Epicyclic gear trains have two degrees of freedom – They require two inputs. Epicyclic gear trains can always be solved by the following two relationships.
Relative angular velocity formula:
Regular gear train formula with Arm stationary
Problem #M5: Gear kinematics: The figure shows an epicyclic gear train. The number of teeth on each gear is as follows:
The input is Gear 2 and its speed is 250 rpm clockwise (positive). Gear 6 is fixed. Determine the speed of the arm and the speed of Gear 4. The drawing is not to scale. (Answers: -114.3 and -357.1 rpm)
Kinematics of Automobile Differential
Considering the Right Wheel, Left Wheel, the Ring Gear and the Drive Shaft as shown in the figure,
Problem # M6 : A pair of spur gears are in mesh with a pitch of 6 are in mesh. The pinion has 18 teeth and rotates at 1800 rpm transmitting 0.5 Hp. Both gears have 20-degree pressure angles. The number of teeth on the gear is 36. Determine the radial and tangential forces on the pinion. Answer (4.2 lbs and 11.7 lbs)
Helical gear forces
Problem # M7: A pair of helical gears transmit 15 KW power and the pinion is rotating at 1000 rpm. The helix angle is 0.50 radians and the normal pressure angle is 0.35 radians. The pitch diameter of the pinion is 70 mm and the pitch diameter of the gear is 210 mm. Determine the tangential, radial, and axial forces between the gear teeth. (Answers: 4092, 1702, 2236 Newtons)
These forces are for pinion and act through the tooth midpoint. Forces acting on the gear are the same but act on different directions.
A pair of straight-tooth bevel gears (as shown in the figure above) are in mesh transmitting 35 hp at 1000 rpm (pinion speed). The gear rotates at 400 rpm. The gear system has a pitch of 6 and a 20-degree pressure angle. The face width is 2 inches and the pinion has 36 teeth. Determine the tangential, radial, and axial forces acting on the pinion. Answers (839 lbs, 283 lbs, 113 lbs).
Worm Gear Kinematics
The velocity ratio of a worm gear set is determined by the number of teeth in gear and worm thread (not the ratio of the pitch diameters).
Nw = Number of threads (single thread =1, double thread =2, etc)
The worm’s lead is
The worm’s axial pitch pa is the same as the gear’s PR circular pitch p.
The worm’s lead angle is the same as the gear’s helix angle
Example: For a speed reduction of 30 fold and a double threaded worm, what should be the number of teeth on a matching worm gear.
Ng = (2) (30) = 60 teeth
The geometric relation for finding worm lead angle
Problem #M9: A worm gear reducer is driven by a 1200 rpm motor. The worm has 3 threads and the gear has 45 teeth. The circular pitch of the gear is p=0.5”, the center distance is 4.5 inches, the normal pressure angle is 20 degrees, and face-width of the gear is b=1 inch. Use a coefficient of sliding friction of 0.029. Determine:
Gear and worm diameters, and worm Lead. (7.16, 1.84, 1.5 in)
The force imbalance continues until the slippage limit is reached.
Problem M #10: A 10”-wide flat belt is used with a driving pulley of diameter 16” and a driven pulley of rim diameter 36” in an open configuration. The center distance between the two pulleys is 15 feet. The friction coefficient between the belt and the pulley is 0.80. The belt speed is required to be 3600 ft/min. The belts are initially tensioned to 544 lbs. Determine the following. (answers are in parentheses)
Belt engagement angle on the smaller pulley (3.03 radians).
Force in belt in the tight side just before slippage. (1000 lbs).