps!

kaiser soze
Joined: 01/01/2012

There are multiple problems with this calculation, starting with the first assumption that maximum piston speed is the same as the crankpin speed. And also the assumption that maximum piston speed occurs when the crankpin is 90 degrees TDC. And there are many more assumptions that are not correct. It just is not a useful calculation.

Rick Pitino
Joined: 21/02/2013

In some disciplines, the terms "moment" and "moment of force" are synonymous with "torque". Other disciplines reserve use of the term "torque" for instances when the angular velocity and/or the moment of inertia of an object are here for more information