Why doesn't tire pressure vary with vehicle payload (much)?
January 20, 2010 11:45 PM Subscribe
Stupid physics question: tire pressure.
Let's say your truck has 4 tires, and they're all pressurized with air to some pressure, say 30 psig. Now say you put an additional 1000 lbs in the truck. Assume for round number's sake, the tire contact area with the ground is 10 square inches.
Why would the tire pressure not go up by (1000 lbs / 4 tires) / 10 in^2 = 25 lbs?
Put another way, does dividing my car weight by the tire pressure give the total tire contact area with the road?
Not homework, I just thought of this filling my car with gas this evening and it's bugging me.
Let's say your truck has 4 tires, and they're all pressurized with air to some pressure, say 30 psig. Now say you put an additional 1000 lbs in the truck. Assume for round number's sake, the tire contact area with the ground is 10 square inches.
Why would the tire pressure not go up by (1000 lbs / 4 tires) / 10 in^2 = 25 lbs?
Put another way, does dividing my car weight by the tire pressure give the total tire contact area with the road?
Not homework, I just thought of this filling my car with gas this evening and it's bugging me.
Best answer: When the tire squishes, the pressure does increase, but just as importantly, so does the contact area. So a new equilibrium is reached where the pressure * the area touching the ground is equal to the new weight.
So let's say your tire pressure is 30psi normally (normally meaning with a quarter of your unladen truck resting on it. So let's say you have a 4000 pound truck, that's 33 sq in of tire touching the ground.
If you now add 250# to that tire, there's a few different options. Maybe now your tire is squashed enough that there's 33 psi and 38 sq in of contact area. Bigger tires will be able to increase the contact area more with less deformation or increase in pressure (less volumetric change is necessary to increase the contact area, AND a similar volumetric change will be a smaller portion of the overall volume.)
If you want a really exaggerated example of this, stand on a tennis ball.
posted by aubilenon at 1:04 AM on January 21, 2010 [1 favorite]
So let's say your tire pressure is 30psi normally (normally meaning with a quarter of your unladen truck resting on it. So let's say you have a 4000 pound truck, that's 33 sq in of tire touching the ground.
If you now add 250# to that tire, there's a few different options. Maybe now your tire is squashed enough that there's 33 psi and 38 sq in of contact area. Bigger tires will be able to increase the contact area more with less deformation or increase in pressure (less volumetric change is necessary to increase the contact area, AND a similar volumetric change will be a smaller portion of the overall volume.)
If you want a really exaggerated example of this, stand on a tennis ball.
posted by aubilenon at 1:04 AM on January 21, 2010 [1 favorite]
Best answer: Off the cuff answer: No, because tire pressure and vehicle weight are different things. I.e., imagine if your tires had zero internal pressure (were flat). That wouldn't mean your truck was weightless. It also wouldn't mean that the tires would have infinite contact area with the road.
(They're different but related: loading the truck will increase the internal tire pressure, but by a small amount.)
Beware of amateur physicists answering this question. If you really want to know, you're best off with a tire engineer who truly understands how tires work. Maybe something along these lines, or this can give you a taste of the complexity behind tires. Ask the tire forum.
(I could make something up about how air pressure is for supporting the tire and maintaining its shape in order to resist deformation from an external load, and it's squishing/compacting deformation that would increase internal tire pressure ... but I'm not a tire engineer.)
(Also note that the book-squishing-a-balloon model is wrong for a tire.)
posted by coffeefilter at 1:07 AM on January 21, 2010
(They're different but related: loading the truck will increase the internal tire pressure, but by a small amount.)
Beware of amateur physicists answering this question. If you really want to know, you're best off with a tire engineer who truly understands how tires work. Maybe something along these lines, or this can give you a taste of the complexity behind tires. Ask the tire forum.
(I could make something up about how air pressure is for supporting the tire and maintaining its shape in order to resist deformation from an external load, and it's squishing/compacting deformation that would increase internal tire pressure ... but I'm not a tire engineer.)
(Also note that the book-squishing-a-balloon model is wrong for a tire.)
posted by coffeefilter at 1:07 AM on January 21, 2010
disclaimer: I am an amateur physicist, so beware.
30 psi x (10sq in. / tire) = 300 lbsf/tire
300 lbsf / tire x 4 tires => 1200 lbs total weight of the car
now add 1000 lbs
1200 lbs + 1000 lbs = 2200 lbs
2200 lbs / 4 tires = 550 lbsf/tire
Now here's the interesting part:
IF the footprint of all the tires continued to be 10 sq in. each
AND the tire sidewall stiffness is zero, contributing nothing to support,
THEN 550 lbsf / 10sq in. => 55 psi.
