## M N

where

W, WF and WR = Total, front and rear weight transfer respectively (N) t = Wheel tract (m)

10.3.7 Lateral (side) force distribution

The total lateral resisting forces generated at all tyre to ground interfaces must equal the centrifugal

Independent suspension (t>J Higid axle tieam suspension

Fig. 10.34(a and b) Comparison of rigid and independent suspension body roll stiffness

Independent suspension (t>J Higid axle tieam suspension

Fig. 10.34(a and b) Comparison of rigid and independent suspension body roll stiffness force acting through the body's centre of gravity. Thus the fore and aft position of the centre of gravity determines the weight distribution between the front and rear wheels and therefore the proportion of cornering force necessary to be generated by their respective tyres.

If Ff and FR are the front and rear tyre to ground cornering forces, then taking moments about FR

Therefore Ff FJ

Therefore Fâ

Fb I

Thus the amount of load and cornering force carried by either the front or rear tyres is proportional to the distance the centre of gravity is from the one or the other axle. Normally there is slightly more weight concentrated at the front half of the vehicle so that greater cornering forces and slip angles are generated at the front wheels compared to the rear.

10.3.8 Comparison of rigid axle beam and independent suspension body roll stiffness (Fig. 10.24)

A comparison between roll stiffness of both rigid axle beam and independent suspension can be derived in the following manner:

Consider the independent suspension (Fig. 10.34(a)). Let the centrifugal force F act through the centre of gravity CG at a height h above the roll centre RC. The overturning couple Fh produced must be equal and opposite to the reaction couple Wtw created by a reduction in the inside wheel reaction â W and a corresponding increase in the outside wheel reaction + W between the effective spring span tw-

If the vertical spring stiffness is S N/m and the vertical deflection at the extremes of the spring span is x m then the angle of body roll 0 degrees can be derived as follows:

tw tw/2

Weight transfer W = x S

Therefore Overturning couple = Fh and Reaction couple = Wt = Sxt

tan 0

but so tan 0

2x tw

2 Fh tStw

This formula shows that the body roll angle is proportional to both centrifugal force F and the couple arm height h but it is inversely proportional to both the spring stiffness k and the square of the spring span tw2, which in this case is the wheel track.

S tw2

A similar analysis can be made for the rigid axle beam suspension (Fig. 10.34(b)), except the spring span then becomes the spring base ts instead of tw. Because the spring span for a rigid axle beam suspension is much smaller than for an independent suspension (tw2 ^ ts2), the independent wide spring span suspension offers considerably more roll resistance than the narrow spring span rigid axle beam suspension and is therefore preferred for cars.

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