Iii also the wheel a makes 10 rpm counter clockwise

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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...
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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.

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