But what is normally seen is for the pressure to remain the same tires to bulge out and increase the footprint. So:
550 lbsf/tire / 30psi => 18.33 sq in per tire will be the new footprint,
once again assuming zero tire wall stiffness
So in a simplistic sense, if you want to keep the same 10sq in footprint, then you have to inflate the tires to 55 psi.
posted by buzzv at 1:48 AM on January 21, 2010
30 psi x (10sq in. / tire) = 300 lbsf/tire
300 lbsf / tire x 4 tires => 1200 lbs total weight of the car
now add 1000 lbs
1200 lbs + 1000 lbs = 2200 lbs
2200 lbs / 4 tires = 550 lbsf/tire
Now here's the interesting part:
IF the footprint of all the tires continued to be 10 sq in. each
AND the tire sidewall stiffness is zero, contributing nothing to support,
THEN 550 lbsf / 10sq in. => 55 psi.
But what is normally seen is for the pressure to remain the same tires to bulge out and increase the footprint. So:
550 lbsf/tire / 30psi => 18.33 sq in per tire will be the new footprint,
once again assuming zero tire wall stiffness
So in a simplistic sense, if you want to keep the same 10sq in footprint, then you have to inflate the tires to 55 psi.
posted by buzzv at 1:48 AM on January 21, 2010
Best answer: The contact patch doesn't really increase when you underinflate a tire given a load.
It might look like the tire spreads out and gets wider, but what you don't see is that the center of the tread buckles up as the outer edges sort of roll over and outwards. That's why underinflated tires wear the outer tread bands in an exaggerated fashion. Due to the construction of the tire, you're not necessarily getting a bigger contact patch when you're underinflated, at least with passenger vehicles.*
As described previously, the structure of a tire isn't like any other balloon, tennis-ball, beach-ball, example that you can give to illustrate the physics of air pressure in a flexible container. A tire is extremely structural, given the steel belt and nylon/kevlar/whatever else ply construction. I weight about 170lb and I can sit on a rim and tire assembly that has 0psi(g) and see basically no shape-change in the tire.
At any rate, the easy answer to your question is that the air doesn't hold the vehicle off the ground. The tire holds the vehicle off the ground, and the air dictates the shape of the tire. The shape of a tire is like an arch whereby it transfers the compressive force from the tread, through the sidewall, to the rim, NOT through the air inside.
*This is true for a solid contact surface like pavement, not snow or sand.
posted by Jon-o at 4:31 AM on January 21, 2010
It might look like the tire spreads out and gets wider, but what you don't see is that the center of the tread buckles up as the outer edges sort of roll over and outwards. That's why underinflated tires wear the outer tread bands in an exaggerated fashion. Due to the construction of the tire, you're not necessarily getting a bigger contact patch when you're underinflated, at least with passenger vehicles.*
As described previously, the structure of a tire isn't like any other balloon, tennis-ball, beach-ball, example that you can give to illustrate the physics of air pressure in a flexible container. A tire is extremely structural, given the steel belt and nylon/kevlar/whatever else ply construction. I weight about 170lb and I can sit on a rim and tire assembly that has 0psi(g) and see basically no shape-change in the tire.
At any rate, the easy answer to your question is that the air doesn't hold the vehicle off the ground. The tire holds the vehicle off the ground, and the air dictates the shape of the tire. The shape of a tire is like an arch whereby it transfers the compressive force from the tread, through the sidewall, to the rim, NOT through the air inside.
*This is true for a solid contact surface like pavement, not snow or sand.
posted by Jon-o at 4:31 AM on January 21, 2010
As an extreme example of the sidewall contribution to holding up the vehicle, look at run-flat tires, the most common type of which relies entirely on the sidewall to support the vehicle even with little or no air pressure in it. Because the sidewall does this so well it can be difficult to know when there is a loss of air and these tire are generally combined with a tire prewssure monitoring system (this may be a government requirement but I could not confirm that). (I am not a physicist or tire engineer, but I had a car with the abominable Michelin PAX system on it and researched the topic a good bit while figuring out the best way to replace it.)
posted by TedW at 5:08 AM on January 21, 2010
posted by TedW at 5:08 AM on January 21, 2010
the easy answer to your question is that the air doesn't hold the vehicle off the ground
This is a rare circumstance, but I think Jon-O is wrong here. Clearly the air is holding the vehicle up; witness the fact that the car drops closer to the ground when the tire is underinflated.
