Figlio del Diavolo wrote:
Try to think of it in terms of moment arms generated around the centerline of crankshaft. With a shorter stroke the moments are not as high. This is the whole source of mechanical ouput of the engine.
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Based on the two different cranks it should be easy to see why it is difficult to get high torque in engines with relatively small strokes from a moment point-of-view.
As a start these two engines have a VERY different displacement, you didn’t catch it because you used two different units, but that F1 engine is a 3000 cc (183 ci) while that NASCAR engine is a 5900 cc (358 ci), close to twice the displacement. Since max torque you can generate from an engine, for a given technological level, is roughly linearly proportional to displacement, it’s evident that the latter, even if not particularly sophisticated, will have higher peak torque, it’s twice as big.
If now you look at a given displacement, you’ll find that the amount of torque generated by an engine isn’t directly influenced by the bore/stroke ratio because even if the stroke is shorter, hence the moment arm is shorter, the force (pressure x area) is larger due to the larger bore and the two effects compensate each other.
Then obviously the short stroke engine will have peak torque at higher rpm but will also have an higher rpm red line, so you just have to let it rev higher, but that doesn’t means it lacks torque, it simply has it at higher rpm. Use the suitable gearbox ratios and you’ll find it all.
There’s actually a possible influence of the bore/stroke ratio on the amount of maximum torque at the crank but it’s an indirect one and only exists while arriving to extreme designs like in F1 where the large bore and the high CR have an effect on the combusting chamber shape hence on the BMEP (that is basically a measure of torque per unit displacement). Anyway that’s a secondary issue non related with the pure kinematics of piston and con rods.
Strangely enough, that myth about long stroke=more torque is something I found only few years ago when I started surfing on the web and mainly coming from US people, I wouldn’t exclude that the imperial units you use (making no clear distinction between torque and work, contrarily to SI units where we use Nm and J respectively) could be one of the cause of the misunderstanding.
Another cause could be US people love for big displacement engines and the fact that consequently that myth is repeated every now and then on car magazines round there bragging about US cars. I wouldn’t also exclude a role is played by the general lack of familiarity with manual gearbox hence with the choice of the right gear ratio for the right situation.
zac510 wrote:
I think you will find that revs x torque only works for hp x lb/ft and with kw/nm. It will not with for nm and hp as that is a cross of imperial and metric.
Just to make it more clear :
Power = revs x torque is the physics law.
If you use the SI units, the ones you should really use for physics calculation if you don’t want to make a multi-million $ mistake (ask NASA...), the ones that will make you really understand what you are doing while applying physics law, then you need torque in Nm and revs in rad/s, you’ll have power in Watt.
If you want to use different units, then you’ll need to adopt conversion factors and these have the tendency to lead to a quantity of mistakes.
For example if you use torque in lb-ft and revs in rpm, you have to divide the result of torque x revs by roughly 5252 to have power in hp (1 hp = 745.7 kW, 1 PS = 1 CV = 735.5 kW) and that gives, as riff_raff said, 792 hp for torque = 260 lbft and rpm = 16000.
Converted in SI units, 260 lb-ft * (0.4536 [lb/kg] * 0.3048 [ft/m] * 9.81 [m/s^2] = 352 Nm. That multiplied for 16000 rpm * 2 * pi / 60= 1675 rad/s gives a power = 590 kW = 792 hp.
