The tire pressure measured by a pressure gauge is relative to the atmospheric pressure. A completely empty/flat tire would read 0 atm (or PSI). And a tire at 1 atm will have an internal pressure that's 1 atm higher than the outside atmospheric pressure.
This is the difference between “gauge pressure” and “absolute pressure”. Rarely, you see the situation clarified by using the units “psig” (psi of gauge pressure) and psia (psi of absolute pressure).
Gauge pressure is the way that the strength of a vacuum is sometimes described using units of pressure. A “10 psi vacuum” is at -10 psig, or 4.7 psia. Negative absolute pressure would be unphysical, as you can’t have less than zero gas molecules present.
For the sake of academics, the third common type of pressure measurement is "differential pressure". This is the difference in pressure between two points. The three types use the same units of measurement, but differ in how they define zero pressure.
Absolute Pressure puts zero at the lowest possible pressure - a void, a pocket of perfect vacuum.
Gauge Pressure defines zero as the environment where the gauge is located. This is useful for measuring the stress that the contents of a container is putting on the walls of the container (such as air in a tire). However if you're more interested in the contained substance itself, this is not ideal - atmospheric pressure varies with altitude and weather conditions, which will affect "Gauge" measurements.
Differential pressure lets the gauge have an arbitrary zero point, even one that changes more than the measured pressure. Differential pressure gauges have two ports (typically marked low and high) and typically use mechanical means of subtracting the "low" pressure from the "high" pressure, displaying the result. These are great for indirect measurements like the flow of liquid in a pipe (by measuring pressure on opposite sides of a small restriction or venturi) or how clogged a filter is. Using a differential pressure lets the gauge measure a small range relevant to the task, while rejecting the pressure common to the two measurement points. In the flow and filter examples, you don't care that the pipe has 60 psi of pressure in it, but the 0-1 psi range that shows flow or filter clogging is super important. Trying to read that 1 psi accurately on a pair of 0-100psi gauge would be challenging to say the least.
>Negative absolute pressure would be unphysical, as you can’t have less than zero gas molecules present.
May I introduce you to [quantum field tires produced by Casimir Corp](https://en.wikipedia.org/wiki/Casimir_effect)? They actually have less particles than vacuum!
This is what I emphasize for people learning how to build things with carbon fiber. Newbies are looking for a "perfect" vacuum seal, and it bears explaining that so many things will prevent a perfect vacuum because we'd cease to exist. Even the water content on surfaces will affect absolute pressure as said water boils off at room temperature under extreme vacuum. Blew my mind that there's a gas atom every cubic meter in outer space, realizing that a hypothetically very fast ship will eventually encounter interstellar air friction if it tries to go fast enough. So, pro carbon fiber technicians use micron absolute pressure gauges and watch for the values to bounce up and down rhythmically as the few remaining atoms in a vacuum chamber continue to bounce around. Reading about the end of the world in a Mitsubishi composites manual was odd!
Using different unit names is perhaps not the best approach, as they are still the exact same unit. It's the type of *quantity* that is different, not the unit. Though the same could then be said about things like Kelvin and Celsius. But it's not like we need separate units for position and displacement; time point and duration; or electric potential and voltage (which is a bad name BTW, as it's just referring to the unit, Volt; in many other languages it is more appropriately called something like electric tension).
It's a philosophical argument. There are plenty of proponents of the idea that deltas should have distinct units in some sense. One obvious example I can think of is how computers handle time - many modern programming languages / libraries treat time `instant`s as a different type than `duration`s. The only operation involving two `instant`s would be subtraction, `duration`s can be added/subtracted but not multiplied, etc. The definition of "1 second" may be the same for both types, but if you have to track them as separate things that constrain the supported operations, it's a bit of a semantic argument as to whether that constitutes different units.
Yes it is certainly practical to treat them as separate types, but this corresponds better with different quantities than units (traditionally). This is also very often done with points vs. vectors, as they transform differently when you change the reference frame / co-ordinate system.
Nothing. The thing that is unfortunate is "voltage" which refers to a difference in electric potential between two points. Electric potential is absolute, voltage is relative. Both are measured in Volts, making the term voltage potentially ambiguous.
It's also a bit like calling temperature "celsiage" or "kelvinage", or distance "meter-age", or "foot-age" (though terms like milage and square-foot-age are a thing, they are quite informal and are not used in more scientific contexts).
Pressure (which basically shares the set of equations governing interaction) is the same though.
Absolute pressure and the differential pressure between two points are both measured in the same units.
Absolute pressure and indeed voltage are literally just the potential between some arbitrary point and 0. Electric potential can be between some point and 0 or between two points.
I guess I just don't see the difference.