As an experiment, I just backed my wife's Jetta onto the concrete driveway apron (to get it off the gravel drive) and did some quick measurements. It looks like the total tire contact area is about 120 square inches. The vehicle should have a curb weight of around 3230#. 3230/120=26.9 pounds/square inch, which is surprisingly close to the inflation pressure. I don't think the sidewalls are doing nearly so much of the load-bearing as Jon-O suggests.
I also believe the tire contact area does get larger when the tires are more heavily loaded. The rough measurements I just took showed the tires' contact patches to be roughly 70% larger in front, under the engine, than in the rear. (Tires were rotated and pressure-checked 6 days ago.)
It seems clear that adding weight to the vehicle increases the weight borne by the air. However, that weight is borne by the entire interior surface area of the tire, pressing outward in all directions, and also pressing on the inside of the rim.
Say I throw a thousand pounds of sand into my Jetta's back seat. Say also, for the sake of simplicity, that the tire contact areas do not change and the tire does not expand. Some more quick guesstimating suggests an interior wheel / tire surface area of 1425 square inches each, or a total of 5700 square inches for the whole vehicle. So that 1000 pounds of additional weight is spread over 5700 square inches; there's a theoretical pressure increase of roughly 0.17 PSI. If the tire is a bit elastic so that it expands a bit under pressure, then the increase is even smaller.
posted by jon1270 at 5:20 AM on January 21, 2010
This is a rare circumstance, but I think Jon-O is wrong here. Clearly the air is holding the vehicle up; witness the fact that the car drops closer to the ground when the tire is underinflated.
As an experiment, I just backed my wife's Jetta onto the concrete driveway apron (to get it off the gravel drive) and did some quick measurements. It looks like the total tire contact area is about 120 square inches. The vehicle should have a curb weight of around 3230#. 3230/120=26.9 pounds/square inch, which is surprisingly close to the inflation pressure. I don't think the sidewalls are doing nearly so much of the load-bearing as Jon-O suggests.
I also believe the tire contact area does get larger when the tires are more heavily loaded. The rough measurements I just took showed the tires' contact patches to be roughly 70% larger in front, under the engine, than in the rear. (Tires were rotated and pressure-checked 6 days ago.)
It seems clear that adding weight to the vehicle increases the weight borne by the air. However, that weight is borne by the entire interior surface area of the tire, pressing outward in all directions, and also pressing on the inside of the rim.
Say I throw a thousand pounds of sand into my Jetta's back seat. Say also, for the sake of simplicity, that the tire contact areas do not change and the tire does not expand. Some more quick guesstimating suggests an interior wheel / tire surface area of 1425 square inches each, or a total of 5700 square inches for the whole vehicle. So that 1000 pounds of additional weight is spread over 5700 square inches; there's a theoretical pressure increase of roughly 0.17 PSI. If the tire is a bit elastic so that it expands a bit under pressure, then the increase is even smaller.
posted by jon1270 at 5:20 AM on January 21, 2010
Jobst Brandt once explained this pretty well with respect to bicycle tires. The overall mechanism by which a tire supports the load isn't any different for car tires.
If you were to measure the forces in the tire sidewall, it's under tension all the time (since it would buckle if you tried to put any significant compression on it), so in that sense the load isn't supported by compression in the sidewall.
But in another sense, you can think of a tire as a pre-tensioned structure, and thought of like that, the load is partly supported by "compression" of that structure (e.g. reduction of pre-tension) where it contacts the ground, and partly by altering the shape of the structure so that the pre-tension acts in a slightly different direction.
posted by FishBike at 6:16 AM on January 21, 2010
If you were to measure the forces in the tire sidewall, it's under tension all the time (since it would buckle if you tried to put any significant compression on it), so in that sense the load isn't supported by compression in the sidewall.
But in another sense, you can think of a tire as a pre-tensioned structure, and thought of like that, the load is partly supported by "compression" of that structure (e.g. reduction of pre-tension) where it contacts the ground, and partly by altering the shape of the structure so that the pre-tension acts in a slightly different direction.
posted by FishBike at 6:16 AM on January 21, 2010
If you want a really exaggerated example of this, stand on a tennis ball.
