Kenworth K270, Body Builder'S Manual

Page: 90 / 96

B-6
Appendix B
Weight Distribution
Use CGf = 246 in equations 1 and 2 to determine how the liftgate weight is distributed to the axles.
Lr =
CGf
WB
X L
322.2
214 X 1455 = 2190.6 lbs (993.6 kg)
Lf = L - Lr 1455 - 2190.6 = -735.6 lbs (-333.6 kg)
This negative weight on the front axle illustrates the difference between the distribution of weight (L) mounted
behind the rear axle versus in front of the rear axle.
• The load carried by the rear axle is greater than the weight of the liftgate itself. Since the weight of
the liftgate (added to the vehicle) cannot be greater than 1,455 lb, the front axle loading is reduced
by a compensating amount (735.6 lb). The combined weight on the front and rear axles is equal to
that of the liftgate.
• Weight added behind the rear axle has the effect of unloading the front axle. The amount of this
front axle load reduction is equal to the “extra” weight added to the rear axle.
• By positioning equipment behind the rear axle, the effective load on the rear axle is more than the
weight of the equipment.
• The farther behind the rear axle the load is mounted, the greater the load on the rear axle. However,
the combined weight, distributed to the front and rear axles (Lf plus Lr), does not exceed the weight
of the liftgate.
In order to get a realistic curb weight, we add weight for a driver and fuel. For purposes of calculation, we use a
standard of 200 lbs. (91 kg) for the driver. Of course, your driver weight will vary.
Using CGf = 10 in equations 1 and 2:
Lr =
CGf
WB
X L
10
214 (200) = 9.34 lbs (4.2 kg)
Lf = L - Lr 200 - 9.34 = 190.6 lbs (86.4 kg)
We calculate the fuel load using 7 lbs per gallon as the weight for diesel fuel.
45 gal x (7 lbs/gal) = 315 lbs (142.8 kg)
Using CGf=73.9 and equations 1 and 2 for the standard tank:
Lr =
CGf
WB
X L
73.9
214 (315) = 108.7 lbs (49.3 kg)
Lf = L - Lr 315 – 108.7 = 206.2 lbs (93.53 kg)