The 1966/67 Dodge Charger Club Home Page. The greatest of all Mopars. The original location on the Internet for the 1966/67 Dodge Charger.

MAIN HOME PAGE > Technical Info TOC > Mathematical Formulas Horsepower/torque equation Liter-to-cubic inch conversion Carburetor Needs MPH to RPM Relationships Finding MPH Finding RPM Finding Gear Ratio Finding Tire Radius Horsepower to Torque Cubic Inch Displacement/Volumetric Efficiency Horsepower/Torque Equation: Bhp = T × RPM/k Where Bhp is Brake Horsepower T is Torque in foot/pounds RPM is engine speed in rounds per minute K is a constant Click Here for the Main Table of Contents Liter to Cubic Inch Conversion 1 Liter = 61.002 c.i. Cubic Inches to Liters: 1 c.i. = 0.01639 liter Relationship between Miles per Hour, (MPH) and Revolutions per Minute, (RPM) Finding it First find the vehicle speed, MPH and the consequent engine RPM operating range: Finding MPH 1) MPH = TIRE RADIUS ÷ 168 × 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 is the MPH at 6,500 RPM with a 4.90 rear axle and a 14 inch radius tire in 4th gear, (1:1)? 14.0 ÷ 168.0 × 6,500.0 ÷ 4.90 ÷ 1.0 _________________ 111 MPH Example: What is the MPH at 6500 RPM with a 4.90 rear axle and a 14 inch radius tire in 3rd gear, (1.34:1)? 14.0 ÷ 168.0 × 6,500.0 ÷ 4.90 ÷ 1.34 _________________ 83 MPH Click Here for the Main Table of Contents Back to the top of this page Finding RPM 2 RPM = 168 × Gear Ratio × MPH ÷ Tire Radius Example: Using the above example, what would the RPM be after shifting from 3rd to 4th gear at 83 MPH? 168.0 × 4.90 × 83.0 ÷ 14.0 ______________ 4,880 RPM Click Here for the Main Table of Contents Back to the top of this page Finding Gear Ratio Gear Ratio = Tire Radius × RPM ÷ 168 ÷ MPH Example: Again, using the above example, what gear ratio is required for 120 MPH at 6,500 RPM? Gear Ratio = 14.0 × 6,500.0 ÷ 168.0 ÷ 120.0 ____________ 4.51 Click Here for the Main Table of Contents Back to the top of this page Finding Tire Radius 4) Tire Radius = 168 × MPH × Gear Ratio ÷ RPM Example: Once again using the above example, what is the tire radius for 110 MPH at 6,500 RPM with a 4.11 gear? 168.0 × 110.0 (MPH) × 4.11 (Gear Ratio) ÷ 6,500.0 (RPM) ___________________________ 11.685 (11.7)inches Note: This would be approximately a 23" diameter tire. Speed should also be taken into account since radius will change somewhat due to centrifugal force. The faster the speed, the larger the tire's diameter will be. Click Here for the Main Table of Contents Back to the top of this page WHAT HP & TORQUE is needed. Three mathematical formulas are given below in determining horsepower needs: Engine horsepower required to reach MPH in quarter mile (HPq): HPq = (0.00426 × MPH) × (0.00426 × MPH) × (0.00426 × MPH) × WEIGHT Note: understates HP required at speeds exceeding 100 MPH. Note: assumes engine HP must be 2 × the HP required at drive wheels. Example: What engine HP is required to achieve 110 MPH in a 3,200 pound vehicle in a 1/4 mile? HPq = (0.00426 × 110) × (0.00426 × 110) × (0.00426 × 110) × 3,200 _____________________________ 329 engineHP Click Here for the Main Table of Contents Back to the top of this page Engine horsepower required to sustain MPH on level ground (HPs): HPs = (MPH ÷ 3) + (WEIGHT ÷ 1,000 × MPH ÷ 10) Note: assumes engine HP must be 2 × the HP required at drive wheels. Example: What engine HP is required to sustain 75 MPH in a 3,600 pound vehicle? HPs = 75 ÷ 3 + (3,600 ÷ 1,000 × 75 ÷ 10) _____________________________ 25 + (3.6 × 7.5) _______________________________________ 52 engine HP Click Here for the Main Table of Contents Back to the top of this page Engine horsepower required to sustain MPH up a grade of G% (HPg): HPg = HPs + (G ÷ 100 × 0.005 × WEIGHT × MPH) Note: Assumes engine HP must be 2 × HP required at drive wheels, Example: What HP to sustain 75 MPH up a 6 % grade in a 3,600 pound vehicle? HPg = HPs + (6 ÷ 100 × 0.005 x3600 × 75) _________________________________ HPs + 81 _________________________________ (3,600 ÷ 10,000 + 0.33) × 75 + 81 _________________________________ 52 + 81 _________________________________________ 133 engine HP Click Here for the Main Table of Contents Back to the top of this page a) Horsepower = TORQUE × RPM ÷ 5,252 Torque = HP × 5,252 ÷ RPM Back to the basics statement: Horsepower comes from torque. Torque comes from the pressure of combustion in the cylinder when 