I've always thought that it was because the frictional force on the tire was increased due to the bulging of the tires increasing the surface area in contact with the road. However, a colleague of mine reminded me that frictional force is independent of surface area. Why then is fuel economy decreased when tire pressure is low?
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$\begingroup$ This is a myth that's been disproved by testing. Tire pressures are set by makers for optimal performance. Either increasing or decreasing away from optimal maker values is both less fuel efficient and more dangerous (as handling and braking suffer). "No less a source than Mythbusters" tested this and found lowering pressure increase fuel consumption. And just about every automotive company and agency has also tried it, but who cares what they say, right ? :-) $\endgroup$– StephenG - Help UkraineCommented Sep 7, 2019 at 5:15
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18$\begingroup$ @StephenG you seem to be arguing against something, but it's not remotely clear what the "this" you're referring to is, and nothing you said after that contradicts the post... $\endgroup$– hobbsCommented Sep 7, 2019 at 8:23
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3$\begingroup$ @StephenG It's no myth that low tire pressure causes poor fuel economy; the question doesn't make any assertion about the other direction. Optimal tire pressures can't be set by tire makers because they depend on (a) the load (vehicle weight) and (b) surface (in short, bumpy roads, such as some unpaved ones, need lower tire pressure for efficiency compared to smooth roads). Passenger vehicles have recommended tire pressures written on the door panel; these don't change when you switch tires to a different brand/model, of the same dimension, on the same rims. $\endgroup$– KazCommented Sep 7, 2019 at 12:19
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$\begingroup$ @StephenG: While car makers set tire pressures for optimal performance, fuel economy seldom if ever factors into their definition of performance. For most mainstream vehicles, ride "comfort" (that is, the detestable "handles like a waterbed" feeling) is the prime factor. If you happen to have a car with a mpg meter (or just keep track of fuel consumption over a period) you can see this for yourself: inflating tires well above the automaker's specs (while staying within the tires' rated pressure) WILL significantly increase fuel economy - and often improves handling, too. $\endgroup$– jamesqfCommented Sep 7, 2019 at 17:30
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$\begingroup$ Seems that a tire that was under inflated would have a net smaller diameter as well and take additional turns to achieve a given distance. Could that be a factor in lower fuel economy as well? $\endgroup$– Michael KarasCommented Sep 7, 2019 at 17:45
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6 Answers
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Lower pressure increases surface contact and increases static friction, and static friction does not involve heat loss, so that is good. But rolling friction is not good and does involve heat loss.
Rolling friction heating is due to the inelastic deformation the rubber of the tire experiences when it is in contact with the road. See this article on rolling resistance from Wikipedia: https://en.wikipedia.org/wiki/Rolling_resistance
When the rubber is in contact with the road for each revolution, it is compressed, it then expands when it leaves the surface. The compression and expansion is not perfectly elastic, thus there is heat loss in the form of friction. The lower the tire pressure, the more rubber that is in contact with the road for each revolution, and the greater the friction heat loss. These increased heat losses add up to lower fuel economy.
Hope this helps.
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1$\begingroup$ The magnitude of the deformation of the tyre and hence hysteresis loss is greater when the pressure in the tyre is lower. $\endgroup$– FarcherCommented Sep 6, 2019 at 22:42
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$\begingroup$ @Farcher I agree, but except for the use of the term "hysteresis loss" I think I'm saying the same thing since I'm saying the deformation is greater at lower pressure. Would you agree? $\endgroup$– Bob DCommented Sep 6, 2019 at 22:47
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$\begingroup$ And this is one reason you get much worse mileage in cold weather: The rubber is stiffer, and much less elastic. $\endgroup$ Commented Sep 7, 2019 at 13:56
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$\begingroup$ @Sherwood Botsford tire pressure drops in cold weather increasing rolling resistance. There are many other reasons for lower mpg in cold weather $\endgroup$– Bob DCommented Sep 7, 2019 at 14:25
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$\begingroup$ Thank you @BobD so if I understand, this is the same phenomenon that would cause rubber to heat when bent? $\endgroup$– jodagCommented Sep 8, 2019 at 16:13
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Static friction does not actually reduce your fuel economy. Without it, your wheels would slip, so it is a good thing.
What causes the loss of fuel economy is rolling resistance. This is due to hysteresis; energy lost due to the cyclic deformation of the tyre as it rolls.
It's not a great analogy but imagine the partially inflated tire as a disc with the axel off centre. When rolled the axel will oscillate up and down - i.e. some energy is lost from being able to contribute to the forward motion.
In your case the axel doesn't bob up and down but energy is lost to reshaping the tite from bulging on one side to bulging on the other with each half revolution.
