Okay, I do need to back up a little and give some credit. This "camshaft" math is what we named Ghetto Math coined by a fellow by the name of Erik K. Sharp guy and it comes from a thread that I was apart of on another board.
Alright. . .now we can proceed. You now need 2 things. A camshaft book for reference and flow numbers off your heads. . . perferablly numbers from your heads but "printed" numbers will work.
We will use a stand lobe that is 230 @.050"
We will use the Canfield 310 head with the following flow numbers:
Lift Int Exh
.300 204 176
.400 266 205
.500 318 237
.600 347 260
Okay we have the following:
(502/2)*5500/1728 = 798.88
798.88/8 cylinders = 99.86 cfm per cylinder
Now lets pick an intake lobe to try. Use the duration @ .50" I will use a cam lobe with 230 @ .050. By dividing the cam duration by the the total degrees of crank rotation of the 4 cycles we get the % of time we have to pack the cylinder for 100% VE
230/720 = .31 (= percent of time to get the air per revolution)
We now use the % of time we have calculated and divide the cylinder CFM requirement by it to come up with the intake port CFM flow that is required to fill our 502 at Max RPM
99.86/.31 = 322 cfm (needed air flow divided by percent of time to get the air in per rev is cfm needed)
So we need 322 cfm and the head at .500" cam lift is 318 CFM. Hey we are right there, but the question is do we run the boat at 5500 rpm all the time?
Okay, I'll stop here for today. Any questions? Please post. Is this helpful? Let me know. No question is to stupid and I will answer to the best of my ability. Blown since you have been good I will give you the equation to convert boost to Barometric pressure: (PSI x 2.035789617) + 29.92
Chris
Where does piston speed come into play?