Engine question (for the engineers!)

The pull you get after 6,000 revs on a 600 inline 4 just doesn't happen on a 600 v-twin, at any rev range.

Or at least that it's how it feels having owned an SV650 and a Thundercat. Could never go back to a v-twin now. Would love to try a triple though next ;D

Really, you need an engine in each configuration, to go with what mood you're in when you open the garage door :-)

I like instantly thumping past traffic and launching out of corners like a catapult on the Falco, but I do occasionally like to scream the tits off Debz' Fazer.

Given the choice the twin always wins for my riding, I really do think they make the best road bikes. An in line four is too smooth, too much like a car engine for me, I want something that feels 'alive' in a bike.

Again with a triple, it feels too smooth and sanitised for me. Bikes for me should feel raw and threatening, not like they're powered by an electric whisk.

The Cross Plane Crank R1 is fun for a four cylinder though, it feels more like a twin, but still has the mental top end rush of a big IL4. Good luck at using all the performance on the road though, unless you go every where in 3rd gear. If you're used to four cylinder litre bikes you really need to have a go on one, they have a very different character and are an incredible bit of kit.

I find twins a tad lazy. On an inline 4 you really have to work the gearbox to keep the revs up. The difference between the two is decreasing though as manufacturers get more top end out of twins and more torque lower in the rev range from 4s.

missing the point slightly here SF......
the inline four may make less torque at a given engine speed, BUT will have the gearing to result in the same torque at the rear wheel for the same road speeds as a twin....BUT the 4 cyl will ultimately rev higher and produce better top eng power (and possibly a better top speed and performance in the top of the rev range).

for the same rear wheel torque the two bikes will be doing approx the same road speed but the 4 will be revving higher, but both bikes will have the same mean piston speed.....

its the resultant torque at the rear wheel you want to assess, not the torque at the crank :-)

Ok, now you're starting to lose me! lol

Thought I was doing ok keeping up thus far! ;D

Quote:

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Originally Posted by Squashed_Fly

So if ultimately, they make about the same power at the top of the rev range, yet V-twins have more usable power/torque lower down the rev range, and what Jaydee says is true about only getting near the top of the rev range 3% of the time on the road, then why don't more people have V-Twin engines where they can use more of the power more of the time?

There must be a good argument somewhere for having to thrash the nuts off something to get it to work as well as something that doesn't need the nuts thrashing off it...?

Or have I missed something in the process? I am trying my best to follow, understand and keep up!

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Errrr...... No!

They make different power and torque curves.

As JD says, the torque can be sorted out by the gearing. What drives you along is torque, not power. The power is the rate of work or energy dissapation so the rate of changes. The top acceleration rate is (all things being equal) determined by the power as is the top speed.

So to demonstrate:

Using Kinetic energy Energy (kE) = 1/2 Mass (M) * velocity (v) squared we can see that as the speed (scalar velocity) (v) doubles, the energy requred to maintain the momentum increases by a factor of four.

From this we can differentiate to say: The rate of gain of kE:

Power (dE/dt) is proportional to, 2 * Acceleration (dv/dt) = Force (N) / Mass(M)

The Force element can be then used to relate to Torque through Torque = Force * Radius, so linking power and torque together on the road.

So power is used to gain momentum. The more power (and torque by inferance) or the lower the Mass the faster you will accelerate, but not neccisaritly go faster because of resistance

Taking wind resistance, the major factor. Assumin the bike is going at a constant speed:

Force of wind (Fd) = Drag factor (c) * velocity (v) squared

and applying it to the equation for potential energy:

Potential energy (pE) = the change in work (delta U)

pE = delta u = integlal (Fd).dx =a + 1/3*c*v cubed.

So you can see as the speed rises the losses due to wind resistance rise dramatically.

So the Power absorbed by the wind is proportional to c*v squared.

The rolling resistance can be assumed to be fairly constatnt throughout.

So, as can be seen. Power is lost though many sources. The engines headline figures are only opart of the story. As JD says. All had to be based at the power and torques applied to the road, not the engine figures as transmission losses can be hideous.

:-?

A nice bit of Newtonian Mechanics and calculus for you in the morning!!!! ;)

Does ya good!!!! ;D

I need a spliff after that!

For one horrible second I was transported back in time to a physics lesson!

Blimey, if that's laymans terms, I'd hate to see you explain something to someone who understands it all!!!!

I was ok at "bigger bangs but less of them" but now totally lost! lol

Gunterpot did spend some time with me explaining the relationship between power and torque, drawing graphs to show why I had a much higher revving powerband (the point where power crosses torque?). Which I kind of understood when I saw it on paper... But now totally lost again! ;D ;D ;D