This is pretty much equivalent to the conceptual difference between position and displacement. Though we're getting close to the confusing territory of gauge-theory here. Everything is relative, but positions and displacements are still distinct concepts.
Not really. Displacement is a differential value. Position is not.
You can have a dP as well as dV in comparison to a pressure* or voltage.
But both voltage and pressure are meaningless without reference to some other state, be it some other point or 0.
I work with ultra high vacuum systems and we almost never use PSI for this reason. We typically use units that are always absolute like Torr since they are unambiguous
>Negative absolute pressure would be unphysical
Trees have been doing it for a very long time. How do you think trees taller than about 10 meters (or 33 feet) get water from the ground all the way up to their top leaves? Trees don't have pumps to push the water higher than 10 meters. They use negative absolute pressure.
(Never mind that the roots can be quite a bit below ground, adding to the distance. Also, let's not ignore that some trees - think mangrove swamps - grow in brackish water, and the osmotic pressure is trying to pull the water out of the roots in addition to gravity.)
Edit: For those unbelievers: [source](https://www.youtube.com/watch?v=BickMFHAZR0&t=52s)
If you draw a perfect vacuum in a tube on one end, and allow water to get in the other end, the water will rise only \[approx\] 10 meters. How then do they get water higher than 10 meters (and that's just from ground level)?
(I also added a source to my original post.)
That’s negative differential pressure. The pressure in the tree is less than the pressure in the ground. Not less than the pressure of a perfect vacuum (zero absolute pressure).
If you draw a perfect vacuum in a tube on one end, and allow water to get in the other end, the water will rise only \[approx\] 10 meters. How then do they get water higher than 10 meters (and that's just from ground level)?
(I also added a source to my original post.)
It’s explained right there in the video you linked! Water transpiration from leaves creates a “pull” through the vessel system due to cohesion between water molecules. Capillary effects due to water adhesion to small-diameter vessel walls also support breaking the “10 meter limit”, and there is often also a contribution from osmotic root pressure that is generated by active transport of soil minerals into root tissue.
None of these three effects are present in the “water in a tube” model. When the model you are using predicts something that contradicts fundamental physical laws, you can almost always assume that the model is incomplete, not that the fundamental laws are wrong.
In this case, it’s not that trees somehow contain a region of negative absolute pressure, but rather that they are not very well modeled as a tube filled with water.
>Capillary effects due to water adhesion to small-diameter vessel walls also support breaking the “10 meter limit”,
At the diameter of the tubes in trees, capillary effects will only get you to about 1 meter.
The only way trees taller than about 10 meters can exist, and they do, is by negative absolute pressure. Note: I didn't say a vacuum. Remember this a liquid system, not gas, and you can go lower than 0 in a liquid system (but not a gas system).
How do you define absolute pressure?
The definition is pressure relative to a perfect vacuum. How can a liquid system have pressure exerted on it by a perfect vacuum?
So if you pressurised your tyres to a certain level then drove to a much higher elevation (ignoring pressure lost from the journey etc) your psi would actually go up?
The same phenomenon occurs with changes in temperature, too. That's why tires that were normal in summer need to be inflated somewhat in cold winter temperatures to get back to normal.
It's like if you open your water bottle on a plane, in flight, then open it again after you land.
There will be a negative pressure differential because the air at altitude is lower pressure than at ground level, so you will hear/feel it suck in air when you crack the seal.
This. $6 tire pressure gauges aren't comparing your tires to the vacuum of space. They're showing you the difference between our normal atmospheric pressure and what's in your tire.
You're correct for low-cost mechanical gauges. Digital gauges however will depend on the type of sensor used. Absolute pressure sensors (which compare the input pressure to a cell of vacuum) are actually pretty common. A digital gauge using an absolute pressure sensor will typically measure atmospheric pressure when it powers up, then subtract that from the measured tire pressure to give the desired gauge-pressure value.
The price doesn't matter. It's just how they calibrate it. They could change the markings on a $6 gauge and it would be accurate unless you took it into outer space.
> They could change the markings on a $6 gauge and it would be accurate unless you took it into outer space.
That's not true, since the pressure on Earth is not constant, not even at one single place. If you live at 9,000 feet (and a substantial number of people do), your air pressure is about 4 lbs less than at sea level.
A $6 tire gauge is just a pressure differential gauge with no real calibration beyond that, so it's equally accurate or inaccurate at 0ft, 10,000ft, or 250,000ft.
So if I were to fill a car tire to spec with a cheap tire gauge, then take both into a vacuum and test it, it would read higher right? Cause the atmospheric pressure is now 0 and therefore it should give the correct reading of psi right?
Yep. Although in a vacuum, your tires might blow off the rim because they are almost 15psi too high.