Great example, but please don't do this. I ripped two tendons this way once.
posted by rokusan at 6:40 AM on January 21, 2010
Great example, but please don't do this. I ripped two tendons this way once.
posted by rokusan at 6:40 AM on January 21, 2010
At any rate, the easy answer to your question is that the air doesn't hold the vehicle off the ground.
Well, that can't be true, because otherwise a flat tire would still hold your car up. Excepting run-flat tires, they don't. A tire's internal structure might contribute some extra support that might actually mean that the pressure would go up.
Think about it this way: The structure of the tire itself exerts some structural pressure itself as it deforms away from its intended shape, kind of like a spring. A spring in a neutral position exerts no pressure. But when you push it, it starts to add pressure.
So when you add more weight to a vehicle, the air in the tire 'wants' to keep the same pressure and just expand the contact patch touching the ground. But, because that would mean deforming the tire, the tire's structure pushes back against that (slightly) therefore increasing the pressure a little bit (but not much)
--
If you're wondering, Why not attach a tire pressure gauge to your tire, then add some weight and see what happens? You probably won't see much change.
posted by delmoi at 7:05 AM on January 21, 2010
Well, that can't be true, because otherwise a flat tire would still hold your car up. Excepting run-flat tires, they don't. A tire's internal structure might contribute some extra support that might actually mean that the pressure would go up.
Think about it this way: The structure of the tire itself exerts some structural pressure itself as it deforms away from its intended shape, kind of like a spring. A spring in a neutral position exerts no pressure. But when you push it, it starts to add pressure.
So when you add more weight to a vehicle, the air in the tire 'wants' to keep the same pressure and just expand the contact patch touching the ground. But, because that would mean deforming the tire, the tire's structure pushes back against that (slightly) therefore increasing the pressure a little bit (but not much)
--
If you're wondering, Why not attach a tire pressure gauge to your tire, then add some weight and see what happens? You probably won't see much change.
posted by delmoi at 7:05 AM on January 21, 2010
pV = nRT
If the volume of the tire is not changing from load, if the gas constant doesn't change when under load, if you're not putting more air into the tire under load, and if the temperature of the gas in the tire does not change under load, then assuming your tires observe the ideal gas law, their pressure p will not change under load.
posted by Blazecock Pileon at 7:06 AM on January 21, 2010 [1 favorite]
If the volume of the tire is not changing from load, if the gas constant doesn't change when under load, if you're not putting more air into the tire under load, and if the temperature of the gas in the tire does not change under load, then assuming your tires observe the ideal gas law, their pressure p will not change under load.
posted by Blazecock Pileon at 7:06 AM on January 21, 2010 [1 favorite]
Response by poster: Ha, I woke up this morning and kicked myself. "Obviously" the contact area would just increase to compensate. Now I like the structural support argument - I'm thinking of one of those popsicle-stick structures they have kids build and then load with weights. A popsicle stick can support quite a bit of weight axially as long as you can keep it from buckling. So now I'm understanding guessing that the air in the tire just keeps the side wall in shape, and the side wall's shape supports the weight. Yes?
posted by ctmf at 7:06 AM on January 21, 2010
posted by ctmf at 7:06 AM on January 21, 2010
the air in the tire just keeps the side wall in shape, and the side wall's shape supports the weight.
The tire without the air supports relatively little. The air without the tire supports essentially nothing. I don't think it makes sense to say one is carrying the car and the other isn't; they are links in a chain.
Notice that trucks meant to carry heavy loads have big, big tires. This is true not only for pickups and dump trucks and other vehicles that need large diameter tires to facilitate driving over off-road-obstacles, but also for semi tractors and trailers which may never see an unpaved surface. If I'm imagining the physics correctly, this is largely because the load is transferred through the air and distributed across the surface area of the entire tire. A big tire has more interior surface area to spread a load out over, so a bigger tire can carry a bigger load.
I think part of the confusion is due to thinking of the load as being inherently vertical / oriented by gravity. There is a vertical load (the vehicle and cargo) in play, but the load is being carried by a compressed gas and thence directed outward against the inside of the gas's container (the tire). From the air's point of view, it's no longer a vertical load; it's pressure from all directions.
posted by jon1270 at 8:26 AM on January 21, 2010
The tire without the air supports relatively little. The air without the tire supports essentially nothing. I don't think it makes sense to say one is carrying the car and the other isn't; they are links in a chain.
