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TABLE 5.1 Solution by Tabulation
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ny + nxy = nx
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(5.16)
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This is best shown by example. Using the planetary train of the previous example, we form Table 5.1, and the equation from the column for gear 5 is n3 N2N4 (n2 n3) = n5 N4N5
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Rearranging and canceling N4, we find n3 1 + N N2 n2 2 = n5 N5 N5 (5.17)
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This is the characteristic equation of the planetary train, as shown in Fig. 5.4. Note that three rotational quantities appear n3, n2, and n5. There must be two input rotations in order to solve for the output. This is easily done when the input rotations and the tooth numbers are inserted. When a positive sense is assigned to counterclockwise and a negative sense to clockwise rotation, the sign of the output rotation indicates its sense of direction. Note that planet 4 was not included in the table (it could have been); however, gear 4 served its purpose by acting as an idler to change a direction of rotation. This is evidenced by the presence of a negative sign in the second row of the column for gear 5. A convenient means of representing a planetary train was shown by Levai [5.4]. Type A of Fig. 5.7a shows an edge view of the planetary train first seen in Fig. 5.4. It and the other 11 configurations represent all possible variations for a planetary train. The equations in Table 5.2 are the characteristic equations of the 12 types. An examination of the equations and their corresponding types reveals that certain ones are identical. Types C and D in Fig. 5.7b are identical because of the arrangement of gears. Whereas in type C the meshes of 2 and 4 and of 7 and 8 are external, the input and output meshes are internal in type D. The same relationship can be seen in types G and H in Fig. 5.7c. Certain pair types are alike in equation form but differ in sign. Compare types E and K and F and L in Fig. 5.7b and d, and B and G (or B and H) in Fig. 5.7a and c. The speed of a planet gear relative to the frame or relative to the arm may be required. If appreciable speeds and forces are involved, this information will facilitate the selection of bearings. Using type A as an example, set up Table 5.3. Row 2 in the column for gear 4 is the speed of gear 4 relative to the arm, and row 3 in the column for gear 4 is its speed relative to the frame.
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Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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MACHINE ELEMENTS IN MOTION
FIGURE 5.7 Twelve variations of planetary trains.
Example 2. Figure 5.8 shows a planetary gear train with input at gear 2. Also, gear 6 is seen to be part of the frame, in which case its rotation is zero. For n2 = 100 r/min clockwise (negative), find output rotation n6. Solution. Gears 2, 4, 5, and 6 and arm 3 form a type B planetary train: n3 1 + Solving for n3 yields n3 1 + 1 1 ( 100) =0 8 8 n3 = For type G: n3 1 Then we solve type G for n6: NN N2N5 + n2 2 5 = n6 N4N6 N4N6 100 r/min 9 N2N5 N N n2 2 6 = n6 N4N6 N4N6
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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