The Physics of Racing
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This is a discussion on The Physics of Racing within the Motorsports Talk forums, part of the Community - Meet other Enthusiasts category; I found, in my archives, a great (yet dated) article that is quite technical, but is explained in easy to ...

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    Geriatric Ginger Mod Rogan's Avatar
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    The Physics of Racing

    I found, in my archives, a great (yet dated) article that is quite technical, but is explained in easy to understand language.
    It was written in 1991, by Brian Beckman, but it all still applies. This makes for a very good read..

    I'll make individual posts for each chapter.


    I will also keep the thread locked, until I get it all posted.. there are 30 parts.

    Thanks for your patience.

    Intro



    Part 1 Weight Transfer
    Part 2 Keeping Your Tyres Stuck to the Ground
    Part 3 Basic Calculations
    Part 4 There Is No Such Thing as Centrifugal Force, Part 4b
    Part 5 Introduction to the Racing Line
    Part 6 Speed and Horsepower
    Part 7 The Traction Budget
    Part 8 Simulating Car Dynamics with a Computer Program
    Part 9 Straights
    Part 10 Grip Angle
    Part 11 Braking
    Part 12 CyberCar, Every Racer's DWIM Car?
    Part 13 Transients (The missing episode)
    Part 14 Why Smoothness?
    Part 15 Bumps In The Road
    Part 16 RARS, A Simple Racing Simulator
    Part 17 "Slow-in, Fast-out!" or, Advanced Analysis of the Racing Line
    Part 18 "Slow-in, Fast-out!" or, Advanced Analysis of the Racing Line, Continued
    Part 19 Space, Time, and Rubber
    Part 20 Four-Point Statics
    Part 21 The Magic Formula: Longitudinal Version
    Part 22 The Magic Formula: Lateral Version
    Part 23 Trail Braking
    Part 24 Combination Slip
    Part 25 Combination Grip
    Part 26 The Driving Wheel, Chapter I
    Part 27 Four-Wheel Weight Transfer
    Part 28 Hazards of Integration
    Part 29 A Magical Trick



    edit: Funny thing.. Heide24 posted a thread about the same article, yesterday LOL I never saw that one.

    Oh well, I'll continue to finish this , so it will be a readable thread, rather than a jump.

    Last edited by Rogan; 08-04-2010 at 10:46 AM.
    Rogan o_0
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    Introduction to PhORS

    Introduction to the PhORS articles
    By Brian Beckman

    "I started this series in 1991 for my local racing club's printed newsletter. The web had just been born, though the Internet was not yet public. Nonetheless, I distributed the articles over the Internet at that time and they become reasonably well known, especially amongst the autocrossing community in the US. The first 13 parts were written in 1991, so they contain some very dated ideas, such as using Scheme for writing simulations. However, the entire series is presented here, as originally written. Perhaps at some later time I will consolidate and update the series, but for now, I am focusing on writing new parts. There are currently a number of 'live threads' in the discussion that I wish to pursue at length.

    My overall goal with the series is to present a fresh outlook on racing physics, understandable to the technically inclined non-specialist. The problems I consider come from a variety of sources. Often, they're motivated by computer simulation, and just as often they arise from competition experiences. Some of the later articles get very technical, but I always try to balance conceptual discussion, which everyone should be able to understand, with mathematical analysis, which might of interest only to specialists, and with numerical results, which, again, should be universally accessible.

    When I first started the series, I purposely avoided the standard reference sources, preferring to figure things out myself from first principles. In the past ten years, a number of superior source books, papers, and programs have become available, and it is no longer sensible for me to avoid them. I've had my fun, now it is time to 'get real.' So, in the later articles, I refer to the well known books by Milliken, Gillespie, Genta, and Carroll Smith; as well as to free simulation packages such as RARS, TORCS, and Racer.

    There is a tremendous amount of activity in racing simulation nowadays that computer hardware is fast enough to permit extremely detailed modelling of racing cars in real time. The realism of Grand-Prix Legends, for instance, was unimaginable in real time in 1991. Despite this growth, I continue to hope that the Physics of Racing series can fulfil its original dual roles of translating racing lore and craft into hardcore physics and of making that physics understandable to real-world working race drivers and teams.
    Rogan o_0
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    Part 1: 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 tyres. But why does weight shift during these manoeuvres? How can weight shift when everything is in the car bolted in and strapped down? Briefly, the reason is that inertia acts through the centre of gravity (CG) of the car, which is above the ground, but adhesive forces act at ground level through the tyre 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 tyres 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 tyres 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 F = ma, where F is a force, m is the mass of the car, and a 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 F and the less m you have, the more a 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 tyres to push forward against the ground, and the ground pushes back. As long as the tyres stay on the car, the ground pushing on them slows the car down.

    Let us continue analysing braking. Weight transfer during accelerating and cornering are mere variations on the theme. We won't consider subtleties such as suspension and tyre deflection yet. These effects are very important, but secondary. The figure shows a car and the forces on it during a "one g" braking manoeuvre. 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 centre is the CG. G is the force of gravity that pulls the car toward the centre 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.

    Lf is the lift force exerted by the ground on the front tyre, and Lr is the lift force on the rear tyre. 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 centre 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 tyre 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 Lf would be the same as Lr. It is always the case that Lf plus Lr equals G, 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. G points down and counteracts the sum of Lf and Lr, which point up.

    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.
    Last edited by Rogan; 08-04-2010 at 08:27 AM.
    Rogan o_0
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    Part 2: Keeping Your Tyres Stuck to the Ground

    The Physics of Racing,
    Part 2: Keeping Your Tyres Stuck to the Ground


    In last month's article, we explained the physics behind weight transfer. That is, we explained why braking shifts weight to the front of the car, accelerating shifts weight to the rear, and cornering shifts weight to the outside of a curve. Weight transfer is a side-effect of the tyres keeping the car from flipping over during manoeuvres. We found out that a one G braking manoeuvre in our 3200 pound example car causes 640 pounds to transfer from the rear tyres to the front tyres. The explanations were given directly in terms of Newton's fundamental laws of Nature.

