Unformatted text preview: e speed of the
gear A relative to the arm C
= NA – N C
and speed of the gear B relative to the arm C,
= NB – N C
Since the gears A and B are meshing directly, therefore they will revolve in opposite directions.
∴ NB – NC
NA – NC
TB Since the arm C is fixed, therefore its speed, N C = 0.
TB If the gear A is fixed, then N A = 0. N B – NC
0 – NC
TB or NB
TB Note : The tabular method is easier and hence mostly used in solving problems on epicyclic gear train. Example 13.4. In an epicyclic gear train, an arm carries
two gears A and B having 36 and 45 teeth respectively. If the arm
rotates at 150 r.p.m. in the anticlockwise direction about the centre
of the gear A which is fixed, determine the speed of gear B. If the
gear A instead of being fixed, makes 300 r.p.m. in the clockwise
direction, what will be the speed of gear B ?
Solution. Given : T A = 36 ; T B = 45 ; N C = 150 r.p.m.
The gear train is shown in Fig. 13.7. Fig. 13.7 Chapter 13 : Gear Trains l 439 We shall solve this example, first by tabular method and then by algebraic method.
1. Tabular method
First of all prepare the table of motions as given below :
Table 13.2. Table of motions.
Revolutions of elements
Step No. Conditions of motion Arm C Gear A Gear B 1. Arm fixed-gear A r otates through + 1
revolution (i.e. 1 rev. anticlockwise) 0 +1 – TA
TB 2. Arm fixed-gear A r otates through + x
revolutions 0 +x – x× TA
TB 3. Add + y revolutions to all elements +y +y +y 4. Total motion +y x +y y –x× TA
TB Speed of gear B when gear A is fixed
Since the speed of arm is 150 r.p.m. anticlockwise, therefore from the fourth row of the table,
y = + 150 r.p.m.
Also the gear A is fixed, therefore
x+y=0 or ∴ Speed of gear B , NB = y – x × x = – y = – 150 r.p.m. 36
= 150 + 150 ×
= + 270 r.p.m.
TB = 270 r.p.m. (anticlockwise) Ans.
Speed of gear B when gear A makes 300 r.p.m. clockwise
Since the gear A makes 300 r.p.m.clockwise, therefor...
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- Fall '08
- gear, Gear Trains, Epicyclic gearing, epicyclic gear train