Some Equations That May Help

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

Note: Tire Radius is distance, in inches, from center of tire to ground. 
Note: Gear Ratio is Rear Axle ratio
Note: Transmission Gear Ratio is the final drive ratio.
Note: All the calculations below are using a TH700R4 auto as an example.

For a list possible transmission gear ratios check here.

Example: What will the MPH be at 2200 RPM with a 3.73 rear axle and 14 inch radius tire (28 inch diameter) in 4th (OD) (.70:1) transmission gear? 

               MPH = 14 ÷ 168 x 2200 ÷ 3.73 ÷ .70 = 70 MPH

Example: in drive (3rd gear) (1:1)? 

               MPH = 14 ÷ 168 x 2200 ÷ 3.73 ÷ 1 = 49 MPH
 

2) RPM = 168 x GEAR RATIO x TRANSMISSION GEAR RATIO x MPH ÷ TIRE RADIUS

Note: I used 70 MPH in the example below.  You can substitute any MPH you would like to use.

Example: What will be the RPM at any given MPH?

               RPM = 168 x 3.73 x .70 x 70 ÷ 14 = 2193 RPM
 

3) GEAR RATIO = (RPM x TIRE RADIUS) ÷ (MPH x 168)

Note: I used my rear gear of 3.73 with a 700R4.  3.73 x .70 = 261 final drive.

Example: What will be the GEAR RATIO at 70 MPH and 2200 RPM with a .70 OD tranny?

               GEAR RATIO = (2200 x 14) ÷ (70 x 168) = 261

NOTE:  In the three equations above, you can use your TIRE DIAMETER if you wish.  Then you would need to substitute the diameter for all TIRE RADIUS numbers.  Also you will need to double the 168 to get 336.  I just used radius (half the diameter) in my equations.  Either way you get the same answer.
 

4) TIRE CHANGE (find new effective gear ratio) = (OLD TIRE DIA. ÷ NEW TIRE DIA.) x GEAR RATIO = EFFECTIVE GEAR RATIO

Example: What would be your new effective rear gear ratio with a tire size change to 27" from 28" with 3.73 rear gears?

               (28 ÷ 27) = 1.037
               1.037 x 3.73 = 3.868 or 3.87
 

5) TIRE DIAMETER = ((SECTION WIDTH x ASPECT RATIO x 2) ÷ 25.4) + RIM DIAMETER = TIRE DIAMETER IN INCHES

Example: What would the diameter of a 275/60R15 tire be?

               Section width = 275 (275 millimeters).  This is how wide the tire is from sidewall to sidewall, not tread width.
               Aspect ratio = 60 (60% of Section width).  This is how tall the sidewall is.

               275 x .60 = 165 millimeters (Sidewall height) (Sidewall height in inches would be: 165 ÷ 25.4 = 6.496 or 6.5 inches)
               165 x 2 = 330 millimeters (Both sidewall added together)
               330 ÷ 25.4 = 12.99 inches (To convert millimeters to inches, divide by 25.4)
               12.99 + 15 = 27.99 or 28 inches in diameter.
 

6) 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 (350 w/.030 overbore)? 

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

7) 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 ccs 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 #6 above, if heads have 64 cc chamber, head gasket is compressed
to 0.040 inch and flat top pistons give 0.025 deck clearance at TDC? 

     CCV = 64 ÷ 16.4 = 3.90 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) ÷ (3.90 + 0.51+ 0.32) = 1+ (44.39 ÷ 4.73) = 10.38 CR
 

8) 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
 

9) Horsepower = TORQUE x RPM ÷ 5252
 

10) 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.

 

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