    This month, we investigate what causes tyres to stay stuck and what causes them to break away and slide. We will find out that you can make a tyre slide either by pushing too hard on it or by causing weight to transfer off the tyre by your control inputs of throttle, brakes, and steering. Conversely, you can cause a sliding tyre to stick again by pushing less hard on it or by transferring weight to it. The rest of this article explains all this in term of (you guessed it) physics.

    This knowledge, coupled with a good "instinct" for weight transfer, can help a driver predict the consequences of all his or her actions and develop good instincts for staying out of trouble, getting out of trouble when it comes, and driving consistently at ten tenths. It is said of Tazio Nuvolari, one of the greatest racing drivers ever, that he knew at all times while driving the weight on each of the four tyres to within a few pounds. He could think, while driving, how the loads would change if he lifted off the throttle or turned the wheel a little more, for example. His knowledge of the physics of racing enabled him to make tiny, accurate adjustments to suit every circumstance, and perhaps to make these adjustments better than his competitors. Of course, he had a very fast brain and phenomenal reflexes, too.

    I am going to ask you to do a few physics "lab" experiments with me to investigate tyre adhesion. You can actually do them, or you can just follow along in your imagination. First, get a tyre and wheel off your car. If you are a serious autocrosser, you probably have a few loose sets in your garage. You can do the experiments with a heavy box or some object that is easier to handle than a tyre, but the numbers you get won't apply directly to tyres, although the principles we investigate will apply.

    Weigh yourself both holding the wheel and not holding it on a bathroom scale. The difference is the weight of the tyre and wheel assembly. In my case, it is 50 pounds (it would be a lot less if I had those $3000 Jongbloed wheels! Any sponsors reading?). Now put the wheel on the ground or on a table and push sideways with your hand against the tyre until it slides. When you push it, push down low near the point where the tyre touches the ground so it doesn't tip over.

    The question is, how hard did you have to push to make the tyre slide? You can find out by putting the bathroom scale between your hand and the tyre when you push. This procedure doesn't give a very accurate reading of the force you need to make the tyre slide, but it gives a rough estimate. In my case, on the concrete walkway in front of my house, I had to push with 85 pounds of force (my neighbours don't bother staring at me any more; they're used to my strange antics). On my linoleum kitchen floor, I only had to push with 60 pounds (but my wife does stare at me when I do this stuff in the house). What do these numbers mean?

    They mean that, on concrete, my tyre gave me 85 / 50 = 1.70 gees of sideways resistance before sliding. On a linoleum race course (ahem!), I would only be able to get 60 / 50 = 1.20G. We have directly experienced the physics of grip with our bare hands. The fact that the tyre resists sliding, up to a point, is called the grip phenomenon. If you could view the interface between the ground and the tyre with a microscope, you would see complex interactions between long-chain rubber molecules bending, stretching, and locking into concrete molecules creating the grip. Tyre researchers look into the detailed workings of tyres at these levels of detail.

    Now, I'm not getting too excited about being able to achieve 1.70G cornering in an autocross. Before I performed this experiment, I frankly expected to see a number below 1G. This rather unbelievable number of 1.70G would certainly not be attainable under driving conditions, but is still a testimony to the rather unbelievable state of tyre technology nowadays. Thirty years ago, engineers believed that one G was theoretically impossible from a tyre. This had all kinds of consequences. It implied, for example, that dragsters could not possibly go faster than 200 miles per hour in a quarter mile: you can go = 198.48 mph if you can keep 1G acceleration all the way down the track. Nowadays, drag racing safety watchdogs are working hard to keep the cars under 300 mph; top fuel dragsters launch at more than 3 gees.

    For the second experiment, try weighing down your tyre with some ballast. I used a couple of dumbbells slung through the centre of the wheel with rope to give me a total weight of 90 pounds. Now, I had to push with 150 pounds of force to move the tyre sideways on concrete. Still about 1.70G. We observe the fundamental law of adhesion: the force required to slide a tyre is proportional to the weight supported by the tyre. When your tyre is on the car, weighed down with the car, you cannot push it sideways simply because you can't push hard enough.

    The force required to slide a tyre is called the adhesive limit of the tyre, or sometimes the stiction, which is a slang combination of "stick" and "friction." This law, in mathematical form, is


    where F is the force with which the tyre resists sliding; is the coefficient of static friction or coefficient of adhesion; and W is the weight or vertical load on the tyre contact patch. Both F and W have the units of force (remember that weight is the force of gravity), so is just a number, a proportionality constant. This equation states that the sideways force a tyre can withstand before sliding is less than or equal to times W. Thus, W is the maximum sideways force the tyre can withstand and is equal to the stiction. We often like to speak of the sideways acceleration the car can achieve, and we can convert the stiction force into acceleration in gees by dividing by W, the weight of the car can thus be measured in gees.

    The coefficient of static friction is not exactly a constant. Under driving conditions, many effects come into play that reduce the stiction of a good autocross tyre to somewhere around 1.10G. These effects are deflection of the tyre, suspension movement, temperature, inflation pressure, and so on. But the proportionality law still holds reasonably true under these conditions. Now you can see that if you are cornering, braking, or accelerating at the limit, which means at the adhesive limit of the tyres, any weight transfer will cause the tyres unloaded by the weight transfer to pass from sticking into sliding.

    Actually, the transition from sticking 'mode' to sliding mode should not be very abrupt in a well-designed tyre. When one speaks of a "forgiving" tyre, one means a tyre that breaks away slowly as it gets more and more force or less and less weight, giving the driver time to correct. Old, hard tyres are, generally speaking, less forgiving than new, soft tyres. Low-profile tyres are less forgiving than high-profile tyres. Slicks are less forgiving than DOT tyres. But these are very broad generalities and tyres must be judged individually, usually by getting some word-of-mouth recommendations or just by trying them out in an autocross. Some tyres are so unforgiving that they break away virtually without warning, leading to driver dramatics usually resulting in a spin. Forgiving tyres are much easier to control and much more fun to drive with.

    "Driving by the seat of your pants" means sensing the slight changes in cornering, braking, and acceleration forces that signal that one or more tyres are about to slide. You can sense these change literally in your seat, but you can also feel changes in steering resistance and in the sounds the tyres make. Generally, tyres 'squeak' when they are nearing the limit, 'squeal' at the limit, and 'squall' over the limit. I find tyre sounds very informative and always listen to them while driving.