Notice that trucks meant to carry heavy loads have big, big tires. This is true not only for pickups and dump trucks and other vehicles that need large diameter tires to facilitate driving over off-road-obstacles, but also for semi tractors and trailers which may never see an unpaved surface. If I'm imagining the physics correctly, this is largely because the load is transferred through the air and distributed across the surface area of the entire tire. A big tire has more interior surface area to spread a load out over, so a bigger tire can carry a bigger load.
I think part of the confusion is due to thinking of the load as being inherently vertical / oriented by gravity. There is a vertical load (the vehicle and cargo) in play, but the load is being carried by a compressed gas and thence directed outward against the inside of the gas's container (the tire). From the air's point of view, it's no longer a vertical load; it's pressure from all directions.
posted by jon1270 at 8:26 AM on January 21, 2010
So now I'm understanding guessing that the air in the tire just keeps the side wall in shape, and the side wall's shape supports the weight. Yes?
This is a little like asking which grain of sand on the beach is the one actually holding you up as you walk, or which water molecule in the ocean is the one you're floating on. The entire system — the tire, the air and the inner tube — gives rise to a force that holds the car up. No single component is responsible.
posted by nebulawindphone at 8:29 AM on January 21, 2010
This is a little like asking which grain of sand on the beach is the one actually holding you up as you walk, or which water molecule in the ocean is the one you're floating on. The entire system — the tire, the air and the inner tube — gives rise to a force that holds the car up. No single component is responsible.
posted by nebulawindphone at 8:29 AM on January 21, 2010
Clearly the air is holding the vehicle up; witness the fact that the car drops closer to the ground when the tire is underinflated.
It's a technicality, but the air is holding the tire in shape, rather than supporting the weight of the vehicle. If you measure tire psi on a lift and then with the car on the ground, the increase in pressure will likely not be measureable with a normal consumer pressure gauge. I've checked.
So, if the pressure doesn't increase with the weight of the car on the tire, then the air isn't bearing any significant load beyond maintaining the structure of the tire.
posted by Jon-o at 9:01 AM on January 21, 2010
It's a technicality, but the air is holding the tire in shape, rather than supporting the weight of the vehicle. If you measure tire psi on a lift and then with the car on the ground, the increase in pressure will likely not be measureable with a normal consumer pressure gauge. I've checked.
So, if the pressure doesn't increase with the weight of the car on the tire, then the air isn't bearing any significant load beyond maintaining the structure of the tire.
posted by Jon-o at 9:01 AM on January 21, 2010
So, if the pressure doesn't increase with the weight of the car on the tire, then the air isn't bearing any significant load beyond maintaining the structure of the tire.
I believe this is incorrect. The weight of the car, spread over the large interior surface of the tires, doesn't count for much -- maybe a little over half of one PSI in the case of my Jetta. That's difficult to measure with typical tire gauges.
posted by jon1270 at 9:49 AM on January 21, 2010
I believe this is incorrect. The weight of the car, spread over the large interior surface of the tires, doesn't count for much -- maybe a little over half of one PSI in the case of my Jetta. That's difficult to measure with typical tire gauges.
posted by jon1270 at 9:49 AM on January 21, 2010
Pointing out the obviously, but there is a fairly simple way to answer this, and that is:
1. Get an accurate tire gauge.
2. Measure tire pressure.
3. Add 1000 lbs to load.
4. Measure again.
Getting a really accurate measurement is going to be the key, as I can tell you from experience that the difference will be more on the order of 1 PSI than 100. None of the tire gauges I happen to have around the house are accurate enough to detect that amount of difference reliably.
posted by flug at 10:08 AM on January 21, 2010
1. Get an accurate tire gauge.
2. Measure tire pressure.
3. Add 1000 lbs to load.
4. Measure again.
Getting a really accurate measurement is going to be the key, as I can tell you from experience that the difference will be more on the order of 1 PSI than 100. None of the tire gauges I happen to have around the house are accurate enough to detect that amount of difference reliably.
posted by flug at 10:08 AM on January 21, 2010
While it didn't come from a physicist per se, the professor of my pavement design class always told us that the force from a vehicle to the pavement was always equal to the tire pressure, regardless of the amount of weight in a vehicle. The tires will deform with additional loading, which does change the V portion of PV=nRT.
posted by hwyengr at 10:53 AM on January 21, 2010
posted by hwyengr at 10:53 AM on January 21, 2010
The tires will deform with additional loading, which does change the V portion of PV=nRT.
