(^710) A Textbook of Engineering Mechanics
We know that the speed of the wheel A, relative to the arm C
=NA – NC
and speed of the wheel B relative to the arm C
=NB – NC
Since the wheel A and B are directly meshing, therefore they will revolve in opposite
directions.
∴
BC
ACNN
NN- A
B
T
T=Moreover, as the arm C is fixed, therefore its speed (NC) is equal to zero. Therefore
B
AN
N- A
B
T
T=Cor. If the wheel A is fixed, therefore its speed (NA) is equal to zero. ThereforeBC
CNN
N- A
B
T
T=orB
CN
N1 A
BT
T=+Note: The tabular method is comparatively easier, and hence mostly used in solving ex-
amples on epicyclic gear trains.
Example 34.6. In an epicyclic gear train, an arm carries two wheels A and B having 36
and 45 teeth respectively. If the arm C rotates at 150 r.p.m. in the clockwise direction about the
centre of the wheel A which is fixed, determine the speed of wheel B. If the wheel A, instead of being
fixed, makes 300 r.p.m. in the anticlockwise direction, what will be the speed of B?
Solution. Given: No. of teeth on wheel A (TA) = 36; No. of teeth on wheel B (TB) = 45 and
speed of arm C (NC) = 150 r.p.m.
We shall solve this example, first by tabular method and then by algebraic method.
Tabular method
First of all prepare the table of motions as given below :Step No. Conditions of motionsRevolutions of
Arm C Wheel A Wheel B- Arm C fixed; wheel A rotates 0 + 1 – A
B
T
through + 1 revolution T- Arm C fixed; wheel A rotates 0 + x – A
B
xT
T
through + x revolutions- Add + y revolutions to all + y + y + y
elements - Total motion (2 + 3) + yx + y –
A
BT
yx
T