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.
What to read after Spiral And Worm Gearing a Treatise On the Principles Dimensions Calculation a?
You can find similar books in the "Read Also" column, or choose other free books by Erik Oberg to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
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.
What to read after Spiral And Worm Gearing a Treatise On the Principles Dimensions Calculation a?
You can find similar books in the "Read Also" column, or choose other free books by Erik Oberg to read onlineMoreLess
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