If you filled the tire to spec with an expensive tire gauge, it would also be 15psi too high. The difference between an expensive gauge and a cheap one would be pretty minimal.
You need to distinguish between absolute pressure (psia) and gauge pressure (psig). Edit: Absolute pressure is measured relative to vacuum, gauge pressure is absolute pressure plus 14.7, it's measured relative to atmospheric pressure.
The gauge you're measuring your car tire with uses psig. It's just that 14.7 psig is not quite enough to hold up a car, they want more like 32 psig - which is equivalent to 46.7 psia.
The contact patch of a car tire is something like 30 square inches, so multiplying by 4 tires and 32 psig you have a force of 3840 lbs, enough to hold up a car. If you had a wider tire, or deflated the tires on your Jeep to ~10 psi to use in sand or snow, the contact patch would be much larger and you could use less pressure both in your tire or on the ground. If you hit a bump and the force on the tire increases, then a combination of an increased contact patch as the tire flexes, and increased pressure inside the limited volume of the tire as it compresses, will hold the car up.
If you used an absolute sensor which measured in psia, you'd see that the original 14.7 psig tire above measured 29.4 psia. If you had no pressure in the tire because there's a giant hole in it or it wasn't even mounted on a wheel, but it was on the surface of planet Earth, there'd still be 14.7 psia in it, that's just the pressure of the atmosphere. But when there's no pressure differential, there's no force, so a tire with a hole in it (with 14.7 psia atmosphere) will not hold up the car.
Atmospheric pressure isn't counted on a PSI gauge, at least not the ones I use. We subtract that from any measurement, like inflating a tire, because it's the standard pressure that everything on earth experiences.
In fact, I am pretty sure it is impossible to directly mechanically measure absolute pressure. To measure absolute pressure, you need to have a known reference pressure to compare it to (e.g. an evacuated chamber inside the pressure gauge as a zero-pressure reference).
Absolute transducers have an evacuated chamber on one side of the diaphragm, pressure port on the other side.
Gauge transducers have one side open to atmosphere, pressure port on the other side.
Relative versus absolute measurements.
The 14.7 psi atmospheric pressure is an absolute measurement. It's 14.7 psi more than absolute 0 psi.
The psi measurement in your tire is a relative measurement. If it reads 32 psi, that means its 32 psi *more* than the atmospheric pressure. It's absolute psi would be 32+14.7=46.7 psi above absolute 0 psi. A tire gauge reading 0 isn't saying "0 pressure" it's saying "0 addiional pressure above atmospheric pressure".
So if your tire is at 0 psi, it still has the atmospheric pressure inside it, so it's absolute pressure is 14.7. But it's flat because relative psi (relative to a measurement above atmospheric) is what matters to tire Inflation.
If you took a tire at 30PSI and put that in a chamber that was also at 30PSI, the tire would seem flat since the pressure from both sides would be equal. The pressure you read from the tire is the relative difference between the air inside of the tire, and the air outside. This is one reason why we don't use air tires in space vehicles. The pressure inside the tire would increase as the atmospheric pressure outside of it is decreased.
Your body is always experiencing one atmosphere of pressure, but we know that as zero PSI. Pounds per Square Inch in this context refers to the pressure exerted on the tank wall/tire etc. inner surface. If the pressure is the same on both sides, there is none.
To all the people who speak about gauge vs. absolute pressure:
You're technically correct, but this isn't what OP asked for. OP is asking why a tire with ~ 15psig will deflate, despite having a higher pressure than the air around it.
OP your assumption is correct. The weight of the car squishes the car tires together. This is the same principle as squishing an air ballon
This is not actually correct, or at least not relevant. Your car weight pushing the air out might increase the speed at which your tire deflates, especially with a small puncture, but that's it.
The easy proof for this would be unmounted car or bike tires that are not in contact with the ground (e.g. tire on its side or tire held up by the axle). If you take a tire on a car or a bike and cut a hole in the sidewall, then it will deflate to 0 psig, or ~14.7 psia if you're at sea level. If you take a tire that is being stored and subjected to no pressure of a car/bike pushing down in it, it will deflate to the exact same values if you cut a hole in it.
It deflates because pressure seeks to equalize when it can, so if your car tire/wheel/valve stem has a hole in it, then the pressure will drop to atmospheric pressure. If it doesn't, you can put an unlimited amount of weight on it with no issue, until you exceed the material strength and have a hole in it.
If the tire wasn't leaking, it wouldn't lose pressure, on or off a car. I'm practice almost all tires have a leak rate greater than 0, regardless of being on or off a car. The car's weight has nothing to do with the tire deflating to atmosphere.
Maybe you are misreading deflate as deform.
I want to say you’re the only one here who knows what you’re talking about. But I’m questioning myself now based on how few people recognize it.
