Relationship of RPM on power to operate valvetrain

[Tom Walker]

Basically, here is my question. Does increasing RPM equate to the valvetrain taking more power to operator it from the standpoint of losing more and more of the force of the valve spring to follow the contour of the closing flank of the camshaft? Just as riding a roller coaster gives you a feeling of weightlessness as you go down the grade, as RPM increase to higher levels, does not the valve springs continue to release less and less force back into the system, until ultimately, you reach a stage of valve float? Any one know of any tests or have any insights. Just curious.

[Tom Walker]

Sorry, did not mean to post twice

[af2]

All I can say is put one spring on and get a breaker bar to turn the engine. Put all 16 on and can almost turn it with a wrench.

[pamotorman]

the valve train friction increases as the RPMs come up or the varicam would not retard the cam timing

[grandsport51]

Basically the valve train doesnt float when it gives up at high revs but is actually valves bouncing off their seats because the springs lost
control . What you are talking about is called valve loft which is deliberately engineered in to bypass lift restrictions

[Circlotron]

Does increasing RPM equate to the valvetrain taking more power to operator it from the standpoint of losing more and more of the force of the valve spring to follow the contour of the closing flank of the camshaft?

Really interesting question. Been thinking about it on and off all day. I think what happens is at very low engine speeds the lifter force on the closing flank is evenly distributed along its length for the whole of the closing cycle. As the rotational speed increases, the force of the lifter on the closing flank gets biased more and more toward the closing ramp end. At the extreme, the tappet would just barely lose contact with the closing flank until the closing ramp ultimately forced deceleration to zero. So in theory all the energy would be recovered except that the friction of the increased load on the closing ramp would reduce this recovered energy somewhat. Probably more than somewhat. My 2c worth.

[Warp Speed]

Yes, as rpm rises, so does the required power to turn it.

[DaveMcLain]

Yes, as rpm rises, so does the required power to turn it.

What does it do when the valvetrain becomes unstable when the spring starts to really surge. Does it tend to take more power to turn than if the spring were under control? It seems like that energy to make the spring oscillate has to come from somewhere or is it not enough to notice?

[PackardV8]

Since our horsepower definition is foot-pound-second, at rest it takes zero horsepower to operate the valvetrain and with increasing RPM the power required to operate the valvetrain could theoretically rise to infinite.

Has anyone ever put a meter on a Spintron and measured the power curve required? That would quantify the answers.

[Warp Speed]

Since our horsepower definition is foot-pound-second, at rest it takes zero horsepower to operate the valvetrain and with increasing RPM the power required to operate the valvetrain could theoretically rise to infinite.

Has anyone ever put a meter on a Spintron and measured the power curve required? That would quantify the answers.

Yes, that is how it is measured.
Power requirement, or torque to turn as it is sometimes referred to, wI'll increase with rpm right until the system loses full control. Way past realistic valve float.
Now, on a firing engine, you also need to add the power it takes to open the exhaust valve under cylinder pressure.

[Cougar5.0]

Does increasing RPM equate to the valvetrain taking more power to operator it from the standpoint of losing more and more of the force of the valve spring to follow the contour of the closing flank of the camshaft?

Really interesting question. Been thinking about it on and off all day. I think what happens is at very low engine speeds the lifter force on the closing flank is evenly distributed along its length for the whole of the closing cycle. As the rotational speed increases, the force of the lifter on the closing flank gets biased more and more toward the closing ramp end. At the extreme, the tappet would just barely lose contact with the closing flank until the closing ramp ultimately forced deceleration to zero. So in theory all the energy would be recovered except that the friction of the increased load on the closing ramp would reduce this recovered energy somewhat. Probably more than somewhat. My 2c worth.

Same here (been thinking about it), I appreciate your attempt to answer the question as asked. It takes energy to compress the spring, but it's always said the spring gives it back when the spring decompresses. But at high RPM, if the spring is expanding almost freely while barely touching the back flank, how much of the energy is being returned to the system? It's a good question.

[Tom Walker]

Thanks for all the replies. Had some doubt I had articulated my question very accurately, but I can tell from the responses that most understood what I was trying to say. Circiotron, I think what you expressed was very helpful. I would suspect that if the force is returned farther down on the closing flank of the cam at high RPM, would not not the geometry of where the lifter reacts with the cam cause it to lose some of its leverage returning force back into the cam?
Again, thanks all for your insights and responses. These are some of the things I think about at night before I fall asleep.

[Frankshaft]

All I can say is put one spring on and get a breaker bar to turn the engine. Put all 16 on and can almost turn it with a wrench.

This. Try installing one side of the engines rockers with 1000+ lb open pressure springs. You would swear your going to snap something off. Alternate between sides, it's effortless

[Tom Walker]

Yes I agree. The lifters on the closing side of the lobe try to propel the camshaft forward, equalizing (mostly), the force required to operate and compress the springs that are in the opening phase. That is what happens at low speed, when we hand crank an engine over with the valvetrain installed.All the spring forces sort of cancel each other out as far as the torque required to turn the camshaft as a whole. My question relates to this, now operate this same engine at 7-8000 RPM. Are the spring forces still equalizing the torque required to turn the camshaft as a whole? I suspect that they are not, and that there is probably somewhat of a linear curve showing increasing torque required to turn the camshaft at these speeds that has nothing to with the increased friction of high RPM, which I agree is increasing,but a different conversation.Thanks for your thoughts, I have no tangible or hard defined answers, just more questions. Probably, should just learn to go to sleep sooner

[af2]

All I can say is put one spring on and get a breaker bar to turn the engine. Put all 16 on and can almost turn it with a wrench.

This. Try installing one side of the engines rockers with 1000+ lb open pressure springs. You would swear your going to snap something off. Alternate between sides, it's effortless

That was my point. It is true.