Iveco DAILY M.Y. 2012, Инструкция для тренажера

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GENERAL SPECIFICATIONS
1-11
D
AILY
Printed 603.95.241 Base - 02/2012
Dimensions and masses
1.15.2 Determining the centre of gravity of the body and payload
Positioning on the longitudinal plane
The examples given below may be used to determine the position of the superstructure’s centre of gravity and the payload.
The technical documentation specific to each model (chassis cab drawing) give the positions permitted with the vehicle in its standard
form. The masses and positioning of the single components of the vehicle are given in the chassis and weight distribution diagram.
166681
respectively L1
=
L
−
W2
⋅
L
W
Figure 1.2
W = Body + payload (kg)
W1 = Reaction to portion of W bearing on front axle (kg)
W2 = Reaction to portion of W bearing on rear axle centre line (kg)
L1 = Distance of the centre of gravity from the rear axle centre line (mm)
L = Wheelbase (mm)
L
1
=
W
1
⋅
L
W
Example of calculation of the load barycentre position
Consider a 40C13 vehicle with a wheelbase of 3,450 mm with:
1. GWW = 4,200 kg (permitted maximum: 1,900 kg on the front and 3,100 kg on the rear)
2. KERB WEIGHT = 1,955 kg (1,340 kg on the front axle, 615 kg on the rear)
The permitted maximum load (body + payload) will be W = 4,200 - 1,955 = 2,245 kg. Let us calculate the position of the center
of gravity in which the maximum permitted on the front axle is achieved. Let us assume an uniform distributed load .
In this case, out of 2,245 kg. W
1
= 1,900 - 1,340 = 560 kg will affect the front axle, while the remaining W
2
= 2,245 - 560 = 1,685
kg will affect the rear axle.
Thus, the following will be obtained:
1. W
1
= 560 kg
2. L = 3450 mm
3. W = 2245 kg
L
1
= W
1
x L / W = 860 mm
The center of gravity of the load (Body + payload) must not be more than 860 mm far from the rear axle; otherwise, the front
axle would be overloaded.