sam85iroc
Materials
- Oct 13, 2011
- 2
Hi all,
I'm new, so nice to meet you! (I hope) and sorry in advance for using the English units of measure
So here it is- we are all familiar with the eq. 84000/rpm=length, but I wanted something more accurate so I did alittle brainstorming and please let me know if I'm off my rocker here...
now, the theory is the valve closing causes a shockwave that reflects up and down the runner at the speed of sound and ideally the valve is just opening when the wave arrives.
speed of sound at 130*F = around 1200fps
crankshaft rotation per full cycle= 720 degrees
example cam intake duration: 250 degrees
example desired torque rpm: 3,333 (for easy calculations)
the amount of time the wave has to travel is the time between the valve's closing in one cycle and it's opening in the next cycle, or "total degrees minus duration" --> 720-250=470 for example
and at 3,333 rpm each degree takes 0.00005sec
thus it takes 0.0235sec to rotate the crank 470*
and at 1200fps, sound can travel 28.2ft in 0.0235sec
so the first harmonic wave would be 14.1ft (wave travels up then back down covering 14.1 * 2 = 28.2
working thus, i crunched the numbers to get the following equation:
x=24a(50/b)
where:
x= ideal runner length
a= 720-cam intake duration
b= target rpm
All good so far??? pracically like the 84k/rpm=lnth but with cam variable..
question 1: ^^^assumes we want the wave to arrive at valve opening.. do we want it there right when the valve opens or do we want the wave to arrive close to the end of valve duration to push in that last bit or at max lift for the highest gain???
question 2: what duration numbers should be used?? typical listed duration value of @0.050in seems quite "open" still.. my gut tells me to use duration @0.006in, at which point the valve is much more "closed"
any input is appreciated. and sorry again for the inches, feet, and F*
I'm new, so nice to meet you! (I hope) and sorry in advance for using the English units of measure
So here it is- we are all familiar with the eq. 84000/rpm=length, but I wanted something more accurate so I did alittle brainstorming and please let me know if I'm off my rocker here...
now, the theory is the valve closing causes a shockwave that reflects up and down the runner at the speed of sound and ideally the valve is just opening when the wave arrives.
speed of sound at 130*F = around 1200fps
crankshaft rotation per full cycle= 720 degrees
example cam intake duration: 250 degrees
example desired torque rpm: 3,333 (for easy calculations)
the amount of time the wave has to travel is the time between the valve's closing in one cycle and it's opening in the next cycle, or "total degrees minus duration" --> 720-250=470 for example
and at 3,333 rpm each degree takes 0.00005sec
thus it takes 0.0235sec to rotate the crank 470*
and at 1200fps, sound can travel 28.2ft in 0.0235sec
so the first harmonic wave would be 14.1ft (wave travels up then back down covering 14.1 * 2 = 28.2
working thus, i crunched the numbers to get the following equation:
x=24a(50/b)
where:
x= ideal runner length
a= 720-cam intake duration
b= target rpm
All good so far??? pracically like the 84k/rpm=lnth but with cam variable..
question 1: ^^^assumes we want the wave to arrive at valve opening.. do we want it there right when the valve opens or do we want the wave to arrive close to the end of valve duration to push in that last bit or at max lift for the highest gain???
question 2: what duration numbers should be used?? typical listed duration value of @0.050in seems quite "open" still.. my gut tells me to use duration @0.006in, at which point the valve is much more "closed"
any input is appreciated. and sorry again for the inches, feet, and F*