Question

5. A single threaded worm rotating at 900 rpm drives a 30-tooth worm gear. The diameter of the worm is 60 mm. The module of the gear is 5 mm. Find (a) helix angle of the worm (b) the speed ratio (c) centre distance and (d) angular speed of the worm gear

Answer 1

N_1 = 900 rpm T_2 = 30 d_1 = 60 mm m = 5 mm helix angle = ? speed ratio = ? center distance = ? angular speed of the worm, year = ? m = d/T m = 60/T_1 T = 60/5 T_1 = 12 m_1 = m_2 m_2 = d_2/T_2 d_2 = 5 times 30 = 150 mm d_2 = 150 mm \r\n p_1 = pi d_1/T_2 p_1 = pi(60)/T_2 p_1 = 15.707 mm tan lambda_1 = p/pi d tan lambda_1 = 15,707/pi 60 lambda = tan^-1(15,707/pi times 60) I lambda_1 = 4,763 I centre distance = m/2[T_1 cos lambda_1 + T_2] = 5/2[12 cos(4.763) + 30] C = 104.896 mm iii angular speed of worm year = pi d_2 N_2/60 = pi times (150) times N_2/60 = pi times 150 times 360/60 W_E = 2827.43 Kd/sec N_2/N_1 = T-1/T_2 rightdoublearrow N_2 = 900 times 12/30 = 360 rpm N_2 = 360