more pressure in springs?
Thread Starter
TECH Apprentice
Joined: Aug 2010
Posts: 333
Likes: 0
From: Mexico
more pressure in springs? using a supercharger or turbocharger, and according to what cam's specs says about valve springs required, would a super or turbocharger ask for more valve spring pressure? or would be just fine with the ones cam requires?
Thread Starter
TECH Apprentice
Joined: Aug 2010
Posts: 333
Likes: 0
From: Mexico
Here's the workup and formula:
boost psi X the area of the intake valve face in square inches.
Area of a circle is pi 3.1415 X (radius of circle X radius of circle)
Or stated as
A = pi(rXr)
If you are running a 2.05" intake valve, then the radius = 1.025"
So the Area = 3.1415 X (1.025" X 1.025")
or
3.1415 X 1.05 = 3.3 (rounded up a tad)
So if you are running 15 psi of boost, then the extra force on the face of the intake valve is 3.3 X 15 = 49.5 pounds.
Since I like a little headroom to grow, I would, in this example, put a spring that was 75-100 lbs. heavy at seat pressure on this motor.
Jim
boost psi X the area of the intake valve face in square inches.
Area of a circle is pi 3.1415 X (radius of circle X radius of circle)
Or stated as
A = pi(rXr)
If you are running a 2.05" intake valve, then the radius = 1.025"
So the Area = 3.1415 X (1.025" X 1.025")
or
3.1415 X 1.05 = 3.3 (rounded up a tad)
So if you are running 15 psi of boost, then the extra force on the face of the intake valve is 3.3 X 15 = 49.5 pounds.
Since I like a little headroom to grow, I would, in this example, put a spring that was 75-100 lbs. heavy at seat pressure on this motor.
Jim
Thread Starter
TECH Apprentice
Joined: Aug 2010
Posts: 333
Likes: 0
From: Mexico
Here's the workup and formula:
boost psi X the area of the intake valve face in square inches.
Area of a circle is pi 3.1415 X (radius of circle X radius of circle)
Or stated as
A = pi(rXr)
If you are running a 2.05" intake valve, then the radius = 1.025"
So the Area = 3.1415 X (1.025" X 1.025")
or
3.1415 X 1.05 = 3.3 (rounded up a tad)
So if you are running 15 psi of boost, then the extra force on the face of the intake valve is 3.3 X 15 = 49.5 pounds.
Since I like a little headroom to grow, I would, in this example, put a spring that was 75-100 lbs. heavy at seat pressure on this motor.
Jim
boost psi X the area of the intake valve face in square inches.
Area of a circle is pi 3.1415 X (radius of circle X radius of circle)
Or stated as
A = pi(rXr)
If you are running a 2.05" intake valve, then the radius = 1.025"
So the Area = 3.1415 X (1.025" X 1.025")
or
3.1415 X 1.05 = 3.3 (rounded up a tad)
So if you are running 15 psi of boost, then the extra force on the face of the intake valve is 3.3 X 15 = 49.5 pounds.
Since I like a little headroom to grow, I would, in this example, put a spring that was 75-100 lbs. heavy at seat pressure on this motor.
Jim
