## This blog illustrates how the position of the centre of gravity of a car can be calculated.

Fig (1) shows the forces on a stationary car. The earth’s gravitational pull (**mg**) acts through the centre of gravity and the reaction (remember: to every action there is an equal and opposite reaction) acts through the contact patches between the tyres and the road. The vectors shown represent the combined reactions at both front wheels (**R _{1}**) and both rear wheels (

**R**).

_{2}We measure the reaction at each wheel on every car we test, and use this information to publish the front/rear weight distribution. We also know the length of the wheelbase, and hence we have enough information to calculate the position of the centre of gravity in a vertical plane when seen from the side.

To do this, we need to know that when an object is stationary three conditions are necessary to ensure that neither linear nor rotational acceleration is taking place:

1. The sum of the vertical forces must be zero.

2. The sum of the horizontal forces must be zero.

3. The clockwise and anti-clockwise moments must be equal in size but opposite in direction.

[*Moment* is just another name for torque ie force times perpendicular distance; the word is used mainly in when the objects being studied are not moving]

These conditions are NOT sufficient to ensure that the object will be stationary, because forces in balance imply either constant speed motion or a state of rest. However, they do form the cornerstone of a great deal of civil engineering calculations because buildings and bridges are objects that don’t move.

Let’s apply the above conditions. There are no horizontal forces, and the vertical forces are known because we measure them, so that we only have to equate the two moments around the centre of gravity caused by the reactions at the wheels.

R1.s_{1} (clockwise) = R2.s_{2} (anti-clockwise) …………………(1)

At this point we’re have two unknowns (s_{1} and s_{2}) and only one equation. One of the interesting results from mathematical theory is the fact that we need as many equations as there are unknowns to enable us to find a solution. This means we have to find another equation linking s_{1} and s_{2}, and we’re in luck because these distances add up to the wheelbase S. Thus we can write:

s_{1} + s_{2} = S……………………………………………………..(2)

At this stage we can assign numbers to the symbols whose values are known. If we take the Porsche 911 GT3 that we tested in the April 2010 issue as an example, we find that m = 1440 kg and S = 2355 mm. The front wheels carry 0,39 per cent of the mass and the rear wheels carry 0,61 per cent.

The weight of the car is equal to 1440 x 9,8 = 14112 N (Mass times g).

The two reaction forces are R_{1} = 0,39 x 14112 = 5504 N and R_{2} = 0,61 x 14112 = 8608 N.

Equation (1) thus becomes:

5504.s_{1} = 8608.s_{2}

and equation (2) becomes:

s_{1} + s_{2} = 2355

from which we deduce that

s_{1} = (8608/5504).s_{2}

or s_{1} = 1,56.s_{2}

We now substitute this value into equation (2) to get:

1,56.s_{2} + s_{2} = 2355

So that s_{2}.(1,56 + 1) = 2355 or s_{2}.(2,56) = 2355

giving s2 = 2355/(2,56) = 918,5 mm.

This means that the centre of mass is 910 mm from the rear wheel centreline.

A similar calculation, utilising a frontal view of the vehicle, can be used to determine the position of the centre of gravity in a transverse plane. However, to calculate the height of this centre one would have to measure either R_{1} or R_{2} when the vehicle is parked at an angle of at least 20 degrees.

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