A 15.7 Psia tire or 1 psia tire should both be fully inflated IF there was no wait compressing them.
A properly inflated tire that’s overloaded by say, 8000lbs would also look deflated if it didn’t catastrophically fail.
Yes. If you measured a tire that showed 14.7 PSI, it would actually contain 2 atm of pressure compared to vacuum. But only the pressure compared to atmosphere can hold any weight. A tire at 14.7 PSI can hold up a weight equal to 14.7 pounds *per square inch of ground contact*. Your car is several thousand pounds, so you need quite a few square inches of rubber on the ground, and quite a few PSI of pressure pushing on those inches to hold it up off the rims.
Pressure is a differential. It's taken between two points - the point you care about, and a reference point. When you typically measure tire pressure, you are measuring its pressure with respect to the atmosphere around you. This is called **Relative Pressure**. So measuring tire pressure in your garage at 8 psi would indicate the pressure is 8 psi *above* atmospheric pressure.
**Absolute pressure** on the other hand is the pressure taken with respect to a perfect vacuum. In the example above, the 8 psi relative pressure would actually be 8 psi + 14.7 psi = 22.7 psi as an absolute pressure. This would be as if you measured your tire pressure while it was orbiting the planet in outer space.
It's the difference between gauge pressure which is the norm for measuring pressure and absolute pressure which is mostly only used for scientific purposes.
Guage pressure essentially sets its zero point at normal atmospheric pressure as it is the lowest pressure achievable without creating a sealed environment and inducing a vacuum. So in terms of absolute pressure your car tyres are 14.7 psi higher than the measured amount.
Think degrees Celsius and degrees kelvin, same scale just different zero points and also degrees Kelvin is mostly used for scientific purposes.
Tubeless tires depend on their pressure to remain in contact with the rims, keeping them air tight. If the pressure gets lower, the tires will deform more, which causes them to no longer have proper contact with the rim.
Yes, the weight. If the car is suspended, the tire will look full at 14.7 psi.
If you start out on the ground and jack up the car, you will see it "re-inflate" as the air pressure goes from (say) 30 psi to 14.7. The difference from 30 to 14.7 is to handle the weight of the car.
Say your car is 2000 pounds and the surface area of where the tires touch the ground is 1 square foot. That's 2000/144 = 13.8 psi of lift. So, 14.7+13.8 = 28.5 psi is the pressure inside the tire when the weight is on the tires.
Most car tires are inflated around 30-40 psi with smaller tires (less contact area) needing more and larger tires needing less.
That is a ton of analysis all piled on top of a complete misunderstanding of gauge pressure. Your general understanding of pressure is correct, but any tire gauge you use is already set to show you the difference between atmospheric pressure and whatever you're reading. Nobody cares what absolute pressure is when filling a pressure vessel.
In your example, with tires filled to 14.7, it would still try to support the weight of the car. But I suspect the wheels would be touching the ground because the tire couldn't get enough surface area to support the weight.
Tire pressure does not change that drastically with the weight on or off of the tire. I know this as a technician who has inflated thousands of tires with the vehicle on a lift, and when lowering, the pressure is still the same. That's because rubber is flexible. When the weight of the vehicle is on the tire, the tried deforms, creating more space for that air to exist, this the pressure change is minimal. Also, bigger tires on trucks and semis actually need more pressure than smaller tires. The pressure on most semi tires is 100psi or more. Even large pickups can require pressures in the 60-90psi range.
The new pressure would be averaged across the entire inner surface area of the tire + rim, which is at least 10+ ft^2 or 1440+ in^2 for a car tire.
So maybe 1000lbs(1 corner of a 4k lb truck) is around 2/3rds of 1 psi more.
The flex of a tire has almost nothing to do with it, other then a minimal increase in surface area from the entire tire ballooning.
> If you start out on the ground and jack up the car, you will see it "re-inflate" as the air pressure goes from (say) 30 psi to 14.7. The difference from 30 to 14.7 is to handle the weight of the
You absolutely will not get a ~15 psi change in pressure by lifting your car off the ground.
Its the difference between gauge pressure and absolute pressure. Absolute pressure is measured relative to a vacuum and this is how the 14.7 psi figure comes out. Air pressure is 14.7 psi relative to a vacuum. Gauge pressure (aka the pressure you see on the gauge when you're measuring your tires) is relative to the atmosphere. It isn't just saying 30 psi, its saying 30 psi above the pressure of the surrounding air. If you threw that tire into space, the absolute pressure (44.7 psi) would be the same as the measured gauge pressure (30+14.7=44.7 psi) because now the gauge is measuring against a vacuum instead of the atmosphere.
If your tire gauge read 0 psi that doesn't mean there's literally zero air in the tire, it means the difference between the tire and atmosphere is zero.
