### Pump Pulley My best guess .. call it a ‘qualified guesstamation’ (based on the very fuzzy drawing) is that the pump has a 10″ Dia pulley mounted. It is driven by a 74″ Dia pulley (250 RPM) giving a pump RPM of 1850 RPM. In the text it states .. “The water from the basin, which overflows from the top into a well is carried by any means desired — commonly a centrifugal pump on account of capacity and low duty— into what is termed the supply tank.

A search online found Belt Driven Centrifugal Pumps which run at 1440 RPM so this is probably close to correct. If I reduce the Pump Drive Pulley from 72″ to 69″ and enlarge the Pump Pulley from 10″ to 12″ then we get 1438 RPM. (I have an Excel spreadsheet set up so I can merrily change the numbers to get the RPM I need).

### Data • $\textbf{HP=45}$
• $\textbf{Pulley Diameter = D}$
• $D = 5"$
• $D = 2.65\text{ mm - O scale}$
• $\textbf{Pulley Circumference = C}$
• $C=3.14'$
• $\textbf{1440 RPM}$
• $\textbf{Pulley Circumferential Velocity}$ $v_{m}$ (Pulley Circumference in Feet x RPM)
• $v_{m}=C_f*RPM$
• $v_{m}=4,524 \text{ FPM}$
• $\textbf{Pulley Power = P}$
• $P = \frac{33000*H}{v_m}$
• $P = \frac{33000*45}{4524}$
• $P = 329$
• $\textbf{Shaft Diameter = d}$
• $d = 3"$
• $d = 1.6\text{ mm - O scale}$
• $\textbf{Belt width = b}$
• $b=4"$
• $b=2.12"\text{ mm - O scale}$
• $\textbf{Face Width = B}$
• $B = \frac{5}{4}b$
• $B = 5"$
• $B=2.65\text{ mm - O scale}$
• $\textbf{Rounding of Pulley Face = s}$
• $s=\frac{1}{20}*b$
• $s=.2"$
• $s=.11\text{ mm - O scale}$
• $\textbf{Rim Edge Thickness = k}$
• $k = 0.08 + \frac{B}{100}$
• $k=.13"$
• $k=.11 \text{ mm - O scale}$
• $\textbf{Pulley Nave Width = w}$
• $w = 0.4 + \frac{d}{6} + \frac{R}{50}$
• $w = 1.02"$
• $w=.54 \text{ mm - O scale}$
• $\textbf{Pulley Nave Length = L}$
• $L = 2.5w$
• $L = 2.55"$
• $L=1.35 \text{ mm - O scale}$
• $\textbf{Number of Arms = N}$
• $N = \frac{1}{2} (5+\frac{R}{b})$
• $N = 3.2$
• $N=\text{4 - Rounding up}$
• $\textbf{Arm Width at Nave = h}$
• $h=0.24+\frac{b}{4}+\frac{R}{10*N}$
• $h=1.39"$
• $h=.74 \text{ mm - O scale}$
• $\textbf{Arm Width at Rim = }h_1$
• $h_1=\frac{2}{3}*h$
• $h_1=.927"$
• $h_1=.5 \text{ mm - O scale}$

### 3D Printing 