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. Torque is measured in Foot/Pounds. Foot/Pounds meaning the force of 1 pound exerted on a lever 1 foot in length as measuered from a pivot point. Example: What torque is required to generate 329 HP at 6,000 RPM? Torque (T) = 329.0 × 5,252.0 ÷ 6,000.0 ______________________ 288.0 foot pounds @ 6,000 RPM Example: What torque is required for 296 HP at 4880 RPM? T = 296.0 × 5,252.0 ÷ 4,880.0 ____________ 319.0 Foot/Pounds @ 4,880 RPM Click Here for the Main Table of Contents Back to the top of this page About Cubic Inch Displacement, Volumetric Efficiency, Combustion Efficiency and CFM: 1) CID = NUMBER OF CYLINDERS × SWEPT VOLUME CID= N × 0.7854 × bore × bore × stroke (all in inches) Example: What is CID of a V8 with a "30 over", 4 inch bore and 3.48 inch stroke? CID = 8.0 × 0.7854 × 4.030 × 4.030 × 3.48 _________________________________ 355.0 Cubic Inch Displacement Click Here for the Main Table of Contents Back to the top of this page VOLUMETRIC EFFICIENCY (VE)= 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. Note: VE above 100% is a supercharging effect and more torque is produced. 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 3) CFM = CUBIC FEET PER MINUTE A measure of air flow into and out of an engine CFM = CID × RPM × VE ÷ 3464 __________ Cubic Feet p/min Example: What CFM is consumed by a 355 CID engine at 4,478 RPM if VE = 105%, (1.05)? 355.0 (CID) × 4,478.0 (RPM) × 1.05 (% VE) ÷ 3,464.0 ______________ 482.0 CFM Example: What CFM by the same engine at 6,400 RPM if VE has fallen to 95% (0.95)? 355.0 (CID) × 6,400.0 (RPM) × 0.95 (% VE) ÷ 3,464.0 _____________ 623.0 CFM Click Here for the Main Table of Contents Back to the top of this page About Compression Ratio, Cubic Inches and Horsepower: 4) CR = COMPRESSION RATIO = CYL. VOLUME @ BDC ÷ CYLINDER VOLUME @ TDC = 1 + (SWEPT VOLUME ÷ VOL @ TDC) = 1 + (0.7854 × BORE × BORE × STROKE) ÷ (CCV + HGV + PDV) Combustion Chamber Volume (CCV) in cubic inches Note: if volume is given in cc’s then divide by 16.4 to get cubic inches. Head Gasket Volume (HGV) in cubic inches = Head gasket compressed × thickness × 0.7854 × bore × bore Piston Deck Volume (PDV) + Piston Dome Effective Volume ___________________________________ (0.7854 × bore × bore × deck to piston distance) + (volume of piston depressions - volume of piston bumps) Example: What is CR of the engine 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 = 72.0 ÷ 16.4 __________________________ 4.39 cubic inches HGV = 0.040 × 0.7854 × 4.030 × 4.030 ___________________ 0.51 CID PDV = 0.025 × 0.7854 × 4.030 × 4.030 + 0.0 - 0.0 ___________________ 0.32 CID CR = 1 + (0.7854 × 4.030 × 4.030 × 3.48) ÷ (4.39 + 0.51+ 0.32) __________________________________ 1 + (44.39 ÷ 5.22) __________________________________ 9.5 Compression Ratio HP = Atmospheric Pressure × CR × VE × CID × RPM ÷ 5,252.0 ÷ 150.8 Example #1: What HP from a 350 CID Torker-level engine @ 6,000 RPM at sea level? 14.7 × 9.5 × 0.95 × 350.0 × 6,000.0 ÷ 5,252.0 ÷ 150.8 _________________ 352 HP Note: Torker VE is typically 100% at Torque peak RPM but 95% at HP peak RPM. Example #2: Effect of a carburetor restrictor plate that causes 1.5 PSI additional manifold vacuum? 14.7 × 9.5 × 0.9 × 350.0 × 6,000.0 ÷ 5,252.0 ÷ 150.8 _______________ 336 HP CID = HP × 5,252.0 × 150.8 ÷ Atmos. Press. ÷ CR ÷ VE ÷ RPM Example #3: What CID is required for 352 HP from a Performer-level engine? 352.0 × 5,252.0 × 150.8 ÷ 14.7 ÷ 8.5 ÷ 0.85 ÷ 5,000.0 _____________ 525.0 CID Note: Performer is 8.5 CR & HP peak is 5,000 RPM @ 85% VE Click Here for the Main Table of Contents Back to the top of this page E-mail address: chargers@3me.com [ HOME | Members' Chargers | Join the ICQ List | Specifications | Chargers 4 Sale | Technical Info | Guestbook ]   Web www.dodge-charger.org Garden Decor We carry a large selection of unique and exquisite garden decor at discount and wholesale prices. You'll find water fountains, garden sculptures, garden furniture, birdhouses, bird feeders, plant stands, wind chimes and a host of garden accents. [ HOME | History | Tech Info | How To | Your Photos | Communications | Products | Services ]

View our Stats Last updated:

All text, images (except for clients' logos throughout site) and forms on all pages of this website ©1995-2010 3rd Millennium Encoding

Ralph M Bohm P.O. Box 500 Bucksport, Maine USA 04416 cell Ph: 207-745-5000