Flat Belt Pulley – II

48 in. dia. Pulley
What the heck. Let’s see what I can come up with for a pulley 48 in. dia. with a 12 in. wide belt. To figure out what belt width depends on how much horsepower is being transmitted, what speed of rotation and so on. I can estimate that the turbine the pulley is for generates 24 h.p. .. shrug .. I have no idea what the rotational speed of the turbine would be so that’s all just nothing but guesses – so .. going with the 48 in. dia pulley and a 12 in. wide belt.
For isolated pulleys the face-width B is taken some-what greater than the width of the belt b; often we take –
B = \frac{5}{4}b = 15
So – that gives me a 15 in. width for the pulley.
-Rim Thickness-
The thickness k of the edge of the rim, or the thickness at the ends of the face-width, may be easily calculated from the formula –
k = 0.08 + \frac{B}{100}

So let’s stick in the 15 in. pulley width …
k = 0.08 + \frac{15"}{100} = .23"

-Rounding-
Rounding may be taken equal to \frac{1}{20} the width of the belt.
So. Rounding = \frac{15"}{20} = \frac{3}{4} in.

-Rim Cross Section-
rim_crossectionPunching in all that data .23 in. thick on the rim with a .75 in. rounding. That gives us a total .98 in. thickness at the center.
-Pulley-Nave-

Pulley-Nave thickness is w = 0.4 + \frac{d}{6} + \frac{R}{50} .. so .. stick in our numbers. The shaft diameter d (.063 in.) is 3 in. and the radius R is 24 in.

The .063 in. dia shaft was what I had on hand, some Evergreen \frac{1}{16}" rod. Sticking our numbers into the formula and we get w = 0.4 + \frac{3}{6} + \frac{24}{50} = 1.38". That gives an OD of 5.76". The length of the nave should not be less than 2.50w = 3.45"

-Number of Arms-

Number of arms is determined by the formula – N = \frac{1}{2} (5+\frac{R}{b}). Insert our numbers and we get N = \frac{1}{2} (5+\frac{24}{12}) = 3.5. Since it’s kind of hard to have fractional arms let’s round that up to 4

-Pulley Arm Width-

The formula h = 0.24 + \frac{b}{4} + \frac{R}{10N}) gives the greater diameter for the pulley-arms. Insert our numbers and we get – h = 0.24 + \frac{12}{4} + \frac{24}{10 X 4}) = 3.84. The width at right angles is determined by the formula – h_1 = \frac{2}{3} X h – insert our numbers and we get h_1 = \frac{2}{3} X 3.84 = 2.56

Section and Rendering
SectionViewHere’s the results of all the number crunching – a nice sectional view and then rendered in Sketchup to make a nice pretty picture.


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