Cylinder Pressure of Combustion Engine

[tos001]

hi, i want to ask about 4 time combustion engine torque.
in my example we simulate a single cylinder gasoline 4 time engine which works 1000-4000 rpms and always have same fuel and air mixture in cylinder and so same burn pressure in each power time. these are the rules of question

now i want to ask that does this cylinder pressure pushes the piston same in different rpms? for example during 1500 rpms and 3000rpms. i ask this because during 3000 rpms piston moves double faster and leaves the burning area faster. so the pressure will effect the piston less because if we apply a force from a constant point to an object our effect will be more if the object is leaving from us at low speed but if the same object leaves from us at high speed our force will be less effective. if i am wrong please correct me. according to this if engine works in high rpms the burn time pressure will effect the piston less because the piston will leave the ignition area more quickly. so in high rpms same cylinder pressure will make less torque.
what do you think about this? is this opinion true

 

[Randy Beikmann]

At different speeds, if the cylinder pressure is the same as a function of crank angle, the work done per stroke will be the same, and so will the torque. The amount of time it takes to complete a stroke does not matter: the work depends on force X distance, not force X time.

 

[tos001]

At different speeds, if the cylinder pressure is the same as a function of crank angle, the work done per stroke will be the same, and so will the torque. The amount of time it takes to complete a stroke does not matter: the work depends on force X distance, not force X time.

mr.Beikmann thanks for your reply but there is a detail you know. the force in cylinder at burn time acts from a constant area this force does not have same speed with the piston. think that you stay constant place and push an object if object is not moving your force will more effective but if object is already moving with high speed and passing near you, your force will not effect as same because of the impulse formula force x time = mass x speed change.. because of the speed of the object your force effect time will be small... F=m x a is valid if the force source has same speed with the object like rockets, plane motors or car tires. according all these if the piston is leaving the burning area faster the time will be small so according to impulse formula effect of the force will be low. please correct me at wrong points thanks

 

[Randy Beikmann]

Your question is a good one. I was essentially talking about work and energy, and you were talking about impulse and momentum.
You're right that an impulse is force X time, and you are right that at a higher speed, the impulse from each stroke is smaller. Therefore the increase in momentum, mass X delta V, is smaller. So at higher speeds, it does take more combustion events to increase the vehicle's speed by the same amount.
How does that fit in with work and kinetic energy? Remember that work is proportional to force X distance, so the kinetic energy increase from each stroke is the same. But kinetic energy is proportional to the square of speed. So again, at faster speeds, it takes more combustion events to increase the velocity by a certain amount: d(Kinetic Energy) is equal to mass X velocity X dVelocity.
So it all agrees, as it must.