Here is another article. In this I will be discussing the effect weight distribution has on a car while it is in a corner. i.e. oversteer or understeer.

As examples, I will use two cars with the large weight biases in either direction.

The greatest weight distribution to the rear that I know of is the Porsche 911 GT2.
Weighing in at 3175 lbs and with a 37/63 weight distribution and 93 inch wheelbase.
The car with the greatest weight distribution to the front that I know is the Volkswagon Scirocco.
Weighing in at 2906 lbs and with a 64/36 weight distribution and a 106 inch wheelbase.

Before we start crunching numbers it is important to know that weight distribution has an important impact on 2 things in cornering. The first is overall turning moment generated by the car in a corner. And the second is the strength that a turning input has due to the position of its COG.

The first step in determining the turning moment produced by a car we simply take the weight distribution and multiply it by the wheelbase to find the position of the COG with respect to the wheels. For the sake of these calculations, I will determine the distance from the rear tires.

For the Porsche 0.37*(93 inch) = 34.41 inch
For the VW 0.64*(106 inch) = 67.84 inch

Next, by subtracting half of the wheelbase we can find out the location of the COG with respect to the center of the car.

For the Porsche (34.41 inch)-(93 inch)/2 = -12.09 inch (or about a foot behind the vehicles center)
For the VW (67.84 inch)-(106 inch)/2 = 14.48 inch (over a foot in front of the vehicles center)

Now, to determine the centripetal force. We multiply the vehicles max lateral G-force by its weight.

For the Porsche (1.10 g's)*(3175 lbs) = 3492.5 lbs
For the VW (0.90 g's)*(2906 lbs) = 2615.4 lbs

When we multiply the centripetal force by the distance away from the center of the car we end up with a torque. If we convert the distance into feet first we can get this torque in units of ft-lbs

For the Porsche (12.09 inch)/12 = 1.0075 ft
(1.0075ft)*(3492.5 lbs) = 3518.69 ft-lbs
For the VW (14.48 inch)/12 = 1.2067 ft
(1.2067ft)*(2615.41 lbs) = 3155.93 ft-lbs

This is the torque the car sees as it is going around a corner right before it loses traction. A car with a 50/50 weight distribution will see none of this torque.

Those are fairly big numbers, for comparison sake, imagine a ten foot pole sticking out from the cars center. For the Porsche that pole would be seeing a 352 lb force trying to spin it into the corner (i.e. oversteer) For the VW, at the end of that ten foot pole it would have a similar 316lb force trying to spin it away from the corner.

Next, the effect weight distribution has on turning input.

First lets determine the distance from the front wheels.

For the Porsche, (93 inch)/2+12.09 inch = 58.59 inch
For an ideal 50/50 Porsche, (93 inch)/2 = 46.5 inch
For the VW, (106 inch)/2-14.48 inch = 38.52 inch
For and ideal 50/50 VW, (106 inch)/2 = 53 inch

These are the moment arms for which the front wheels apply their turning moment. The longer it is, the greater the turning moment the front wheels produce.

Comparing the Porsche to an ideal Porsche, its moment arm is 26% greater than an ideal moment arm. This means that not only will the car have a tendency to turn in, but due to the COG location its front wheels will be 26% more sensitive to steering input. This gives the car a very sensitive magical tuning feel, and it also makes spinning out even easier to accomplish

Comparing the VW to an ideal version of itself, its moment arm is 27.3% smaller then the ideal moment arm. Also meaning that the Scirocco will be 27.3% less sensitive to steering input. This also amplifies the problem of understeer we calculated earlier.

And there you have it, you now have a further understanding of the important effect weight distribution has on a vehicle.

The Effect of Weight Distribution on Turning Moments

About the Author