Giving me good vibrations

by Simon Hargreaves

Bikes make lots of vibrations, some good and some bad. Here's why...

Anyone with fingers knows bikes vibrate, sometimes strongly enough to make picking your nose impossible. But how we perceive vibration depends on its type, pattern, frequency and cause.

Bikes are subject to different types of vibration. A bumpy road creates random forced vibration at medium to high frequency (around 5-35 Hz depending on vehicle speed). It's perceived as unpleasant, as is head buffeting caused by wind turbulence. Engineers and aerodynamicists try to minimise these bad vibrations.

But we usually talk about engine vibration. There are many sources: apart from the reciprocating mass of the piston and the crank's rotational mass, there's also combustion vibration, various rocking couples, vibration generated by cams and other mechanical friction, by changing gas pressure, and the component drag of oil.

But the vibes that matter come from whirling lumps of metal. There are good and bad vibrations, and their magnitude and type is related to engine design and configuration.

Broadly, there are two 'types' of engine vibration caused by reciprocating masses: primary and secondary. Primary vibration occurs at engine speed, secondary at twice engine speed. Primary vibration is caused by a piston assembly accelerating up and down on a crank, creating an inertia force equal to its mass times its acceleration. With no opposing force, it's unbalanced. To balance it and reduce vibration, singles (or 360° parallel twins, with pistons moving as one) use a balance shaft or shafts, or a vestigial piston, or a weighted, crank-mounted disc. Each counter-balances by providing opposition to the inertial force.

In many multicylinder engines, primary balance is achieved by using pistons reciprocating alternately. So 90° V-twins, 180° parallel twins, flat twins, 120° inline triples, 90° V4s and inline fours and sixes have primary balance. But anyone who's ridden any of them will know none are vibe-free.

This is either or partly because of uneven firing intervals (e.g. a 90° V-twin) or because balancing a reciprocating piston with an alternate piston along a crank generates a rocking couple (flat and parallel twins, triples, V4s). It's these two types of low-frequency vibration we generally like; they're character (a BMW boxer twin, with its rocking from side to side, the off-beat thump of a Ducati). Although they too can be countered; Triumph and Yamaha triples are balanced but need balancer shafts to counter an extreme rocking couple that would otherwise limit engine performance.

But secondary vibration, caused by the uneven acceleration of pistons through their stroke, is unpleasant (it's worse in inline fours because pistons move in pairs). It's usually a high frequency vibration through the bars at cruising speed – if an inline four is turning at 6000rpm in top, the secondary vibes will be in the 200Hz region, which is exactly the right frequency for 'white finger' – the tingling numbness in fingers after a long ride at a steady speed. Within packaging constraints there's little can be done (multiple balance shafts, fitting infinite-length con-rods, or using a Wankel rotary), which is why manufacturers try to damp vibes on inline four tourers with heavy bar-end weights and massive footrest assemblies.

It doesn't always work. As anyone with fingers knows.

pittsy
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It is said that the speed of a piston will be zero twice per Rev. Seems logical. One moment it is whizzing upwards, next moment whizzing downwards. So at some point it must have been travelling at zero speed. That point is tdc and bdc. Again, all seems logical.

But hang on. Tdc and bdc are defined by a line. At zero degrees. A line that has no width. So if we keep zooming in, so to speak, to check the piston speed as the crank moves ever closer to zero degrees, it never actually seems to exist at zero degrees for any time. So that's a bit strange. If there's no time, then how can it happen? How can the piston speed ever be zero? And the other strange thing is, it seems that this is independent of rpm. There's no doubt that the piston speed increases with rpm, but the infinitely thin line defining tdc and therefore zero piston speed is just as infinitely thin at high rpm as it is at low rpm.

So, just how long is a piston at tdc? How long is it travelling at zero speed?

Conundrum.

malauder
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I think all the time the piston is experiencing acceleration, it is never at one speed for any amount of time. In fact, as it's accelerating hardest at TDC & BDC, then it is at zero speed for less time than any other speed!

Half way between TDC & BDC the acceleration must be zero, as it goes between positive and negative, so the piston is at constant speed. But for how long? Well, for as long as the acceleration is zero - which is no time at all!

Yep - a conundrum all right!

pittsy
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I guess that is the very nature of acceleratin....to not be at any speed for any amount of time. Doesn't sound quite as crazy thought of like that. But when one stops (pun) to ponder on the implications it all starts to get a bit weird. Well it does in my head anyway. Lol.

unconventional rebel
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I find it odd that a tyre must be stationary at the edge even when moving at speed, otherwise it would be wheel-spinning. Meanwhile then hub is never still (while the bike is being ridden obviously..) but is attached to the stationary bit of the tyre.

Is the same thing going on?

malauder
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Entirely true UR. As you say - the bit of the tyre on the road has a zero horizontal speed, whereas by the time it is at the top of the wheel it is moving twice the speed of the motorcycle (or any other vehicle for that matter).
In the case of a train - there is always some part of the wheel below the track (on the inside) which is going backwards!

unconventional rebel
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OK, so why is it going backwards?

malauder
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All the parts of the wheel above the rail are going forwards, the wheel in contact with the rail is stationary (assuming it's not skidding) and so the wheel below the rail contact point has to be going backwards because of the rotation. Easier to explain with a diagram!
Here's a technical link - with a diagram near the bottom
http://math.stackexchange.com/questions/721935/vector-valued-function-wh...

unconventional rebel
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Took a bit of brain stretching - but got there in the end, thanks. Just need to wait for it to come up in a quiz...

pittsy
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Is it helpful to think of (or better still, watch) a caterpillar tracked vehicle? Pick on a section of the track and watch it as the vehicle is moving along. When the section being watched contacts the ground it appears to stay still, which it does with respect to the ground. But once it reaches the end and is picked up by the the end wheel and over the top you can see it moving forwards twice as fast as the vehicle. It has to or it wouldn't ever get to the front again. Weird!

Looked at another way, can we think of a single wheel (motorcycle rear wheel for example) as being like a caterpillar track of infinitely short (zero) centre distance?

Dunno, but it helps me to think that way!

unconventional rebel
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It helped with the stationary bit, but not the bit going backwards...

pittsy
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unconventional rebel wrote:
It helped with the stationary bit, but not the bit going backwards...

Yes, sorry, the caterpillar track was my way of reconciling what is happening to a wheel, but it maybe doesn't illustrate the going backwards bit.

I'm definitely no expert so this is me just thinking out loud. Does it help to think of the rear wheel of a motorcycle as an infinite number of levers arranged in succession in a circle? So if you were to freeze frame you'd see one of these levers going from the ground vertically up to the axle centre. This lever is what moves the bike forwards. There just happens to be an infinite succession of them, so the axle remains the same distance off the ground and the force always stays parallel to the ground (forgetting tyre distortions etc).

If that is right then if we look at the train wheel in the same way, then because the wheel has a lip or flange on it, which means the outer diameter is below "ground" if you imagine the lever extending to the outer diameter then if there was just a lever and not a wheel the end of the lever will move backwards relative to the ground.

Probably a terrible attempt at an explanation!