(A) Edelbrock Performer RPM aluminum heads or other aftermarket heads of similar design.
(B) Edelbrock Victor Jr. aluminum heads or other aftermarket heads of similar design.

Guidelines For Selecting The Edelbrock Total Power Package:
The following equations and rules apply only to four-cycle engines powered by pump gasoline. The equations have been simplified for ease of understanding. Answers will be approximate but generally will be close enough for use as a guideline.

About Mile per Hour and Revolutions per Minute:
First find the vehicle speed, MPH and the consequent engine RPM operating range:
1) MPH = TIRE RADIUS ?? 168 X ENGINE RPM ?? GEAR RATIO

Note: Tire Radius is distance, in inches, from center of tire to ground.
Note: Gear Ratio is Rear Axle ratio multiplied by Transmission Gear ratio.
Example: What MPH at 6500 RPM with a 4.9 rear axle and 14 inch radius tire in 4th (1:1) gear?

MPH = 14 ?? 168 x 6500 ?? 4.90 ?? 1 = 111 MPH
Example: in 3rd gear (1.34)?

MPH = 14 ?? 168 x 6500 ?? 4.90 ?? 1.34 = 83 MPH

2) RPM = 168 x GEAR RATIO x MPH ?? TIRE RADIUS

Example: For the case in #1, what will be the RPM after shift from 3rd to 4th gear at 83 MPH?

RPM = 168 x 4.90 x 83 ?? 14 = 4880 RPM

3) GEAR RATIO = TIRE RADIUS x RPM ?? 168 ?? MPH

Example: For the case in #1, what Gear Ratio is required for 120 MPH at 6500 RPM?

GR = 14 x 6500 ?? 168 ?? 120 = 4.51

4) TIRE RADIUS = 168 x MPH x GEAR RATIO ?? RPM

Example: For the case in #1, what tire radius for 110 MPH but at 6000 RPM with a 4.11 gear?

168 x 110 x 4.11 ?? 6000 = 12.7 inches
Note: Approximately a 25" diameter tire. Remember that the tire radius will be less during hard acceleration than when the vehicle is standing still. Also, radius will be greater at high speed due to tire expansion from centrifugal force.

WHAT HP & TORQUE is needed: Equations #5, #6, and #7 show how to compute the engine horsepower needed for three different applications.
5) Engine horsepower required to reach MPH in quarter mile (HPq):

HPq = (0.00426 x MPH) x (0.00426 x MPH) x (0.00426 x MPH) x WEIGHT

Note: understates HP required at speeds exceeding 100 MPH.
Note: assumes engine HP must be 2 x the HP required at drive wheels.
Example: What engine HP is required to achieve 110 MPH in a 3200 pound vehicle in 1/4 mile?

HPq = (0.00426 x 110) x (0.00426 x 110) x (0.00426 x 110) x 3200 = 329 engine HP

6) Engine horsepower required to sustain MPH on level ground (HPs):

HPs = (MPH ?? 3) + (WEIGHT ?? 1,000 x MPH ?? 10)
Note: assumes engine HP must be 2 x the HP required at drive wheels
Example: What engine HP is required to sustain 75 MPH in a 3600 pound vehicle?

HPs = 75 ?? 3 + (3600 ?? 1,000 x 75 ??10) = 25 + (3.6 x 7.5) = 52 engine HP

7) Engine horsepower required to sustain MPH up a grade of G% (HPg):

HPg = HPs + (G ?? 100 x 0.005 x WEIGHT x MPH)
Note: Assumes engine HP must be 2x HP required at drive wheels, calculate HPs with #6.
Example: What HP to sustain 75 MPH up a 6 % grade in a 3600 pound vehicle?