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2$\begingroup$ Paul, static friction does not reduce fuel economy and is, as you say, needed to accelerate your car. But sliding friction and rolling friction do reduce your fuel economy. These forms of friction produce heat. And heat takes energy away from performing useful work (like taking your car from A to B). So you need more energy (gas) to go from A to B with sliding or rolling friction than without. $\endgroup$– Bob DCommented Sep 6, 2019 at 22:11
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3$\begingroup$ If I may suggest, add the word "Static" to friction at the start of your answer. Bob $\endgroup$– Bob DCommented Sep 6, 2019 at 22:42
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$\begingroup$ @BobD: Which sliding friction are you thinking of for this problem? If the wheel is slipping (like driving in on a slick surface), then the lower your friction coefficient, the higher your power costs are: E.g., for a fixed air resistance force F and viscous ground friction coefficent mu, power loss at the ground contact scales as P=mu*(v_slip)^2, and the slip velocity to provide the forward force is v_slip = F/mu, so power loss scales as P = F^2/mu: the higher mu is, the closer the wheel is to not slipping, and the less power is lost to the ground friction. $\endgroup$– RLHCommented Sep 9, 2019 at 0:14
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$\begingroup$ @RLH I am not thinking about sliding friction at all in this problem. Only about rolling friction. I only mentioned sliding friction because that is another potential friction loss. But the answer to the OP's question only involves rolling friction. $\endgroup$– Bob DCommented Sep 9, 2019 at 0:29
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I have a slightly different intuition on this point than the other answers, which may help.
Imagine you are riding a skateboard or roller skates (with very hard wheels) across a soft rubber surface. In this case, it is not the wheel that deforms, but the surface:
The wheel/surface interface deforms by exactly the right amount to where the load on the wheel is balanced by the upward force of the surface.
Now, for your wheel to advance by the next increment, it has to go "uphill" to get out of the dent that it has caused itself in the surface. You can think of the wheel's progress as being a series of "climbs" out of the dip, followed by the surface collapsing vertically again under the weight. Of course in reality, your progress will not take place in this manner, you will be pushing a zone of compressed rubber in front of the wheel, but the result is the same. If you have ever tried to ride a skateboard over grass or push a pushchair on a beach, you will have experienced this problem.
For a car (and most other wheel/surface interfaces) the wheel is softer than the surface, so the wheel deforms, not the surface:
This is why it is so much more efficient for bicycles to have larger diameter wheels; the flat-spot on the bottom of the wheel can be as big as necessary, but the angle between the flat-spot and the next millimetre of tyre will be correspondingly smaller, the bigger the wheel:
Both of these circles are contacting the black line with the same size flat-spot, but the red circle is deformed from being a circle by far more than the blue one.
Also, the "hill" that the smaller wheel has climb is far steeper than the larger wheel, as shown by the angles A and B.
Finally, I think it's a bit abstract to think about fuel-consumption. You should just imagine how much harder it would be to 1) Ride a bicycle with flat tyres, 2) Push a car with flat tyres. It would take more energy. In your car, the energy comes from your petrol-tank, hence, you get a lower mpg figure when your tyres are under-inflated.
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$\begingroup$ I like this analogy but it's incomplete. You also have to imagine that the back half of the valley (i.e the part you are not "riding up") is "pushing the tire up the hill". It is releasing the stored energy from when it was compressed just a moment earlier. But it is not perfect so it doesn't' release as much energy as it absorbed. All the while like you said the hill is effectively "absorbing" energy as the wheel "rolls up it". It is the difference between the energy put into the floor and the energy put back by the floor that account for this loss. It is called hysteresis loss. $\endgroup$– tir38Commented Jul 16, 2023 at 2:39
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When using pneumatic tires, the "rolling friction" represents the effort required to continuously reshape the tire. The lower the pressure in the tire, the greater the amount of deformation as it rolls, and thus the more work will be required to achieve such deformation. If one were to inflate tires sufficiently that they would barely deform as they rolled, this friction would be greatly reduced (thus improving fuel economy) but the performance and handling advantages of pneumatic tires would be sacrificed.
If one were driving a car with absolutely rigid wheels, then every little bump in the road surface would cause the force between the wheel and the ground to be increase as the car hits the bump, and then decrease as soon as the wheel is over. Because stopping force is limited by the breaks as well as traction, momentary increases in traction won't improve it, but momentary decreases in traction will reduce it. Pneumatic tires minimize this problem because the contact area can be continuously reshaped to minimize the increase in force when hitting a bump, and thus the inevitable reduction in force afterward.
The reason that tire pressures are stated on vehicles rather than tires is that the amount of pressure needed to limit deformation to the optimal level will vary depending upon a vehicle's weight. Ideally, tires would supply a table of recommended inflation pressures based upon vehicle weight, but most vehicles are intended for use with a relatively narrow range of tire sizes, and tires of a given size will tend to have similar inflation recommendations. The exact optimal pressure for different manufacturers' tires won't be identical, but most tires will perform reasonably well over a sufficiently large range of pressures that a pressure chosen for a typical tire will usually work reasonably well--even if not quite optimally--for others tire of similar size.
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Friction being independent of surface area only applies to smooth surfaces. Roads are rough: the tire is constantly being deformed by irregularities in the road. You can see this in real time if your car has a fairly sensitive fuel economy meter: all else being equal, the rougher the road, the worse the fuel economy.
For a practical demonstration, consider two extremes: the high rolling resistance of wide, low pressure tires used by off-road vehicles (and wannabees), and the very low rolling resistance of the inflexible steel wheel on steel rail of trains.
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The street moves with constant linear speed under the tire, but the outside of the tire moves with constant angular speed. The more tire has contact with the road at one point of time, the more discrepancy between moving parts of the tire and the moving road occurs. This discrepancy is to some degree resolved by ruboff. It causes loss of energy and material.