Percent of lost lift due to 58 degree lifter angle?
- Thread starter gregsdart
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Percentage?!?! No
My Comp solid Magnum cam w/.525 advertised theoretical ratio came in at .510. Edelbrock head and Comp roller rockers. I don't remember pushrod length. Very good geometry!
My Comp solid Magnum cam w/.525 advertised theoretical ratio came in at .510. Edelbrock head and Comp roller rockers. I don't remember pushrod length. Very good geometry!
B3422w5
Well-Known Member
Usually about 30 in an LA motor with a typical 550/600 lift flat tappet regards deflection
I'm going to make an educated guess????
More lift = more loss at valve?
B3422w5
Well-Known Member
Yea.
If you have ever had a 700 lift roller( I used to in my W5 motor) its pretty creepy watching the valve train rolling the motor over to check lash ... lol...... looks like no way it should stay together at and Rpm above what a tractor runs at.
And that was with Jesel stuff
nm9stheham
Well-Known Member
To get a close number:
- Take the cosine of the AVERAGE pushrod to lifter angle through the lift range
- multiply by the lift at the lifter
- then subtract that number from the lift at the lifter.
Example for a peak lift :
- If the AVERAGE pushrod to lifter angle through the lift range is 14 degrees, then cosine(14*) = .9703
- Multiply by the lift at the lifter: .9703 * .300" = .2911"
- Loss of lift at top of pushrod = .300 - .2911 = .0089"
The above is ONLY for the pushrod-to-lifter angle loss.
Edit to add: For the loss of lift at the valve, multiply the above by the rocker ratio. In this example, where the valve lift is nominally .450", the loss at the valve would be 1.5 * .0089" = .0134". Of course, that assumes the rocker is a true 1.5 ratio.....
As the lift increases, the pushrod to lifter angle increases and that increases the % loss of lift due to the cosine function. So, to be more accurate, you could measure the angle for each 1/3 or 1/4 of the lift range and compute the loss of lift for each range and add them up.
Measuring at the valve tip and working backwards to get this number is not accurate, since there is another loss in lift due to the lift variations inherent in the rocker angles as they move through their range.
Here is an interesting history of why the lifter bank angle ended up where it did in the LA: http://www.cranecams.com/pdf-tech-tips/chrysler-sm-block152-153.pdf
- Take the cosine of the AVERAGE pushrod to lifter angle through the lift range
- multiply by the lift at the lifter
- then subtract that number from the lift at the lifter.
Example for a peak lift :
- If the AVERAGE pushrod to lifter angle through the lift range is 14 degrees, then cosine(14*) = .9703
- Multiply by the lift at the lifter: .9703 * .300" = .2911"
- Loss of lift at top of pushrod = .300 - .2911 = .0089"
The above is ONLY for the pushrod-to-lifter angle loss.
Edit to add: For the loss of lift at the valve, multiply the above by the rocker ratio. In this example, where the valve lift is nominally .450", the loss at the valve would be 1.5 * .0089" = .0134". Of course, that assumes the rocker is a true 1.5 ratio.....
As the lift increases, the pushrod to lifter angle increases and that increases the % loss of lift due to the cosine function. So, to be more accurate, you could measure the angle for each 1/3 or 1/4 of the lift range and compute the loss of lift for each range and add them up.
Measuring at the valve tip and working backwards to get this number is not accurate, since there is another loss in lift due to the lift variations inherent in the rocker angles as they move through their range.
Here is an interesting history of why the lifter bank angle ended up where it did in the LA: http://www.cranecams.com/pdf-tech-tips/chrysler-sm-block152-153.pdf
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nm9stheham
Well-Known Member
Read it, didn't see why, but could see "Cost Savings" as a factor.
I don't think lannon that much lift on my future W5 build. But I can certainly understand where your coming from.
I'll be keeping lift at a max of ,650 or less.
MOPAROFFICIAL
Oogliboogli
Average street motor with .500-.550 lift loses around .020+lift
The no.4 and no.3 exhaust push rods....they can lose more depending on rockers, lifter height or plain head casting being the reason...still looking into it.. I'll have to post a pic of my own to show this.
