--- title: "Worm gear drive numerical" subject: "Theory of Mechanism and Machine I" chapter: 5 key: "semester_6/theory_of_mechanism_and_machine_i:worm-gear-lead-angle-efficiency" source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/worm-gear-lead-angle-efficiency site: BCT Engineering --- # Worm gear drive numerical Theory of Mechanism and Machine I · Chapter 5: Bevel, Helical and Worm Gears - Asked in: year not given, 6 marks - Video: [WORM GEARS - Forces and Speed Relations in Just Under 15 Minutes!](https://www.youtube.com/watch?v=7cNR4YJ1cnI) — Less Boring Lectures - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/worm-gear-lead-angle-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-5 ## Question A triple-start worm of axial module 6 mm and diametral quotient (q = d/m) of 10 drives a worm wheel of 45 teeth. Find the lead, the lead angle, the pitch circle diameter of the worm and wheel, the centre distance and the velocity ratio. Estimate the efficiency of the drive for a coefficient of friction of 0.05 and a normal pressure angle of 20 degrees. Also state two advantages and two disadvantages of worm gearing. ## Short answer Lead L=56.55 mm, lead angle λ=16.70°, worm pitch diameter dw=60 mm, wheel dg=270 mm, centre distance C=165.0 mm, velocity ratio=15, efficiency η≈83.6%. Advantages: high reduction in single stage, smooth quiet operation, self-locking possible. Disadvantages: low efficiency due to sliding, heat generation, needs bronze wheel and good lubrication, large axial thrust. ## Answer #### Worm drive relations For a worm with $z_1$ starts (threads), axial pitch $p_x = \pi m$, and diameter $d_w = qm$: - Lead $L = z_1p_x$. - Lead angle $\tan\lambda = \dfrac{L}{\pi d_w} = \dfrac{z_1}{q}$. - Wheel pitch diameter $d_g = mz_2$; the wheel helix angle equals $\lambda$ (for 90 degree shafts, with the same hand). - Centre distance $C = \dfrac{d_w + d_g}{2}$; velocity ratio $= z_2/z_1$. #### Calculation Axial pitch: $p_x = \pi \times 6 = 18.850$ mm. Lead: $L = 3 \times 18.850 = 56.55$ mm. $$ \lambda = \tan^{-1}\frac{z_1}{q} = \tan^{-1}\frac{3}{10} = 16.70^\circ $$ Worm diameter: $d_w = qm = 10 \times 6 = 60$ mm. Wheel diameter: $d_g = 6 \times 45 = 270$ mm. $$ C = \frac{60 + 270}{2} = 165.0\ \text{mm} $$ $$ \text{Velocity ratio} = \frac{z_2}{z_1} = \frac{45}{3} = 15 $$ #### Efficiency $$ \eta = \frac{\cos\phi_n - \mu\tan\lambda}{\cos\phi_n + \mu\cot\lambda} = \frac{\cos 20^\circ - 0.05 \times 0.3000}{\cos 20^\circ + 0.05/0.3000} = \frac{0.9247}{1.1064} $$ $$ \eta = 83.6\ \% $$ #### Advantages and disadvantages | Advantages | Disadvantages | |---|---| | Very high speed reduction in one stage | Low efficiency (sliding contact), heat generation | | Smooth and quiet operation | Needs bronze wheel and good lubrication, costly | | Can be self-locking (small lead angle) | Large axial thrust on the worm | **Answer: $L = 56.55$ mm, $\lambda = 16.70^\circ$, $d_w = 60$ mm, $d_g = 270$ mm, $C = 165.0$ mm, VR $= 15$, $\eta \approx 83.6\ \%$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a worm with z1 starts and diameter quotient q, the lead angle satisfies: A. tan(lambda) = q/z1 B. sin(lambda) = z1/q C. tan(lambda) = z1/q D. tan(lambda) = z1 m Answer: C. tan(lambda) = L/(pi d_w) = z1 p_x/(pi q m) = z1/q. Here tan(lambda) = 3/10, so lambda = 16.70°. (Key point: Lead and lead angle) 2. For a triple-start worm of axial module 6 mm, the lead is: A. 18.85 mm B. 6 mm C. 56.55 mm D. 113.1 mm Answer: C. Axial pitch = pi x 6 = 18.850 mm, and lead = 3 x 18.850 = 56.55 mm. (Key point: Lead value) 3. With d_w = qm = 60 mm, wheel of 45 teeth (d_g = 270 mm) and 3 starts, the centre distance and velocity ratio are: A. 330 mm and 15 B. 165 mm and 15 C. 165 mm and 45 D. 210 mm and 3 Answer: B. C = (60 + 270)/2 = 165 mm and VR = z2/z1 = 45/3 = 15. (Key point: Centre distance and velocity ratio) 4. Which statement fits the worm drive here (eta about 83.6%)? A. Efficiency is high because contact is rolling B. No lubrication or special wheel material is needed C. Efficiency rises as the friction coefficient rises D. Efficiency is reduced by sliding contact, and a bronze wheel and good lubrication are needed Answer: D. Worm gears have sliding contact, so lower efficiency and heat. They need a bronze wheel and good lubrication, but give very high reduction in one stage. 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