The Physics of Racing

Weight Transfer

Most autocrossers and race drivers learn early in their careers the importance of balancing a car. Learning to do it consistently and automatically is one essential part of becoming a truly good driver. While the skills for balancing a car are commonly taught in drivers' schools, the rationale behind them is not usually adequately explained. That rationale comes from simple physics. Understanding the physics of driving not only helps one be a better driver, but increases one's enjoyment of driving as well. If you know the deep reasons why you ought to do certain things you will remember the things better and move faster toward complete internalization of the skills.

Balancing a car is controlling weight transfer using throttle, brakes, and steering. This article explains the physics of weight transfer. You will often hear instructors and drivers say that applying the brakes shifts weight to the front of a car and can induce oversteer. Likewise, accelerating shifts weight to the rear, inducing understeer, and cornering shifts weight to the opposite side, unloading the inside tires. But why does weight shift during these maneuvers? How can weight shift when everything is in the car bolted in and strapped down? Briefly, the reason is that inertia acts through the center of gravity (CG) of the car, which is above the ground, but adhesive forces act at ground level through the tire contact patches. The effects of weight transfer are proportional to the height of the CG off the ground. A flatter car, one with a lower CG, handles better and quicker because weight transfer is not so drastic as it is in a high car.

The rest of this article explains how inertia and adhesive forces give rise to weight transfer through Newton's laws. The article begins with the elements and works up to some simple equations that you can use to calculate weight transfer in any car knowing only the wheelbase, the height of the CG, the static weight distribution, and the track, or distance between the tires across the car. These numbers are reported in shop manuals and most journalistic reviews of cars.

Most people remember Newton's laws from school physics. These are fundamental laws that apply to all large things in the universe, such as cars. In the context of our racing application, they are:

The first law: a car in straight-line motion at a constant speed will keep such motion until acted on by an external force. The only reason a car in neutral will not coast forever is that friction, an external force, gradually slows the car down. Friction comes from the tires on the ground and the air flowing over the car. The tendency of a car to keep moving the way it is moving is the inertia of the car, and this tendency is concentrated at the CG point.

The second law: When a force is applied to a car, the change in motion is proportional to the force divided by the mass of the car. This law is expressed by the famous equation , where is a force, is the mass of the car, and is the acceleration, or change in motion, of the car. A larger force causes quicker changes in motion, and a heavier car reacts more slowly to forces. Newton's second law explains why quick cars are powerful and lightweight. The more and the less you have, the more you can get.

The third law: Every force on a car by another object, such as the ground, is matched by an equal and opposite force on the object by the car. When you apply the brakes, you cause the tires to push forward against the ground, and the ground pushes back. As long as the tires stay on the car, the ground pushing on them slows the car down.

Let us continue analyzing braking. Weight transfer during accelerating and cornering are mere variations on the theme. We won't consider subtleties such as suspension and tire deflection yet. These effects are very important, but secondary. The figure shows a car and the forces on it during a ``one g'' braking maneuver. One g means that the total braking force equals the weight of the car, say, in pounds.

In this figure, the black and white ``pie plate'' in the center is the CG. is the force of gravity that pulls the car toward the center of the Earth. This is the weight of the car; weight is just another word for the force of gravity. It is a fact of Nature, only fully explained by Albert Einstein, that gravitational forces act through the CG of an object, just like inertia. This fact can be explained at deeper levels, but such an explanation would take us too far off the subject of weight transfer.

is the lift force exerted by the ground on the front tire, and is the lift force on the rear tire. These lift forces are as real as the ones that keep an airplane in the air, and they keep the car from falling through the ground to the center of the Earth.

We don't often notice the forces that the ground exerts on objects because they are so ordinary, but they are at the essence of car dynamics. The reason is that the magnitude of these forces determine the ability of a tire to stick, and imbalances between the front and rear lift forces account for understeer and oversteer. The figure only shows forces on the car, not forces on the ground and the CG of the Earth. Newton's third law requires that these equal and opposite forces exist, but we are only concerned about how the ground and the Earth's gravity affect the car.

