I think that for any engine/motor (electric or combustion) you can flatten the torque curve by sacrificing peak power. This is essentially the trade-off between straight line performance (which depends mainly on peak power) and driveability (which depends largely on the 'flatness' of the torque curve).
Not necessarily T.W., see the dyno charts linked below.. & compare the two Kawasaki machines..
Indeed T-C, here is a dyno plot of a Hirth engine, another 2T 3cyl, & comparatively softly tuned, for flight..
Sorry JAW, I must take issue with that. Although a rising torque curve will result in higher power (same engine, same average torque) . . .
Your final paragraph gg, hits the nail on the head - naturally, sporty engines get sporty gear sets..gruntguru wrote: ↑Wed Jul 19, 2017 5:34 amSorry JAW, I must take issue with that. Although a rising torque curve will result in higher power (same engine, same average torque) . . .
For single gear acceleration, the flat torque curve will get the overtaking done sooner since acceleration at the beginning of the manouvre is more important than acceleration in the middle or at the end. (Again this assumes average torque is the same for both cases.)
The rising-torque engine will provide better acceleration through the gears provided the ratios are close enough to keep the revs in the region where it has more torque than the other engine.
manolis wrote: ↑Tue Jul 18, 2017 4:28 pmHello J.A.W.
“A rising torque curve enables an exponential increase in acceleration, & that's handy for overtaking..”
A rising torque curve enables a linearly proportional acceleration.
More correctly: the ratio of the force pushing forwards a car or motorcycle, to the torque provided by the engine, is constant and depends on the selected gear ratio in the gearbox.
In the RoadLoad DOS program at http://www.pattakon.com/pattakonEduc.htm the basis for the calculations is the torque curve and the overall gear ratios:
The torque curve at the bottom of the above image is the same with the "engine force" pushing forwards the motorcycle (top of the image).
The acceleration results by subtructing from the "engine force" the overall resistance force, and then by dividing with the total mass of the motorcycle.
Aren't we say the same thing with diferent words?
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