B20 Vtec power.

[ewendc2r]

I hate to say it, but the comments about your condescending approach to 'offering opinion' is spot on, there isn't a need for people to act god, if a question is asked it'd be great if a reasonable informative response was posted..

I only say this because you've done it in response to me before, and I couldn't figure out in what way it could be construed as positive input..

Oh, and I'm pretty sure (after some thought) that string is technically spot on.

/breathe.

[joyride]

i didnt read all of the above because its way too technical for me.

tinkerbell/string: i know both of you guys have a wealth of knowledge and experience; its good that you are both trying to come to the same conclusion but please do not try and argue with each other directly. you know PM is always available to discuss it there.

cheers guys. its good to see OH still has some 'thinkers' onboard

[Benson]

So why do people run forged rods in their turbo motors with >15psi through the motor and cap the revs to 7200rpm? i.e EVO's

IF what your saying is right string, then people dont need forged rods

[aaronng]

You guys forgot that compression and elongation forces on the rod bolt affect the metal differently. As long as your are generating more combustion force than stock, that means the other piston in the intake cycle would experience a stronger elongation force when it is being pulled down by the crankshaft. The metal in the rod bolt can eventually experience fatigue failure after sufficient cycles if the force exceeds the amount needed to slightly stretch the metal in the rod bolt.

[tinkerbell]

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?

As long as your are generating more combustion force than stock, that means the other piston in the intake cycle would experience a stronger elongation force when it is being pulled down by the crankshaft. The metal in the rod bolt can eventually experience fatigue failure after sufficient cycles if the force exceeds the amount needed to slightly stretch the metal in the rod bolt.

thanks for answering my question so simply! so, more power does lead to more stress, despite RPM...

[string]

So why do people run forged rods in their turbo motors with >15psi through the motor and cap the revs to 7200rpm? i.e EVO's

IF what your saying is right string, then people dont need forged rods

I haven't discussed the strength of the rods in compression because I'm not qualified to make comments - I don't know anything about the specific materials in question. You can't avoid a higher torque combustion event from pushing down on the crankshaft (through the connecting rod) harder - if you're making more power than a stock cast rod can handle in compression (or handle in the event of detonation) then you'll have to go for a stronger item.

As long as your are generating more combustion force than stock, that means the other piston in the intake cycle would experience a stronger elongation force when it is being pulled down by the crankshaft.

Consider two engines, one 10% more torquey than the other but otherwise identical. They are both connected to a load which holds the engine speed constant (one load is 10% larger obviously). One engine is putting out 10% more power but both are operating at the exact same speed.

In both cases, the rods and pistons are moving at identical rates. If two objects have equal positions at all points in time, they also have equal velocities and equal accelerations. Since they have equal accelerations, we know that from F=ma, they must both be receiving the exact same force, even though one engine is producing 10% more power than the other.

[aaronng]

Consider two engines, one 10% more torquey than the other but otherwise identical. They are both connected to a load which holds the engine speed constant (one load is 10% larger obviously). One engine is putting out 10% more power but both are operating at the exact same speed.

In both cases, the rods and pistons are moving at identical rates. If two objects have equal positions at all points in time, they also have equal velocities and equal accelerations. Since they have equal accelerations, we know that from F=ma, they must both be receiving the exact same force, even though one engine is producing 10% more power than the other.

If both engines are at the same speed, and both have equal acceleration and hence equal force, then both engines are producing the same amount of power. The more "powerful" engine is however using less throttle to keep the engine spinning at the same speed as the weaker engine.

[string]

If both engines are at the same speed, and both have equal acceleration and hence equal force, then both engines are producing the same amount of power. The more "powerful" engine is however using less throttle to keep the engine spinning at the same speed as the weaker engine.

Both are at the same speed and thus have equal piston accelerations. One engine has 10% more torque than the other and neither has angular acceleration - of course they're not producing the same power.

Both engines can effectively be thought to be at full throttle - the speeds of both engines are mediated by the choice of load (10% larger to counter 10% more power).

[aaronng]

Both are at the same speed and thus have equal piston accelerations. One engine has 10% more torque than the other and neither has angular acceleration - of course they're not producing the same power.

Both engines can effectively be thought to be at full throttle - the speeds of both engines are mediated by the choice of load (10% larger to counter 10% more power).

Torque = Force x Vector distance

where Vector distance is the horizontal distance between the point at which you are measuring engine torque (crankshaft) and the point where the force is applied (piston-rod pin)

Since we assume that the engines are identical, then the horizontal axis distance from pin to crankshaft is the same for both engines and the force is the same for both engines, hence the torque must also be the same.