    So, to keep your tyres stuck to the ground, be aware that accelerating gives the front tyres less stiction and the rear tyres more, that braking gives the front tyre more stiction and the rear tyres less, and that cornering gives the inside tyres less stiction and the outside tyres more. These facts are due to the combination of weight transfer and the grip phenomenon. Finally, drive smoothly, that is, translate your awareness into gentle control inputs that always keep appropriate tyres stuck at the right times. This is the essential knowledge required for car control, and, of course, is much easier said than done. Later articles will use the knowledge we have accumulated so far to explain understeer, oversteer, and chassis set-up.
    Last edited by Rogan; 08-04-2010 at 08:34 AM.
    Rogan o_0
    '96 Dodge 2500 CTD @ 40psi - over 700 lb/ft TQ, 7" stack, and 5speed! - SOLD
    '01 Dodge 2500 CTD 6-holed hand-shaker - 3850# dual disk - 900 lb/ft - SOLD
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    Part 3: Basic Calculations

    The Physics of Racing,
    Part 3: Basic Calculations
    Brian Beckman

    In the last two articles, we plunged right into some relatively complex issues, namely weight transfer and tyre adhesion. This month, we regroup and review some of the basic units and dimensions needed to do dynamical calculations. Eventually, we can work up to equations sufficient for a full-blown computer simulation of car dynamics. The equations can then be 'doctored' so that the computer simulation will run fast enough to be the core of an autocross computer game. Eventually, we might direct this series of articles to show how to build such a game in a typical microcomputer programming language such as C or BASIC, or perhaps even my personal favourite, LISP. All of this is in keeping with the spirit of the series, the Physics of Racing, because so much of physics today involves computing. Software design and programming are essential skills of the modern physicist, so much so that many of us become involved in computing full time.

    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, v, then the distance d it travels in time t is d = vt 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

    d = 15 mph times 1 sec times 22/15 = 22 feet


    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.

    Every time you plough 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 d = at2 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 Gs. One G 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,

    d (feet) = 16a (Gs) t (seconds)2


    roughly. So, a car accelerating from a standing start at G, 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 G 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, a = G, and we get


    which is about 13 seconds. Not too unreasonable! A real car will not be able to keep up full G 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 wheel spin, 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 3.06 x 330 = 1010 foot-pounds 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 centre of the rear wheels (this is a lot of torque!). The distance from the centre 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 3100 foot-pounds / 1.08 feet = 2870 Now, at rest, the car has about 50/50 weight distribution, so there is about 1600 pounds of load on the rear tyres. You will remember from last month's article on tyre adhesion that the tyres 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 tyre contact patch at about 1600 pounds. The tyres 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 F = ma (part 1) acceleration of about G, or half the weight of the car. The main reason a car will accelerate with only G 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 tyres. 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 tyres 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. And now you know how to do back-of-the-envelope calculations, too.
    Last edited by Rogan; 08-04-2010 at 08:33 AM.
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    Part 4: There Is No Such Thing as Centrifugal Force

    The Physics of Racing,
    Part 4: There Is No Such Thing as Centrifugal Force
    Brian Beckman


    One often hears of "centrifugal force." This is the apparent force that throws you to the outside of a turn during cornering. If there is anything loose in the car, it will immediately slide to the right in a left hand turn, and vice versa. Perhaps you have experienced what happened to me once. I had omitted to remove an empty Pepsi can hidden under the passenger seat. During a particularly aggressive run (something for which I am not unknown), this can came loose, fluttered around the ****pit for a while, and eventually flew out the passenger window in the middle of a hard left hand corner.

    I shall attempt to convince you, in this month's article, that centrifugal force is a fiction, and a consequence of the fact first noticed just over three hundred years ago by Newton that objects tend to continue moving in a straight line unless acted on by an external force.

    When you turn the steering wheel, you are trying to get the front tyres to push a little sideways on the ground, which then pushes back, by Newton's third law. When the ground pushes back, it causes a little sideways acceleration. This sideways acceleration is a change in the sideways velocity. The acceleration is proportional to the sideways force, and inversely proportional to the mass of the car, by Newton's second law. The sideways acceleration thus causes the car to veer a little sideways, which is what you wanted when you turned the wheel. If you keep the steering and throttle at constant positions, you will continue to go mostly forwards and a little sideways until you end up where you started. In other words, you will go in a circle. When driving through a sweeper, you are going part way around a circle. If you take skid pad lessons (highly recommended), you will go around in circles all day.

    If you turn the steering wheel a little more, you will go in a tighter circle, and the sideways force needed to keep you going is greater. If you go around the same circle but faster, the necessary force is greater. If you try to go around too fast, the adhesive limit of the tyres will be exceeded, they will slide, and you will not stick to the circular path-you will not "make it."

    From the discussion above, we can see that in order to turn right, for example, a force, pointing to the right, must act on the car that veers it away from the straight line it naturally tries to follow. If the force stays constant, the car will go in a circle. From the point of view of the car, the force always points to the right. From a point of view outside the car, at rest with respect to the ground, however, the force points toward the centre of the circle. From this point of view, although the force is constant in magnitude, it changes direction, going around and around as the car turns, always pointing at the geometrical centre of the circle. This force is called centripetal, from the Greek for "centre seeking." The point of view on the ground is privileged, since objects at rest from this point of view feel no net forces. Physicists call this special point of view an inertial frame of reference. The forces measured in an inertial frame are, in a sense, more correct than those measured by a physicist riding in the car. Forces measured inside the car are biased by the centripetal force.

    Inside the car, all objects, such as the driver, feel the natural inertial tendency to continue moving in a straight line. The driver receives a centripetal force from the car through the seat and the belts. If you don't have good restraints, you may find yourself pushing with your knee against the door and tugging on the controls in order to get the centripetal force you need to go in a circle with the car. It took me a long time to overcome the habit of tugging on the car in order to stay put in it. I used to come home with bruises on my left knee from pushing hard against the door during an autocross. I found that a tight five-point harness helped me to overcome this unnecessary habit. With it, I no longer think about body position while driving - I can concentrate on trying to be smooth and fast. As a result, I use the wheel and the gearshift lever for steering and shifting rather than for helping me stay put in the car!

