Braking causes Lf to be greater than Lr. Literally, the "rear end gets light," as one often hears racers say. Consider the front and rear braking forces, Bf and Br, in the diagram. They push backwards on the tyres, which push on the wheels, which push on the suspension parts, which push on the rest of the car, slowing it down. But these forces are acting at ground level, not at the level of the CG. The braking forces are indirectly slowing down the car by pushing at ground level, while the inertia of the car is 'trying' to keep it moving forward as a unit at the CG level.

The braking forces create a rotating tendency, or torque, about the CG. Imagine pulling a table cloth out from under some glasses and candelabra. These objects would have a tendency to tip or rotate over, and the tendency is greater for taller objects and is greater the harder you pull on the cloth. The rotational tendency of a car under braking is due to identical physics.

The braking torque acts in such a way as to put the car up on its nose. Since the car does not actually go up on its nose (we hope), some other forces must be counteracting that tendency, by Newton's first law. G cannot be doing it since it passes right through the centre of gravity. The only forces that can counteract that tendency are the lift forces, and the only way they can do so is for Lf to become greater than Lr. Literally, the ground pushes up harder on the front tyres during braking to try to keep the car from tipping forward.

By how much does Lf exceed Lr? The braking torque is proportional to the sum of the braking forces and to the height of the CG. Let's say that height is 20 inches. The counterbalancing torque resisting the braking torque is proportional to Lf and half the wheelbase (in a car with 50-50 weight distribution), minus Lr times half the wheelbase since Lr is helping the braking forces upend the car. Lf has a lot of work to do: it must resist the torques of both the braking forces and the lift on the rear tyres. Let's say the wheelbase is 100 inches. Since we are braking at one g, the braking forces equal G, say, 3200 pounds. All this is summarized in the following equations:

3200 lbs times 20 inches = Lf times 50 inches - Lr times 50 inches

Lf + Lr = 3200 lbs (this is always true)

With the help of a little algebra, we can find out that

Lf = 1600 + 3200 / 5 = 2240 lbs,  Lr = 1600 - 3200 / 5 = 960 lbs

Thus, by braking at one g in our example car, we add 640 pounds of load to the front tyres and take 640 pounds off the rears! This is very pronounced weight transfer.

By doing a similar analysis for a more general car with CG height of h, wheelbase w, weight G, static weight distribution d expressed as a fraction of weight in the front, and braking with force B, we can show that

Lf = dG + Bh / w,  Lr = (1 - d)G - Bh / w

These equations can be used to calculate weight transfer during acceleration by treating acceleration force as negative braking force. If you have acceleration figures in gees, say from a G-analyst or other device, just multiply them by the weight of the car to get acceleration forces (Newton's second law!). Weight transfer during cornering can be analysed in a similar way, where the track of the car replaces the wheelbase and d is always 50% (unless you account for the weight of the driver). Those of you with science or engineering backgrounds may enjoy deriving these equations for yourselves. The equations for a car doing a combination of braking and cornering, as in a trail braking manoeuvre, are much more complicated and require some mathematical tricks to derive.

Now you know why weight transfer happens. The next topic that comes to mind is the physics of tyre adhesion, which explains how weight transfer can lead to understeer and oversteer conditions.