[string]

Since we assume that the engines are identical, then the horizontal axis distance from piston to crankshaft is the same for both engines and the force is the same for both engines, hence the torque must also be the same.

The force on the combustion stroke piston is NOT the same. The example states that one engine has 10% more torque than the other = 10% more force down on the combustion stroke piston.

The remaining 3 pistons (who's motion is determined entirely by the crankshaft, unlike the combustion piston which has an unbalanced [without a load] force) and rods of both engines share the same position function, and thus velocities, accelerations and therefore forces from the crankshaft.

[aaronng]

The force on the combustion stroke piston is NOT the same. The example states that one engine has 10% more torque than the other = 10% more force down on the combustion stroke piston.

The remaining 3 pistons (who's motion is determined entirely by the crankshaft, unlike the combustion piston which has an unbalanced [without a load] force) and rods of both engines share the same position function, and thus velocities, accelerations and therefore forces from the crankshaft.

So if you say that the 3 remaining pistons have motion determine by the crankshaft, and one engine's combustion piston is making 10% more force, then how can you assume that the other 3 pistons have the same force as the weaker engine? The crankshaft is directly driven by the combustion piston, so if you have 10% more force, that will also exert 10% more force on the other 3 pistons.

P = T x RPM / 5252

If you keep the two engines at the same constant velocity (constant RPM), then both engines would be making the same torque at the same RPM, otherwise one engine would have more power and accelerate more than the other engine (as per above equation). Even if you had both engines with identical velocities and accelerating at the same rate (increasing RPM), then both engines would also have to be exerting the same amount of torque, assuming that both engines are identical in terms of friction/losses because power has to be identical between both engines so that acceleration is identical.

F = m x a
T = F x r
Substituting gives T = m x a x r

Therefore if you assume that m (mass), a (acceleration) and r (horizontal vector) are identical for both engines, then T (torque) must be the same as well. If you say one engine has 10% more torque, then that violates the 2nd law of motion since the force has to be 10% higher as well.

[string]

P = T x RPM / 5252

If you keep the two engines at the same constant velocity (constant RPM), then both engines would be making the same torque at the same RPM, otherwise one engine would have more power

Exactly... One engine does "have more power" than the other - the entire context of this discussion, glad you're with it.

and accelerate more than the other engine (as per above equation).

The 10% more powerful engine has a 10% larger load. The loads absorb energy such that both engines are at constant angular velocity. Read the example again.

Even if you had both engines with identical velocities and accelerating at the same rate (increasing RPM), then both engines would also have to be exerting the same amount of torque, assuming that both engines are identical in terms of friction/losses because power has to be identical between both engines so that acceleration is identical.

Again, the loads are not identical. The 10% more powerful engine is flowing 10% more power to the load. Angular acceleration is zero in both cases.

F = m x a
T = F x r
Substituting gives T = m x a x r

Therefore if you assume that m (mass), a (acceleration) and r (horizontal vector) are identical for both engines, then T (torque) must be the same as well. If you say one engine has 10% more torque, then that violates the 2nd law of motion since the force has to be 10% higher as well.

The torque at the load is 10% higher with the more powerful engine. Since the magnitude of the load is 10% larger (as cited in the example), it is in the same steady state as the smaller load. Disregarding your bastardization of the equations, the force is 10% larger - on the combustion rod and on the load.

So if you say that the 3 remaining pistons have motion determine by the crankshaft, and one engine's combustion piston is making 10% more force, then how can you assume that the other 3 pistons have the same force as the weaker engine?

You have two engines spinning at the same rate. One engine is producing 10% more power - it's combustion events give a 10% greater net torque on the crank. There is no angular acceleration of the engine - the remaining 3 pistons have motion determined entirely by the crank - the crankshafts in both cases must be providing the exact same force. The only way for the forces to be different in the more powerful engine is if it were spinning at a different rate - which it isn't.

The crankshaft is directly driven by the combustion piston, so if you have 10% more force, that will also exert 10% more force on the other 3 pistons.

Incorrect. The load will be driven with 10% more force - the pistons require only as much force as necessary to facilitate their motion - a motion which is identical to the weaker engine.


The only way to increase the force on the remaining 3 pistons is to increase the angular velocity of the engine. I specifically crafted this example with equal engine speed, but non-equal power flow to gracefully simplify the concept.