    The 'forces' that the driver and other objects inside the car feel are actually centripetal. The term centrifugal, or "centre fleeing," refers to the inertial tendency to resist the centripetal force and to continue going straight. If the centripetal force is constant in magnitude, the centrifugal tendency will be constant. There is no such thing as centrifugal force (although it is a convenient fiction for the purpose of some calculations).

    Let's figure out exactly how much sideways acceleration is needed to keep a car going at speed v in a circle of radius r. We can then convert this into force using Newton's second law, and then figure out how fast we can go in a circle before exceeding the adhesive limit-in other words, we can derive maximum cornering speed. For the following discussion, it will be helpful for you to draw little back-of-the-envelope pictures (I'm leaving them out, giving our editor a rest from transcribing my graphics into the newsletter).

    Consider a very short interval of time, far less than a second. Call it dt (d stands for "delta," a Greek letter mathematicians use as shorthand for "tiny increment"). In time dt, let us say we go forward a distance dx and sideways a distance ds. The forward component of the velocity of the car is approximately v = dx / dt. At the beginning of the time interval dt, the car has no sideways velocity. At the end, it has sideways velocity ds / dt. In the time dt, the car has thus had a change in sideways velocity of ds / dt. Acceleration is, precisely, the change in velocity over a certain time, divided by the time; just as velocity is the change in position over a certain time, divided by the time. Thus, the sideways acceleration is


    How is ds related to r, the radius of the circle? If we go forward by a fraction f of the radius of the circle, we must go sideways by exactly the same fraction of dx to stay on the circle. This means that ds = f dx. The fraction f is, however, nothing but dx / r. By this reasoning, we get the relation


    We can substitute this expression for ds into the expression for a, and remembering that v = dx / dt, we get the final result


    This equation simply says quantitatively what we wrote before: that the acceleration (and the force) needed to keep to a circular line increases with the velocity and increases as the radius gets smaller.

    What was not appreciated before we went through this derivation is that the necessary acceleration increases as the square of the velocity. This means that the centripetal force your tyres must give you for you to make it through a sweeper is very sensitive to your speed. If you go just a little bit too fast, you might as well go much too fast - you're not going to make it.
    Last edited by Rogan; 08-04-2010 at 08:46 AM.
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    Geriatric Ginger Mod Rogan's Avatar
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    The following table shows the maximum speed that can be achieved in turns of various radii for various sideways accelerations. This table shows the value of the expression


    which is the solution of a = v2 / r for v, the velocity. The conversion factor 15/22 converts v from feet per second to miles per hour, and 32.1 converts a from gees to feet per second squared. We covered these conversion factors in part 3 of this series.

    Table 1. Speed (Miles Per Hour)
    Code:
    Acceleration 	        Radius (Feet)
    (Gees) 	                50.00 	100.0 	150.0 	200.0 	500.0
    0.25 	                13.66 	19.31 	23.66 	27.32 	43.19
    0.50 	                19.31 	27.32 	33.45 	38.63 	61.08
    0.75 	                23.66 	33.45 	40.97 	47.31 	74.81
    1.00 	                27.32 	38.63 	47.31 	54.63 	86.38
    1.25 	                30.54 	43.19 	52.90 	61.08 	96.57
    1.50 	                33.45 	47.31 	57.94 	66.91 	105.79
    1.75 	                36.13 	51.10 	62.59 	72.27 	114.27
    2.00 	                38.63 	54.63 	66.91 	77.26 	122.16

    For autocrossing, the columns for 50 and 100 feet and the row for 1.00G are most germane. The table tells us that to achieve 1.00G sideways acceleration in a corner of 50 foot radius (this kind of corner is all too common in autocross), a driver must not go faster than 27.32 miles per hour. To go 30 mph, 1.25G is required, which is probably not within the capability of an autocross tyre at this speed. There is not much subjective difference between 27 and 30 mph, but the objective difference is usually between making a controlled run and spinning badly.

    The absolute fastest way to go through a corner is to be just over the limit near the exit, in a controlled slide. To do this, however, you must be pointed in just such a way that when the car breaks loose and slides to the exit of the corner it will be pointed straight down the optimal racing line at the exit when it "hooks up" again. You can smoothly add throttle during this manoeuvre and be really moving out of the corner. But you must do it smoothly. It takes a long time to learn this, and probably a lifetime to perfect it, but it feels absolutely triumphal when done right. I have not figured out how to drive through a sweeper, except for the exit, at anything greater than the limiting velocity because sweepers are just too long to slide around. If anyone (Ayrton Senna, perhaps?) knows how, please tell me!

    The chain of reasoning we have just gone through was first discovered by Newton and Leibniz, working independently. It is, in fact, a derivation in differential calculus, the mathematics of very small quantities. Newton keeps popping up. He was perhaps the greatest of all physicists, having discovered the laws of motion, the law of gravity, and calculus, among other things such as the fact that white light is made up of multiple colours mixed together.

    It is an excellent diagnostic exercise to drive a car around a circle marked with cones or chalk and gently to increase the speed until the car slides. If the front breaks away first, your car has natural understeer, and if the rear slides first, it has natural oversteer. You can use this information for chassis tuning. Of course, this is only to be done in safe circumstances, on a rented skid pad or your own private parking lot. The police will gleefully give you a ticket if they catch you doing this in the wrong places.
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    Part 5: Introduction to the Racing Line

    Part 5: Introduction to the Racing Line

    This month, we analyze the best way to go through a corner. "Best" means in the least time, at the greatest average speed. We ask "what is the shape of the driving line through the corner that gives the best time?" and "what are the times for some other lines, say hugging the outside or the inside of the corner?" Given the answers to these questions, we go on to ask "what shape does a corner have to be before the driving line I choose doesn't make any time difference?" The answer is a little surprising.

    The analysis presented here is the simplest I could come up with, and yet is still quite complicated. My calculations went through about thirty steps before I got the answer. Don't worry, I won't drag you through the mathematics; I just sketch out the analysis, trying to focus on the basic principles. Anyone who would read through thirty formulas would probably just as soon derive them for him or herself.