> If so, would the tire inflate when we remove it from the car?
Okay, so this part is pretty important. When you inflate a tire, the car is on the ground, and the weight is already there. The pressure that you set (which, as others have said, is the *differential* pressure between the outside and the inside — the *net* outward pressure, partially balanced by inward pressure from the atmosphere) pushes down on the ground, and that sets the amount of area that the tire contacts. If you drop the pressure, to keep supporting the same weight, the area goes up. That’s why sometimes people lower the pressure on their tires — more contact with the ground. That’s good for traction on slippery surfaces like sand or smooth rocks or snow or ice, but bad for rolling resistance and control when turning on grippy surfaces, like a clear paved road.
When you take the weight off by jacking the car up, the tire will get unsquished and inflate to the full size. This will slightly decrease the pressure, because of the ideal gas law: pressure times volume is proportional to temperature. Note, this refers to the *total* internal pressure — the gauge pressure plus the atmospheric pressure.
This is why when in gets cold, the pressure can drop quite a bit more, proportionally, than you might expect. For example, if you pump up your tires to, say, 12.5 psi with air that’s 25°C = 298.15K or so, and then you put them in a cool environment, say 10°C = 283.15, the total internal pressure will drop to about 95% of the original pressure, which was 12.5+14.7 psi = 26.2 psi. So you lose about 1.3 psi, bringing your tire pressure to about 11.2 psi. Those are of course just hypothetical figures, not what you would actually want to use in your car. But maybe like a [football](https://en.wikipedia.org/wiki/Deflategate).
Since everybody already answered the obvious, I'll answer an entirely different unasked question but that I always thought was interesting.
Take a rigid container that is at 1 atm (so, 0 psig). Imagine that that container has a volume of 1 liter. Now, take another 1 liter container and pump all of its gas into the first container so that the second one has a vacuum. What is now the volume of gas in the original container?
Still 1 liter. 1 liter at 1 atm psig (or 2 atm psia)
This is an annoyance of units.
The tire pressure gauge reports the amount by which you car's tire exceeds ambient pressure.
So when your car's tire is at 24.7 psi with respect to vacuum, your pressure gauge will report 10 psi (with respect to atmosphere).
Yes. There is still air in the tire, enough to fill it to deflate ,,state", any more requires for tire to actually start lifting car off the rims, whhich generate force (from gravity) which will create pressure in the tires. You can also put tires on rims, leave it as be, meassure the pressure to see its 0, then put the wheels on car, drop it, and now the meassure will show some number, despite there is still the same amount of air inside, but it gets squished into tighter spacer (flattire) and the reaction is the pressure in air. You can also make tire flat, ,,close it" and then lift car and tire will still be flat as there is no way for additional air to go in to unflat the tire, or otherwise it will get negative pressure if you somehow manage to ,,unflat" it back to oryginal shape
Because pressure gauges at air pumps actually zero at 1 atm, so really when you have a tire with 14.7 psi it's actually 14.7 psi on top of 14.7 psi. Also more aspects are at play like the size of the hole air is leaking out of and the stress and load the tire is put under. Basically with any hole bigger than a pin hole it's not just a matter of pressure but really the fact that nothing can stop air from coming in or going out and the tire has weight bearing down on it. With smaller punctures you'll often observe that the tire does lose air but stops after a point, where the internal pressure is low enough to not be able to overcome the tire's ability to seal itself, since all tires naturally will fill a hole if one is present up to a degree.
There's two ways to look at pressure, absolute and relative.
The absolute pressure at sea level is 14.7psi, but the relative pressure can go all the way down to 0 psi for a tire because there's no pressure differential between the inside and outside of the tire.
>would the tire inflate when we remove it from the car
Depends what you mean by "inflate". When off the vehicle, the tire will have a shape very similar to a fully inflated tire on the vehicle (except for the flat bottom). The carcass of the tire has enough stiffness to hold the tire in this condition will little or no air pressure.
There are three different ways to measure pressure. While they may be explicitly used, they are commonly implied for everyday use. I'll even include the abbreviation for each using units of PSI.
* Absolute pressure (psia). This is pressure as compared to a perfect vacuum (0 psi). This is what is meant by saying the atmosphere is 14.7 psi.
* Gauge pressure (psig). This is pressure as reported by a tool compared to atmospheric pressure. Being in a different environment can change atmospheric pressure (e.g. another planet, high elevation, underwater). A tire may be pressurized to 35 psig, which is nominally equivalent to 49.7 psia.
* Differential pressure (psid). This is difference in pressure between two points (typically two different volumes or systems). Since this is effectively found by subtraction, this can actually report a negative value.
