Explanation:
Gear Train:
Sometimes, two or more gears are made to mesh with each other to transmit power from one shaft to another. Such a combination is called a gear train.
They can be of many types:
Epicyclic gear train:
Simple Gear Train:
Compound Gear Train:
Reverted Gear train:
The figure shows an epicyclic gear train with gears A and B having 40 and 50 teeth, respectively. If the arm C rotates at 200 rpm in anticlockwise direction about the centre of gear A which is fixed, then speed of revolution of gear B in rpm would be
Concept:
\(\frac{N_1}{N_2}~=~\frac{T_2}{T_1}\)
where N1 = rpm of gear 1, N2 = rpm of gear 2, T1 = Number of teeth in gear 1, T2 = Number of teeth in gear 2
Calculation:
Given:
Gear ‘A’ is stationary and Arm ‘C’ rotates at 200 rpm in an anticlockwise direction about the centre of gear A
Therefore, N_{C} = 200 rpm, NA = 0
Now, to solve such a problem first we follow the following steps
Motion  Arm  A (40) 
B (50) 
Arm is Fixed (Let A rotate with x rpm in CW direction) 
0  +x (CW)  \(\frac{T_A}{T_B}x~=~\frac{40}{50}x~(CCW)\) 
The arm rotates with +y rpm in CW  y  y + x (CW)  \(y~~\frac{4}{5}x\) (CCW) 
Now, by applying all the conditions, we have
A is fixed hence NA = 0
x + y = 0
x =  y (1)
Also, the rpm of Arm C is 200
N_{C} = 100
y = 200
x =  200 rpm
The speed of B is
N_{B} = 200  0.8(200) = 360 rpm
Explanation
Degree of freedom(DOF) – Degree of freedom of plane mechanism is defined as the number of inputs or independent coordinates needed to define the configuration or position of all the links of mechanism with respect to a fixed link.
Kutzback equation of degree of freedom,
DOF = 3 × (L – 1) – 2j – h
Where L = number of links in the mechanism, j = Binary joint or lower pair, h = higher Pair
DOF of the epicyclic gear train
L = 4
j = 3 Between ( 1 and 2, 1 and 4, 3and 4)
h = 1 Between 2 and 3
DOF = 3 × (4 – 1) – 2 × 3 – 1 = 2
Alternate Method
Degree of freedom is the number of inputs required to produce a unique output.
In Epicyclic gear train, there are two inputs required to produce a unique output i.e. one at arm and other at sun gear or at ring gear.
Input 1:
If sun is fixed, then input is given to Ring gear Or If ring gear is fixed, then input is given to sun gear.
Input 2:
Input is given to Arm.
Concept:
Calculation:
Given:
TB = 100; TA = 25
Step No. 
Conditions of motion 
Revolutions of elements 

Arm C 
Gear A 
Gear B 

1.
2.
3.
4. 
Arm fixedgear A rotates though + 1 revolution (i.e. 1 rev. anticlockwise) Arm fixedgear A rotates through + x revolutions Add + y revolutions to all elements
Total motion 
0
0
+y
+y 
+ 1
+x
+y
x + y 
\( \frac{{{T_A}}}{{{T_B}}}\)
\( x \times \frac{{{T_A}}}{{{T_B}}}\)
+y
\(y  x \times \frac{{{T_A}}}{{{T_B}}}\) 
y = 1; NB = 0
\(\begin{array}{l} {N_B} = y  x \times \frac{{{T_A}}}{{{T_B}}} = 1  x \times \frac{{25}}{{100}} = 1  \frac{x}{4} = 0 \Rightarrow x = 4\\ {N_A} = x + y = 4 + 1 = 5 \end{array}\)
With respect to the epicyclic gear train shown in the figure below, A has 75 teeth and B has 25 teeth; A is fixed and arm C makes 5 revolutions:
The number of revolutions made by B is
Concept:
The velocity ratio between two mating gears is given by
\(\frac{N_2}{N_1}=\frac{T_1}{T_2}\)
Where ω is the angular velocity and T is the number of teeth.
Calculation:
The given gear train is an epicyclic gear train
Sr. No. 
Condition of motion 
Revolutions of elements 

Arm C 
Gear A 
Gear B 

1 
Arm fixed, gear A rotates through + 1 (CCW) 
0 
+1 
\(  \frac{{{T_A}}}{{{T_B}}}\) 
2 
Arm fixed, gear A rotates through + x revolutions 
0 
+x 
\(  x \times \frac{{{T_A}}}{{{T_B}}}\) 
3 
Add +y revolutions to all elements 
+y 
+y 
+y 
4 
Total Motion 
+y 
x + y 
\(y  x\frac{{{T_A}}}{{{T_B}}}\) 
Given TA = 75, TB = 25, NC = y = 5
NA = 0 = x + y ⇒ x = y
\({{\rm{N}}_{\rm{B}}}{\rm{\;}} = {\rm{\;y\;}}{\rm{\;x}}\frac{{{T_A}}}{{{T_B}}} = {\rm{y}} + {\rm{y}}\frac{{{{\rm{T}}_{\rm{A}}}}}{{{{\rm{T}}_{\rm{B}}}}} = 5 + 5 \times \frac{{75}}{{25}} = 20\)
Concept:
Calculation:
Given:
T_{B} = 100; T_{A} = 25
Step No. 
Conditions of motion 
Revolutions of elements 