    There are several simplifying assumptions I make to get through the analysis. First of all, I consider the corner in isolation; as an abstract entity lifted out of the rest of a course. The actual best driving line through a corner depends on what comes before it and after it. You usually want to optimize exit speed if the corner leads onto a straight. You might not apex if another corner is coming up. You may be forced into an unfavorable entrance by a prior curve or slalom.

    Speaking of road courses, you will hear drivers say things like "you have to do such-and-such in turn six to be on line for turn ten and the front straight." In other words, actions in any one spot carry consequences pretty much all the way around. The ultimate drivers figure out the line for the entire course and drive it as a unit, taking a Zen-like approach. When learning, it is probably best to start out optimising each kind of corner in isolation, then work up to combinations of two corners, three corners, and so on. In my own driving, there are certain kinds of three corner combinations I know, but mostly I work in twos. I have a long way to go.

    It is not feasible to analyse an actual course in an exact, mathematical way. In other words, although science can provide general principles and hints, finding the line is, in practice, an art. For me, it is one of the most fun parts of racing.

    Other simplifying assumptions I make are that the car can either accelerate, brake, or corner at constant speed, with abrupt transitions between behaviours. Thus, the lines I analyse are splices of accelerating, braking, and cornering phases. A real car can, must, and should do these things in combination and with smooth transitions between phases. It is, in fact, possible to do an exact, mathematical analysis with a more realistic car that transitions smoothly, but it is much more difficult than the splice-type analysis and does not provide enough more quantitative insight to justify its extra complexity for this article.

    Our corner is the following ninety-degree right-hander:


    This figure actually represents a family of corners with any constant width, any radius, and short straights before and after. First, we go through the entire analysis with a particular corner of 75 foot radius and 30 foot width, then we end up with times for corners of various radii and widths.

    Let us define the following parameters:
    r = radius of corner centre line = 75 feet
    W = width of course = 30 feet
    ro = radius of outer edge = r + W = 90 feet
    ri = radius of inner edge = r - W = 60 feet

    Now, when we drive this corner, we must keep the tyres on the course, otherwise we get a lot of cone penalties (or go into the weeds). It is easiest (though not so realistic) to do the analysis considering the path of the centre of gravity of the car rather than the paths of each wheel. So, we define an effective course, narrower than the real course, down which we may drive the centre of the car.

    w = width of car = 6 feet
    Ro = effective outer radius = ro - w = 87 feet
    Ri = effective inner radius = ri + w = 63 feet
    X = effective width of course = W - w = 24 feet

    This course is indicated by the labels and the thick radius lines in the figure.

    From last month's article, we know that for a fixed centripetal acceleration, the maximum driving speed increases as the square root of the radius. So, if we drive the largest possible circle through the effective corner, starting at the outside of the entrance straight, going all the way to the inside in the middle of the corner (the apex), and ending up at the outside of the exit straight, we can corner at the maximum speed. Such a line is shown in the figure as the thick circle labeled "line m." This is a simplified version of the classic racing line through the corner. Line m reaches the apex at the geometrical centre of the circle, whereas the classic racing line reaches an apex after the geometrical centre - a late apex - because it assumes we are accelerating out of the corner and must therefore have a continuously increasing radius in the second half and a slightly tighter radius in the first half to prepare for the acceleration. But, we continue analyzing the geometrically perfect line because it is relatively easy. The figure shows also Line i, the inside line, which come up the inside of the entrance straight, corners on the inside, and goes down the inside of the exit straight; and Line o, the outside line, which comes up the outside, corners on the outside, and exits on the outside.

    One might argue that there are certain advantages of line i over line m. Line i is considerably shorter than Line m, and although we have to go slower through the corner part, we have less total distance to cover and might get through faster. Also, we can accelerate on part of the entrance chute and all the way on the exit chute, while we have to drive line m at constant speed. Let's find out how much time it takes to get through lines i and m. We include line o for completeness, even though it looks bad because it is both slower and longer than m.

    If we assume a maximum centripetal acceleration of 1.10g, which is just within the capability of autocross tyres, we get the following speeds for the cornering phases of Lines i, o, and m:
    Cornering Speed (mph)
    Code:
    Line i 	Line o 	Line m
    32.16 	37.79 	48.78
    vi 	vo 	vm
    Line m is all cornering, so we can easily calculate the time to drive it once we know the radius, labelled k in the figure. A geometrical analysis results in

    k = 3.414(Ro - 0.707Ri) = 145 feet

    and the time is


    For line i, we accelerate for a bit, brake until we reach 32.16 mph, corner at that speed, and then accelerate on the exit. Let's assume, to keep the comparison fair, that we have timing lights at the beginning and end of line m and that we can begin driving line i at 48.78 mph, the same speed that we can drive line m. Let us also assume that the car can accelerate at g and brake at 1g. Our driving plan for line i results in the following velocity profile:


    Because we can begin by accelerating, we start beating line m a little. We have to brake hard to make the corner. Finally, although we accelerate on the exit, we don't quite come up to 48.78 mph, the exit speed for line m. But, we don't care about exit speed, only time through the corner. Using the velocity profile above, we can calculate the time for line i, call it ti, to be 4.08 seconds. Line i loses by 9/10ths of a second. It is a fair margin to lose an autocross by this much over a whole course, but this analysis shows we can lose it in just one typical corner! In this case, line i is a catastrophic mistake. Incidentally, line o takes 4.24 seconds = to.
    Last edited by Rogan; 08-04-2010 at 09:15 AM.
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    What if the corner were tighter or of greater radius? The following table shows some times for 30 foot wide corners of various radii:
    Code:
    radius 	30.00 	45.00 	60.00 	75.00 	90.00 	95.00
    to 	3.99 	4.06 	4.15 	4.24 	4.35 	4.38
    ti 	3.94 	3.94 	4.00 	4.08 	4.17 	4.21
    tm 	2.64 	2.83 	3.01 	3.18 	3.34 	3.39
    margin 	1.30 	1.11 	1.01 	0.90 	0.83 	0.82
    Line i never beats line m even though that as the radius increases, the margin of loss decreases. The trend is intuitive because corners of greater radius are also longer and the extra speed in line m over line i is less. The margin is greatest for tight corners because the width is a greater fraction of the length and the speed differential is greater.