HPg = HPs + (6 ?? 100 x 0.005 x 3600 x 75) = HPs + 81 = (3600 ?? 10,000 + 0.33) x 75 + 81 = 52 + 81 = 133 engine HP

8a) Horsepower = TORQUE x RPM ?? 5252
8b) Torque = HP x 5252 ?? RPM

Horsepower comes from torque. Torque comes from the pressure of combustion in the cylinder because combustion pressure causes the piston to turn the crankshaft which is measured as torque. The trick is to generate high enough pressure on each stroke and to do it often enough (RPM) to produce the horsepower needed.
Example: What torque is required to generate 329 HP at 6000 RPM?

T = 329 x 5252 ?? 6000 = 288 foot pounds @ 6000 RPM

Example: What torque is required for 296 HP at 4880 RPM?

T = 296 x 5252 ?? 4880 = 319 foot pounds @ 4880

About Cubic Inches, Volumetric Efficiency, Combustion Efficiency and CFM:
9) CID = NUMBER OF CYLINDERS x SWEPT VOLUME

CID = N x 0.7854 x bore x bore x stroke (all in inches)
Example: What is CID of a V8 with a "30 over", 4 inch bore and 3.48 inch stroke?

CID = 8 x 0.7854 x 4.030 x 4.030 x 3.48 = 355 cu. inches

10) VE = VOLUMETRIC EFFICIENCY = Engine Actual Air Intake ?? CID:

If VE is less than 1 (or 100%) the amount and quality of charge in the cylinder is reduced so less torque is produced. VE above 100% is a supercharging effect and more torque is produced.
Power Level: Stock Performer Torker II Perf. RPM Victor Jr. Victor

Peak VE% 60-80 75-90 90-100 95-105 105-115 110-122

11) CE = COMBUSTION EFFICIENCY = How well the energy in the fuel is converted into crankshaft torque. Affected by; air/fuel ratio, ignition timing, charge mixing and other factors.

 Condition      Best Power    Best Economy    Lean Misfire



 Air/Fuel Ratio   12-12.5      14.5-15.5           17

12) CFM = CUBIC FEET PER MINUTE
A measure of air flow into and out of an engine (CFM = CID x RPM x VE ?? 3464).
Example: What CFM is consumed by a 355 CID engine at 4478 RPM if VE = 105% (1.05)?

CFM = 355 x 4478 x 1.05 ?? 3464 = 482 CFM

Example: What CFM by the same engine at 6400 RPM if VE has fallen to 95% (0.9)?

CFM = 355 x 6400 x 0.95 ?? 3464 = 623 CFM

About Compression Ratio, Cubic Inches and Horsepower:
13) CR = COMPRESSION RATIO = CYL. VOLUME @ BDC ?? CYLINDER VOLUME @ TDC

= 1 + (SWEPT VOLUME ?? VOL @ TDC)
= 1+ (0.7854 x BORE x BORE x STROKE) ?? (CCV + HGV + PDV)
CCV = Combustion Chamber Volume, in cubic inches
Note: if volume is given in cc‚Äôs then ?? 16.4 to get cubic inches.
HGV = Head Gasket Volume, in cubic inches,
= Head gasket compressed thickness x 0.7854 x bore x bore
PDV = (Piston Deck Volume) + (Piston Dome Effective Volume)
= (0.7854 x bore x bore x deck to piston distance) + (volume of piston depressions - volume of piston bumps)
Example: What is CR of the engine in #9 if heads have 72 cc chamber, head gasket is compressed to 0.040 inch and flat top pistons give 0.025 deck clearance at TDC?
CCV = 72 ?? 16.4 = 4.39 cubic inches
HGV = 0.040 x 0.7854 x 4.030 x 4.030 = 0.51 c.i.
PDV = 0.025 x 0.7854 x 4.030 x 4.030+ 0- 0 = 0.32 c.i.
CR = 1+ (0.7854 x 4.030 x 4.030 x 3.48 ?? (4.39 + 0.51+ 0.32) = 1+ (44.39 ?? 5.22) = 9.5 CR

14) HP = Atmos. Press. x CR x VE x CID x RPM ?? 5252 ?? 150.8

Example: What HP from a 350 CID Torker-level engine @ 6000 RPM at sea level?

