1-10 GENERAL SPECIFICATIONS E UROCARGO Euro 4/5 Figure 1.1 refers to vehicles with 2 axles. Figure 1.1 Positioning on the longitudinal plane 91517 or L 1 = L − W 2 ⋅ L L 1 = W 1 ⋅ L W W W = Outfit + useful load (kg) W 1 = Portion of W weighing on the front axle (kg) W 2 = Portion of W weighing on the centreline of the rear axle or on the centreline of the bogie (kg) L 1 = Distance of the center of gravity from the centreline of the rear axle or from the centreline ofthe bogie (mm) L = Calculation wheelbase (mm) Computational example of the position of the centre of gravity of the load We consider an ML120E18/P vehicle, wheelbase 4185 mm, with 1.
GVW = 12000kg (maximum permissible 4400kg on front and 8480kg on rear) 2. kerb weight = 4220kg (2760kg on front axle and 1460kg on rear) The maximum permissible load (outfit + useful load) is W=12000-4220=7780 kg. We obtain the position of the centre of gravity where the maximum permissible load on the front axle is reached. We hypothesize a uniform load distribution. In this case, of the 7780 kg, W1=4400-2760=1640 kg will go on the front axle and the remainder W2=7780-1640=6140 kg on the rear axle. So we will have 1. W 1 = 1640 kg 2. L = 4185 mm 3.
W = 7780 kg L 1 = W 1 x L / W= 882 mm The centre of gravity of the load (outfit + useful load) can be no more than 882mm from the rear axle, otherwise there would be an overload on the front axle. Dimensions and weights Base - January 2008 Print 603.93.431