    How about for various widths? The following table shows times for a 75 foot radius corner of several widths:
    Code:
    width 	10.00 	30.00 	50.00 	70.00 	90.00>
    to 	2.68 	4.24 	5.47 	6.50 	7.41
    ti 	2.62 	4.08 	5.32 	6.45 	7.51
    tm 	2.46 	3.18 	3.77 	4.27 	4.73
    margin 	0.16 	0.90 	1.55 	2.18 	2.79
    The wider the course, the greater the margin of loss. This is, again, intuitive since on a wide course, line m is a really large circle through even a very tight corner. Note that line o becomes better than line i for wide courses. This is because the speed differential between lines o and i is very great for wide courses. The most notable fact is that line m beats line i by 0.16 seconds even on a course that is only four feet wider than the car! You really must "use up the whole course."

    So, the answer is, under the assumptions made, that the inside line is never better than the classic racing line. For the splice-type car behaviour assumed, I conjecture that no line is faster than line m.

    We have gone through a simplified kind of variational analysis. Variational analysis is used in all branches of physics, especially mechanics and optics. It is possible, in fact, to express all theories of physics, even the most arcane, in variational form, and many physicists find this form very appealing. It is also possible to use variational analysis to write a computer program that finds an approximately perfect line through a complete, realistic course.
    Last edited by Rogan; 08-04-2010 at 09:18 AM.
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    Part 6: Speed and Horsepower

    The Physics of Racing,
    Part 6: Speed and Horsepower

    The title of this month's article consists of two words dear to every racer's heart. This month, we do some "back of the envelope" calculations to investigate the basic physics of speed and horsepower (the "back of the envelope" style of calculating was covered in part 3 of this series).

    How much horsepower does it take to go a certain speed? At first blush, a physicist might be tempted to say "none," because he or she remembers Newton's first law, by which an object moving at a constant speed in a straight line continues so moving forever, even to the end of the Universe, unless acted on by an external force. Everyone knows, however, that it is necessary to keep your foot on the gas to keep a car moving at a constant speed. Keeping your foot on the gas means that you are making the engine apply a backward force to the ground, which applies a reaction force forward on the car, to keep the car moving. In fact, we know a few numbers from our car's shop manual. A late model Corvette, for example, has a top speed of about 150 miles per hour and about 240 hp. This means that if you keep your foot all the way down, using up all 240 hp, you can eventually go 150 mph. It takes a while to get there. In this car, you can get to 60 mph in about 6 seconds (if you don't spin the drive wheels), to 100 mph in about 15 seconds, and 150 in about a minute.

    All this seems to contradict Newton's first law. What is going on? An automobile moving at constant speed in a straight line on level ground is, in fact, acted on by a number of external forces that tend to slow it down. Without these forces, the car would coast forever as guaranteed by Newton's first law. You must counteract these forces with the engine, which indirectly creates a reaction force that keeps the car going. When the car is going at a constant speed, the net force on the car, that is, the speeding-up forces minus the slowing-down forces, is zero.

    The most important external, slowing-down force is air resistance or drag. The second most important force is friction between the tyres and the ground, the so-called rolling resistance. Both these forces are called resistance because they always act to oppose the forward motion of the car in whatever direction it is going. Another physical effect that slows a car down is internal friction in the drive train and wheel bearings. Acting internally, these forces cannot slow the car. However, they push backwards on the tyres, which push forward on the ground, which pushes back by Newton's third law, slowing the car down. The internal friction forces are opposed by external reaction forces, which act as slight braking forces, slowing the car. So, Newton and the Universe are safe; everything is working as it should.

    How big are the resistance forces, and what role does horsepower play? The physics of air resistance is very complex and an area of vigorous research today. Most of this research is done by the aerospace industry, which is technologically very closely related to the automobile industry, especially when it comes to racing. We'll slog through some arithmetic here to come up with a table that shows how much horsepower it takes to sustain speed. Those who don't have the stomach to go through the math can skim the next few paragraphs.

    We cannot derive equations for air resistance here. We'll just look them up. My source is Fluid Mechanics, by L. D. Landau and E. M. Lifshi.tz, two eminent Russian physicists. They give the following approximate formula:


    The factors in this equation are the following:
    Cd = coefficient of friction, a factor depending on the shape of a car and determined by experiment; for a late model Corvette it is about 0.30;
    A = frontal area of the car; for a Corvette, it is about 20 square feet;
    = Greek letter rho, density of air, which we calculate below;
    v = speed of the car.

    Let us calculate the density of air using "back of the envelope" methods. We know that air is about 79% Nitrogen and 21% Oxygen. We can look up the fact that Nitrogen has a molecular weight of about 28 and Oxygen has a molecular weight of about 32. What is molecular weight? It is the mass (not the weight, despite the name) of 22.4 litres of gas. It is a number of historical convention, just like feet and inches, and doesn't have any real science behind it. So, we figure that air has an average molecular weight of


    I admit to using a calculator to do this calculation, against the spirit of the "back of the envelope" style. So sue me.

    We need to convert 1.29 gm/l to pounds of mass per cubic foot so that we can do the force calculations in familiar, if not convenient, units. It is worthwhile to note, as an aside, that a great deal of the difficulty of doing calculations in the physics of racing has to do with the traditional units of feet, miles, and pounds we use. The metric system makes all such calculations vastly simpler. Napoleon Bonaparte wanted to convert the world the metric system (mostly so his own soldiers could do artillery calculations quickly in their heads) but it is still not in common use in America nearly 200 years later!

    Again, we look up the conversion factors. My source is Engineering Formulas by Kurt Gieck, but they can be looked up in almost any encyclopaedia or dictionary. There are 1000 litres in a cubic meter, which in turn contains 35.51 cubic feet. Also, a pound-mass contains 453.6 grams. These figures give us, for the density of air


    This says that a cubic foot of air weighs 8 hundredths of a pound, and so it does! Air is much more massive than it seems, until you are moving quickly through it, that is.

