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Im specifically referring to motorcycle engines, but for arguments sake a car engine wouldn't matter either

Say 100Nm was measured at the wheel using a rolling dyno, then based on the reduction ratios(say 5) from wheel to sprocket to gearbox to the clutch, would the torque produced at the crankshaft be less? 100/5 = 20Nm

Assuming no losses, the power produced at the engine would be the same at the wheel but since there are multiple gear ratios, the torque is increased whilst the angular speed has reduced. So if the crankshaft was turning at 10000 rpm... with an overall gear ratio of 5. The wheels are turning at 2000rpm?

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Generally, most road vehicles are geared so that the wheel speed is less than engine speed and with top gear usually being around about 1:1. Ratios which produce wheel speed greater than engine speed are sometimes called overdrives. On older cars with relatively few selectable gears these are sometimes fitted as a separate module to the gearbox and selected by a separate lever or switch not unlike the hi/lo ratio selector in some 4wd transmissions.

Conventionally when we talk about reduction ratios that refers to a reduction in speed not torque.

So in your example with reduction ration of 5:1 the torque at the wheels would be 500Nm not 20Nm.

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  • $\begingroup$ 1:1 is the typical top gear ratio through the transmission, but don't forget there is another ratio typically at least 2:1 at the differential. You would be hard pressed to find an example of a real world road vehicle where the overall gear ratio was less than 1:1 even in overdrive. $\endgroup$ – agentp Jul 23 '17 at 3:09
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assuming equal diameter (wheel and crankshaft) the torque is highest on the engine, with some losses it arrives less to the wheels. But assuming no losses , the torque is mainly affected by the diameter so u will have to convert the torque to a linear value unaffected by the radius of the shaft 1 rpm = 1*2*pi*shaft Radius*60 m/s so now this value is fully transmitted to the wheels assuming no losses then u reconvert it to RPM using the wheels diameter.

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  • $\begingroup$ Im not sure I follow...could you provide an example with numbers? What's confusing me is that you said assume the crankshaft and wheel are of equal diameter...but the diameter of a crankshaft is much smaller than the rear wheel. $\endgroup$ – james Jul 22 '17 at 10:20
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I would propose another approach, let's calculate the speed at which a vehicle should go while the engine is at $1000RPM$ so that torque at the wheel is equal to the torque at the engine. This depends only on the diameter of the wheel.

A typical car tire have a radius of is $25cm$. The circumference is around $PI \over 2$ meters. So when the engine is at $1000RPM$, in order for the wheels to rotate at the same speed and thus having the same torque, the car speed will have to be $1570m/min$ or $94Km/h$.

Since usually a motorcycle have bigger wheels, and also low torque, fast engines, it seems highly unlikely than torque at the wheel will ever be lower than that of the engine, even with the taller gear.

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