kitabel
Automotive
- Dec 14, 2005
- 20
I've seen many comments that any reduction in tappet, pushrod or rocker short side lever weight has almost no effect on determining spring rate and RPM.
This must, IMHO, be due to the amplified movement of the long-side bits (valve, collar, locks, lash cap, long lever) by the rocker ratio. Although no longer common, fairly competitive engines were once built with 1:1 rockers (H-D knucklehead, BSA Gold Star), and in these cases the short and long side parts must have equal effect, yes?
What I cannot grasp is what relative value to assign to the differences between short and long side parts based on a rocker ratio of more than 1:1, let's say 2:1 to keep the math easy.
1. The long side parts travel 2X the distance, have 2X the mean and maximum velocity, and accelerate at 2X the rate of the short side parts.
Are these factors cumulative? Duplicative? Exponential?
2. To balance this, the short side parts have 2X the spring load at all points.
If common wisdom (short side parts less important) prevails, clearly factors 1. & 2. are not equal and self-cancelling, with 1. higher, but how much?
IMHO it can't be a direct addition, subtraction or proportion but must be a (fractional) exponent, or the values would not match on both sides for a 1:1 rocker.
Possible answer #1 is to ignore acceleration in 1., multiply speed and distance to get (rocker ratio^2), then divide by ratio to compensate for spring load.
This leaves the relative value of a gram of short side weight as proportionate to long side weight X (1/rocker ratio), or 50% in this case. A 1.5:1 rocker would yield 66.7%, etc.
Possible answer #2 is to use all factors in 1., multiply acceleration, speed and distance to get (rocker ratio^3), then divide by ratio to compensate for spring load.
This leaves the relative value of a gram of short side weight as proportionate to long side weight X (1/rocker ratio^2), or 25% in this case.
A 1.5:1 rocker would yield 1/(1.5^2) or 44.4%, etc.
Too easy?
Partial exponents rather than squares, like ^1.5?
Better idea?
This must, IMHO, be due to the amplified movement of the long-side bits (valve, collar, locks, lash cap, long lever) by the rocker ratio. Although no longer common, fairly competitive engines were once built with 1:1 rockers (H-D knucklehead, BSA Gold Star), and in these cases the short and long side parts must have equal effect, yes?
What I cannot grasp is what relative value to assign to the differences between short and long side parts based on a rocker ratio of more than 1:1, let's say 2:1 to keep the math easy.
1. The long side parts travel 2X the distance, have 2X the mean and maximum velocity, and accelerate at 2X the rate of the short side parts.
Are these factors cumulative? Duplicative? Exponential?
2. To balance this, the short side parts have 2X the spring load at all points.
If common wisdom (short side parts less important) prevails, clearly factors 1. & 2. are not equal and self-cancelling, with 1. higher, but how much?
IMHO it can't be a direct addition, subtraction or proportion but must be a (fractional) exponent, or the values would not match on both sides for a 1:1 rocker.
Possible answer #1 is to ignore acceleration in 1., multiply speed and distance to get (rocker ratio^2), then divide by ratio to compensate for spring load.
This leaves the relative value of a gram of short side weight as proportionate to long side weight X (1/rocker ratio), or 50% in this case. A 1.5:1 rocker would yield 66.7%, etc.
Possible answer #2 is to use all factors in 1., multiply acceleration, speed and distance to get (rocker ratio^3), then divide by ratio to compensate for spring load.
This leaves the relative value of a gram of short side weight as proportionate to long side weight X (1/rocker ratio^2), or 25% in this case.
A 1.5:1 rocker would yield 1/(1.5^2) or 44.4%, etc.
Too easy?
Partial exponents rather than squares, like ^1.5?
Better idea?
