Conversions

Convert Ft-Lbs to kg-cm
Enter Ft-Lbs Here



Result-01 = kg-cm
Result-03 = oz-in
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1 ft = 30.48 cm
1 lb = .453592 kgs
=> (30.48 cm)*(.453592 kg) = 13.825495454615 kg-cm
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1 ft = 12 inches
1 lb = 16 ounces
1 ft-lb = (12 inches)*(16 ounces) = 192 in-oz
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Convert RPM to Ft/sec & mi/hr
Enter Diameter of Wheel in inches Here
Enter RPM Here

Result-01 = Ft/sec
Result-02 = Mi/Hr
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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
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Convert mi/hr to ft/sec & RPM
Enter Miles per Hour Here
Enter Diameter of Wheel Here in Inches

Result-01 = Ft/sec
Result-02 = RPM
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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
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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
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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
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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
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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)
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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
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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
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