Changing shape doesn't necessarily mean changing volume, although if there are measurements to determine if one observes a change in volume, I'm curious to know what gets reported. The ideal gas law seems pretty well understood, regardless of what other arguments are forwarded.
posted by Blazecock Pileon at 11:13 AM on January 21, 2010
Changing shape doesn't necessarily mean changing volume, although if there are measurements to determine if one observes a change in volume, I'm curious to know what gets reported. The ideal gas law seems pretty well understood, regardless of what other arguments are forwarded.
posted by Blazecock Pileon at 11:13 AM on January 21, 2010
Like I said in the previous thread, I've measured this with a digital gauge accurate within a tenth of a psi. I found that the tire pressure was the same loaded or unloaded. If lowering the weight of a vehicle on to the tires isn't going to increase the pressure even a tenth of a psi, adding a hundred pounds to the car isn't going to make even the slightest difference.
Driving your car for ten minutes will change the tire pressure more than adding weight.
posted by Jon-o at 11:31 AM on January 21, 2010
Driving your car for ten minutes will change the tire pressure more than adding weight.
posted by Jon-o at 11:31 AM on January 21, 2010
That's a handy data point, Jon, but I don't think it follows that the tire is carrying the weight and the air isn't.
posted by jon1270 at 11:42 AM on January 21, 2010
posted by jon1270 at 11:42 AM on January 21, 2010
That's a handy data point, Jon, but I don't think it follows that the tire is carrying the weight and the air isn't.
No, but does show that the load isn't being supported by compression of the air in the tire. Unlike, say, a rigid cylinder with the load sitting on top of a piston in that cylinder, where any increase in the load turns directly into compression of the air and an increase in pressure.
I still find it helpful to try to visualize this in terms of what holds the wheel rim up off the ground. It's not hard to imagine how the vehicle weight gets through the suspension as far as the wheel rim, but what supports the rim with all that weight on it?
It's not the pressure of the air pushing on the rim, because it pushes on the rim equally in all directions. Even if putting additional load on the tire did compress it and increase the air pressure, that pressure is still applied equally in all directions to the circular rim, so it has no net effect.
What does have an effect is the sidewalls of the tire, pulling (or pushing, depending on how you look at it) UN-equally in all directions on the rim. And that pull (or push) is unequal because the shape of the tire is not symmetrical when it's under load, it bulges outward where it contacts the ground, causing the sidewalls to pull (or push) on the rim from more of an angle there. That results in more sideways forces on the rim, and less up-and-down force, over that section where the bulge is.
The air pressure is what causes the sidewalls to exert force on the rim all the way around, and the bulge at the bottom is what causes that force to be unbalanced slightly, in the right direction to hold the rim up off the ground.
(It says as much in the link I posted earlier, but thought expanding on it right here in the thread might be useful.)
posted by FishBike at 12:41 PM on January 21, 2010 [1 favorite]
No, but does show that the load isn't being supported by compression of the air in the tire. Unlike, say, a rigid cylinder with the load sitting on top of a piston in that cylinder, where any increase in the load turns directly into compression of the air and an increase in pressure.
I still find it helpful to try to visualize this in terms of what holds the wheel rim up off the ground. It's not hard to imagine how the vehicle weight gets through the suspension as far as the wheel rim, but what supports the rim with all that weight on it?
It's not the pressure of the air pushing on the rim, because it pushes on the rim equally in all directions. Even if putting additional load on the tire did compress it and increase the air pressure, that pressure is still applied equally in all directions to the circular rim, so it has no net effect.
What does have an effect is the sidewalls of the tire, pulling (or pushing, depending on how you look at it) UN-equally in all directions on the rim. And that pull (or push) is unequal because the shape of the tire is not symmetrical when it's under load, it bulges outward where it contacts the ground, causing the sidewalls to pull (or push) on the rim from more of an angle there. That results in more sideways forces on the rim, and less up-and-down force, over that section where the bulge is.
The air pressure is what causes the sidewalls to exert force on the rim all the way around, and the bulge at the bottom is what causes that force to be unbalanced slightly, in the right direction to hold the rim up off the ground.
(It says as much in the link I posted earlier, but thought expanding on it right here in the thread might be useful.)
posted by FishBike at 12:41 PM on January 21, 2010 [1 favorite]
No, but does show that the load isn't being supported by compression of the air in the tire.
I would say it shows that the volume of space enclosed by the tire hasn't changed much. See ideal gas law above. I suspect that the change in air pressure, while small, is not zero, and still feel that the internal area of the tire is related to its load-carrying ability.
