Planetary Gearing Design And Efficiency
Planetary Gearing Design And Efficiency
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P.
Next to be considered are the bearings of the planet gears. Here the load on the bearings is 78-78 Ib. , and the speed of the bearings is 419-05 r. P. M. The surface speed at the bearings is 0-262 X 1 > 419-05 = 109-791 ft. Per min. Also, the bearing surface is 1 X 2 =3 2-25 sq. Ins. ; hence the bearing pressure is 78-78/2-25 = 35-01 Ib. Per sq. Inch. By equation (5), therefore, the frictional resistance is 0-224 Ib. Per sq. Inch, and the total frictional resistance is 0-224 X 2-25 = 0-504 I...b. Hence, the power absorbed in friction is equal to (0-504 x 109-79)/33000 = 0-0017 h. P.
The power required to overcome the friction between the teeth of the sun gear B and the planet gear C is given by the formula, (2Nn\iF sec a)/(33000 X P X 12), which is equal to (2 x 419-05 X 28 x 0-01 x 39-39 X 1-0642)/(33000 X 10 x 12) = 0-0025 h. P.
Similarly, the power required to overcome friction between the teeth of the planet gear C and the annulus D is (2Nn\nF 3 sec a)/(33000 X P X 12), and this is (2 x 419-05 x 28 X 0-01 X 2127-06 X 1-0642) /(33000 X 10 x 12) = 0-1341 h.
What to read after Planetary Gearing Design And Efficiency?
You can find similar books in the "Read Also" column, or choose other free books by H W Herbert Wilfrid Hardy to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
P.
Next to be considered are the bearings of the planet gears. Here the load on the bearings is 78-78 Ib. , and the speed of the bearings is 419-05 r. P. M. The surface speed at the bearings is 0-262 X 1 > 419-05 = 109-791 ft. Per min. Also, the bearing surface is 1 X 2 =3 2-25 sq. Ins. ; hence the bearing pressure is 78-78/2-25 = 35-01 Ib. Per sq. Inch. By equation (5), therefore, the frictional resistance is 0-224 Ib. Per sq. Inch, and the total frictional resistance is 0-224 X 2-25 = 0-504 I...b. Hence, the power absorbed in friction is equal to (0-504 x 109-79)/33000 = 0-0017 h. P.
The power required to overcome the friction between the teeth of the sun gear B and the planet gear C is given by the formula, (2Nn\iF sec a)/(33000 X P X 12), which is equal to (2 x 419-05 X 28 x 0-01 x 39-39 X 1-0642)/(33000 X 10 x 12) = 0-0025 h. P.
Similarly, the power required to overcome friction between the teeth of the planet gear C and the annulus D is (2Nn\nF 3 sec a)/(33000 X P X 12), and this is (2 x 419-05 x 28 X 0-01 X 2127-06 X 1-0642) /(33000 X 10 x 12) = 0-1341 h.
What to read after Planetary Gearing Design And Efficiency?
You can find similar books in the "Read Also" column, or choose other free books by H W Herbert Wilfrid Hardy to read onlineMoreLess
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