Some Helpful Equations

About Mile per Hour and Revolutions per Minute
First find the vehicle speed, MPH and the consequent engine RPM operating range:

Formula for MPH
MPH = TIRE RADIUS ÷ 168 x ENGINE RPM ÷ 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

Note: Tire Radius is distance, in inches, from center of wheel to the top of the tire.

Note: Gear Ratio is Rear Axle ratio divided by Transmission Gear ratio.


Formula for RPM
RPM = 168 x GEAR RATIO x MPH ÷ TIRE RADIUS

Example: Using the first example, 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




Formula for Gear Ratio
GEAR RATIO = TIRE RADIUS x RPM ÷ 168 ÷ MPH

Example: Using the first example, what Gear Ratio is required for 120 MPH at 6500 RPM?

GR = 14 x 6500 ÷ 168 ÷ 120 = 4.51


Formula for Tire Radius
TIRE RADIUS = 168 x MPH x GEAR RATIO ÷ RPM

Example: Using the first example, 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.



Computing HP & Torque

Formula for HPQ
HPq = (0.00426 x MPH)3 x WEIGHT


HPq = Engine horsepower required to reach MPH in quarter mile

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) x3200 = 329 engine HP


Formula for HP and Torque
HP = Torque x rpm ÷ 5252 • Torque = HP x 5252 ÷ RPM


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, Flow Bench Conversion and CFM

Formula for VE
VE = (CFM x 3456) ÷ (CID x RPM)


If VE (volumetric efficiency) 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


Flow Bench Conversion Factor
VE = (CFM x 3456) ÷ (CID x RPM)


Typically flow bench values are given for a pressure drop of 28 in H2O. To convert flow figures from a different pressure drop to 28 in H2O use the formula above.

Example: You have flow figures of 152 cfm at 10 in H2O. What if the same head was flowed at 28 in H2O?

CFM H2O = 152 x √(28÷10) = 254 cfm


Formula for CID (cubic inch displacement)
CID = NUMBER OF CYLINDERS x SWEPT VOLUME


Note: 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


Formula for CFM
CFM = CUBIC FEET PER MINUTE


A measure of air flow into and out of an engine (CFM = CID x RPM x VE ÷ 3456). Example: What CFM is consumed by a 355 CID engine at 4500 RPM if VE = 105% (1.05)?

CFM = 355 x 4500 x 1.05 ÷ 3456 = 485 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 ÷ 3456 = 625 CFM


About Compression Ratio

Formula for CR
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’s 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



About Fuel System Flow

Formula for Injector Size per Hp
LBS/HR = HP ÷ 16


In general if you use BSFC = .50 and an acceptable duration of 100% then use above formula for an eight cylinder engine.

Note: lbs/hr. = ((BSFC ÷ 8 ) x HP) ÷ Peak Inj. Duration

Example: You plan to make 448 hp. What size injector should you use?

lbs/hr = 448 ÷ 16 = 32


Injector Conversion - lbs/hr to cc/min
CC/MIN = LBS/HR X 9.71 OR LBS/HR = CC/MIN X .103


Example: How many cc/min. does a 32 lb/hr. injector flow?

cc/min = 32 x 9.71 = 311 cc/min


Pump Flow Conversion Factors
GPM = lbs/hr ÷ 369.8


Used to convert lbs/hr to gallons per minute (of gasoline).

Note: typically GPM = HP ÷ 740

Example: Your engine uses 221 lbs. per hour of fuel. What are the fuel system requirements?

GPM = 221 ÷ 369.8 = 0.60 gal/min




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 “HO” 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’s 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 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.

Last Modified: 7-13-05 JCH