Suspension. The crane track is laid level, laterally. Therefore
the difference of elevation between points c and b, or h and a, is 3.193 feet
multiplied by 1.00 percent, which is equal to the difference between points c
and e or a and f. See A of Figure 10 for the elevations of these points.
Frame corner "a" is suspended 0.0319 feet multiplied by 2, and the sill at
point A of Elevation in Figure 10 is suspended an identical amount, or 0.0638
feet. The suspension at point B, a pivot point, equals the suspension at
point A. The truck intermediate equalizers, in all probability, are not
exactly balanced; therefore, assume point D descends and point C rises.
Vertical Position. The vertical position of point C is now 0.1277
feet above normal; it follows that point E is the same amount above normal.
The drive truck assembly No. 2 being heavier than the idler truck assembly No.
2, will cause point F to rise 0.1277 feet above point E. Point F will now be
0.2554 (3 1/16 inches) feet above point G, the normal position. The flange is
1-1/8 inches in depth, which leaves a clearance (3-1/16 - 1-1/8) = 1-15/16
inches (0.1615 feet) between the top of the rail and the flange on each wheel
of idler truck assembly No. 2, assuming that the idler truck assembly remains
balanced because of friction. This clearance will permit the truck assembly
to swivel and cause derailment. Distance (S) of the wheels of one truck
assembly above the rail is determined by:
S ' R [radian (arc sin
& arc sin
)] (g) [ ]
in which g is the rate of grade on the inside rail, and w is the total number
R1 = 120
= 32 (from equation 9)
= 100 (Radian 1 deg 49 min 47 sec) (0.01) (8)
= 0.2554 or 3-1/16 inch
Balance. If the wheels or idler truck assembly No. 2 do not remain
balanced, which is the most likely condition, the distance (S) which one
wheel, or pair of wheels, will be above the rail is twice the amount shown by
equation (9) and the formula for the worst, most likely condition will be: