chap09 - 34.298 ψ = = ° Ans 29 3 3 2 40 teeth 1.25...

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105 Chapter 9 Worms and Worm Gears 9.1 A worm having 4 teeth and a lead of 1.0 in drives a worm gear at a velocity ratio of 7.5. Determine the pitch diameters of the worm and worm gear for a center distance of 1.75 in. 2 1 in 4 teeth 0.25 in/tooth x p N = = = l ( 29 ( 29 3 2 3 2 7.5 4 teeth 30 teeth N N ϖ ϖ = = = 3 3 2 1.194 in x R N p π = = 2 3 1.75 0.556 in R R = - = Ans. 9.2 Specify a suitable worm and worm gear combination for a velocity ratio of 60 and a center distance of 6.50 in using an axial pitch of 0.500 in/tooth. Use 2 1 tooth N = , ( 29 ( 29 3 2 3 2 60 1 teeth 60 teeth N N ϖ ϖ = = = 3 3 2 4.775 in x R N p π = = 2 3 6.500 1.725 in R R = - = Ans. 9.3 A triple-threaded worm drives a worm gear having 40 teeth. The axial pitch is 1.25 in and the pitch diameter of the worm is 1.75 in. Calculate the lead and lead angle of the worm. Find the helix angle and pitch diameter of the worm gear. ( 29 2 3 teeth 1.25 in/tooth 3.750 in x N p = = = l Ans. ( 29 ( 29 1 1 2 tan 2 tan 3.750 in 1.75 in 34.298 R λ π π -
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Unformatted text preview: 34.298 ψ = = ° Ans. ( 29 3 3 2 40 teeth 1.25 in/tooth 2 7.957 in x R N p = = = Ans. 9.4 A triple threaded worm with a lead angle of 20˚ and an axial pitch of 0.400 in/tooth drives a worm gear with a velocity reduction of 15 to 1. Determine for the worm gear: ( a ) the number of teeth, ( b ) the pitch diameter, and ( c ) the helix angle. ( d ) Determine the pitch diameter of the worm. ( e ) Compute the center distance. ( 29 2 3 teeth 0.400 in/tooth 1.200 in x N p = = = l ( 29 2 2 tan 1.200 in 2 tan 20 0.524 7 in R = = ° = l Ans. ( 29 ( 29 3 2 3 2 15 0.524 7 in 7.871 in R R = = = Ans. 3 3 2 123.6 teeth x N R p = = Ans. 20.0 = = ° Ans. 2 3 0.525 in 7.871 in 8.396 in R R + = + = Ans. [Page intentionally blank.] 106...
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