HP = 14.7 x 9.5 x 0.95 x 350 x 6000 ?? 5252 ?? 150.8 = 352 HP
Note: Torker VE is typically 100% at Torque peak RPM but 95% at HP peak RPM.
Example: Effect of a carb restrictor plate that causes 1.5 PSI additional manifold vacuum?

HP = 14.7 x 9.5 x 0.9 x 350 x 6000 ?? 5252 ?? 150.8 = 336 HP

15) CID = HP x 5252 x 150.8 ?? Atmos. Press. ?? CR ?? VE ?? RPM

Example: What CID required for 352 HP from a Performer-level engine?

CID = 352 x 5252 x 150.8 ?? 14.7 ?? 8.5 ?? 0.85 ?? 5000 = 525 c.i.
Note: Performer is 8.5 CR & HP peak is 5000 RPM @ 85% VE

CFM RULES...
CFM and Carburetors:
Carburetors are rated by CFM (cubic feet per minute) capacity. 4V carburetors are rated at 1.5 inches (Hg) of pressure drop (manifold vacuum) and 2V carburetors at 3 inches (Hg). Rule: For maximum performance, select a carburetor that is rated higher than the engine CFM requirement. Use 110% to 130% higher on single-plane manifolds . Example: If the engine needs 590 CFM, select a carburetor rated in the range of 650 to 770 CFM for a single-plane manifold. A 750 would be right. An 850 probably would cause driveability problems at lower RPM. A 1050 probably would cause actual loss of HP below 4500 RPM. For dual-plane manifolds use 120% to 150 % higher.

CFM and Manifolds:
Manifolds must be sized to match the application. Because manifolds are made for specific engines, select manifolds based on the RPM range.

CFM and Camshafts:
With the proper carburetor and manifold it is possible to select a cam that loses 5% to15% of the potential HP. These losses come from the wrong lift and duration which try to create air flow that does not match the air flow characteristics of the carburetor, manifold, head and exhaust so volumetric efficiency is reduced. An increase in camshaft lobe duration of 10 degrees will move the HP peak up 500 RPM but watch out, it may lose too much HP at lower RPM.

CFM and Cylinder Heads:
Usually, cylinder heads are the limiting component in the whole air flow chain. That is why installing only a large carburetor or a long cam in a stock engine does not work. When it is not possible to replace the cylinder heads because of cost, a better matching carburetor, manifold, cam and exhaust can increase HP of most stock engines by 10 to 15 points. To break 100% Volumetric Efficiency, however, better cylinder heads or OEM ‚ÄúHO‚Äù level engines are usually needed.

CFM and Exhaust:
An engine must exhaust burned gases before it can intake the next fresh charge. Cast iron, log style manifolds hamper the exhaust process. Tube style exhaust systems are preferred. But headers are often too big; especially for Performer and Performer RPM levels. Improving an engine‚Äôs Volumetric Efficiency depends on high exhaust gas velocity to scavenge the cylinder but this will not happen if the exhaust valve dumps into a big header pipe. On the newer computer controlled vehicles it is also important to ensure that all emissions control devices, and especially the O2 sensor, still work as intended.

CFM and Engine Control:
Spark timing must be matched to Volumetric Efficiency because VE indicates the quantity and quality of charge in each cylinder on each stroke of the engine. Different engine families require distinctly different spark advance profiles. And even engines of equal CID but different CR require their own unique spark advance profiles. Rule: Expect 0.1% to 0.5% loss in Torque for each 1 degree error in spark timing advanced or retarded from best timing. Also, detonation will occur with spark advanced only 3 degrees to 5 degrees over best timing and detonation will cause 1% to 10% torque loss, immediately, and engine damage if allowed to persist.