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Old Jan 15, 2008 | 09:39 PM
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Default Rocker Arm Weight

The stock LS1 (GM 10214664) rocker arm without its trunnion (as shown in the accompanying attachment) weighs 2.90 ounces (82.21 grams). I would like to have some idea of the relative rotational inertia of various LS rocker arms. I am particularlly interested in the Yella Terra Ultra Lite 6645. Anyone who has an aftermarket rocker that can be disassembled, can weigh it at their local Post Office. Thanks.
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Old Jan 16, 2008 | 05:12 AM
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The weight is a nice think to have, but won't tell you the inertial properties. A rocker can weigh more but have less inertia. If someone wanted to spend the time to measure and model the rockers in a CAD program, you could figure it out.
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Old Jan 16, 2008 | 12:38 PM
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Originally Posted by vettenuts
The weight is a nice think to have, but won't tell you the inertial properties. A rocker can weigh more but have less inertia.
It is also possible that a lighter rocker could have higher inertia. We know that the stock rockers are very good. If a rocker is much heavier than the stocker you should have serious doubts that it has lower inertia.
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Old Jan 16, 2008 | 05:44 PM
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From: Little Rhody
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Originally Posted by Gary Z
It is also possible that a lighter rocker could have higher inertia. We know that the stock rockers are very good. If a rocker is much heavier than the stocker you should have serious doubts that it has lower inertia.
That's exactly my point, and I don't disagree, but I don't like basing it solely on weight.
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Old Jan 17, 2008 | 03:18 PM
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Vettenuts - If you or anyone else wants to post true rocker inertia data, I would be delighted to see it. Because I expect that such data will not be forthcoming, I am proposing a simple alternative that should closely correlate with the true value. How about we find how much the rocker weights differ and THEN decide whether the difference is significant?

-Gary
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Old Jan 18, 2008 | 05:07 AM
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Gary,

I have thought about taking measurements many times and plugging a few of the rocker designs into a CAD program then checking their strength/stiffness with some FEA. Never seem to find the spare time to do it. I have been trying to get this information from the vendors for a couple of years.

Maybe the other measurement you should consider makong is tip weight with the roller in it. You could measure that with the center supported as installed. I guess you peaked my interest because you are one of the only guys I have seen in since I have been on this forum trying to sort this out with some technical information. I have a set of Yella Terra's on the work bench right now, I could take a tip measurement next week after my business trip. We have a high precision "grams" scale at work.
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Old Jan 18, 2008 | 07:52 AM
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Originally Posted by vettenuts
Gary,

I have thought about taking measurements many times and plugging a few of the rocker designs into a CAD program then checking their strength/stiffness with some FEA. Never seem to find the spare time to do it. I have been trying to get this information from the vendors for a couple of years.

Maybe the other measurement you should consider makong is tip weight with the roller in it. You could measure that with the center supported as installed. I guess you peaked my interest because you are one of the only guys I have seen in since I have been on this forum trying to sort this out with some technical information. I have a set of Yella Terra's on the work bench right now, I could take a tip measurement next week after my business trip. We have a high precision "grams" scale at work.
This is one of the inputs in Engine Analyzer Pro, so I've been looking for this info for a long time also (> 10 years). Same as you've reported, no luck from anyone other than what I've been able to measure myself.

I have taken the tip weight measurement you identify here. I have a medical scale that's accurate to a tenth of a gram. I have a calibration weight that I test accuracy with to ensure weight measurements are correct. Here's what I came up with:

Test condition: The rocker arm trunnion supported and the tip of the rocker on the scale. Scale zeroed prior to measurement.

Stock rocker: 8.2 grams
Crane 1.8 roller rocker: 22.8 grams
Harland Sharp non-adjustable 1.7 roller rocker: 22.5 grams

I would be interested in the tip weight of the Yella Terra's, I assume your using 1.7 ratio.