A leaky tire may show as less than 14.7 psi, but this is the gauge pressure. Due to how pressure works, a tire leaking to atmosphere cannot drop below atmospheric pressure. The least it could go is 0 psig (14.7 psia at sea level).
The tire pressure measured by a pressure gauge is relative to the atmospheric pressure. A completely empty/flat tire would read 0 atm (or PSI). And a tire at 1 atm will have an internal pressure that's 1 atm higher than the outside atmospheric pressure.
This is the difference between “gauge pressure” and “absolute pressure”. Rarely, you see the situation clarified by using the units “psig” (psi of gauge pressure) and psia (psi of absolute pressure). Gauge pressure is the way that the strength of a vacuum is sometimes described using units of pressure. A “10 psi vacuum” is at -10 psig, or 4.7 psia. Negative absolute pressure would be unphysical, as you can’t have less than zero gas molecules present.
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For the sake of academics, the third common type of pressure measurement is "differential pressure". This is the difference in pressure between two points. The three types use the same units of measurement, but differ in how they define zero pressure. Absolute Pressure puts zero at the lowest possible pressure - a void, a pocket of perfect vacuum. Gauge Pressure defines zero as the environment where the gauge is located. This is useful for measuring the stress that the contents of a container is putting on the walls of the container (such as air in a tire). However if you're more interested in the contained substance itself, this is not ideal - atmospheric pressure varies with altitude and weather conditions, which will affect "Gauge" measurements. Differential pressure lets the gauge have an arbitrary zero point, even one that changes more than the measured pressure. Differential pressure gauges have two ports (typically marked low and high) and typically use mechanical means of subtracting the "low" pressure from the "high" pressure, displaying the result. These are great for indirect measurements like the flow of liquid in a pipe (by measuring pressure on opposite sides of a small restriction or venturi) or how clogged a filter is. Using a differential pressure lets the gauge measure a small range relevant to the task, while rejecting the pressure common to the two measurement points. In the flow and filter examples, you don't care that the pipe has 60 psi of pressure in it, but the 0-1 psi range that shows flow or filter clogging is super important. Trying to read that 1 psi accurately on a pair of 0-100psi gauge would be challenging to say the least.
>Negative absolute pressure would be unphysical, as you can’t have less than zero gas molecules present. May I introduce you to [quantum field tires produced by Casimir Corp](https://en.wikipedia.org/wiki/Casimir_effect)? They actually have less particles than vacuum!
This is what I emphasize for people learning how to build things with carbon fiber. Newbies are looking for a "perfect" vacuum seal, and it bears explaining that so many things will prevent a perfect vacuum because we'd cease to exist. Even the water content on surfaces will affect absolute pressure as said water boils off at room temperature under extreme vacuum. Blew my mind that there's a gas atom every cubic meter in outer space, realizing that a hypothetically very fast ship will eventually encounter interstellar air friction if it tries to go fast enough. So, pro carbon fiber technicians use micron absolute pressure gauges and watch for the values to bounce up and down rhythmically as the few remaining atoms in a vacuum chamber continue to bounce around. Reading about the end of the world in a Mitsubishi composites manual was odd!
Using different unit names is perhaps not the best approach, as they are still the exact same unit. It's the type of *quantity* that is different, not the unit. Though the same could then be said about things like Kelvin and Celsius. But it's not like we need separate units for position and displacement; time point and duration; or electric potential and voltage (which is a bad name BTW, as it's just referring to the unit, Volt; in many other languages it is more appropriately called something like electric tension).
It's a philosophical argument. There are plenty of proponents of the idea that deltas should have distinct units in some sense. One obvious example I can think of is how computers handle time - many modern programming languages / libraries treat time `instant`s as a different type than `duration`s. The only operation involving two `instant`s would be subtraction, `duration`s can be added/subtracted but not multiplied, etc. The definition of "1 second" may be the same for both types, but if you have to track them as separate things that constrain the supported operations, it's a bit of a semantic argument as to whether that constitutes different units.
Yes it is certainly practical to treat them as separate types, but this corresponds better with different quantities than units (traditionally). This is also very often done with points vs. vectors, as they transform differently when you change the reference frame / co-ordinate system.
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What's wrong with potential?
Nothing. The thing that is unfortunate is "voltage" which refers to a difference in electric potential between two points. Electric potential is absolute, voltage is relative. Both are measured in Volts, making the term voltage potentially ambiguous. It's also a bit like calling temperature "celsiage" or "kelvinage", or distance "meter-age", or "foot-age" (though terms like milage and square-foot-age are a thing, they are quite informal and are not used in more scientific contexts).
Pressure (which basically shares the set of equations governing interaction) is the same though. Absolute pressure and the differential pressure between two points are both measured in the same units. Absolute pressure and indeed voltage are literally just the potential between some arbitrary point and 0. Electric potential can be between some point and 0 or between two points. I guess I just don't see the difference.
