Vehicle Resistive Forces
- Author:
- Timo Budarz
![[url=https://pixabay.com/en/railway-station-gleise-1363771/]"Train Yard"[/url] by fotolehrling is in the [url=http://creativecommons.org/publicdomain/zero/1.0/]Public Domain, CC0[/url]
While you might think rail yard when you see this picture, I often think uncompromised efficiency when I see trains. As we'll discuss in this chapter, they are the most efficient means of transporting goods, cargo and large numbers of passengers ever invented. Sadly the rail network in the US is in decline.](https://www.geogebra.org/resource/GdrQwJMd/w6K4kN87k4kgQdSB/material-GdrQwJMd.png)
There are three main resistive forces that vehicles encounter as they move along. Fuel or electric energy must be spent to overcome the resistive forces in order to move a vehicle along the road or track or waterway. It is instructive to look at these forces and consider the different modes of transportation. We will see that trains are by far the most efficient mode to transport materials or cargo. A train can transport one ton (2000 pounds) of cargo over 450 miles on one gallon of fuel! In fact, a nationwide rail network recently reported an average of 436 miles/ton for all its trains in operation. Try doing that in a Toyota Prius. Hitch a trailer onto the hybrid Toyota and fill the tank with a gallon of gasoline while pulling 2000 lbs and you might go 30 miles if the wind is in your favor. Of course this will be much worse in a pickup truck. On the other hand, you might haul 10,000 lbs (5 tons) in a large pickup and make 10 miles per gallon in absolutely perfect conditions. Hauling 5 tons 10 miles is the same as hauling one ton 50 miles in the way it's computed - but that's still nine times less efficient than an average train.
The downside of trains for commuting, which can sometimes turn the numbers around is that they are not always filled with passengers, and even when they are, humans don't weigh much compared with typical cargo carried by freight trains. That being the case, if you calculate energy burned per passenger mile or kilometer it doesn't look as good for trains since most of the weight of the train is the train itself, even when filled with passengers. On the other hand, when transporting cargo the train weighs much less than its cargo. This problem can be solved for passenger transport. While nobody is asking for my solution to the problem, I'd make very light weight commuter trains with aerodynamic profiles. But being an avid cyclist and motorcyclist, I feel the same way about cars. They are too big and heavy when you consider their common use - carrying a single passenger to work with a briefcase or backpack. But people will often argue that safety and comfort play heavily (pun intended) into the design, so I understand the dilemma. It would in fact be just as safe, but only if everyone rode in similarly light vehicles. And that could never happen overnight.
The resistive forces that all vehicles must overcome are: Bearing forces, rolling resistance (if there are wheels) and aerodynamic (or hydrodynamic for boats) drag. Each of these has a different dependence on the speed of the vehicle, and therefore at different speeds, different terms dominate. Let's have a look at each of these factors.
Resistive Forces
Bearing Resistance: When a vehicle has moving parts like axles and wheels and gears, it will always take a force to move them. This force is typically independent of speed. The major one we will consider which applies to cars as well as trains and trucks is bearing friction. The bearing friction is like the friction term that we described in the chapter on sliding friction. The bearing friction is
where there is a coefficient associated with the bearings time the normal force. If we had used a coefficient of friction it would have been just like sliding friction. Typical bearing coefficient values are 0.001-0.005.
Rolling Resistance: Rolling resistance is due to the deformation of a wheel or tire as it contacts the surface on which is rolls. The surface too might deform, and this plays into the resistance. It is always best if the surface and the wheel are equally hard or soft. Trains accomplish this by having tracks made of the same material as the wheels. A very different scenario is a dune buggy in the sand. The buggy will roll best with very low tire pressure such that the tires are equivalently soft to the sand on which they roll.
In all cases, rolling resistance is speed dependent. To a decent approximation it is linear in speed. This is the form we will use. The force required to overcome rolling resistance is
It still depends on the normal force, but now there is another pair of unit-less coefficients called the rolling resistance coefficients that describe how rapidly the force builds with speed. The zero and one on the coefficients should be thought of as zeroth order (no speed dependence) and first order (linear in v). I wrote the speed as a ratio to one m/s so that the coefficients can both remain unit-less, which is the most natural thing to do. In some texts you will see the first order coefficient in units of s/m, and the equation written without the 1m/s in the denominator. This accomplishes the same requirement of having units that agree on both sides of the equation. Typical values of are 0.01 to 0.02 and for are 0.0001 to 0.0002 for rubber tires on roads. The linear coefficient is about 10-30x smaller for trains on tracks and the zeroth order coefficient is negligible (meaning assumed zero) for a steel train wheel on a steel track.