Arm C 
Gear A 
Gear B 

1.
2.
3.
4. 
Arm fixedgear A rotates though + 1 revolution (i.e. 1 rev. anticlockwise) Arm fixedgear A rotates through + x revolutions Add + y revolutions to all elements
Total motion 
0
0
+y
+y 
+ 1
+x
+y
x + y 
\( \frac{{{T_A}}}{{{T_B}}}\)
\( x \times \frac{{{T_A}}}{{{T_B}}}\)
+y
\(y  x \times \frac{{{T_A}}}{{{T_B}}}\) 
y = 1; N_{B} = 0
\(\begin{array}{l} {N_B} = y  x \times \frac{{{T_A}}}{{{T_B}}} = 1  x \times \frac{{25}}{{100}} = 1  \frac{x}{4} = 0 \Rightarrow x = 4\\ {N_A} = x + y = 4 + 1 = 5 \end{array}\)
In an epicyclic gear train, shown in the figure, the outer ring gear is fixed, while the sun gear rotates counterclockwise at 100 rpm. Let the number of teeth on the sun, planet and outer gears to be 50, 25, and 100, respectively. The ratio of magnitudes of angular velocity of the planet gear to the angular velocity of the carrier arm is ________.
Concept:
Let S is suffix for sun, P for planet and D for outer ring.
N_{D} = 0
N_{S} = 100 (Taking CW as positive)
T_{S} = 50, T_{P} = 25, T_{D} = 100
Motion 
Arm 
Sun (S) 
Planet (P) 
Ring (D) 
Arm fixed 
0 
1 
\(\frac{{  50}}{{25}}\) 
\(  \left( {\frac{{50}}{{25}} \times \frac{{25}}{{100}}} \right)\) 
x 
0 
x 
\(\frac{{  x50}}{{25}}\) 
\( x\left( {\frac{{50}}{{25}}} \right)\left( {\frac{{25}}{{100}}} \right)\) 
y 
y 
x + y 
\(\frac{{y  x \times 50}}{{25}}\) 
\(y  x\left( {\frac{{50}}{{25}}} \right)\left( {\frac{{25}}{{100}}} \right)\) 
Now y + x = 100 (i)
\(y  x\left( {\frac{{50}}{{25}}} \right)\left( {\frac{{25}}{{100}}} \right) = 0 \Rightarrow y  \frac{x}{2} = 0\) (ii)
From (i) & (ii)
\(x = \frac{{  200}}{3},\;y = \frac{{  100}}{3}\)
\({N_P} = y  2x = \frac{{  100}}{3} + \frac{{400}}{3} = 100\)
\({N_{arm}} = y = \frac{{  100}}{3}\)
\(\frac{{{N_P}}}{{{N_{arm}}}} = \frac{{100}}{{\frac{{  100}}{3}}} =  3 = 3\left( {magnitude} \right)\)
An epicyclic gear train is shown in the figure below. The number of teeth on the gears A, B and D are 20, 30 and 20, respectively. Gear C has 80 teeth on the inner surface and 100 teeth on the outer surface. If the carrier arm AB is fixed and the sun gear A rotates at 300 rpm in the clockwise direction, then the rpm of D in the clockwise direction is
The speed of the gears in epicyclic gear train can be analyzed from the following table,
If the external toothed gear is meshing with the internal toothed gears then the direction of the velocity remains same, and the velocity ratio between two mating gears is given by
\(\frac{N_2}{N _1}=\frac{T_1}{T_2}\)
Where N is the angular velocity in rpm and T is the number of teeth
Conditions of motion 
Revolutions of elements 

