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. 1912 Excerpt: ... balancing force, the latter must pass through A and also through the stationary point bd, so that its direction is known. The rector polygon on the right gives its magnitude, thus P2 is to be as shown. Example 2. Find the resistance P2 which must be produced by the crank pin of an engine, Fig. 75, to balance the force P, on the piston, Fig. 75 In this case both P, and P2 may be regarded as forces on b and these will intersect at bc, also their resultant must pass through the centre bd and also through bc and is thus known in direction and position. In the diagram to the right draw P, = A B to scale to represent the pressure on the piston, draw A C parallel to P and B C to P2 then P2 is given by the length B C. The moment of P2 on the crank shaft is OD X P2, and it may readily be shown by geometry that this is equal to P, X Oac, or the turning effect on the crank shaft due to P, is found in magnitude, direction and sense by simply transferring P, to ac. Let the force P3 act normal to a, to find its magnitude. Here P3 and P, intersect at H and their resultant passes through H and bd. Draw the triangle EFG making EF equal to the known value of P, and FG and EG parallel respectively to P3 D and bd--H. Then FG = P3 and EG equals the resultant force P' of P, and P3 and the force P3 is called the crank effort, being the force passing through the crank pih normal to a which just balances the steam pressure P,. Example 3. The direction of pressure between the teeth of a pair of gears is AB, Fig. 76, the pitch circles of which are shown, to find the relation between P, and P2. In this case AB meets P, in A and P2 in B, join A--ad and B--bd. The forces acting on a are now P, and the force P3 due to the gear b and the resultant P of these must pass through ad so that ...

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