### Frame V2 and Footings

I addressed the problem with the thin(ish) angle support beams by making hem a casting. I then added rubber blocks between them and the case and finally moving them to sit over the horizontal structure of the frame. This made sense both from 3D printing and in real life as this would be much stronger.

The concrete footings were basically ‘eyeballed’ .. there is enough space under the supported coal breaker for a conveyor belt. These were printed on my FDM printer and the layering actually adds to the look of a concrete casting .. IMO.

### Footing Height

When I had printed the footings and test fitted the frame on top I realized that some sort of jig would help align the footings. In the end I made a bit of ‘ground’ with 1mm slots that the bottom of the footings would fit into.

This ended up with a 17.777 mm height from the bottom of the frame/top of the footing to the ‘ground’.

### Pulley Dia

With the frame height now known I can now determine the pulley diameter.

From the center of the pulley shaft to the ground is 28.72 mm. If I make the radius of the pulley 27 mm this gives a 1.72 mm clearance between the face of the pulley and the ground – a little over 3″ full-size.

Previously on page 3 I calculated that using a 32.5″ dia pulley it would require a belt 20″ wide while a 64″ dia pulley would require a 10″ wide belt. If I used the 27 mm dimension shown here this would be 51″ full-size … or a 102″ dia pulley. I am curious to see how the belt width calculates. This is one of the reasons I put all of the calculations down on page 3 so I could do future calculations.

Pulley Diameter : 102″
Pulley Circumference : 320.4″
$v_m = 7090$
$P = 33000 * 45/7090 = 209.4$
Belt Width : 2.5″

Umm. 2.5″ .. this would be a really thin pulley .. in O scale only 1.32 mm wide. Not happy with that. A 6 ft dia / 72 inch pulley running at the stated HP and RPM has a P of 787.82 and would require a 9 in belt. In O scale this would be 0.187″/7.38 mm .. much better .. it would seem.

What has to be remembered is that the belt from this pulley runs to another ‘drive’ pulley which we can assume is on the engine/motor shaft. This pulley is under the same design constraints. If this drive pulley is the same diameter as the driven pulley then fine .. the 9″ belt width will work. If though the drive pulley is smaller than the belt has to be wider and the driven pulley simply has to match the width of the drive pulley.  If the drive pulley were 32″ dia then the belt width would be 20″.

Note: This Coal Breaker runs at 100 rpm. Suppose I have a steam engine which runs at 250 rpm. How would this affect the drive and driven pulley?

• Drive Pulley 32.5″ dia (I had previously run calculations for this dia pulley)
$H = \text{Pulley Horsepower}=45$
$RPM = \text{Steam Engine}=250$
$v_m = \text{Circumferential Velocity}=2095$
$P = \text{Circumferential Power}=33000 * 45/2095 = 709$
So. For $P = 709$ with riveted joints requires a 8″ wide belt. A single laced leather belt would need to be 14″ wide.
• Driven Pulley 80″ dia
$H = \text{Pulley Horsepower}=45$
$RPM = \text{Steam Engine}=100$
$v_m = \text{Circumferential Velocity}=2094$ (note that Circumferential Velocity is the same for both pulleys which means that …)
$P = 709$ so belt width is the same.

I went into great detail for Flat Belt Pulley design and the following is taken from that.

Pulley Width –
B = Face Width
b = Belt Width
$B = \frac{5}{4}b$
For the riveted joint 8″ wide belt we would need a 10″ wide pulley and for the single laced leather belt 14″ wide we would need a 17.5″ wide pulley. I have no problem with designing for the 10″ belt.

– Rim Edge Thickness –
k = Rim Thickness
B = Face Width = 10″
$k = 0.08 + \frac{B}{100}$
$k = 0.08 + \frac{10}{100}=0.18"$
I included this just for the fun of it. In O scale/1:48 that 0.18″ would be 0.00375″/.1 mm .. way too thin. This needs to be at minimum .3 mm and better at a .5 mm minimum.

– Rounding at Pulley Face –
s = Rounding at the pulley face
b = Belt Width = 10″
$s=\frac{1}{20}*b$
$s=\frac{1}{20}*10=0.5"$
.5″ in O scale/1:48 will be 0.0104″/0.265 mm .. again too thin. So this too is pretty much just “for the fun of it”.

– Pulley Nave –
d = pulley shaft dia = 1/8″ O scale = 6″ Full size
$w = \text{Thickness of a pulley nave}=0.4 + \frac{d}{6} + \frac{R}{50}$
$L = \text{Length of pulley nave}=2.50w$

32″ Drive Pulley:

• $w=1.72"=.036"/0.91 \text{ mm} \text{ O scale}$
• $L=4.375"=0.091"/2.315 \text{ mm} \text{ O scale}$

80″ Breaker Pulley:
$w=2.2" \ L=5.5=0.0.115"/2.91 \text{ mm} \text{ O scale}$

• $w=2.2"=..046"/1.168 \text{ mm} \text{ O scale}$
• $L=5.5"=0.114"/2.9 \text{ mm} \text{ O scale}$

– Number of Arms –
b = Belt Width = 8″
$N = \frac{1}{2} (5+\frac{R}{b})$
32″ Drive Pulley : $N = \frac{1}{2} (5+\frac{16}{8}=3.5=4)$
80″ Breaker Pulley : $N = \frac{1}{2} (5+\frac{40}{8})=5$

– Pulley Arm Widths –
b = Belt Width
N = Number of arms
$h = \text{width of arm with plane of pulley and taken at nave} = 0.24 + \frac{b}{4} + \frac{R}{10N})$
$h_1 = \text{width of arm at rim} = \frac{2}{3} X h$

32″ Drive pulley :

• $h = \text{width of arm with plane of pulley and taken at nave} = 0.24 + \frac{8}{4} + \frac{16}{40})=2.64=0.055"/1.397\text{ mm in O scale}$
• $h_1 = \text{width of arm at rim} = \frac{2}{3} X 2.64=1.76"=0.037"/0.93\text{ mm in O scale}$

80″ Breaker Pulley :

• $h = \text{width of arm with plane of pulley and taken at nave} = 0.24 + \frac{8}{4} + \frac{40}{50})=3.04=0.063"/1.6\text{ mm in O scale}$
• $h_1 = \text{width of arm at rim} = \frac{2}{3} X 3.04=2.03=0.042"/1.074\text{ mm in O scale}$

Note: These numbers are fun but mostly not applicable except as a general guide. In the end I can only design everything within the constraints of the 3D printing process.