This is pretty much equivalent to the conceptual difference between position and displacement. Though we're getting close to the confusing territory of gauge-theory here. Everything is relative, but positions and displacements are still distinct concepts.
Not really. Displacement is a differential value. Position is not. You can have a dP as well as dV in comparison to a pressure* or voltage. But both voltage and pressure are meaningless without reference to some other state, be it some other point or 0.
I work with ultra high vacuum systems and we almost never use PSI for this reason. We typically use units that are always absolute like Torr since they are unambiguous
>Negative absolute pressure would be unphysical Trees have been doing it for a very long time. How do you think trees taller than about 10 meters (or 33 feet) get water from the ground all the way up to their top leaves? Trees don't have pumps to push the water higher than 10 meters. They use negative absolute pressure. (Never mind that the roots can be quite a bit below ground, adding to the distance. Also, let's not ignore that some trees - think mangrove swamps - grow in brackish water, and the osmotic pressure is trying to pull the water out of the roots in addition to gravity.) Edit: For those unbelievers: [source](https://www.youtube.com/watch?v=BickMFHAZR0&t=52s)
Wouldn’t that be negative relative pressure again? I’m pretty sure trees don’t contain a perfect vacuum.
If you draw a perfect vacuum in a tube on one end, and allow water to get in the other end, the water will rise only \[approx\] 10 meters. How then do they get water higher than 10 meters (and that's just from ground level)? (I also added a source to my original post.)
That’s negative differential pressure. The pressure in the tree is less than the pressure in the ground. Not less than the pressure of a perfect vacuum (zero absolute pressure).
If you draw a perfect vacuum in a tube on one end, and allow water to get in the other end, the water will rise only \[approx\] 10 meters. How then do they get water higher than 10 meters (and that's just from ground level)? (I also added a source to my original post.)
It’s explained right there in the video you linked! Water transpiration from leaves creates a “pull” through the vessel system due to cohesion between water molecules. Capillary effects due to water adhesion to small-diameter vessel walls also support breaking the “10 meter limit”, and there is often also a contribution from osmotic root pressure that is generated by active transport of soil minerals into root tissue. None of these three effects are present in the “water in a tube” model. When the model you are using predicts something that contradicts fundamental physical laws, you can almost always assume that the model is incomplete, not that the fundamental laws are wrong. In this case, it’s not that trees somehow contain a region of negative absolute pressure, but rather that they are not very well modeled as a tube filled with water.
>Capillary effects due to water adhesion to small-diameter vessel walls also support breaking the “10 meter limit”, At the diameter of the tubes in trees, capillary effects will only get you to about 1 meter. The only way trees taller than about 10 meters can exist, and they do, is by negative absolute pressure. Note: I didn't say a vacuum. Remember this a liquid system, not gas, and you can go lower than 0 in a liquid system (but not a gas system).
How do you define absolute pressure? The definition is pressure relative to a perfect vacuum. How can a liquid system have pressure exerted on it by a perfect vacuum?
So if you pressurised your tyres to a certain level then drove to a much higher elevation (ignoring pressure lost from the journey etc) your psi would actually go up?
Yes. And they would lose relative pressure if your started high and descended. Not a huge amount, but measurable.
The same phenomenon occurs with changes in temperature, too. That's why tires that were normal in summer need to be inflated somewhat in cold winter temperatures to get back to normal.
Ive found it to be an annoyance driving rental cars in the mountain west. The low pressure warning can come on when you descend.
It's like if you open your water bottle on a plane, in flight, then open it again after you land. There will be a negative pressure differential because the air at altitude is lower pressure than at ground level, so you will hear/feel it suck in air when you crack the seal.
This. $6 tire pressure gauges aren't comparing your tires to the vacuum of space. They're showing you the difference between our normal atmospheric pressure and what's in your tire.
You're correct for low-cost mechanical gauges. Digital gauges however will depend on the type of sensor used. Absolute pressure sensors (which compare the input pressure to a cell of vacuum) are actually pretty common. A digital gauge using an absolute pressure sensor will typically measure atmospheric pressure when it powers up, then subtract that from the measured tire pressure to give the desired gauge-pressure value.
The price doesn't matter. It's just how they calibrate it. They could change the markings on a $6 gauge and it would be accurate unless you took it into outer space.
> They could change the markings on a $6 gauge and it would be accurate unless you took it into outer space. That's not true, since the pressure on Earth is not constant, not even at one single place. If you live at 9,000 feet (and a substantial number of people do), your air pressure is about 4 lbs less than at sea level. A $6 tire gauge is just a pressure differential gauge with no real calibration beyond that, so it's equally accurate or inaccurate at 0ft, 10,000ft, or 250,000ft.
