# 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- Punching 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 Here’s the results of all the number crunching – a nice sectional view and then rendered in Sketchup to make a nice pretty picture.