That said, the rest of what you wrote makes sense. I suppose I owe Jon-O an apology; I can see how one would say, correctly, that the tire supports the car while the air supports the tire. (Sorry, Jon.)
posted by jon1270 at 2:52 PM on January 21, 2010
I would say it shows that the volume of space enclosed by the tire hasn't changed much. See ideal gas law above. I suspect that the change in air pressure, while small, is not zero, and still feel that the internal area of the tire is related to its load-carrying ability.
That said, the rest of what you wrote makes sense. I suppose I owe Jon-O an apology; I can see how one would say, correctly, that the tire supports the car while the air supports the tire. (Sorry, Jon.)
posted by jon1270 at 2:52 PM on January 21, 2010
No worries!
I had problems with this at first, also. I couldn't understand how the air doesn't carry any of the load until someone explained to me, with pictures and animations, that even though there's 30psi of air in the tires, air is so not-dense that the oxygen, water, and nitrogen molecules aren't ever touching each other enough for the weight of the car to transfer any force through the air. It's not like a liquid that readily transfers hydraulic pressure. So, low molecular density combined with the relatively small tire deformation results in a situation where the pressure doesn't increase when there's weight on the wheel.
posted by Jon-o at 3:30 PM on January 21, 2010
I had problems with this at first, also. I couldn't understand how the air doesn't carry any of the load until someone explained to me, with pictures and animations, that even though there's 30psi of air in the tires, air is so not-dense that the oxygen, water, and nitrogen molecules aren't ever touching each other enough for the weight of the car to transfer any force through the air. It's not like a liquid that readily transfers hydraulic pressure. So, low molecular density combined with the relatively small tire deformation results in a situation where the pressure doesn't increase when there's weight on the wheel.
posted by Jon-o at 3:30 PM on January 21, 2010
Best answer: FishBike: The air pressure is what causes the sidewalls to exert force on the rim all the way around, and the bulge at the bottom is what causes that force to be unbalanced slightly, in the right direction to hold the rim up off the ground.
Your explanation relates to the transmission of force from the tire to the rim. It does not explain the transmission of force from from the ground to the tire. Yes, the medium of transmission of force is the tire casing, not the air, but that doesn't address the OP's question about air pressure.
We know that in a properly inflated tire that the force from the ground to the tire is even across the entire patch since that is the definition of a properly inflated tire. The opposite and equal force of the air on the inside of the tire pressing down is evenly distributed across the tire patch and perpendicular to the patch. Those two forces, the ground pushing up on the outside of the tire and the air pushing down on the inside of the tire must be equal or otherwise the elastic rubber would not be flat on the ground. Therefore the size of the patch must be weight divided by the tire pressure.
The force from the ground is transmitted to the tire patch and then transmitted up the sidewall by the elastic forces of the rubber between the tread and sidewall.
When you add weight to the car, the tire patch simply increases in size by an amount equal to the extra weight divided by the tire pressure. This small deflection of the tire does not significantly change the volume of the tire and therefore does not change the air pressure in the tire.
posted by JackFlash at 3:44 PM on January 21, 2010
Your explanation relates to the transmission of force from the tire to the rim. It does not explain the transmission of force from from the ground to the tire. Yes, the medium of transmission of force is the tire casing, not the air, but that doesn't address the OP's question about air pressure.
We know that in a properly inflated tire that the force from the ground to the tire is even across the entire patch since that is the definition of a properly inflated tire. The opposite and equal force of the air on the inside of the tire pressing down is evenly distributed across the tire patch and perpendicular to the patch. Those two forces, the ground pushing up on the outside of the tire and the air pushing down on the inside of the tire must be equal or otherwise the elastic rubber would not be flat on the ground. Therefore the size of the patch must be weight divided by the tire pressure.
The force from the ground is transmitted to the tire patch and then transmitted up the sidewall by the elastic forces of the rubber between the tread and sidewall.
When you add weight to the car, the tire patch simply increases in size by an amount equal to the extra weight divided by the tire pressure. This small deflection of the tire does not significantly change the volume of the tire and therefore does not change the air pressure in the tire.
posted by JackFlash at 3:44 PM on January 21, 2010
air is so not-dense that the oxygen, water, and nitrogen molecules aren't ever touching each other enough for the weight of the car to transfer any force through the air.
