Percent of lost lift due to 58 degree lifter angle?

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