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Lightweight Electric/Hybrid Vehicle Design

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<strong>Design</strong> for optimum body-structural and running-gear performance efficiency 227<br />

tabular approach to gravimetric analysis is combined with moment equations to determine axle<br />

loadings; the numbers locating the main assemblies on the vehicle correspond with the first column<br />

of the table and go on to include front wing (5), front suspension (6), power unit (7) and on to<br />

(51)/(52) as shown. The alignment charts are nomograms which represent the weight of a given<br />

material as a function of the surface area or length and help in the estimation of body structural<br />

weight, a surprisingly large percentage of the total on conventional cars. Successful prediction of<br />

body weight distribution is key to the early fixing of centre-of-gravity position upon which<br />

performance prediction depends.<br />

The importance of vehicle weight is made clear by Taborek 9 who points out that a key relationship<br />

in performance considerations is that tractive force P is equal to load on the driving axle W times<br />

tyre-to-ground adhesion coefficient μ. But, while W for a static vehicle can easily be found by a<br />

moment calculation, the dynamic axle reactions of a moving vehicle are more difficult to work<br />

out. The dynamic weight transfer:<br />

dW = (H/L)(T d /r −fW) and P = (T e RE)/r<br />

where T d is torque at drive axle, T e torque at engine, R total reduction ratio, E transmission efficiency<br />

and r rolling radius. H is height of vehicle centre of gravity (CG) above ground, L the wheelbase<br />

and f the symbol for rolling resistance coefficient.<br />

With front and rear-wheel drive, dynamic front axle load = W(L r + fH)/(L + μH) where L r is the<br />

distance from CG to rear axle and μ is adhesion coefficient; this leads to the maximum transferable<br />

tractive force as<br />

P f,max = μW[(L r + fH)/(L + μH)]<br />

the square-bracketed term being the weight distribution factor dW/W = w so that W df = W wf P max =<br />

μW wf and P r.max = μWdr<br />

With four-wheel-drive, P 4.max = μW and w 4 = 1; otherwise the chart at (c) can be used to determine<br />

the relative distributions. The weight distributions derived above are dimensionless, being derived<br />

from ratios, and the chart also contains only dimensionless factors, with the weight of the vehicle<br />

being eliminated by dividing the tractive force equation by W, giving P max /W = μw.<br />

The dotted line shows an example for a vehicle with h/L = 0.35 and L f /L = 0.45 defining its CG<br />

position, for which it is required to find maximum transferable tractive force and the maximum<br />

engine torque for a low gear reduction factor of RE/r = 7.3 – given also μ = 0.6 for the adhesion<br />

coefficient. Starting in the lower left quadrant, the intersection of the m = 0.6 line cuts curves for<br />

front and rear wheel drive at h/L = 0.35. Horizontal projections from those points similarly lead to<br />

the P max /W scale. Values thus obtained give the first solution while projecting into the top left<br />

quadrant gives the second. For a vehicle weight of 4000 kg, P max = 1620 kg and T e.max = 224 kNm<br />

(front drive) or 840 kg and 112 kNm (rear drive).<br />

In predicting performance, Fig. 8.16, this data can be used in a chart, such as at (a), to obtain<br />

available vehicle speeds on different gradients. The chart plots maximum values of tractive force<br />

at the wheel, as functions of speed, in different gears. Also on the chart, gradient slopes are plotted<br />

against vehicle weight. The 0% line is drawn at a slope designating a rolling resistance of 35.4 lb/<br />

ton (80 kg/tonne). Propelling force P lines are obtained by graphical subtraction of wind resistance<br />

from maximum tractive force, in this method due to Goldschmidt and Hadar of the Israel Institute<br />

of Metals. Dotted lines illustrate worked examples for vehicles weighing 1910 and 1875 lb<br />

respectively. The first is shown to be able to climb a slope of 10% at 36 mph reducing to 9% at the<br />

same speed if that speed is maintained. The second climbs an 8% slope at 43.5 and 38.5 mph in

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