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Miles per Hour vs. Revolutions per Minute

Note: Tire Radius is distance, in inches, from center of tire to ground.
Gear Ratio is Rear Axle ratio multiplied by Transmission Gear ratio.

1) MPH = TIRE RADIUS ÷ 168 X ENGINE RPM ÷ GEAR RATIO

Example: What is the speed in 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 2: How about the speed 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: Now, what will be the RPM after shifting 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: What Gear Ratio is required to turn 6500 RPM at 120 MPH?

GR = 14 x 6500 ÷ 168 ÷ 120 = 4.51

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

Example: What tire radius is required for 110 MPH and 6000 RPM using a 4.11 gear?

168 x 110 x 4.11 ÷ 6000 = 12.7 inches

12.7 x 2 = approx. 25 inches

TIRE SIZE = RADIUS x 2

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.

HP vs. 1/4 mile

5) Engine horsepower required to reach a certain MPH through a 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.
This also 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 through the 1/4 mile?

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

HP vs. MPH

6) The engine horsepower required to sustain a certain 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 is required 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 8a: What torque is required to generate 329 HP at 6000 RPM?

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

Example 8b: What torque is required for 296 HP at 4880 RPM?

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

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 = Actual Engine 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.

Good intake manifolds generally have a peak VE in the range between 85-120.

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

12) CFM = CUBIC FEET PER MINUTE

CFM is a measure of air flow into and out of an engine:

CFM = CID x RPM x VE ÷ 3464

Example 1: 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 2: 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

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

Example 1: How many  HP can a 350 CID engine with VE of 95, running 6000 RPM, at sea level put out?

HP = 14.7 x 9.5 x 0.95 x 350 x 6000 ÷ 5252 ÷ 150.8 = 352 HP

Note: VE is 100% at Torque peak RPM but 95% at HP peak RPM.

Example 2: What is the effect on HP, if a carb restrictor plate that causes 1.5 PSI additional manifold vacuum, is used?

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: Engine used 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 specific application. Because manifolds are made for specific engines and RPM ranges, select manifolds based on the RPM range in which you are trying to make power. There is usually a trade off or comprimise that you must decide on, especially with older (used) intakes. Either you want low to midrange or you want midrange to top end power. But with recent advances in intake manifold technology, you can pretty much have the best of both worlds in one intake, if you chose your new intake wisely.

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 (and usually will) lose too much HP at lower RPM, if you aren't careful. Don't overcam and engine with too little compression or vice-versa. You will end up with a "doggy" engine that doesn't perform well. 

CFM and Cylinder Heads

Usually, cylinder heads are the number one 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 or beef up 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 HP or more depending on the engine. To break 100% Volumetric Efficiency, however, better cylinder heads, work done on heads to increase overall flow, or OEM “High Output” / "High Performance" 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 mild street car 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 too big of a 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. 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 continue running.