Spiral And Worm Gearing a Treatise On the Principles Dimensions Calculation a

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This may not be enough to allow for 178 WORM GEARING a worm shaft of sufficient strength; besides it would give a very narrow face to the worm-gear. We, therefore, probably prefer to use a double -threaded worm, the pitch diameter of which i- X 2 will be = 7. 42 inches. The face of the worm-gear will then be f X 7. 42 = 4-95 inches, or, say, 5 inches.
Assuming a worm-gear of 28 inches pitch diameter, if inch pitch, and 50 teeth, the worm shaft will be running 5 X -/- = 125 R. P. M. , which give
...s a speed of rubbing of n -. , * 12 X cos8f deg.
= 247 feet per minute.
Referring to the table, "Safe Load on Worm-gear Teeth, " we find for a speed of 250 feet per minute an allowable load of 371 pounds per unit of product (pitch X width of tooth). The total allowable load in this case will be37iXi|X5= 3246 pounds. This load at 14 inches radius gives a turning moment of 14 X 3246 = 45, 444 inch-pounds, while only 42, 000 inch- pounds is required.
If the above machine were applied for lowering packages in- stead of elevating same, as previously assumed, the gearing would have to lock while running at a full speed of 247 feet per minute, at which speed we would not have a friction coefficient of more than 0.


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