The Practical Mechanic; Comprising a Clear Exposition of the Principles and Practice of Mechanism, with Their Application to the Industrial Arts

The Practical Mechanic; Comprising a Clear Exposition of the Principles and Practice of Mechanism, with Their Application to the Industrial Arts

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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1879 edition. Excerpt: ...ins. 3. Required the diameter of a pinion to make 12$ revolutions in the same time as one of 32 ins. diameter making 26. 32x26 = 66.56 ins. 12.5 4. A shaft, having 22 revolutions per minute, is to drive another shaft at the rate of 15, the distance between the two shafts upon the line of centres is 45 inches; what should be the diameter of the wheels? Then, 1st, 22 + 15:22::45:26.75 = inches in the radius of the pinion. 2d. 22 +15:15::45:18.24 = inches in the radius of the spur. 5. A driving shaft, having 16 revolutions per minute, is to drive a shaft 81 revolutions per minute, the motion to be communicated by two geared wheels and two pulleys, with an intermediate shaft; the driving wheel is to contain 54 teeth, and the driving pulley upon the driven shaft is to be 25 inches in diameter; required the number of teeth in the driven wheel, and the diameter of the driven pulley. Let the driven wheel have a velocity of y/ i6 x 81 = 36, a mean proportional between the extreme velocities 16 and 81. Then, lst. 36:16::54:24 = teeth in the driven wheel. 2d. 81:36::25:11.11 = inches diameter of the driven pulley. 6. If, as in the preceding case, the whole number of revolutions of the driving shaft, the number of teeth in its wheel, and the diameters of the pulleys are given, what are the revolutions of the shafts? Then, 1st. 18:16::54: 48 = revolutions of the intermediate shaft. 2d. 15: 4S::25: 80 = revolutions of the driven shaft. To compute the Diameter of a Wheel for a given Pitch and Number of Teeth. Rote.--Multiply the diameter in the following table for the number of teeth by the pitch, and the product will give the diameter at the pitch circle. Example.--What is the diameter of a wheel to contain 48 teeth of 2.5 ins. pitch? 15.29 X 2.5 = 38.225 ishow more

Product details

  • Paperback | 100 pages
  • 189 x 246 x 5mm | 195g
  • Rarebooksclub.com
  • Miami Fl, United States
  • English
  • black & white illustrations
  • 1236530012
  • 9781236530011