# Automotive Vehicles - Physics

## Physics In Automobiles and Trucks

There are many aspects of vehicles where the laws of Physic apply. We will try to cover most of the main ones here.

## Movement Forward

In order for any vehicle to move forward, it must provide some Thrust. Thrust is necessary to first cause acceleration (increase in speed / velocity) according to Newton's laws of motion. Additional Thrust is required to overcome Drag forces, and that component is necessary even once the vehicle is traveling at a constant speed.

Drag exists from two different sources. Aerodynamic Drag is due to having to push air out of the way of where the vehicle needs to go, and also due to the turbulence generally caused behind it as the air has to re-fill that space). Mechanical Drag is due to all the moving mechanisms in the vehicle that have frictional losses, most specifically the wheel bearings, but is actually nearly entirely due to the action of the tires on the road surface.

In nearly all actual situations, all the other causes of mechanical drag factors can be ignored, and actually considered instead as fixed losses, and just the Tire Resistance considered, regarding the Mechanical Drag. It is often called Tire Drag as a result.

It has been experimentally determined that the total Tire Resistance Drag is nearly proportional to the weight of the vehicle, and is slightly affected by vehicle speed. It is also affected by the air pressure in the tires and the temperature of the tires. In general, with modern synthetic rubber tires, inflated to the specified pressures, the total Tire Resistance Drag is generally around 1.2% to 1.4% of total vehicle weight at 30 mph, and around 1.6% to 2.0% of total vehicle weight at 70 mph, once the tires are driven at that speed for half an hour or so such that they get up to normal operating temperature (often around 240°F).

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For example, a standard 3,000 pound car therefore has around 48 to 60 pounds of Tire Resistance Drag at 70 mph.

Nearly all of this drag arises from the fact that the tire sidewalls have to bend / flex each time they cross under the axle. At 70 mph, every part of the tread of every tire has to flex like that around 15 times every second! THAT is the cause of the tires heating up to very high operating temperatures. When a tire is under-inflated, it flexes MORE every time, which causes even more heating of the tire, and THAT is usually the cause of a low-pressured tire "blowing out" at speed, because the tire got far hotter than the design temperature.

By the way, in contrast, a rigid wheel, which does NOT flex like that, such as a railroad car wheel, only has around 1/9 the rolling resistance at speed! In cars, no one would tolerate how hard the ride would be! But people have told me that they inflate their car tires to 45 PSI or 50 PSI, in an attempt to reduce this loss. Yes, that can work, but the ride inside that vehicle can become unbearable on rough roads, AND the benefit is generally still only about 2% or 3%. Not even close to even one mile-per-gallon improvement in fuel mileage! THAT is why no one else does that! Along with not wanting to constantly stress the tires far beyond the pressures that the tire manufacturers had intended in their products!

The other large contribution to Drag is Aerodynamic Drag. This drag is directly proportional to the frontal area of the vehicle, and very close to being exactly proportional to the SQUARE of the speed difference between the air and the vehicle. It is also strongly affected by the shape of the vehicle, so there is a Drag Coefficient Factor which is included. Aerodynamic Drag is also proportional to the density of the air, so high in the mountains it is less than near sea level.

Here are some numbers regarding the air resistance of a moving vehicle. This is a slightly more general approach, which is used for aircraft, trains, ships and sometimes cars and trucks. It is actually the same thing. A parameter called the Dynamic Pressure must be multiplied by the frontal area of the vehicle and by a factor called a shape coefficient (CD (or aerodynamic coefficient of drag) to get the actual total air-resistance force. Streamlined vehicles have a lower CD, as low as 0.2, while extremely boxy vehicles can have coefficients nearly 1.0. These are all easily determined values. The result is a value (at 60 mph speed) of around 120 to 170 pounds of air resistance force, for a mid-sized sedan-style car.

The Dynamic Pressure is related to the Momentum in the air and is simply the product of the mass-flow of the air times the speed. In the example below, it is (1/415 slug/cu ft * 88 f/s) * (88 f/s) which is 18.6 pounds of force (A pound of force is equal to one slug-ft/sec2 or mass times acceleration as Newton said) (60 mph is the same as 88 ft/sec).

Interestingly, some respected texts include a factor of 1/2 in that calculation, showing it as 1/2 * m * v2 instead of the correct m * v2 presented here. They apparently make a mistake in thinking that the amount is an amount of (kinetic) ENERGY rather than the correct MOMENTUM figure. If you see a factor of 1/2 in that calculation (Even Mark's Handbook for Mechanical Engineers presents it incorrectly!), remember to make that correction!

