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.
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
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.
275 x .60 = 165 millimeters (Sidewall height) (Sidewall height in inches
would be: 165 ÷ 25.4 = 6.496 or 6.5 inches)
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)
Example: What is CR of the engine
in #6 above, if heads have 64 cc chamber, head gasket is compressed
CCV = 64
÷ 16.4 = 3.90 cubic inches
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.
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