The no.4 and no.3 exhaust push rods....they can lose more depending on rockers, lifter height or plain head casting being the reason...still looking into it.. I'll have to post a pic of my own to show this.
B3422w5
Well-Known Member
Loved that motor( W5)
Modern cylinder head in the Detroit area has a very good cnc program for that head.
I sold my car with the 422w5 in it about 7 years ago to a guy in New Jersey. He has been 9.60's at 140 with it on 002/003 leafs. Currently weighs about 3220
The heads were hand ported and finally became sprinklers leaking. Modern CNC'd a new set of Blue plug W5 heads he found and the combo was down only about 15 horse from before( with way more integrity) with that CNC job.
They do very nice work
nm9stheham
Well-Known Member
Yes, I assume they just moved the A right on over to the LA with minimal block changes.... less retooling, etc.
T56MaxTorq
Well-Known Member
For a magnum engine, I measured .500 lift at the valve on a .531 cam with Hughes rollers and checking lifters with a starrett dial.
nm9stheham
Well-Known Member
So for the OP...did you want to know the loss of lift due just to the lifter angle?
moper
Well-Known Member
I think the loss plots as a curve. It's not a percentage, it's a function. The higher the lift, and/or the higher the rocker ratio the more loss there is. I usually account for .025-.030 when I'm planning things out but I don't normally expect to use lifts higher than .600, or rocker ratios beyond 1.5.
B3 Would be a better resource for the math...lol.
B3 Would be a better resource for the math...lol.
That was the intent. Next, we have to look at the rockers and how they deliver that lift to the valve. Other factors I see are lift loss due to parts flex and compression. I see at least .010 lift loss just changing from checking springs to roller cam springs on m y drag motor.
I think you have something there, but a basic percentage will give us a lot of information to use. IMHO the curve would be fairly close to the percentage. Plotting it out at the top of the pushrods, then again at the valve with race springs in place would be interesting. Then look at a spintron test for the real facts
moper
Well-Known Member
I think the problem is a function is not a constant. The actual number of lost lift will change based on how much total each lobe has, plus the ratio of the rocker and length of the pushrod itself. So it won;t be a percentage that can be applied universally. That's why I tend to stick with ".025-.030"...lol.
nm9stheham
Well-Known Member
If you have some pushrod to lifter angles, or pushrod angles from vertical or horizontal, I can give you some numbers. I don't have an engine top open right now to measure them myself and just guessed in the example above.
The final number is a tad sensitive to the actual angles we are dealing with and since those pushrod angles are going to vary with different rockers and lift amounts, the angles at different lifts would be nice. Working with stock rocker up to .325" lifter lift might be one place to start for a generic number. I'd ideally work up a table of loss for .250", .325" and .400" lifter lift (to cover all common valve lifts) , using angle values taken at approximately .1" lift increments.
At the rocker, it is a whole 'nother story.
The final number is a tad sensitive to the actual angles we are dealing with and since those pushrod angles are going to vary with different rockers and lift amounts, the angles at different lifts would be nice. Working with stock rocker up to .325" lifter lift might be one place to start for a generic number. I'd ideally work up a table of loss for .250", .325" and .400" lifter lift (to cover all common valve lifts) , using angle values taken at approximately .1" lift increments.
At the rocker, it is a whole 'nother story.
yellow rose
Overnight Sensation
I'd like to here what B3 has to say about how rocker geometry affects net lift loss. I know on my last set up I only lost about .008 with 150 on the seat and IIRC 425ish over the nose.
Nm9 has it about right. There is no percentage, it depends on the pushrod angles which change throughout the lifter and rocker travels. The cam base circle will change the starting point angle, but the top of the lobe will produce about the same angle with a given lifter. The lifter seat height will have an effect too.
Now consider the crappy rocker designs out there, and that will also affect the angles. Add in my geometry correction on the LA stuff, and it changes yet again. It's a lot of stressing over something you can't do much about. Besides, the max lift loss isn't nearly as important as the area lost from the rocker geometry being out in left field.
Yes, they still have the 59 degree lifter angle.
yellow rose
Overnight Sensation
That's what I figured. Having the geometry off compounds area lost but doesn't really change net loss.
Did I get that correct?
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