Here are the values for one of my Corvettes, a fairly aerodynamic car. Dynamic pressure (only depends on speed and not the type of vehicle) at 60 mph is around 18.6 pounds per square foot. (at 70 mph, it's 25.3 psf). The frontal area (from GM) is 19.0 square feet, the aerodynamic coefficient of drag is 0.330. That aerodynamic drag coefficient is relatively constant at different speeds, except at extremely slow speeds where there is little turbulence. So, for my car at 60 mph, the Aerodynamic drag is 19.0 * .330 * 18.6 or about 116.7 pounds. (at 70 mph it is about 158.9 pounds, and at 40 mph, 51.9 pounds). The Tire Resistance drag is (at 60 mph) around 0.015 of the vehicle weight, and is dependent on the type of tires, their inflation, temperature, speed and other things. Since my Corvette weighs around 3200 pounds, this gives about 48 pounds at 60 mph. This makes the Total Drag as 116.7 + 48 or 164.7 pounds at 60 mph (and 218.9 pounds at 70 mph) (and 83.9 pounds at 40 mph).

THIS is the total vehicle Drag that exists at those various speeds. Therefore, the vehicle has to provide forward Thrust of exactly that amount, if the vehicle is to remain at a constant speed. In other words, at 60 mph, no wind, flat level surface, etc, my Corvette needs to constantly produce about 165 pounds of forward thrust, to STAY at 60 mph.

Vehicles generally provide that Thrust by turning drive wheels. As long as there is sufficient traction between the tires and roadway, the engine can provide that necessary Thrust by rotating the drive wheels.

People generally talk about HORSEPOWER rather than Thrust with vehicles, but Thrust is generally described for aircraft and watercraft. They are closely related, and the choice was made because one or the other was more suitable for the specific applications. Thrust times speed gives ft-lb/sec, and simply by dividing by 550 (the number of ft-lb/sec in one horsepower), we can get horsepower! In the case above, at 60 mph (88 ft/sec), we have 165 lb * 88 ft/sec / 550 or 26.4 horsepower. THAT is actually all the power the Corvette needs to stay at a constant 60 mph speed! (All the extra horsepower that cars have is needed for acceleration, to be discussed below.)

## Production of Thrust

Much of this subject is covered in the related page on Physics in an Automotive Engine, linked at the end of this web-page. However, there are some subjects of Physics between when torque is developed inside an engine and it getting to become a forward force for the vehicle.

A moderately powerful automobile engine can produce around 200 pound-feet of torque (on fairly heavy usage of the gas pedal!) Let's say we could directly apply that torque to the driving wheels of the vehicle. Common tires on cars have roughly a one-foot effective radius. (A tire's measured diameter or radius is greater than this, but the tire is flattened at the bottom, and the actual effectual radius, the distance between the wheel axle center and the road, is less, as suggested.) Therefore, this would mean that we could provide a "maximum rearward force" by the tire tread on the pavement, of around 200 pounds force (when the engine is really working hard!)

Newton told us that force equals mass times acceleration. If we are considering a vehicle with a (mass equivalent) weight of 3200 pounds, then we would have 200 = 3200 * acceleration, or acceleration would be 1/16 G (the acceleration due to gravity). This would be a pitiful acceleration! Starting from a traffic light, after one (long) block (660 feet) we would have a speed given by v2 = 2 * A * D or v2 = 2* (1/16 * 32 ft/sec/sec) * 660 or v2 = 2600 or v = 51 ft/second! That would be an impressive 35 miles per hour! (after all out full-bore acceleration for a full city block!)

And THAT would only be if the engine could produce its maximum torque over the whole range of engine speed, which it cannot! Figure HALF that, or around 17 mph absolute maximum speed a block after a traffic light!

SO!

We use GEARING to MULTIPLY TORQUE. First, rear-wheel-drive vehicles have a DIFFERENTIAL, which has several functions, but one of which is a torque multiplier of around 3:1 to 4:1. Commonly, cars with automatic transmissions have around 3.25 ratio.

This would help, and our traffic light full-gas-pedal acceleration figured above would now give us about 31 mph, better but still pitiful.

SO! All cars also have a transmission, which has several different gearings that are used. First gear generally has another 3:1 gearing ratio. A side benefit from this is that the engine can be used over a narrower speed range, so that it can have a consistent greater torque output. The result of these two additional advantages result in the vehicle described above being able to get up to over 60 mph in that city block after the traffic light. (A rather obvious violation of Municipal Laws, so don't do that at home!)

As a side note, a Corvette can (the author says 'allegedly') get up to about 100 mph in a quarter mile (two long city blocks) of hard acceleration. Using the formulas above, we would have v2 = (145 ft/sec)2 or 21000 = 2 * A * D. A quarter mile is 1320 feet, so that is 2640 * A, so A is around 8 ft/sec/sec, for an average acceleration of around 0.25 G. Near the start, the acceleration is greater, due to the torque multiplying of transmission gears.

