This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 00 r.p.m. clockwise, therefore from the fourth row of the
table
y = – 100
...(iii)
Also the wheel A makes 10 r.p.m. counter clockwise, therefore from the fourth row of the
table,
x + y = 10 or x = 10 – y = 10 + 100 = 110
...(iv)
∴ Speed of wheel B = y + x × 64 26
TA TD
×
= – 100 + 110 ×
×
= – 100 +105.4 r.p.m.
28 62
TC TB = + 5.4 r.p.m. = 5.4 r.p.m. counter clockwise Ans.
Example 13.9. In an epicyclic gear of the ‘sun and planet’ type shown
in Fig. 13.13, the pitch circle diameter of the internally toothed ring is to be
224 mm and the module 4 mm. When the ring D is stationary, the spider A,
which carries three planet wheels C of equal size, is to make one revolution in
the same sense as the sunwheel B for every five revolutions of the driving
spindle carrying the sunwheel B. Determine suitable numbers of teeth for all
the wheels.
Fig. 13.13
Solution. Given : dD = 224 mm ; m = 4 mm ;
NA = NB / 5
Let T B , T C a nd T D b e the number of teeth on the sun wheel B ,
planet wheels C and the internally toothed ring D. The table of motions is given below : Table 13.8. Table of motions.
Revolutions of elements
Step No.
1. 2. 3.
4. Conditions of motion
Spider A fixed, sun wheel
B r otates through + 1
revolution ( i . e . 1 rev.
anticlockwise)
Spider A fixed, sun wheel
B r otates through + x
revolutions
Add + y revolutions to all
elements
Total motion
Spider A Sun wheel B Planet wheel C 0 +1 – 0 +x – x× +y +y +y +y x +y y – x× TB
TC Internal gear D
– TB
TC TB TC
T
× =– B
TC TD
TD
– x× TB
TD +y
TB
TC y – x× TB
TD 448 l Theory of Machines We know that when the sun
wheel B makes + 5 revolutions, the spider A makes + 1 revolution. Therefore
from the fourth row of the table,
y = + 1 ; and x + y = + 5
∴ x=5–y=5–1=4 Main rotor Drive shaft
Cockpit
Tail boom Since the internally toothed ring
D is stationary, therefore from the fourth
row of the table, y – x× TB
=0
TD T
1– 4× B =0
TD or ∴ TB 1
=
TD 4 Tail rotor Landing skids Engine, transmission fuel, etc. Power transmission in a helicopter is essentially through
gear train...
View
Full
Document
This note was uploaded on 02/13/2014 for the course MIE 301 taught by Professor Celghorn during the Fall '08 term at University of Toronto Toronto.
 Fall '08
 CELGHORN

Click to edit the document details