1. The angle at which the teeth are inclined to the normal of the axis of rotation is called _______________
a) pitch angle
b) lead angle
c) normal angle
d) joint angle
Explanation: Lead angle is the angle at which the teeth are inclined to the normal of the axis of rotation. Lead angle of the worm gear is same as the helix angle of the helical gear. Thus, Ψ2 = λ1.
2. What is the velocity ratio of worm gears?
a) (lπ)/d2
b) (πd2)/l
c) l/(πd2)
d) d2/(lπ)
Explanation: Velocity ratio in worm gears is given as the ratio of the angle turned by the gear to the angle turned by the worm.
Thus, velocity ratio = (2l/d2)/(2π) = l/(πd2).
3. What is the centre distance for the worm gear?
a) (mn/2)(T1 cotλ1 – T2)
b) (mn/2)(T2 cotλ1 + T1)
c) (mn/2)(T2 cotλ1 – T1)
d) (mn/2)(T1 cotλ1 + T2)
Explanation: C = (mn/2)(T1 cotλ1 + T2)
This equation can be derived by using the formula for centre distance of a helical gear which is given as C = (mn/2) ((T1/cos Ψ1) + (T2/cos Ψ2)
As, Ψ2 = λ1, Ψ1 = 90° – λ1.
4. What is the formula to calculate maximum efficiency of a worm gear?
a) (1+sinø)/(1-sinø)
b) (1-sinø)/(1+sinø)
c) (tan(λ1-ø))/tan λ1
d) (tan(λ1+ø))/tan λ1
Explanation: The maximum efficiency of the worm gear is given to be (1-sinø)/(1+sinø), whereas the efficiency of the worm gear is given as (tan(λ1-ø))/tan λ1
5. Find the helix angle of the worm if the pitch of the worm gear is 12 mm and the pitch diameter is 50 mm.
a) 8.687°
b) 11.231°
c) 9.212°
d) 10.319°
Explanation: tan λ1 = Lead / Pitch circumference = 2p/πd1 = 24/50π = 0.1528
λ1 = 8.687°.
6. Find the speed of the gear if the worm is a three start worm rotating at 500 rpm. The gear has 20 teeth.
a) 125 rpm
b) 100 rpm
c) 75 rpm
d) 50 rpm
Explanation: N1/N2 = T2/T1
500/N2 = 20/3
N2 = 75 rpm
Thus, the gear rotates at a speed of 75 rpm.
7. For a two start worm gear having a pitch of 20 mm and a lead angle 12°, find the centre distance if the larger gear has 25 teeth.
a) 148.22 mm
b) 124.93 mm
c) 121.19 mm
d) 109.53 mm
Explanation: C = (mn/2)(T1 cotλ1 + T2)
Therefore, C = (pn/2π)(T1 cotλ1 + T2) = 109.53 mm
8. Calculate the lead angle of the worm gear for maximum efficiency if θ = 90° and the coefficient of friction is 0.05.
a) 48.21°
b) 42.23°
c) 43.57°
d) 46.43°
Explanation: µ = 0.05°; ø = tan-1(0.05) = 2.862°; θ = 90°
For maximum efficiency, Ψ1 = (θ+ ø)/2 = 92.862/2 = 46.43°
Ψ1 = 90° – λ1 = 46.43°
λ1 = 90° – 46.43° = 43.57°.
9. Find the maximum efficiency if the lead angle is given to be 10° and the coefficient of friction is 0.07.
a) 79.82%
b) 72.23%
c) 76.29%
d) 70.72%
Explanation: λ1 = 10°, ø = tan-1(0.07) = 4°
Efficiency = tan(λ1)/ tan(λ1+ ø) = 0.7072 = 70.72%.
10. Calculate the maximum efficiency of the worm gears which have a friction angle of 0.06.
a) 88.71%
b) 83.23%
c) 89.91%
d) 86.49%
Explanation: ø = tan-1(0.06) = 3.43°
Maximum efficiency = (1-sin ø)/(1+sin ø) = 0.8871 = 88.71%