The balancing of engines by Dalby William Ernest 1874-1918
Author:Dalby, William Ernest, 1874-1918
Language: eng
Format: epub, pdf
Tags: Steam-engines, Balancing of machinery
Publisher: London : E. Arnold
Published: 1920-03-25T05:00:00+00:00
Equation (9) becomes—
x\ = -1195 ± x/1'428 -f 2-643 therefore—
x\ = + -823
the negative value being untenable, since x\ must be real. Therefore, retaining the positive sign, Xi = + '908.
FIG. 110.
M.-607
M-6'6'7
FIG. 111. From equation (7), Art. 96—
2 X '908
= --551
Then—
1 = ±\/l - -908 2 = ±'421
2 = ±X/1-'551 2 = ±'834
The three assumptions made include —
7/1 is equal in magnitude and opposite in sign to 7/4
yi » » „ 2/3
The individual signs of y\ and 7/2 are to be determined from the general equations. From equation (5) —
MI and M 2 are positive, a\ and #2 are of the same sign ; therefore 7/1 must be opposite in sign to y%. Arranging the results—
Xi = +'908 7/1 = +'421, giving the direction of No. 1 crank
X2 = --551 y a = -'834, „ „ „ No. 2 „
. T3 = --551 7 /3 = +-834, „ „ „ No. 3 „
34 = +-908 7/4 =--421, „ „ „ No. 4 „
From equation (4), Art. 96 —
Mi : M 3 = cca : a* = '551 : '90S = -607 : 1 This is also the ratio M 4 : M 3
And-
MI = M.i = 1 by hypothesis
Therefore — •
M, = M :{ = -G07
Figs. 110 and 111 show the centre lines and crank angles, set out in their proper relative positions. The crank angles, in degrees, are added between the successive cranks and between the cranks Nos. 1 and 4, and 3 and 2.
Case II. Unsymmetrical Engine. — Given that /3 = 100° and that 8 = 90°, find the angles Ti, 72, the ratio of the reciprocating masses and the cylinder centre lines so that the engine may be in balance for primary and secondary forces and primary couples.
The following is a convenient semi-graphical way of solving this problem. Calculate the values of 7 2 and 7i from equation (11) of the previous article. Next calculate the value of MI from equation (12). Now draw an end view of the crank angles, numbering them as in Fig. Ill, and set out the two sides of the
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