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find the speed of B ; and 4. If the arm G makes 100 r.p.m. clockwise
and wheel A makes 10 r.p.m. counter clockwise ; find the speed of
Solution. Given : T C = 28 ; T D = 26 ; T E = T F = 18
1. Sketch the arrangement
The arrangement is shown in Fig. 13.12.
2. Number of teeth on wheels A and B
Let TA = Number of teeth on wheel A , and
TB = Number of teeth on wheel B .
If dA , dB , dC , dD , dE and dF are the pitch circle diameters of wheels A , B , C, D, E and F
respectively, then from the geometry of Fig. 13.12,
dA = dC + 2 dE
dB = dD + 2 dF
Since the number of teeth are proportional to their pitch circle diameters, for the same
TA = T C + 2 T E = 28 + 2 × 18 = 64
TB = T D + 2 T F = 26 + 2 × 18 = 62
3. Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A is fixed
First of all, the table of motions is drawn as given below :
Table 13.7. Table of motions.
Revolutions of elements
No. Conditions of
motion Arm Wheel
A 1. Arm fixed- wheel A
rotates through + 1
revolution (i.e. 1 rev.
anticlockwise) 0 2. Arm fixed-wheel A
rotates through + x
revolutions 0 +x + x× 3. Add + y revolutions
to all elements +y +y +y 4. Total motion +y x+y +1 Wheel
E + TA
wheel C-D – TA TE
TE y + x× – x× TA
TC y–x× + TA TD
TC TF TA
TE Wheel F + TA TD TF
TC TF TB =+
+ x× TA TD
TC TF +y
TC Wheel B y + x× TA TD
TC TF TA TD
TC TB + x× TA TD
TC TB +y y + x× TA TD
TC TB Chapter 13 : Gear Trains 447 l Since the arm G makes 100 r.p.m. clockwise, therefore from the fourth row of the table,
y = – 100 ...(i) Also, the wheel A is fixed, therefore from the fourth row of the table,
x = – y = 100
∴ Speed of wheel B = y + x × ...(ii) 64 26
= – 100 + 100 ×
= – 100 + 95.8 r.p.m.
TC TB = – 4.2 r.p.m. = 4.2 r.p.m. clockwise Ans.
4. Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A makes 10 r.p.m. counter
Since the arm G m akes 1...
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- Fall '08
- gear, Gear Trains, Epicyclic gearing, epicyclic gear train