Valve Rotation

[Alvor315]

At the moment I'm looking at the original 170 1.5 ratio rockers. There is evidence on the surface that pushes on the valves that the valves spin. It also seems that there is an angle cut to the rocker to valve contact surface, my guess is that it is put there just for the reason of spinning the valve. The combination of the angled rocker surface and valve sweep would impart a torque to just one side of the valve stem, and thus rotate it. I have heard in passing before that the valves rotate to reduce wear, and it makes sense that a non-roller rocker arm can cause the valve to spin, but if you were to switch to roller-rockers are they also designed to spin the valves?

It seems to me that in placing roller-tips on rockers it would significantly reduce the friction found in sweeping the rocker arm across the top of the valve stem. (Hence the purpose of roller-tipped rockers) However, in reducing this friction, how is the valve then rotated?

I haven't been able to find anything on the internet about the fact that the valves are rotated. Is this a closely guarded secret?

 

[XPC66]

Yeah at reasonable revs, the valves rotate. It's one of the things that contributed to the subscription that giving the engine a good workout now and again would avoid problems (carbon buildup on the valves, bore ridges, etc)

 

[Alvor315]

I think I may have divined the answer. It took gobs of black magic and toad warts or something, but I think I have the answer.

Tell me what you guys think of this hypothesis:

With a roller-rocker when the valve stem is pushed down all the force is transfered through the valve keepers and to the springs, but in order for the springs to compress they must rotate slightly with the retainers. For an example of this, think of when a slinky is pulled apart; the slinky will try to rotate in hand. Think of that principle in reverse with respect to the valve springs. Because the bottom of the springs are fixed in position with the friction between them and the head. When the force transfered by the rockers is released the friction between the valve springs and the head is significantly reduced and the springs rotate at the bottom returning to their static position. Since the mass of the springs is less than that of the valves, the valves remain relatively stationary while the spring rotates back to equilibrium. Thus the valve and spring have both rotated. :roll:

(takes a breath)

Considering this phenomenon it explains why the valves would not rotate at lower rpm. The release of pressure from the rocker arms must exceed the rate at which the springs can apply pressure to the rocker arms. Thus the valves do not rotate until the engine reaches higher rpm. This theory also fits non-roller rockers because the spring pressure must halt the the rotation of the valves in order for them to not rotate at low rpm.

What do you guys think? Does that sound plausible?

Should this hypothesis be true it also allows for another quality of roller-rocker arm design that I have been brooding over. If the dragging friction over the top of the valve stem is not what causes the valve to rotate, then rocker arms can be designed to begin their stroke above the 0* mark and end their stroke below the 0* mark, thus greatly reducing the valve stem sweep. According to my calculations, the sweep of a 1.6:1 ratio rocker on our engines that is opening a valve to 0.600 inches, and is starting from a position exactly at 0* must sweep 0.122 inches across the the valve stem. A rocker of the same dimensions starting from a position of roughly 12 degrees will sweep a displacement of roughly 0.0307 inches to the 0* position and the same distance to the -12* position. Since the rocker sweeps the same distance both above and below the 0* mark (though in opposite directions) the total sweep distance is 0.0614 inches, which is roughly half of the distance swept by the rocker with a 0* starting position.

Have I missed anything? Any input would be much appreciated.

Allan

 

[rhetor]

Allan,

This would seem like less friction, but it seems to me like the same amount of pressure over a smaller area of travel. Accelerated wear.

I believe that the lifters are offest from the cam lobes to induce a spin, but the valve rotation is like you had mentioned: due to spring action.

Look up exhaust valve rotators. See what they look like?

 

[Alvor315]

I have seen positive valve rotators

http://www.tpub.com/content/construction/14264/css/14264_94.htm

but what I'm talking about is the the phenomenon that occurs when the engine reaches sufficient rpm to cause the valves to rotate.

