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Application Information In many applications, Joyce Bevel Gear jacks are more efficient and faster than worm gear driven jacks. To determine the suitabilityof a Bevel Gear jack for your application, use the steps below to calculate load, travel speed and duty cycle. Step 1 - Determine load in pounds. Step 2 - Determine velocity in feet per minute. Step 3 - Determine duty cycle in terms of minutes operation (time on) and minutes resting (time off). Step 4 - Calculate PVPV=Load in pounds x velocity in feet per minute 1000 Step 5 - Calculate TT= cooling time (from chart on P. 95) x T ime on Time off Step 6 - Plot the point on the graphs for the calculated values. If the point falls below the line the application is satisfactory. If it isabove the line, recalculate T for the next size jack and plot the new values for the next larger size jack. Note: T must be calculated for the size jack corresponding to the graph, since each jack has a different cooling time. Step 7 - Calculate horsepower requiredR.P.M. = velocity in feet per minute x 12 x input turns per 1" travel (from chart on P. 95) Horsepower = Pinion tor que (fr om char t) x Load x RPM A 5000 pound load must be raised 30 inches in 15 seconds. The load remains in position for 2 minutes. It is then lowered and 63,025 Sample remainslowered for 30 seconds. The cycle begins again. Determine the appropriate bevel gear jack and calculate the horsepower required. Step 1 - Load = 5000 pounds Step 2 - Velocity = 30 inches in 15 seconds = 10 f.p.m. Step 3 - Duty Cycle =Time on = 15 sec. up + 15 sec. down = 30 sec. = .5 min. Time off = 2 min. up + 30 sec. down = 2 min. 30 sec. = 2.5 min. off Step 4 - PV = 5000 x 10 = 501000 Step 5 - T = 38 (for BG 150) x .5 = 7.62.5 Step 6 - The point for PV, 50, and T, 7.6, falls below the line for BG 150-D and above the line for BG 150-S, therefore BG 150-D isappropriate. Step 7 - RPM = 10 x 12 x 5.38 = 645.6Horsepower = .066 (5000) x 646 = 3.3863,025
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