Then the CFM needs to be 421.14. If not this will put the blower forcing air through a restriction. "Dirty" air will be a result.

Chris

zt260 110 octane 13 lbs of boost WOW!!!!!

AFR claims they flow at 425CFM depending on lift. Looks like we are in the ballpark.

zt , good luck with it , will be waiting for yourfinal report .......mine is done and being dressed with marine parts this minute . haven't seen final dyno numbers or $$$$$ amount yet ......bob !!!

Fortunately the cam will be chose last. So what cam specs are we talking about? Recommendations anyone???

To make the 1200HP you want, the engine will need to 120% VE at 6500. Cam will be in the .900 lift range and duration in the 290 range @.050. at .400" it will still need to be in the 130 range. Seat duration will be 344.

Chris

He's using a blower and will need to make about 650HP N/A and then use 13 pounds boost will be close to his goal. He doesn't need that big of a cam or rpm.

Marty,
Converting boost to Barometric pressure I then calculated the camshaft. The "throat" area of the intake valve lap will be a restriction, therefore the camshaft will have to over compensate.

Chris

When you double atmospheric pressure ie:14.7 lbs of boost, you @double hp, subtracting out some effeciency factors. He ought to be able to easily make 1200 hp with 13 lbs of boost with a motor that makes 650 NA.

I run 13 lbs of boost with a Procharger and use Canfield 355's, In. flow at .700 is 397, with a .655 lift cam, LSA 114.

Hey guys, it sure is fun jumping on the supercharger threads isn't it?

That engine should easily make 650 HP naturally aspirated. To know how much boost you need, you have to know what the intake air temperature will be after the intercooler, which means you have to know the cooling effectiveness of the core. I don't know the cooling effectiveness of the core in this case, but I can guarantee you will need more than 13 psi to make 1200 HP.

How do I know? Let's assume that the intercooler is 100% effective and the supercharged air is the same temperature as ambient air (impossible but bear with me). Let's also assume that it costs no HP to spin the twin blowers (also impossible).

The density ratio would then be:

(14.7 + 13)/14.7 = 1.88

The new HP level would be:

650 X 1.88 = 1222 HP

That's the best theoretical (and impossible) HP you could get. But since the intake air temperature will be hotter than ambient air, the true air density ratio might only be 1.75. And you have to deduct the power taken off the crank to spin the blowers. So the actual HP might be:

(650 X 1.75) - 40 = 1100 HP

As I said, we don't know the cooling effectiveness, but my guess is you would need 14.7 psi and 90% cooling effectiveness to reach 1200 HP with a 650 HP base engine. You can't beat the Ideal gas law!