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. 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 just the Tire Resistance considered, regarding the Mechanical Drag.
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).
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 has to bend / flex each time it crosses 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!
The other large contribution to Drag is Aerodynamic Drag. This drag is directly proportional to the frontal area of the vehicle, and 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.
The Dynamic Pressure is simply the product of the mass-density of the air times the square of the speed divided by two. In the example below, it is 1/2 * (88 f/s)2 * 1/415 slug/cu ft which is 18.6 pounds of force.
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 25.3 psf). The frontal area (from GM) is 19.0 square feet, the aerodynamic coefficient of drag is 0.330. The aerodynamic drag coefficient is relatively constant at different speeds. 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 51.9 + 32 or 83.9 pounds at 40 mph).
THIS is the total 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 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.)
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!)
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 accelation 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!)
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 coeffient factor, and cause the driving tires to spin REAL FAST! (NOT that the author has ever done that!)
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.
(future sections)
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C Johnson, Physicist, Physics Degree from Univ of Chicago