Arm 
Gear A 
Gear B 
Gear C 
Gear D 

Arm is fixed, wheel A rotates +1 revolutions 
0 
+1 
\( \frac{{{T_A}}}{{{T_B}}}\) 
\( \frac{{{T_A}}}{{{T_B}}} \times \frac{{{T_B}}}{{{T_{Ci}}}} =  \frac{{{T_A}}}{{{T_{Ci}}}}\) 
\(\frac{{{T_A}}}{{{T_{Ci}}}} \times \frac{{{T_{Co}}}}{{{T_D}}} = \frac{{{T_A}}}{{{T_D}}}\frac{{{T_{Co}}}}{{{T_{ci}}}}\) 
Arm is fixed, wheel A is rotated trough +x revolution 
0 
+x 
\( x\frac{{{T_A}}}{{{T_B}}}\) 
\( x\frac{{{T_A}}}{{{T_{Ci}}}}\) 
\(x\frac{{{T_A}}}{{{T_D}}}\frac{{{T_{Co}}}}{{{T_{ci}}}}\) 
Add +y revolution to all 
y 
Y + x 
\(y  x\frac{{{T_A}}}{{{T_B}}}\) 
\(y  x\frac{{{T_A}}}{{{T_C}}}\) 
\(y + x\frac{{{T_A}}}{{{T_D}}}\frac{{{T_{Co}}}}{{{T_{ci}}}}\) 
Calculation:
Given: T_{A} = 20, T_{B} = 30, T_{D} = 20, T_{Ci} = 80, T_{Co} = 100, N_{Arm} = 0, N_{A} = 300 rpm, N_{D} =?
N_{Arm} = 0 = y
N_{A} = 300 rpm = y + x ⇒ x = 300
\({N_D} = y + x\frac{{{T_A}}}{{{T_D}}}\frac{{{T_{Co}}}}{{{T_{ci}}}} = x\frac{{{T_A}}}{{{T_D}}}\frac{{{T_{Co}}}}{{{T_{ci}}}} = 300 \times \frac{{20}}{{20}} \times \frac{{100}}{{80}} = 375\;rpm\)
Mistake: Take care of the inner and outer teeth of gear C.
Step no 
Arm 
Pinion 
Gear 
Fixed arm x rotate y 
0 0 y 
+ 1 + x X + y 
T_{P}/T_{G}  x T_{P}/T_{G} y – x T_{P}/T_{G} 
N_{G} = 0, (Given)
\(y  \frac{x}{4} = 0\),
y = x/4; x=4y
for one revolution of arm: y = 1; x = 4 rev.
N_{P}=X+y
∴ NP = 5 revolution.
For the epicyclic gear arrangement shown in the figure, ω_{2} = 100 rad/s clockwise (CW) and ω_{arm} = 80 rad/s counter clockwise (CCW). The angular velocity ω_{5 }(in rad/s) is
Answer: 3 Concept: Step 1. Make a table.
Step 2. Fill the first row.
Step 3: Fill the second row.
Step 4:
Step 5: Assume clockwise rotation (CW) as positive (+ve) and counterclockwise (CCW) as negative (ve). Calculation: Given T_{2} = 20, T_{3} = 24, T_{4} = 32, T_{5} = 80 \({\omega _2} = 100\) rad/sec, \({\omega _{arm}} = \;  80\) rad/sec
\(\therefore x + y = 180\;and\;y = \;  80\) ⇒ \(x  80 = 100\) ⇒\(x = 180.\) Now, \({N_5} =  \frac{1}{3} \times x + y\) ⇒ \(  \frac{1}{3} \times 180  80\) ⇒ \(  140\) rad/sec. (ve signifies CCW) 
Statement (I): In an epicyclic gear train, the size of the gearbox is smaller than that of the spur gearbox for the same horsepower and the same velocity ratio.
Statement (II): In an epicyclic gearbox, more than one pair of gear pinion contacts always exist, whereas it is not so in spur gearbox.A epicyclic (planetary) gearbox gives high torque transmission with good stiffness and low noise, in a more compact footprint than other gearbox types.
Since planetary gears mesh with the sun gear and ring gear at several locations, more teeth are engaged to drive the load, compared to a conventional gear and pinion mesh. Therefore, for the same load, planetary gearing requires smaller gears although in greater number than a standard piniontogear reduction.
More than one pair of contact between pinion and gear results in less load per tooth of gear while in case of spur gearbox, the gears are having single contact and hence for epicyclic gears, the gearbox size is small for the same horsepower.
Therefore, statement II) gives exact explanation for statement I).Statement (I): The epicyclic gear train has a central gear and an epicyclic gear which produces epicyclic motion being moved by a crank arm.
Statement (II): The arm contains the bearings for the epicyclic gear to maintain two gears in mesh.An epicyclic gear train also known as planetary gear consists of two gears mounted so that the center of one gear revolves around the center of the other. Epicyclic gear sets essentially consists of two gears with their axes or shafts connected to an arm or carrier. If the arm is fixed, the gear train is simple and gear A can drive gear B or vice versa, but if gear A is fixed and the arm is rotated about the axis of gear A , then the gear B is forced to rotate upon and around gear.
The epicyclic gear trains may be simple or compound. Compound epicyclic gear trains are called planetary gear train.
So, from above explanation Statement I) is correct as it gives the definition of epicyclic gear set.
Moreover, arm has bearing at ends which connect the end of arm with centre of gears maintaining two gears in mesh. Statement II) is also correct but does not explain statement I).
Applications: The epicyclic gear trains are useful for transmitting high velocity ratios with gears of moderate size in a comparatively lesser space. These are used in the back gear of lathe, differential gears of the automobiles, hoists, pulley blocks, wrist watches etc.