Lifter Load

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cavaliers1323
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Lifter Load

Post by cavaliers1323 »

With a 1.6 rocker, P5249464 springs, and crane 753901 cam I calculate my lifter load to be 390.72# change that to 1.7 rocker and I get 415.14# HOWEVER I read a bit more and come up on this;

"The actual load on the lifter foot is the sum of the static spring pressure, the frictional loading in the valve train, and the inertial loading of the mass of the valve train."

Anyone mind helping deciphering this http://www.eng.wayne.edu/page.php?id=758
dwg86
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Stroker Displacement: 4.6
Vehicle Year: 2003
Vehicle Make: Jeep
Vehicle Model: Wrangler
Location: Atlanta, Ga

Re: Lifter Load

Post by dwg86 »

It shouldn't be that much. With 270#/ inch spring rate and 120# installed...To calculate your spring rate:
Installed # + (lift x spring rate)=? For example if you had a .500 lift cam using the mopar springs... 120+(270x.500)=? Multiply .500x270=135 So... 120+135=255# at max lift.

There is no way I could even try to figure out that page that you listed :frustrated:
cavaliers1323
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Re: Lifter Load

Post by cavaliers1323 »

The formula you supplied is for open pressure, the lift on the cam I mentioned is 0.46" which gives me 244.2# open. The formula for lifter load is lifted load= open pressure x rocker ratio. With a 1.6:1 that's 390.72#, with a 1.7:1 that's 415.14# Please double check my math, but I believe it to be correct.

I guess my my question is, are my numbers accurate or do I need to equate things like ramp, mass, speed, etc? Making my number way off.
cavaliers1323
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Re: Lifter Load

Post by cavaliers1323 »

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