MIL-HDBK-1005/6
Example of Vertical Alignment. A horizontally curved track shall
3.17.1
be placed on level grade. Derailment, with attendant danger of overturning,
is probable because of the following conditions.
a) Points on the rails directly under the corners of the crane
traversing the curved track would be at different elevations and not in the
same plane. Because the frame and legs are a rigid structure, the sills and
girders cannot appreciably conform to other than a plane surface.
b) In understanding the geometry of this problem, consider the
crane as simply a rigid rectangular frame. The higher inside corner and the
lower outside corner of this frame (always diagonally opposite) will remain on
the rails at all times. Of the other two corners, the one nearer the center
of gravity will be forced to the rail, thus suspending the fourth corner
(diagonally opposite) above the rail by twice the amount of the difference in
elevation at points on the rails under laterally opposite corners.
c) An identical suspension occurs at the corresponding point on
the sill of the actual crane traversing such track. The suspension of one
truck above the rail will be the amount of suspension at the sill multiplied
by N' where:
(1)
N' = total number of pivot points (at one corner) + 1
2
(2)
N' = total number of wheels of crane
8
Application. The following hypothetical case illustrates the
3.17.2
application of this procedure. From A of Figure 10:
$ = arc sin 18.8716
(3)
100.00
$ = 10 deg 52 min 40 sec
N = arc sin 18.8716
(4)
120.00
N = 9 deg 02 min 53 sec
)2 = $ - N = 1 deg 49 min 47 sec
(5)
Arc cb and arc ha = (100.00)(radian 1 deg 49 min 47 sec)
= 3.193 ft.
(6)
)1 = 2N = 18 deg 05 min 46 sec
(7)
Arc bh = (100.00)(radian 18 deg 05 min 46 sec)
= 31.584 ft.
(8)
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