### Coal Breaker Pulley

This is the pulley on the Coal Breaker itself. I am designing for the larger breaker which requires 45 hp. I have a page on crushers down to 10 hp on my Coal Breaker page. You can the assortment of breakers from 10 HP with a 10 ton per hour capacity up to the 45 HP breaker with a 83 ton per hour capacity.

The pulley is 64″ in diameter with a 12-1/2″ face for a 10″ belt. It has five arms and an opening for a 3″ shaft. I changed the opening to 6″ to match the shafting on my Coal Breaker. This was calculated as you can see below.

3D printing : I am putting progress on this on the bottom of the page.

### Data

• $\textbf{HP=45}$
• $\textbf{Pulley Diameter = D}$
• $D = 64"$
• $D = 33.87\text{ mm - O scale}$
• $\textbf{Pulley Circumference = C}$
• $C=16.755'$
• $\textbf{100 RPM}$
• $\textbf{Pulley Circumferential Velocity}$ $v_{m}$ (Pulley Circumference in Feet x RPM)
• $v_{m}=C_f*RPM$
• $v_{m}=1689 \text{ FPM}$
• $\textbf{Pulley Power = P}$
• $P = \frac{33000*H}{v_m}$
• $P = \frac{33000*45}{1689}$
• $P = 879$
• $\textbf{Shaft Diameter = d}$
• $d = 6"$
• $d = 3.18\text{ mm - O scale}$
• $\textbf{Belt width = b}$
• $b=10"$
• $b=5.29"\text{ mm - O scale}$
• $\textbf{Face Width = B}$
• $B = \frac{5}{4}b$
• $B = 12\frac{1}{2}"$
• $B=6.61\text{ mm - O scale}$
• $\textbf{Rounding of Pulley Face = s}$
• $s=\frac{1}{20}*b$
• $s=\frac{1}{20}*10$
• $s=.5"$
• $s=.265\text{ mm - O scale}$
• $\textbf{Rim Edge Thickness = k}$
• $k = 0.08 + \frac{B}{100}$
• $k = 0.08 + \frac{12.5}{100}$
• $k=.205"$
• $k=.11 \text{ mm - O scale}$
• $\textbf{Pulley Nave Width = w}$
• $w = 0.4 + \frac{d}{6} + \frac{R}{50}$
• $w = 0.4 + \frac{6}{6} + \frac{32}{50}$
• $w = 2.2"$
• $w=1.17 \text{ mm - O scale}$
• $\textbf{Pulley Nave Length = L}$
• $L = 2.5w$
• $L = 5.5"$
• $L=2.92 \text{ mm - O scale}$
• $\textbf{Number of Arms = N}$
• $N = \frac{1}{2} (5+\frac{R}{b})$
• $N = \frac{1}{2} (5+\frac{32}{10})$
• $N = 4.1$
• $N=\text{5 - Rounding up}$
• $\textbf{Arm Width at Nave = h}$
• $h=0.24+\frac{b}{4}+\frac{R}{10*N}$
• $h=0.24+\frac{10}{4}+\frac{32}{10*5}$
• $h=0.24+2.5+.64$
• $h=3.54"$
• $h=1.88 \text{ mm - O scale}$
• $\textbf{Arm Width at Rim = }h_1$
• $h_1=\frac{2}{3}*h$
• $h_1=\frac{2}{3}*3.38$
• $h_1=2.36"$
• $h_1=1.25 \text{ mm - O scale}$

### 3D Printing

• Plan: These calculations were generated for a “Real-World” pulley. To 3D print these in 1:48 requires some modifications. The Rim Edge Thickness is shown as .205″ which scales to 0.11 mm .. way too thin. I increased this to 0.3 mm .. which should be ok. The same with the arms .. except I will just say .. “may” be ok. If there are problems .. which would be related to simply the thickness of the parts I will enlarge them until the pulley prints without problems.
• Results – Test Print #1: The first test print actually went pretty well. I printed it flat without any tilt .. sometimes you CAN break the rules. I did not want any support scars on the face of the pulley.
• There was sag between the supports as I had too few spaced too far apart. You can see on the far side where the print pulled away from supports. The hole in the center was a tight fit on a 1/8″ rod .. looking at the CAD I found I had forgotten to pad that hole a bit so for the next print I enlarged it to 0.128″ dia.
• I added the supports manually for this test. The next one I will use Chitubox to add supports