Question 8.9: A rack is driven by pinion shown in Figure 8–46. The pinion ...

A rack is driven by pinion shown in Figure 8–46. The pinion rotates at 125 rpm, has 24 teeth, and a diametral pitch of 6.

a. Determine the pitch diameter of the pinion.
b. Determine the dimension from the pitch line to the back of the rack, B.
c. Calculate the distance of the pinion center line to the back of the rack.
d. Determine the linear velocity of the rack.
e. How long would it take to move a rack that has a length of 20 ft?
f. How many revolutions would the pinion turn in moving the rack 20 ft?

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Given: n_{pinion} = 125 rpm; Pinion P_d = 6; N_{pinion} = 24; Length of rack = L = 20 ft
a. D_p = N_p/P_d = 24/6 = 4.000 in
b. Distance from the pitch line to the back of the rack: B = 1.333 in [From Table 8–10]

c. Distance from back of the rack to the pinion centerline: [Call this value B-C]
B-C = B + Pinion radius = B + D_p/2 = 1.333 in + (4.000 in)/2 = 3.333 in

d. Linear velocity of the rack = V_{rack} = rω = (D_p/2)(n_p)

V_{rack} = (2.000 in)(125 rev/min)(2π rad/rev)(1.0 ft/12 in) = 130.9 ft/min = 130.9 fpm

e. Time to move rack 20 ft: Using V = s/t, then t = s/V

 

t=\frac{s}{V}=\frac{20ft}{130.9ft/min}.\frac{60sec}{min}=9.167 sec

 

f. Number of pinion revolutions, θ_P, to move rack 20 ft:
From Equation 8–37:

S_{RACK}=\frac {D_P}{2} θ_P

Then

θ_P=\frac {S_{RACK}}{D_P/2}=\frac{20ft}{2.0 in}.\frac{12 in}{ft}=120 rad.\frac{1 rev}{2k}= 19.09rev

 

TABLE 8–10 Example rack specifications
Diametral pitch Pitch line to back (B) Overall thickness Face width Nominal length [ft]
64 0.109 0.125 0.125 2
48 0.104 0.125 0.125 2
32 0.156 0.187 0.187 4
24 0.208 0.250 0.25 4
20 0.450 0.500 0.5 6
16 0.688 0.750 0.75 6
12 0.917 1.000 1 6
10 1.150 1.250 1.25 6
8 1.375 1.500 1.5 6
6 1.333 1.500 2 6
5 1.300 1.500 2.5 6
4 1.750 2.000 3.5 6

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