If the car were standing still or coasting, and its weight distribution were 50-50, then would be the same as . It is always the case that plus equals , the weight of the car. Why? Because of Newton's first law. The car is not changing its motion in the vertical direction, at least as long as it doesn't get airborne, so the total sum of all forces in the vertical direction must be zero. points down and counteracts the sum of and , which point up.

Braking causes to be greater than . Literally, the ``rear end gets light,'' as one often hears racers say. Consider the front and rear braking forces, and , in the diagram. They push backwards on the tires, 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. cannot be doing it since it passes right through the cetner of gravity. The only forces that can counteract that tendency are the lift forces, and the only way they can do so is for to become greater than . Literally, the ground pushes up harder on the front tires during braking to try to keep the car from tipping forward.

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

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

Thus, by braking at one g in our example car, we add 640 pounds of load to the front tires 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 , wheelbase , weight , static weight distribution expressed as a fraction of weight in the front, and braking with force , we can show that

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 analyzed in a similar way, where the track of the car replaces the wheelbase and 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 maneuver, 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 tire adhesion, which explains how weight transfer can lead to understeer and oversteer conditions.

Basic Calculations

Physics is the science of measurement. Perhaps you have heard of highly abstract branches of physics such as quantum mechanics and relativity, in which exotic mathematics is in the forefront. But when theories are taken to the laboratory (or the race course) for testing, all the mathematics must boil down to quantities that can be measured. In racing, the fundamental quantities are distance, time, and mass. This month, we will review basic equations that will enable you to do quick calculations in your head while cooling off between runs. It is very valuable to develop a skill for estimating quantities quickly, and I will show you how.

Equations that don't involve mass are called kinematic. The first kinematic equation relates speed, time, and distance. If a car is moving at a constant speed or velocity, , then the distance it travels in time is or velocity times time. This equation really expresses nothing more than the definition of velocity.

If we are to do mental calculations, the first hurdle we must jump comes from the fact that we usually measure speed in miles per hour (mph), but distance in feet and time in seconds. So, we must modify our equation with a conversion factor, like this

If you ``cancel out'' the units parts of this equation, you will see that you get feet on both the left and right hand sides, as is appropriate, since equality is required of any equation. The conversion factor is 5280/3600, which happens to equal 22/15. Let's do a few quick examples. How far does a car go in one second (remember, say, ``one-one-thousand, two-one-thousand,'' etc. to yourself to count off seconds)? At fifteen mph, we can see that we go or about 1 and a half car lengths for a 14 and 2/3 foot car like a late-model Corvette. So, at 30 mph, a second is three car lengths and at 60 mph it is six. If you lose an autocross by 1 second (and you'll be pretty good if you can do that with all the good drivers in our region), you're losing by somewhere between 3 and 6 car lengths! This is because the average speed in an autocross is between 30 and 60 mph.

Everytime you plow a little or get a little sideways, just visualize your competition overtaking you by a car length or so. One of the reasons autocross is such a difficult sport, but also such a pure sport, from the driver's standpoint, is that you can't make up this time. If you blow a corner in a road race, you may have a few laps in which to make it up. But to win an autocross against good competition, you must drive nearly perfectly. The driver who makes the fewest mistakes usually wins!

The next kinematic equation involves acceleration. It so happens that the distance covered by a car at constant acceleration from a standing start is given by or 1/2 times the acceleration times the time, squared. What conversions will help us do mental calculations with this equation? Usually, we like to measure acceleration in s. One happens to be 32.1 feet per second squared. Fortunately, we don't have to deal with miles and hours here, so our equation becomes, roughly. So, a car accelerating from a standing start at , which is a typical number for a good, stock sports car, will go 8 feet in 1 second. Not very far! However, this picks up rapidly. In two seconds, the car will go 32 feet, or over two car lengths.

