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Porter governor speed range and lift

Short answer

The minimum speed (r=150 mm) is 143.7 rpm, maximum (r=200 mm) is 154.9 rpm, so the range is 11.2 rpm. Sleeve lift is 2(√(0.3²-0.15²)-√(0.3²-0.2²))=72.4 mm. Sensitiveness=(154.9-143.7)/((143.7+154.9)/2)=0.0750.

A Porter governor has equal arms each 300 mm long, pivoted on the axis of rotation. Each ball has a mass of 5 kg and the central sleeve load has a mass of 25 kg. The radius of rotation of the balls is 150 mm when the governor is at its lowest position and 200 mm at the highest position. Neglecting friction, find the range of speed and the lift of the sleeve. Also find the sensitiveness of the governor.

Data: l=0.3l = 0.3 m (arms and links equal, hung from the axis), m=5m = 5 kg, M=25M = 25 kg, r1=0.15r_1 = 0.15 m, r2=0.2r_2 = 0.2 m.

Equilibrium relation

For a Porter governor with arms and links of equal length meeting the axis (α=β\alpha = \beta, so q=tan⁡β/tan⁡α=1q = \tan\beta/\tan\alpha = 1):

ω2=(m+M) gm h,h=l2−r2\omega^2 = \frac{(m+M)\,g}{m\,h}, \qquad h = \sqrt{l^2 - r^2} 
         O (pivot on axis)
        /|\
       / | \   arms l
      /  |h \
     B   |   B   balls m at radius r
      \  |  /   links l
       \ | /
        [M]  sleeve load

Lowest position (r1=0.15r_1 = 0.15 m)

h1=0.32−0.152=0.2598 mh_1 = \sqrt{0.3^2 - 0.15^2} = 0.2598\ \text{m} ω12=(5+25)×9.815×0.2598=226.55⇒ω1=15.05 rad/s,N1=143.7 rpm\omega_1^2 = \frac{(5+25)\times 9.81}{5 \times 0.2598} = 226.55 \Rightarrow \omega_1 = 15.05\ \text{rad/s},\quad N_1 = 143.7\ \text{rpm}

Highest position (r2=0.2r_2 = 0.2 m)

h2=0.32−0.22=0.2236 mh_2 = \sqrt{0.3^2 - 0.2^2} = 0.2236\ \text{m} ω22=30×9.815×0.2236=263.23⇒ω2=16.22 rad/s,N2=154.9 rpm\omega_2^2 = \frac{30 \times 9.81}{5 \times 0.2236} = 263.23 \Rightarrow \omega_2 = 16.22\ \text{rad/s},\quad N_2 = 154.9\ \text{rpm}

Range, lift and sensitiveness

Range=N2−N1=154.9−143.7=11.2 rpm\text{Range} = N_2 - N_1 = 154.9 - 143.7 = 11.2\ \text{rpm} Lift of sleeve=2(h1−h2)=2(0.2598−0.2236)=0.0724 m=72.4 mm\text{Lift of sleeve} = 2(h_1 - h_2) = 2(0.2598 - 0.2236) = 0.0724\ \text{m} = 72.4\ \text{mm} Sensitiveness=N2−N1Nmean=11.2149.3=0.0750\text{Sensitiveness} = \frac{N_2 - N_1}{N_{mean}} = \frac{11.2}{149.3} = 0.0750

Answer: Speed range =143.7= 143.7 to 154.9154.9 rpm (11.2 rpm); sleeve lift =72.4= 72.4 mm; sensitiveness =0.0750= 0.0750.

✓ Answer checked by Claude Opus 5.5 (high effort)

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