About the sweep distance. Pressure is force per unit area and the area of that force never changes. The pressure is due to the force on the area of the rocker wheel contact surface. Similar to a tire contact pattern, the entirety of the force is applied to that same area. Since the wheel roles across the top of the valve stem, the lateral friction forces are negligible, but the lateral forces as a component of the geometry of the valve are still present. Consider a long isosceles triangle with the smallest angle at the bottom and you are standing at the top. You are forced to walk from side to side on this triangle. You want to walk the shortest distance from side to side to keep it from tipping don't you? The shorter the sweep distance, the less wear on the valve guides, and probably several other parts as well.

 

[Alvor315]

Another thought on minimal valve sweep, is that if the valves to rotate the wear from a small sweep would be minimal and uniform as well, because the roller extends beyond the end of the the valve stem on either side and then the entire area of the valve top would be covered as the valve rotated.

 

[rhetor]

Alvor, this is true and a good point.

however, the roller tip rolls on a pin. The smaller the degree of rotation, the more likely to get wear on a small part of the pin or roller (in the vector of the force.) you would end up with an oblong hole.

I'm guessing that the pin and roller are hard enough that they will not see a substantial wear in either case.

keep in mind, all materials, no matter how rigid, will deflect under a load.

This is most likely a moot point though.

Just to be clear, you are saying that a roller tip rocker should have a perpendicular position when the valve has reached half of its downward travel, correct? And that by your observations the current rockers are perpendicular when the valve is fully seated?

 

[Alvor315]

Yes those are my observations. From measuring the original rocker, the center of the push rod, the fulcrum and the valve-stem contact surface all lie within the same plane. What I'm suggesting is raising the the valve-stem contact surface by about 12 degrees on our rocker arms. That would allow the sweep distance to be cut in half.

As for the pins, I believe the same type of motion, that I hypothesize is happening to the valves also happens to roller rockers. That is, when the revolutions reach a certain point, the release of pressure on the roller wheel is such that the roller wheel will not rotate due to contact with the valve stem, thus again, minimizing wear. If this doesn't happen, then roller rockers as they are used now should be example enough that the roller pins don't fatally wear.

Allan

 

[rhetor]

This makes sense.

You can do a simple physics calculation using inertia and a spring constant. I am not the best at this stuff, but i believe that since the spring is returning the valve, which returns the rocker, pushrod, and lifter to the lobe on the cam, you need to calculate acceleration.

The weight of the pushrod, valve and keepers, as well as the lifter and the friction/suction of the oil will have to be considered. You can actually just ignore all this, do a quick calc and get relative numbers.

The spring should give a constant acceleration to the valve throughout the rpm range. You also have an acceleration in the opposite direction from the cam lobe. Once your acceleration from the cam lobe overcomes the acceleration from the spring, you get valve float.

So based on this logic, you do see less and less pressure as the rpm's come up. There will indeed be a point at which the friction from the keepers or rollers, etc., will be less than the force causing the spin. Thats the point that it will spin.

Now, it is my understanding that when considering the roller sweeping across the valve stem, you will have the same total duration of force and the same amount of force in every situation, provided the valve moves the same distance. If you were to take small snap shots of the action, you would see a series of sequential force applications. The total duration of the force will be concentrated on a smaller area with less sweep. Although there is no room for this theoretically, you will see a longer "dwell" in the same area. This is just a guess though. Guesses are useless until tested.

 

[XPC66]

Valve float is more to do with the spring's installed height, installed load, spring mass and it's subsequent resonant frequency. Redline is just the resonant frequency x 120/7.5


The opening and closing acceleration is pretty much a sinusodial event using redline, displacement and angular acceleration. When I calculate my spring selections I throw in all the masses including even the shims.

One of the first thing you should do when testing the rockers is to establish the contact patch on the valve tip by coating them with marking blue and looking for a symetrical wear pattern, otherwise it will cause uneven side loading. It's something I do religiously to ensure even wear on the tips & guides and the gradual rotation of the valves themselves.

The right hand rule always comes into play,. Just like a front wheel drive will torque steer (due to introduced bump steer from unequel length drive shafts) so the poppit valves will rotate when the angular force of the rocker is applied. To do this the rotation torque must exceed the spring preload. That preload effectively reduces as the frequency of the spring action increases, thus as the engine revs rise.