B20 Vtec power.

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imagine a nitrous B20B.

stock RPM limit.

combustion goes bang, rod goes down.

where are the other three rods?

there is another piston/rod going down too, but it is being pulled down by the crank with no force on top.... (NB - this rod and piston has inertia)

imagine what is happening to the rod bolts on that rod?

are the forces on those rod bolts going to be the same if the forces on the other piston are higher due to the higher combustion pressure?

If the forces pulling the other rod down were somehow greater with the higher power, then you would be accelerating the piston faster (since mass is obviously conserved) - which of course does not happen in a state of constant engine speed.

The forces on the remaining non-powered rods are determined by the angular speed of the engine, not the torque onto the crank-shaft. For a given engine speed: the physical nature of the rod and piston assembly determines how much force is required to overcome it's state; the reverse is not true: increased torque on the crank pushes the piston harder. If you want to balance the equation, increase the "load" on the engine - which involves increasing the angular acceleration of the load - the exact expected result!

The forces on the rod can be examined one at a time and then summed to get the final result. Comparing two otherwise identical engines at the same operating speed means you can cancel out the inertial loads as they are equal in each case.

Summary: You can push an object a given distance in a given time with only one constant force. You can't push twice as hard and have it take just as long. (I.e. you can't push/pull a rod harder yet have it move at the same speed).