Last edited by 405HP_Z06; Jan 18, 2008 at 09:57 AM.
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Old Jan 18, 2008 | 09:42 AM
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From: Little Rhody
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Yes I am. Do you have a photo or anything of your setup? That way we can try and get an apples to apples comparison as I will try to duplicate it.
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Old Jan 18, 2008 | 09:57 AM
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Guys, for the purpose of estimating rotational inertia, the so-called “tip weight” really is useless. Think about it. Any tip weight could be reduced to zero by ADDING mass to the pushrod-side of the rocker. A tug-boat rocker arm could have zero tip weight. Tip weight bears no relation to “equivalent tip mass”.

Conversely, the comparative weights of the rocker beams (minus trunnions), might be very useful information. In particular, we can conclude that any aluminum roller-tip rocker beam that weighs as much or more than the stock rocker beam will have higher than stock rotational inertia.

-Gary

Last edited by Gary Z; Jan 18, 2008 at 10:16 AM.
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Old Jan 18, 2008 | 09:59 AM
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Originally Posted by vettenuts
Yes I am. Do you have a photo or anything of your setup? That way we can try and get an apples to apples comparison as I will try to duplicate it.

I don't off hand, but I can take some. I'll do it tonight and post them.
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Old Jan 18, 2008 | 10:07 AM
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Originally Posted by Gary Z
Guys, for the purpose of estimating rotational inertia, the so-called “tip weight” really is useless. Think about it. Any tip weight could be reduced to zero by ADDING mass to the pushrod-side of the rocker. A balanced rocker arm could have very high inertia and zero tip weight. Tip weight bears no relation to “equivalent tip mass”.
Hi Gary,
Help me understand this, I'm not seeing what your seeing. Example: If I had a balanced rocker arm with the following characteristics:

Rocker minus trunnion: 100 grams
Assumption for sake of arguement: Balanced beam with equal weight on both sides equates to 50 grams on each side.

If I add weight to the pushrod side, say 10 grams, I now have an unbalanced rocker arm with 60 grams on one side and 50 grams on the other.

So, 50 grams is 50 grams on the valve side of the rocker. How could this additional 10 grams cancel out the 50 grams for a net of zero?

Originally Posted by Gary Z
Conversely, the comparative weights of the rocker beams (minus the trunnion), might be very useful information. In particular, we can conclude that any aluminum roller-rocker beam that weighs as much or more than the stock rocker beam will have higher than stock inertia.

-Gary
Agree, we can make this assumption; however, what is the real number. Is it 5% more, 10% more, or 50% more. The design of the stock rocker provides for a very low moment of inertia. We know that aftermarket rockers provide a higher moment of inertia, but how much?
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Old Jan 18, 2008 | 10:35 AM
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Originally Posted by 405HP_Z06
We know that aftermarket rockers provide a higher moment of inertia...
I think this is probably true of most aftermarket rockers but I don't know it is true of all.
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Old Jan 18, 2008 | 10:45 AM
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405 - Maybe I'm not seeing what you're seeing. A rocker can be balanced (tip weight reduced to zero) by adding weight (mass) to the pushrod side of the rocker. Adding mass anywhere increases the rotational inertia.

-Gary
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Old Jan 18, 2008 | 10:49 AM
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Originally Posted by Gary Z
I think this is probably true of most aftermarket rockers but I don't know it is true of all.
Agree, I was making a generalization. I have not seen any substantial evidence that ANY aftermarket rocker arm has a lower moment of inertia as compared to a stock rocker. Unless I'm totally off base, I think we are trying to determine the real inertia data for stock and aftermarket LS1 rocker arms?
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Old Jan 18, 2008 | 10:52 AM
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Originally Posted by Gary Z
Guys, for the purpose of estimating rotational inertia, the so-called “tip weight” really is useless.