    Let's finish off our equation for air resistance. We want to fill in all the numbers except for speed, v, using the Corvette as an example car so that we can calculate the force of air resistance for a variety of speeds. We get


    We want, at the end, to have v in miles per hour, but we need v in feet per seconds for the calculations to come out right. We recall that there are 22 feet per second for every 15 miles per hour, giving us


    Now (this gets confusing, and it wouldn't be if we were using the metric system), a pound mass is a phoney unit. A lb-mass is concocted to have a weight of 1 pound under the action of the Earth's gravity. Pounds are a unit of force or weight, not of mass. We want our force of air resistance in pounds of force, so we have to divide lb-mass ft / sec2 by 32.1, numerically equal to the acceleration of Earth's gravity in ft / sec2, to get pounds of force. You just have to know these things. This was a lot of work, but it's over now. We finally get


    Let's calculate a few numbers. The following table gives the force of air resistance for a number of interesting speeds:
    Code:
    v (mph) 	15 	30 	60 	90 	120 	150
    F (pounds) 	3.60 	14.5 	58.0 	130 	232 	362
    We can see that the force of air resistance goes up rapidly with speed, until we need over 350 pounds of constant force just to overcome drag at 150 miles per hour. We can now show where horsepower comes in.
    Last edited by Rogan; 08-04-2010 at 09:26 AM.
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    Horsepower is a measure of power, which is a technical term in physics. It measures the amount of work that a force does as it acts over time. Work is another technical term in physics. It measures the actual effect of a force in moving an object over a distance. If we move an object one foot by applying a force of one pound, we are said to be doing one foot-pound of work. If it takes us one second to move the object, we have exerted one foot-pound per second of power. A horsepower is 550 foot-pounds per second. It is another one of those historical units that Napoleon hated and that has no reasonable origin in science.

    We can expend one horsepower by exerting 550 pounds of force to move an object 1 foot in 1 second, or by exerting 1 pound of force to move an object 550 feet in 1 second, or by exerting 1 pound of force to move an object 1 foot in 0.001818 seconds, and so on. All these actions take the same amount of power. Incidentally, a horsepower happens to be equal also to 745 watts. So, if you burn about 8 light bulbs in your house, someone somewhere is expending at least one horsepower (and probably more like four or five) in electrical forces to keep all that going for you, and you pay for the service at the end of the month!.

    All this means that to find out how much horsepower it takes to overcome air resistance at any speed, we need to multiply the force of air resistance by speed (in feet per second, converted from miles per hour), and divide by 550, to convert foot-lb/sec to horsepower. The formula is


    and we get the following numbers from the formula for a few interesting speeds.
    Code:
    v (mph) 	30 	55 	65 	90 	120 	150 	200
    F (pounds) 	14.5 	48.7 	68.0 	130 	232 	362 	644
    horsepower 	1.16 	7.14 	11.8 	31.3 	74.2 	145 	344
    I put 55 mph and 65 mph in this table to show why some people think that the 55 mph national speed limit saves gasoline. It only requires about 7 hp to overcome drag at 55 mph, while it requires almost 12 hp to overcome drag at 65. Fuel consumption is approximately proportional to horsepower expended.

    More interesting to the racer is the fact that it takes 145 hp to overcome drag at 150 mph. We know that our Corvette example car has about 240 hp, so about 95 hp must be going into overcoming rolling resistance and the slight braking forces arising from internal friction in the drive train and wheel bearings. Race cars capable of going 200 mph usually have at least 650 hp, about 350 of which goes into overcoming air resistance. It is probably possible to go 200 mph with a car in the 450-500 hp range, but such a car would have very good aerodynamics; expensive, low-friction internal parts; and low rolling resistance tyres, which are designed to have the smallest possible contact patch like high performance bicycle tyres, and are therefore not good for handling.
    Last edited by Rogan; 08-04-2010 at 09:29 AM.
    Rogan o_0
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  13. #12
    Geriatric Ginger Mod Rogan's Avatar
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    Part 7: The Traction Budget

    The Physics of Racing,
    Part 7: The Traction Budget

    This month, we introduce the traction budget. This is a way of thinking about the traction available for car control under various conditions. It can help you make decisions about driving style, the right line around a course, and diagnosing handling problems. We introduce a diagramming technique for visualizing the traction budget and combine this with a well-known visualization tool, the "circle of traction," also known as the circle of friction. So this month's article is about tools, conceptual and visual, for thinking about some aspects of the physics of racing.

    To introduce the traction budget, we first need to visualize a tyre in contact with the ground. Figure 1 shows how the bottom surface of a tyre might look if we could see that surface by looking down from above. In other words, this figure shows an imaginary "X-ray" view of the bottom surface of a tyre. For the rest of the discussion, we will always imagine that we view the tyre this way. From this point of view, "up" on the diagram corresponds to forward forces and motion of the tyre and the car, "down" corresponds to backward forces and motion, "left" corresponds to leftward forces and motion, and "right" on the diagram corresponds to rightward forces and motion.

    The bottom surface of a tyre viewed from the top as though with "X-ray vision."

    The figure shows a shaded, elliptical region, where the tyre presses against the ground. All the interaction between the tyre and the ground takes place in this contact patch: that part of the tyre that touches the ground. As the tyre rolls, one bunch of tyre molecules after another move into the contact patch. But the patch itself more-or-less keeps the same shape, size, and position relative to the axis of rotation of the tyre and the car as a whole. We can use this fact to develop a simplified view of the interaction between tyre and ground. This simplified view lets us quickly and easily do approximate calculations good within a few percent. (A full-blown, mathematical analysis requires tyre coordinates that roll with the tyre, ground coordinates fixed on the ground, car coordinates fixed to the car, and many complicated equations relating these coordinate systems; the last few percent of accuracy in a mathematical model of tyre-ground interaction involves a great deal more complexity.)