Gah, I'm going to get myself in trouble again if I'm not careful, but I don't think this is a good way to put it. The fact that air molecules aren't 'touching' is immaterial to whether they support weight. Fishbike mentioned the idea of weight supported by compressed gas in a rigid cylinder -- those gas molecules aren't "touching" either, but the compressed gas can certainly support weight.
posted by jon1270 at 6:01 PM on January 21, 2010
Gah, I'm going to get myself in trouble again if I'm not careful, but I don't think this is a good way to put it. The fact that air molecules aren't 'touching' is immaterial to whether they support weight. Fishbike mentioned the idea of weight supported by compressed gas in a rigid cylinder -- those gas molecules aren't "touching" either, but the compressed gas can certainly support weight.
posted by jon1270 at 6:01 PM on January 21, 2010
Now is the time to deploy my timeless and favorite fallback,
"Hey man, I'm just a mechanic, not an engineer."
posted by Jon-o at 6:34 PM on January 21, 2010
"Hey man, I'm just a mechanic, not an engineer."
posted by Jon-o at 6:34 PM on January 21, 2010
Response by poster: Here's why I don't think internal area of the tire is related much (though, if I were an engineer I wouldn't have asked the original question):
If I go out and let some air out of my tire, the pressure goes down. If the internal area of the tire was the factor that mattered, well, that hasn't changed, and neither has the weight of the car, so tire pressure should be unaffected. That's not what happens.
Best answers marked are not necessarily the ones I believe most correct, just the ones that made me go "aha."
posted by ctmf at 11:43 PM on January 21, 2010
If I go out and let some air out of my tire, the pressure goes down. If the internal area of the tire was the factor that mattered, well, that hasn't changed, and neither has the weight of the car, so tire pressure should be unaffected. That's not what happens.
Best answers marked are not necessarily the ones I believe most correct, just the ones that made me go "aha."
posted by ctmf at 11:43 PM on January 21, 2010
Yeah, I'm not an engineer (or molecular physicist) either.
Thanks for a fun set of ideas to play with. This is deceptively challenging stuff to think about.
posted by jon1270 at 2:18 AM on January 22, 2010
Thanks for a fun set of ideas to play with. This is deceptively challenging stuff to think about.
posted by jon1270 at 2:18 AM on January 22, 2010
I haven't had much time to think about it but please ignore my earlier off the cuff thought that internal surface area matters. Volume matters but you couldn't measure it easily. I think the original proposition of pressure times contact surface are related is correct, at least assuming a sufficiently elastic rubber. How you take into account the changing contact area due to the increased load I don't know.
posted by caddis at 3:39 AM on January 22, 2010
posted by caddis at 3:39 AM on January 22, 2010
I don't know if anyone's still reading this, but oh well.
posted by delmoi at 3:11 PM on January 30, 2010
If the volume of the tire is not changing from load, if the gas constant doesn't change when under load, if you're not putting more air into the tire under load, and if the temperature of the gas in the tire does not change under load, then assuming your tires observe the ideal gas law, their pressure p will not change under load.Causality goes both ways in that equation; an increase in pressure will also increase temperature. And of course the volume does change. The volume gets smaller, and therefore the pressure goes up.
Changing shape doesn't necessarily mean changing volume,Not true, changing the shape definitely changes the volume, as long as the surface area doesn't change. A sphere has most volume per surface area, and the less 'spherical' a shape, the less volume given unchanging surface area.
It's not the pressure of the air pushing on the rim, because it pushes on the rim equally in all directions. Even if putting additional load on the tire did compress it and increase the air pressure, that pressure is still applied equally in all directions to the circular rim, so it has no net effect.It's precisely the resistance to deformation that holds the rim up. If the rim were to come down to the ground, the volume in the tire would be smaller, and pressure would go up even more.
I still find it helpful to try to visualize this in terms of what holds the wheel rim up off the ground. It's not hard to imagine how the vehicle weight gets through the suspension as far as the wheel rim, but what supports the rim with all that weight on it?Only if you want to confuse yourself. Think about a balloon that's only half full sitting on a shelf. If you add more air, it becomes more spherical. Why? Certainly not due to shape of the rubber. Rather it's because the air actually does push more on the deformed areas. Same thing with a tire. The rim is just as much hanging off the top of the tire as it is sitting on the bottom of it.
posted by delmoi at 3:11 PM on January 30, 2010
This thread is closed to new comments.
posted by phrontist at 12:42 AM on January 21, 2010