Conversions I wrote this short python program Convert mi/hr to RPM given diameter of wheel in inches Enter mi/hr Here Enter Diameter of wheel in inches Hello World i am an ai Robot.! Result-01 = RPM 672.2710474632273 = rb8 = Answer 60 = rb6 30 = rb7 Where is the wizard? He is sitting by the pool.! = rb2 ************ 1 ft = 30.48 cm 1 lb = .453592 kgs => (30.48 cm)*(.453592 kg) = 13.825495454615 kg-cm ********* 1 ft = 12 inches 1 lb = 16 ounces 1 ft-lb = (12 inches)*(16 ounces) = 192 in-oz *********** This is David's python program Convert Ft-Lbs to kg-cm Enter ft-lbs here Enter not used input Result-01 = kg-cm Result-02 = not used Hello World i am an ai Robot.! = rb4 = Answer 672.2710474632273 = rb8 = Answer Where is the wizard? He is sitting by the pool.! = rb2 ************ Circumference of wheel = pi*D D = Diameter of wheel in inches => 1 Revolution = (pi*D)/12 feet/rev 1 minute = 60 seconds => rev/sec => RPM/60 ft/sec = (pi*D)/12)*(RPM/60) => (rev/sec)*(ft/rev) 88 ft/sec = 60 mi/hr => to convert ft/sec to mi/hr multiply by (60/88) => RPM to Mi/hr = (pi*D*RPM)/88 ********* Convert Miles per Hour to Ft/sec and RPM Enter Miles per Hour Here Enter Diameter of Wheel Here in Inches Result-01 = Ft/sec Result-02 = RPM ************ Circumference of wheel = pi*D D = Diameter of wheel in inches => 1 Revolution = (pi*D)/12 feet/rev 1 minute = 60 seconds => rev/sec => RPM/60 ft/sec = (pi*D)/12)*(RPM/60) => (rev/sec)*(ft/rev) = ft/sec RPM = ((ft/sec)/(pi*D))*12*60 88 ft/sec = 60 mi/hr => to convert ft/sec to mi/hr multiply by (60/88) => RPM to Mi/hr = (pi*D*RPM)/88 ********* Electric Motors => Torque = (HP*5250)/RPM Enter Horse Power of Motor Here Enter RPM Here Enter Diameter of Pulley in inches Here Result-01 = Torque Result-02 = Lbs it can lift with the pulley ************ Torque = (HP*5250)/RPM = ft-lbs Three inch pulley => radius = 1.5 inches = 1.5/12 foot If the Torque = 1.52 ft-lbs which is standard for a half horse power motor, Then the motor can lift 1.52/(1.5/12) = 12.16 lbs ********* Pulleys => What is the RPM of the second Pulley Enter Diameter in inches of the Pulley on the Motor Enter RPM of the Motor Here Enter Diameter of the second Pulley in inches Here Result-01 = RPM of the second Pulley ************ RPM of the motor = 1725 Diameter of motor Pulley = 3 inches 3 inch Pulley = 3*pi = distance = 9.42477 inches = 1 rev Second Pulley diameter = 4 inches = 4*pi = 12.56636 inches = 1 rev Second Pulley RPM = (3/4)*(RPM of the motor) ********* Speed of car by measuring skid marks. Enter Length of the Skid Marks in feet Enter Friction of rubber against road surface Result-01 = acceleration = -u*g => 2*a*x Result-02 = V = SqRt(2*a*x) = ft/sec Result-02 = V = (SqRt(2*a*x))*(60/88) = mi/hr ************ It turns out that the acceleration depends only on the friction constant used. a = u*g where u = friction constant and g = 32 = gravity. Then V = SqRt(2*a*x) = Ft/Sec where x = length of skid marks. => V = (SqRt(2*a*x))*(60/88) = Mi/Hr ********* ************