Jackshaft Drive Pulley

You can see how poor the original drawing was. I have colored the Jackshaft Drive Pulley in red to highlight it in the drawing. It’s size is to under-drive the Jackshaft Pulley slower than the 250 RPM of the Steam engine.

D=30"
C=94.25"C=7.85'
v_{m}=C_f*RPMv_{m}=7.85*250v_{m}=1962.5

45 HP : P=\frac{33000*H}{v_m}P=\frac{33000*45}{1963}P=757
20 HP : P=\frac{33000*H}{v_m}P=\frac{33000*20}{1963}P=336

From the ‘Belt Width Table‘ we can see that at 45 Hp the belt — b — needs to be 9″ wide and at 20 HP only 4″ wide. Pulley width — B — is taken as B=\frac{5}{4}b .. pulley widths are 11\frac{1}{8}” and 5” respectively.

Once again, running calculations for a five-jig Stewart Washery at 100 HP

100 HP : P=\frac{33000*H}{v_m}P=\frac{33000*100}{1963}P=1681

This requires a 18″ wide belt or a 22.5″ wide pulley. This is (again) well within the error of me trying to trace a small, blurry image. Since I am modeling a single-jig I am happy to go with the 11\frac{1}{8}” wide pulley calculated for 45 HP.

Data

  • \textbf{HP=45}
  • \textbf{Pulley Diameter = D}
    • D = 30"
    • D = 15.88\text{ mm - O scale}
  • \textbf{Pulley Circumference = C}
    • C=7.85'
  • \textbf{250 RPM}
  • \textbf{Pulley Circumferential Velocity} v_{m} (Pulley Circumference in Feet x RPM)
    • v_{m}=C_f*RPM
    • v_{m}=1964 \text{ FPM}
  • \textbf{Pulley Power = P}
    • P = \frac{33000*H}{v_m}
    • P = \frac{33000*45}{1963}
    • P = 757
  • \textbf{Shaft Diameter = d}
    • d = 3"
    • d = 1.59\text{ mm - O scale}
  • \textbf{Belt width = b}
    • b=9"
    • b=4.77"\text{ mm - O scale}
  • \textbf{Face Width = B}
    • B = \frac{5}{4}b
    • B = 11\frac{1}{4}"
    • B=5.96\text{ mm - O scale}
  • \textbf{Rounding of Pulley Face = s}
    • s=\frac{1}{20}*b
    • s=\frac{1}{20}*9
    • s=.45"
    • s=.24\text{ mm - O scale}
  • \textbf{Rim Edge Thickness = k}
    • k = 0.08 + \frac{B}{100}
    • k = 0.08 + \frac{11.25}{100}
    • k=.193"
    • k=.11 \text{ mm - O scale}
  • \textbf{Pulley Nave Width = w}
    • w = 0.4 + \frac{d}{6} + \frac{R}{50}
    • w = 0.4 + \frac{3}{6} + \frac{15}{50}
    • w = 1.2"
    • w=.64 \text{ mm - O scale}
  • \textbf{Pulley Nave Length = L}
    • L = 2.5w
    • L = 3"
    • L=1.59 \text{ mm - O scale}
  • \textbf{Number of Arms = N}
    • N = \frac{1}{2} (5+\frac{R}{b})
    • N = 3.33
    • N=\text{4 - Rounding up}
  • \textbf{Arm Width at Nave = h}
    • h=0.24+\frac{b}{4}+\frac{R}{10*N}
    • h=2.865"
    • h=1.52 \text{ mm - O scale}
  • \textbf{Arm Width at Rim = }h_1
    • h_1=\frac{2}{3}*h
    • h_1=1.91"
    • h_1=1.02 \text{ mm - O scale}

3D Printing

3D CAD for O scale model : The dimensions shown are full-size in inches, 1:48 in mm with adjustments made for printing with mm inclosed in parentheses.

  • The rim (k) was calculated using k = 0.08 + \frac{B}{100} at .1 mm – which is way too small. I enlarged this to .3 mm as the minimum thickness for a resin printer.
  • The shaft hole diameter was enlarged  to 1.67 mm to clear the 1/16″ shaft (resin shrinks when curing).
  • I have concerns with the size of the arms as they are 1.01 mm at the rim – h_1=\frac{2}{3}*h and 1.52 mm at the Nave – h=0.24+\frac{b}{4}+\frac{R}{10*N} This is the width .. the depth being half that.
  • This is then the starting point using the sizes generated by the formulas. If the print fails then I will adjust as necessary.
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