Just to prove to you that this isn't crazy, let's answer the question ``How long will it take a car accelerating at to do the quarter mile?'' We invert the equation above (recall your high school algebra), to get and we plug in the numbers: the quarter mile equals 1320 feet, , and we get which is about 13 seconds. Not too unreasonable! A real car will not be able to keep up full acceleration for a quarter mile due to air resistance and reduced torque in the higher gears. This explains why real (stock) sports cars do the quarter mile in 14 or 15 seconds.

The more interesting result is the fact that it takes a full second to go the first 8 feet. So, we can see that the launch is critical in an autocross. With excessive wheelspin, which robs you of acceleration, you can lose a whole second right at the start. Just visualize your competition pulling 8 feet ahead instantly, and that margin grows because they are `hooked up' better.

For doing these mental calculations, it is helpful to memorize a few squares. 8 squared is 64, 10 squared is 100, 11 squared is 121, 12 squared is 144, 13 squared is 169, and so on. You can then estimate square roots in your head with acceptable precision.

Finally, let's examine how engine torque becomes force at the drive wheels and finally acceleration. For this examination, we will need to know the mass of the car. Any equation in physics that involves mass is called dynamic, as opposed to kinematic. Let's say we have a Corvette that weighs 3200 pounds and produces 330 foot-pounds of torque at the crankshaft. The Corvette's automatic transmission has a first gear ratio of 3.06 (the auto is the trick set up for vettes-just ask Roger Johnson or Mark Thornton). A transmission is nothing but a set of circular, rotating levers, and the gear ratio is the leverage, multiplying the torque of the engine. So, at the output of the transmission, we have of torque. The differential is a further lever-multiplier, in the case of the Corvette by a factor of 3.07, yielding 3100 foot pounds at the center of the rear wheels (this is a lot of torque!). The distance from the center of the wheel to the ground is about 13 inches, or 1.08 feet, so the maximum force that the engine can put to the ground in a rearward direction (causing the ground to push back forward-remember part 1 of this series!) in first gear is Now, at rest, the car has about 50/50 weight distribution, so there is about 1600 pounds of load on the rear tires. You will remember from last month's article on tire adhesion that the tires cannot respond with a forward force much greater than the weight that is on them, so they simply will spin if you stomp on the throttle, asking them to give you 2870 pounds of force.

We can now see why it is important to squeeeeeeeze the throttle gently when launching. In the very first instant of a launch, your goal as a driver is to get the engine up to where it is pushing on the tire contact patch at about 1600 pounds. The tires will squeal or hiss just a little when you get this right. Not so coincidentally, this will give you a forward force of about 1600 pounds, for an (part 1) acceleration of about , or half the weight of the car. The main reason a car will accelerate with only to start with is that half of the weight is on the front wheels and is unavailable to increase the stiction of the rear, driving tires. Immediately, however, there will be some weight transfer to the rear. Remembering part 1 of this series again, you can estimate that about 320 pounds will be transferred to the rear immediately. You can now ask the tires to give you a little more, and you can gently push on the throttle. Within a second or so, you can be at full throttle, putting all that torque to work for a beautiful hole shot!

In a rear drive car, weight transfer acts to make the driving wheels capable of withstanding greater forward loads. In a front drive car, weight transfer works against acceleration, so you have to be even more gentle on the throttle if you have a lot of power. An all-wheel drive car puts all the wheels to work delivering force to the ground and is theoretically the best.

Technical people call this style of calculating ``back of the envelope,'' which is a somewhat picturesque reference to the habit we have of writing equations and numbers on any piece of paper that happens to be handy. You do it without calculators or slide rules or abacuses. You do it in the garage or the pits. It is not exactly precise, but gives you a rough idea, say within 10 or 20 percent, of the forces and accelerations at work.