I should note here that there are many models for rolling resistance used in practice. Which model you use depends on how precise you want to be. The simplest models show rolling resistance to be a constant like friction and bearing force. A better model like we are using shows linear dependence on speed. For road-going vehicles the most precise models have both linear dependence on speed as well as higher order terms - the most precise of which depends on speed to the 2.5 power. That equation is called the Stuttgart equation named after the city in Germany where it was developed. In Germany, vehicles travel very quickly on public roads, and at those higher speeds it becomes clear that the speed dependence is not linear.
Aerodynamic Drag: We discussed drag in earlier sections of this chapter. I hope you recall that it builds with the square of the speed. As a reminder, the magnitude of the air drag is given by:
At usual transport speeds, this squared dependence tends to dominate even in trains and trucks which have quite large bearing and rolling resistance forces due to very large normal forces. It is worth mentioning here that for freight trains with empty flat bed cars between full cars, the drag is much higher. Therefore, the arrangement of the train cars is essential to their efficiency. Such ordering considerations don't exist in other transport modes. Of course it is well known that driving close behind a semi truck on the freeway will save you gas while not stealing any fuel from the semi. In fact the semi will save some fuel with you tail-gating.
In this sense, the combined fuel efficiency of a semi and a car traveling together will be much better if they are closely arranged without any gaps, just as is the case for train cars. The same is true of drafting in a cycling peloton - the name given to a large group of cyclists racing or riding together. Both leader and followers benefit, but the benefit in all of these cases is much more pronounced for the follower.
Surface Gradient: In earlier sections we discussed the force with which gravity tries to pull an object down an incline. That force is
If a vehicle is climbing a hill, this presents another resistive force to be overcome. If the vehicle is descending a hill, this force will obviously help the vehicle along rather than hindering its progress. It is conventional to consider the gradient positive when the road is uphill and negative when it's downhill. With this convention in mind we should write this expression as so that when the angle is positive (meaning uphill) the force opposes the vehicle, and when the angle is negative (meaning downhill) the force will help the vehicle along.
If the total value of all of the forces on the vehicle goes positive, as it may with a steep enough hill, then brakes would have to be applied to prevent descending too fast. Since road surfaces rarely exceed 10% grades, or 5.7o, we don't really have to be concerned about the normal force in the previous expressions changing much due to the cosine term since the cosine of that angle is 0.995, but with steep gradients or for very high precision, we should make that modification.
Combined Resistive Forces
Taken together and seen on a plot, it is clear that depending on speed and weight and size of the mode of transport, the various factors can be more or less significant.
Vehicle Resistive Forces Plotted
Power
Power is a topic that rightfully follows from discussions of energy, which is a later chapter. But right now it seems like power would be relevant to discuss since we often discuss vehicle engines in terms of the power they produce. So consider this a preview of a later discussion on power.
One way in which power could be used right now is to try to calculate the top speed of a vehicle. If I took a car to a long stretch of flat road and put the gas to the floor and just waited until speed stopped rising, that would correspond to a situation where the force propelling the car forward (due to the engine/drivetrain) would match the resistive forces opposing the motion of the car. Overall that would mean that the net force on the car is zero. That leads to zero acceleration, and a constant velocity. That is the condition at top speed.
Power P is the product of speed and force. It's really a dot product, but in the current context it is correct to just write P=Fv. The power required to go a certain speed v while opposing resistive forces F is just the product of the two. Power is measured in watts, but 746W=1hp, so if you like horsepower you can convert that way.
At very high speeds, usually air drag dominates the resistive forces. So if we just use air drag for F (which is quadratic in speed v), you can see that multiplying by another v makes power cubic in v, or
A cubic rate of growth is quick. It implies that for a vehicle to double its top speed (so v becomes 2v), it takes 23=8 times more power. That's why it is so hard for any car to reach 200 mph while it's rather trivial to reach 100 mph in almost any car.
If I'm a cyclist and want to be able to ride at 10% higher speed, or 1.1v where v is my current speed, and if I wonder how much more power it will take, all I have to do is take the factor of 1.1 and cube it. Since 1.13=1.33, it means it takes 33% (or one third) more power to do this. So if my current maximum cruising speed is 25mph on my bike, and I wish to go 10% faster (26.5mph), I need 1/3 more power. In reality it might take 300W (watts) to go 25mph. So to cruise at 26.5mph it will take 1.33x300W=400W! That is no small increase. For an athlete who has been training for a while and can sustain 300W, making 400W may just never happen. In that case, the best option is to work on the other factors in the drag equation... like the frontal area, which can be altered by posture on the bike, and the drag coefficient, which is altered by design of the bike, clothing, and posture of the rider.