tre of the pin on which the connecting rod is to move, and the centre of the crank pin, for the length of the connecting rod. This method is given in Nicholson's Operative Mechanic and Mechanist, page 170; but it is incorrect, and when the length is so taken, the connecting rod will be found too long. TO CONSTRUCT AN ECCENTRIC WHEEL. From the centre of the shaft O (Fig. 9) take O P equal to half the length of the stroke which you intend the wheel to work; and from P as a centre, with any radius greater than P D, describe a circle, and this circle will represent the required wheel. For every circle, drawn from the centre P, will work the same length of stroke, whatever may be its radius; as, whatever you increase the distance of the circumference of the circle from the centre of motion on the one side, you will have a corresponding increase on the opposite side equal to it. Thus, suppose an eccentric wheel to work a stroke of 18 inches is required, the diameter of the shaft being 6 inches; and if 2 inches be the thickness of metal necessary for keying it on to the shaft, then set off, from O to P, 9 inches; and 9+ 5 = 14 inches, the radius of the wheel required. Formula. Let S represent the space the end A is moved through by the eccentric wheel, and s the space the slide moves. Then, A B x 8 BC XS; and this equation, solved for A B, B C, S, and s, gives the following: Ex. 1.-Given the length of the stroke of the slide = 8 inches, the length of the arm B C 4 inches, and the distance of the centre of the eccentric wheel from the centre of the shaft 10 inches; required the length of the arm A B. eccentricity 6 inches, to find the length of the arm B C. The crank is used for converting a rectilinear into a rotary motion. In the crank, as applied in the steam engine, the effect which is produced, is to the effect, were the force to act perpendicularly on the crank all the way round, as twice the diameter of a circle is to the circumference; in consequence of which, many practical men have considered that there is a corresponding loss of power by using a crank; without ever considering that the piston, or moving power, only moves through twice the diameter of the crank's orbit while the crank moves through its whole circumference. For here the same principle holds good as in all other mechanical contrivances, viz. the power multiplied by the space which it passes over, is equal to the weight or resistance multiplied by the space which it passes over. On the effective Leverage of a Crank. Let the circle A B C D represent the orbit of the crank, OD any position of the crank below the horizontal line, OA any position above it. Then, to find the effective leverage of the crank when in the position O D, produce the line P D, which represents the connecting rod; and from T B F F the centre O draw OE perpendicular to it; then OE will represent the effective leverage, which may be determined by trigonometry. For, in the triangle OD P, we have the length of the crank O D, the length of the connecting rod DP, and the angle DOP which the crank makes with a vertical line passing through O, to find the angle OD P. And by Part I. Art. 40, Cor. 3, the effect of a force acting obliquely against a lever is as the sine of the angle of direction which the force makes with the lever; and if the length of the crank be multiplied by the sine of this angle, the product is the effective leverage. Or it may be done by the following rule: Rule. Multiply the sine of the angle which the crank makes with the vertical line by the length of the crank, and divide this product by the length of the connecting rod, and the quotient will give the sine of the angle which the connecting rod makes with the vertical line. This angle being found, we may find the angle E DO, which is the supplement of the angle which the connecting rod makes with the crank. For, LE DO=LDPO+LDO P. And if the sine of the angle EDO be multiplied by the length of the crank in inches, the product will give the effective leverage in inches. Ex.-Given the length of the connecting rod 5 feet, the length of the crank 1 foot, and the angle which it makes with the vertical line 60 degrees, to find the effective leverage. The natural sine of 60°, the angle which the crank makes with the vertical line, •86603. angle which the connecting rod makes with the vertical line, which is, by a table of sines, 9° 38'. LEDO=2DPO+4DOP = 9°58′+60° = 69° 58′, the angle which the connecting rod makes with the crank, the natural sine of which is '93949, the effective leverage; or 93949 x 12 11.27388 inches. When the crank is in any position O B above the horizontal line, draw O F perpendicular to B P', and it will represent the effective leverage of the crank; and to find the angle which the connecting rod makes with the crank, we have the angle B O C which the crank makes with the vertical line A C, the length of the connecting rod, and the length of the crank given. But, in this case, we have 4 BOC 4 OBP + 4O P/ B. .. LOB PLBOC-LO P/ B. And if the sine of the angle O B P' be multiplied by the length of the crank in inches, the product is the effective leverage in inches. Ex.-If the connecting rod be 6 times the length of the crank, and the angle which the crank makes with the vertical line 51 degrees, The sine of 51° 78261; and 782616·13043 natural sine of the angle which the connecting rod makes with the vertical line. The corresponding arc is 74 degrees. = = = LOBP LBOC-LO P/B 514° 74° 44°, the angle which the connecting rod makes with the crank, the natural sine of which is 69466 the effective leverage; and 69466 x 12 8.33592 inches. × = ON THE GOVERNOR. In Art. 138, Part I. it was shewn that the periodic time varies as the square root of the altitude, or the altitude varies as the square of the periodic time, let the radius of the base be what it may. And since the number of revo lutions is inversely as the time, the altitude will be inversely as the square of the number of revolutions; and it is known that the distance between the point of suspension and the plane in which the centres of the balls move, is equal to the length of a pendulum which would make two vibrations, viz. one backward and one forward, in the same time that the balls perform one revolution. Now, the length of a seconds pendulum, or a pendulum which makes 60 vibrations in a minute, is 39 inches; therefore, if we consider the altitude of a conical pendu lum to be 39 inches, the corresponding number of revolutions in a minute is 30, this being known. Then, as the square of the number of revolutions is to the square of 30, so is 394 to the altitude required. Ex-What is the distance between the point of suspen sion of a conical pendulum and the plane of revolution, when it makes 25 revolutions per minute? 35212.5 25° 30°: 39.125; 625 tude required. 56.34 inches, the alti From this we have the following rule:-Divide 35212-5 by the square of the number of revolutions per minute, and the quotient is the altitude of the conical pendulum. Ex-If the length of the arms of a governor or conical pendulum be 30 inches from the point of suspension to the centre of the ball, and the pendulum revolves 40 times in a minute, what will be the diameter of the circle described by the centre of the ball? 35212.5 402 22 inches, the altitude or distance between the point of suspension and the plane of revolution. And, to find the radius of the circle described;-From the square of the length of the arm subtract the square of the altitude, and the square root of the remainder will be the radius required. Thus, (30o 22o) =20.4 inches pearly, the radius of the circle described by the ball. |