Solving Gearbox Ratio for 12" Wheel Travel @ 2mph

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SUMMARY

The discussion focuses on calculating the gearbox ratio necessary for a 12-inch wheel to achieve a speed of 2 mph. The correct approach involves determining the wheel's circumference using the formula Pi multiplied by the diameter, resulting in a circumference of approximately 3.14 feet. To achieve 2 mph, the wheel must rotate at approximately 56.02 RPM, which translates to a gearbox output requirement of around 56 RPM. The final formula for calculating the required RPM is derived from converting the speed in mph to revolutions per minute.

PREREQUISITES
  • Understanding of basic physics principles related to rotational motion
  • Familiarity with the mathematical constant Pi (π)
  • Knowledge of speed conversion from miles per hour to feet per hour
  • Basic skills in algebra for manipulating equations
NEXT STEPS
  • Research gearbox design principles and ratios
  • Learn about calculating wheel circumference and its applications
  • Explore speed and RPM conversion techniques in mechanical systems
  • Investigate the impact of different wheel sizes on gearbox requirements
USEFUL FOR

Engineers, mechanical designers, hobbyists working on robotics or vehicle design, and anyone involved in optimizing gearbox ratios for specific speed requirements.

SevenToFive
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I need to figure out a gearbox ratio to get a 12 inch wheel to travel at 2mph. If I remember right the equation should be 1foot * 3.14*gearbox output rpm*60 then all divided by 5280 to get the final mph. Am I on the right path? Then I need a gearbox output around 58rpm.

Thanks to all of you who reply.
 
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SevenToFive said:
I need to figure out a gearbox ratio to get a 12 inch wheel to travel at 2mph. If I remember right the equation should be 1foot * 3.14*gearbox output rpm*60 then all divided by 5280 to get the final mph. Am I on the right path? Then I need a gearbox output around 58rpm.
You're close to the "right path." Just one little adjustment.
 
Do it backwards.
12” diam = 1 ft wheel, at 2mph.
wheel circumference = Pi * diam = Pi feet.
2 mph = 2 * 5280 feet per hour = 10560 ft per hour
revs per hour = 10560 / Pi = 3361.35
revs per min = 10560 / ( 60 * Pi ) = 56.02 RPM
 
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