-Gary
Gary,

While not the end all be all, in my opinion it provides another puzzle piece since a rocker that has a heavy tip weight will likely have more inertia as well and I don't think its totally useless information in the scheme of things. Your example of no tip weight would likely find a home in a rocker that has the adjustment on the pushrod side of the rocker rather than the fulcrum, as it could end up with either negative tip weight or zero/small but in reality provide a large rotational mass to accelerate. Its another simple measurement that can be taken quickly to add to the information pool.
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Old Jan 18, 2008 | 10:57 AM
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Originally Posted by Gary Z
405 - Maybe I'm not seeing what you're seeing. A rocker can be balanced (tip weight reduced to zero) by adding weight (mass) to the pushrod side of the rocker. Adding mass anywhere increases the rotational inertia.

-Gary
Okay, so in the above example adding the additional 10 grams to the pushrod side of the rocker would increase the rotational inertia and not negate the weight at the valve tip side?

To compute the inertia of a rocker arm, both sides of the rocker are part of the equation. However, in practice weight at the pushrod side of the rocker is less of an impact to valve train control than weight at the valve side of the rocker arm.

Originally Posted by Gary Z
405 - Maybe I'm not seeing what you're seeing. A rocker can be balanced (tip weight reduced to zero) by adding weight (mass) to the pushrod side of the rocker.....
Not trying to be difficult or disagreeable, but I don't understand this statement.
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Old Jan 18, 2008 | 12:57 PM
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Originally Posted by 405HP_Z06
Not trying to be difficult or disagreeable, but I don't understand this statement.
I am trying to explain why the tip weight measurements you describe in Post #7 are unrelated to rocker arm rotational inertia. The tip weight you describe can be zero. Rocker inertia can not be zero.
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Old Jan 18, 2008 | 02:40 PM
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Originally Posted by Gary Z
The tip weight you describe can be zero.
How is this? If the tip weight was zero, than there wouldn't be a mass to accelerate so inertia would be zero. I must have missed where you explained this.....
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Old Jan 18, 2008 | 03:06 PM
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Rotary inertia is a function of both mass (weight) and radius (distance from center of rotation). For example, you could have a balanced see saw that has zero mass when weighed on the end, but the rotary inertia would be high especially as radius from the center increases. Hope that helps.
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Old Jan 18, 2008 | 04:49 PM
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Originally Posted by vettenuts
Rotary inertia is a function of both mass (weight) and radius (distance from center of rotation). For example, you could have a balanced see saw that has zero mass when weighed on the end, but the rotary inertia would be high especially as radius from the center increases. Hope that helps.
That helps, but I do not agree that the mass of the rocker arm from the axis of rotation to the valve tip end of the rocker has no bearing on the moment of inertia.

The axis of rotation for a stock rocker arm and aftermarket rocker arm is the same distance. The weight of the valve end of a stock rocker arm from the axis of rotation to the rocker arm tip is 8 grams. The weight of the valve end of a Crane rocker arm from the axis of rotation to the rocker arm tip is 22 grams. Knowing the axis of rotation is the same, the weight of the object changes the moment of inertia. The object with more mass will have a higher moment of inertia. i.e. the Crane rocker arm.

Mass is a measure of how difficult it is to get something to move in a straight line, or to change an object's straight-line motion. The more mass something has, the harder it is to start it moving, or to stop it once it starts. Similarly, the moment of inertia of an object is a measure of how difficult it is to start it spinning, or to alter an object's spinning motion. The moment of inertia depends on the mass of an object, but it also depends on how that mass is distributed relative to the axis of rotation: an object where the mass is concentrated close to the axis of rotation is easier to spin than an object of identical mass with the mass concentrated far from the axis of rotation.
The moment of inertia of an object depends on where the axis of rotation is. The moment of inertia can be found by breaking up the object into little pieces, multiplying the mass of each little piece by the square of the distance it is from the axis of rotation, and adding all these products up


Reference: http://physics.bu.edu/~duffy/py105/Torque.html
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