### 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"$$C=7.85'$
$v_{m}=C_f*RPM$$v_{m}=7.85*250$$v_{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.