"Axle ratio" is the common term for the reduction in the gear ratio of most implementations of mechanical differential drive axles in cars, trucks and tractors. These drive axle mechanisms receive driving torque from a driveshaft or transmission output, multiply it by the axle ratio, and transmit the final driving force to the axle shafts, and thus to the drive wheels. The higher the numerical ratio, the higher the output torque, but the lower the speed, in an inverse proportion. If torque doubles, the speed will be half of the driveshaft parameters.

- Skill level:
- Easy

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## Instructions

- 1
Define the tractor axle application: In this case, a payloader tractor is used to jump train tracks in order to move railroad freight cars into place for unloading. It has been determined that this rail car movement is most efficient and manageable at about 3mph. If the tractor has 6-foot-high tires, a diesel engine with a maximum 200-horsepower (hp) rating at 1800 RPM (revolutions-per-minute), and a 12 : 1 reduction transmission, you can calculate the axle ratio that will deliver the most power to the wheels at the 3-mph operating speed.

- 2
Calculate the diesel engine's torque at its 1800 RPM peak power speed. Using the motor equation, Torque = hp X 5252/RPM = 200hp X 5252/1800 RPM = 583.56 foot-pounds.

- 3
Calculate the drive axle speed (in RPM) necessary to achieve 3-mph. If the drive wheel is 6-feet high, then the circumference = 6 feet X pi = 6 X 3.1416 = 18.85 feet; 3mph = 3mph X 5280 feet/mile/60 min/hr = 264 feet per minute; if each turn of the wheel provides 18.85 feet of movement, then the wheel needs to turn 264 feet/min/18.85 feet/revolution = 14 RPM axle speed to provide the 3-mph speed.

- 4
Calculate the final axle drive ratio needed to deliver the 3-mph speed: If the engine speed is 1800 RPM at peak power output, then the transmission output is 1800 RPM/12; the reduction transmission ratio = 150 RPM driveshaft speed; the axle ratio = 150 RPM driveshaft speed/14 RPM axle (wheel) speed or 10.714 : 1. Axle torque will be 583.56 X 12 X 10.714 = 75,027 foot-pounds, and 75,027 foot-pounds/3-foot wheel radius = 11344 Kilogram drive thrust available to push heavy freight cars.

- 1
Define the rally car operational scenario: An open class rally car has a 4-speed transmission with a final 1 : 1 output ratio. The engine reaches its peak power at 5800 RPM and has a redline of 6600 RPM. If it is statistically determined that the car spends most of its race time centred at 89mph in top gear, and tire diameter is 23 inches, you can calculate the optimal axle ratio to keep the car racing at maximum power most of the time.

- 2
Determine the axle RPM at the car's 89mph optimal range with the 23-inch diameter tires: Tire circumference = 23 X pi = 23 X 3.1416 = 72.26 inches/12 inches/foot = 6.021 feet; 89mph X 5280/3600 seconds/hr = 130.53 feet/second; 130.53-feet/sec speed/6.021 feet per revolution = 21.68 revolutions per second.

- 3
Calculate the axle ratio that allows the engine to operate at its peak power speed of 5800 RPM at 89mph: 5800 RPM/60 seconds/minute = 96.67 revolutions/second. If the engine requires 96.67 revolutions to turn the rear wheels 21.68 revolutions, then the axle ratio = 96.67/21.68 = 4.46 : 1.

#### Tips and warnings

- Always select axle ratios that allow peak power to be delivered at the intended operational speed.
- Automotive engines that continually operate at maximum power output wear out or blow up quickly, while heavy duty diesel engines are intended to operate at maximum output continually.