Inertial torque magnitude

By Kevin Ash - 15/09/2011

Yamaha‘s 2009 R1‘s innovative cross-plane crankshaft gave us a new term: inertial torque (and plenty of crackpot theories to explain it). Now the R1 has traction control it's a reminder that this bike is unique.

A full assessment of how the cross-plane crank works can be found here: Yamaha Cross-plane crank analysis so here I‘ve gone 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, 5Nm), try adding a couple of noughts and you‘ll be closer...

A brief resumé first of the inertial torque phenomenon: 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 - hence the name.

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 (the inertia one piston is gaining is cancelled out by the inertia another is losing), resulting 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 badly with the concept), the inertial torque effect has been known since Victorian times. 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. But even twin cylinder steam traction and stationary engines in the 19th century, then steam railway engines in the 20th, were designed so the inertial torque effects cancelled each other out, by setting crankpins at 90 degrees to each other.

To work out how large this is in modern motorcycles, I‘ll use a GSX-R1000 engine as an example, revving at 12,000rpm, and to make the calculations much easier I‘ll assume 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.

This calculation is only to get an idea of the scale. It can be done more accurately using the Simple Harmonic Motion technique, but that‘s more difficult to follow and understand what‘s going on, so I‘m using more basic Newtonian physics instead.

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 π x the stroke, 57.3mm on the GSX-R. That‘s a peak speed (v) of π x 0.0573 x 200 = 36m/s (metres per second) That‘s 0-80mph (130kph) 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)

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

= 5040N (N is Newtons, and 9.8 Newtons is the same as 1kg, so a piston experiences up to half a ton force as it accelerates).

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 = 5040 x 0.02865

= 144Nm.

That‘s a massive 106lb.ft (14.6kgm, 144Nm) of torque, more than the engine makes in the usual way. And this is only from one piston... In other words, spinning at 12,000rpm, the inertial torque in a GSX-R1000 engine peaks at 424lb.ft (58.5kgm, 575Nm), and that‘s a conservative figure. Calculate more accurately, reva bit harder and it could be 50 per cent higher again.

It‘s a huge force but we don‘t feel it directly because it changes direction every quarter turn of the crank, slowing it down then speeding it up as the torque flips from positive to negative. But it‘s there in a normal four and it interferes with the delivery of the engine‘s real torque.

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