    You will recall that forces on the tyre from the ground are required to make a car change either its speed of motion or its direction of motion. Thinking of the X-ray vision picture, forces pointing up are required to make the car accelerate, forces pointing down are required to make it brake, and forces pointing right and left are required to make the car turn. Consider forward acceleration, for a moment. The engine applies a torque to the axle. This torque becomes a force, pointing backwards (down, on the diagram), that the tyre applies to the ground. By Newton's third law, the ground applies an equal and opposite force, therefore pointing forward (up), on the contact patch. This force is transmitted back to the car, accelerating it forward. It is easy to get confused with all this backward and forward action and reaction. Remember to think only about the forces on the tyre and to ignore the forces on the ground, which point the opposite way.

    You will also recall that a tyre has a limited ability to stick to the ground. Apply a force that is too large, and the tyre slides. The maximum force that a tyre can take depends on the weight applied to the tyre: F W where F is the force on the tyre, is the coefficient of adhesion (and depends on tyre compound, ground characteristics, temperature, humidity, phase of the moon, etc.), and W is the weight or load on the tyre.

    By Newton's second law, the weight on the tyre depends on the fraction of the car's mass that the tyre must support and the acceleration of gravity, g = 32.1 ft / sec2. The fraction of the car's mass that the tyre must support depends on geometrical factors such as the wheelbase and the height of the centre of gravity. It also depends on the acceleration of the car, which completely accounts for weight transfer.

    It is critical to separate the geometrical, or kinematic, aspects of weight transfer from the mass of the car. Imagine two cars with the same geometry but different masses (weights). In a one g braking manoeuvre, the same fraction of each car's total weight will be transferred to the front. In the example of Part 1 of this series, we calculated a 20% weight transfer during one g braking because the height of the CG was 20% of the wheelbase. This weight transfer will be the same 20% in a 3500 pound, stock Corvette as in a 2200 pound, tube-frame, Trans-Am Corvette so long as the geometry (wheelbase, CG height, etc.) of the two cars is the same. Although the actual weight, in pounds, will be different in the two cases, the fractions of the cars' total weight will be equal.

    Separating kinematics from mass, then, we have for the weight W = f(a)mg where f(a) is the fraction of the car's mass the tyre must support and also accounts for weight transfer, m is the car's mass, and g is the acceleration of gravity.

    Finally, by Newton's second law again, the acceleration of the tyre due to the force F applied to it is a = F / f(a)m We can now combine the expressions above to discover a fascinating fact:

    a = F / f(a)m amax

    The maximum acceleration a tyre can take is g, a constant, independent of the mass of the car! While the maximum force a tyre can take depends very much on the current vertical load or weight on the tyre, the acceleration of that tyre does not depend on the current weight. If a tyre can take one g before sliding, it can take it on a lightweight car as well as on a heavy car, and it can take it under load as well as when lightly loaded. We hinted at this fact in Part 2, but the analysis above hopefully gives some deeper insight into it. We note that amax being constant is only approximately true, because changes slightly as tyre load varies, but this is a second-order effect (covered in a later article).

    So, in an approximate way, we can consider the available acceleration from a tyre independently of details of weight transfer. The tyre will give you so many gees and that's that. This is the essential idea of the traction budget. What you do with your budget is your affair. If you have a tyre that will give you one g, you can use it for accelerating, braking, cornering, or some combination, but you cannot use more than your budget or you will slide. The front-back component of the budget measures accelerating and braking, and the right-left component measures cornering acceleration. The front-back component, call it ay, combines with the left-right component, ax, not by adding, but by the Pythagorean formula:


    Rather than trying to deal with this formula, there is a convenient, visual representation of the traction budget in the circle of traction. Figure 2 shows the circle. It is oriented in the same way as the X-ray view of the contact patch, Figure 1, so that up is forward and right is rightward. The circular boundary represents the limits of the traction budget, and every point inside the circle represents a particular choice of how you spend your budget. A point near the top of the circle represents pure, forward acceleration, a point near the bottom represents pure braking. A point near the right boundary, with no up or down component, represents pure rightward cornering acceleration. Other points represent Pythagorean combinations of cornering and forward or backward acceleration.

    The beauty of this representation is that the effects of weight transfer are factored out. So the circle remains approximately the same no matter what the load on a tyre.

    The Circle of Traction.
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  14. #13
    Geriatric Ginger Mod Rogan's Avatar
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    In racing, of course, we try to spend our budget so as to stay as close to the limit, i.e. the circular boundary, as possible. In street driving, we try to stay well inside the limit so that we have lots of traction available to react to unforeseen circumstances.

    I have emphasized that the circle is only an approximate representation of the truth. It is probably close enough to make a computer driving simulation that feels right (I'm pretty sure that "Hard Drivin'" and other such games use it). As mentioned, tyre loads do cause slight, dynamic variations. Car characteristics also give rise to variations. Imagine a car with slippery tyres in the back and sticky tyres in the front. Such a car will tend to oversteer by sliding. Its traction budget will not look like a circle. Figure 3 gives an indication of what the traction budget for the whole car might look like (we have been discussing the budget of a single tyre up to this point, but the same notions apply to the whole car). In Figure 3, there is a large traction circle for the sticky front tyres and a small circle for the slippery rear tyres. Under acceleration, the slippery rears dominate the combined traction budget because of weight transfer. Under braking, the sticky fronts dominate. The combined traction budget looks something like an egg, flattened at top and wide in the middle. Under braking, the traction available for cornering is considerably greater than the traction available during acceleration because the sticky fronts are working. So, although this poorly handling car tends to oversteer by sliding the rear, it also tends to understeer during acceleration because the slippery rears will not follow the steering front tyres very effectively.

    A traction budget diagram for a poorly handling car.

    The traction budget is a versatile and simple technique for analysing and visualizing car handling. The same technique can be applied to developing driver's skills, planning the line around a course, and diagnosing handling problems.
    Rogan o_0
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  15. #14
    Geriatric Ginger Mod Rogan's Avatar
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    more later..
    Rogan o_0
    '96 Dodge 2500 CTD @ 40psi - over 700 lb/ft TQ, 7" stack, and 5speed! - SOLD
    '01 Dodge 2500 CTD 6-holed hand-shaker - 3850# dual disk - 900 lb/ft - SOLD
    '97 Dodge 3500 CTD DUALLY built Auto - 40psi boosties - 750 lb/ft

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