Another side note: There is a limit to how much Thrust one can apply with tires. There is a static coefficient of friction which depends on the type of road surface, rain, snow, tire temperature, and other things. In general, the static coefficient of friction for tires on standard dry roadway is around 0.7. That means that the weight on those driving wheels (the downward force) times 0.7 gives the maximum tire friction possible without the tires breaking free and spinning. Most cars have around half the vehicle weight on the driving wheels, which means that the maximum forward thrust available from those tires is around half of 0.7 G, or around 0.35 G. People familiar with sports cars (and other crazies) are aware that, in low gears, where torque multiplication is greatest, it is possible to apply enough engine torque to the tires to overcome this static friction coefficient factor, and cause the driving tires to spin REAL FAST! (NOT that the author has ever done that!)

OK, continuing! On a rear wheel drive car, there are two additional effects which enter the picture. First, the torque of the driveshaft from the engine acts to try to twist the rear axle. Engine manufacturers seem to have all agreed to have their engines rotate in the same direction, so this torque always has the effect of trying to lift some body weight off the passenger rear tire. This is generally a relatively minor effect, but it still tends to cause THAT tire to break loose on extreme acceleration, and once it has broken loose, there is less torque trying to break the driver's rear tire loose. So if you have a MODERATELY powerful engine, you may have wondered why only one rear tire tends to use up its tread very fast. That is why. If you have an extremely powerful engine and you also have a limited slip differential, then both rear tires can and do break free at the same time.

The second additional effect is that extreme torque sent to the rear wheels has the effect of trying to twist the rear axle as a reaction effect (Newton, again). Many hot cars from the 1950s and 1960s which had leaf spring rear suspensions had this as a serious problem, and dragsters nearly all do. The solution to that can be seen on most dragsters, of an extra torque arm attached to the rear axle and going forward, where it is connected by a drag link to the frame of the vehicle. That structure ensures that the rear axle cannot twist at all, and it greatly reduces wheel spin to a bare minimum.

Yet another side note: On wet roadways, the static coefficient of friction can be around 0.2. Really worn tires (like slicks) only have around 0.5 SCF. On icy roads, the SCF can be as low as 0.06. This all means that far less engine torque (or braking torque) will break the tires free from the pavement, which explains why cars sometimes skid and slide, and drivers are sometimes caught unprepared when the vehicle does such things.

Further, once a tire breaks free of the pavement, a DIFFERENT coefficient of friction applies, one called the DYNAMIC coefficient of friction. For essentially all pairs of materials, the Dynamic coefficient is a lower number than the Static. This results in two behaviors which you may have noticed. One is that once tire spin is initiated, such as in doing a donut, less engine power is required to continue the spinning. The other is that when you are on a wet or icy or snow covered road, once traction has broken free, the vehicle is very difficult to get back under control, until the vehicle speed has greatly reduced. There are ways for a talented driver to improve this last situation, such as by applying the parking brake (only rear wheels) while NOT applying the regular brakes, and aiming the front wheels in the direction of the current slide, to try to get the front (steering) wheels spinning and hopefully again in (Static) traction with the pavement, but few drivers know how to do such things. Long ago (when I was young and ignorant), I found myself driving at night on an Interstate in Michigan at around 80 mph on what I did not realize was black ice. Basically the same as a skating rink. Miles ahead I saw red flashing lights so I LIGHTLY touched the brake pedal, only to see the speedometer immediately drop to zero! But I was still moving at 80 mph, but now sliding rather than driving. Fortunately, at 3 am, there were no other vehicles around, so the fact that my car started gradually rotating was not an immediate crisis. But to be sliding down an interstate highway at probably 75 mph, sideways, is quite distressing! I was quite concerned that I might encounter a sudden dry spot of highway, where the tires might all suddenly grab and I might then do a multiple rollover like in the movies. Fortunately that did not occur. Also fortunate was that I continued to slide on the lane areas of the highway, as I was quite aware that if I slide sideways and encountered a bridge support, it probably would not be fun. In any case, I knew the things described above and immediately applied the parking brake, to keep the rear wheels from turning, while doing everything I could to encourage the front wheels to start spinning. Very slowly, my headlights started turning back around toward the front, and I tried to steer the wheels to be as much along the highway as possible. Once the car was again mostly pointed in the right direction, the front wheels DID start spinning and I could again steer. The parking brake was left on, to have at least a slight effect of slowing the car down, now while I was actually able to aim it to stay in one lane. The whole experience scared the daylights out of me and taught me what black ice was! Once I got the car speed down to about 20 mph, I released the parking brake and could then again see the vehicle speed (which is measured by the rotation of the rear tires). I eventually got close to all the red flashing lights and saw around two dozen cars and semis which had had extremely bad damage, including many rollovers. I stayed driving at 20 mph for quite a while that night, until I was absolutely sure that I was well past the black ice area.

(future sections)

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This presentation was first placed on the Internet in April 2006.

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C Johnson, Theoretical Physicist, Physics Degree from Univ of Chicago