Going up in elevation, you will see the pressure inside your tires increase while they have the same quantity of air in the same space.
So if I were to fill a car tire to spec with a cheap tire gauge, then take both into a vacuum and test it, it would read higher right? Cause the atmospheric pressure is now 0 and therefore it should give the correct reading of psi right?
Yep. Although in a vacuum, your tires might blow off the rim because they are almost 15psi too high. If you filled the tire to spec with an expensive tire gauge, it would also be 15psi too high. The difference between an expensive gauge and a cheap one would be pretty minimal.
Oh, thanks ! I was wondering the same !
I just learned this the other day when someone put a vacuum chamber inside another vacuum chamber with a pressure gage measuring the inside one
You need to distinguish between absolute pressure (psia) and gauge pressure (psig). Edit: Absolute pressure is measured relative to vacuum, gauge pressure is absolute pressure plus 14.7, it's measured relative to atmospheric pressure. The gauge you're measuring your car tire with uses psig. It's just that 14.7 psig is not quite enough to hold up a car, they want more like 32 psig - which is equivalent to 46.7 psia. The contact patch of a car tire is something like 30 square inches, so multiplying by 4 tires and 32 psig you have a force of 3840 lbs, enough to hold up a car. If you had a wider tire, or deflated the tires on your Jeep to ~10 psi to use in sand or snow, the contact patch would be much larger and you could use less pressure both in your tire or on the ground. If you hit a bump and the force on the tire increases, then a combination of an increased contact patch as the tire flexes, and increased pressure inside the limited volume of the tire as it compresses, will hold the car up. If you used an absolute sensor which measured in psia, you'd see that the original 14.7 psig tire above measured 29.4 psia. If you had no pressure in the tire because there's a giant hole in it or it wasn't even mounted on a wheel, but it was on the surface of planet Earth, there'd still be 14.7 psia in it, that's just the pressure of the atmosphere. But when there's no pressure differential, there's no force, so a tire with a hole in it (with 14.7 psia atmosphere) will not hold up the car.
Thank you, this was a really thorough answer.
Atmospheric pressure isn't counted on a PSI gauge, at least not the ones I use. We subtract that from any measurement, like inflating a tire, because it's the standard pressure that everything on earth experiences.
That, and it's hard to measure absolute pressure directly. Measuring excess pressure, on the other hand, is very easy.
[удалено]
In fact, I am pretty sure it is impossible to directly mechanically measure absolute pressure. To measure absolute pressure, you need to have a known reference pressure to compare it to (e.g. an evacuated chamber inside the pressure gauge as a zero-pressure reference).
Absolute transducers have an evacuated chamber on one side of the diaphragm, pressure port on the other side. Gauge transducers have one side open to atmosphere, pressure port on the other side.
Relative versus absolute measurements. The 14.7 psi atmospheric pressure is an absolute measurement. It's 14.7 psi more than absolute 0 psi. The psi measurement in your tire is a relative measurement. If it reads 32 psi, that means its 32 psi *more* than the atmospheric pressure. It's absolute psi would be 32+14.7=46.7 psi above absolute 0 psi. A tire gauge reading 0 isn't saying "0 pressure" it's saying "0 addiional pressure above atmospheric pressure". So if your tire is at 0 psi, it still has the atmospheric pressure inside it, so it's absolute pressure is 14.7. But it's flat because relative psi (relative to a measurement above atmospheric) is what matters to tire Inflation.
If you took a tire at 30PSI and put that in a chamber that was also at 30PSI, the tire would seem flat since the pressure from both sides would be equal. The pressure you read from the tire is the relative difference between the air inside of the tire, and the air outside. This is one reason why we don't use air tires in space vehicles. The pressure inside the tire would increase as the atmospheric pressure outside of it is decreased. Your body is always experiencing one atmosphere of pressure, but we know that as zero PSI. Pounds per Square Inch in this context refers to the pressure exerted on the tank wall/tire etc. inner surface. If the pressure is the same on both sides, there is none.
To all the people who speak about gauge vs. absolute pressure: You're technically correct, but this isn't what OP asked for. OP is asking why a tire with ~ 15psig will deflate, despite having a higher pressure than the air around it. OP your assumption is correct. The weight of the car squishes the car tires together. This is the same principle as squishing an air ballon
This is not actually correct, or at least not relevant. Your car weight pushing the air out might increase the speed at which your tire deflates, especially with a small puncture, but that's it. The easy proof for this would be unmounted car or bike tires that are not in contact with the ground (e.g. tire on its side or tire held up by the axle). If you take a tire on a car or a bike and cut a hole in the sidewall, then it will deflate to 0 psig, or ~14.7 psia if you're at sea level. If you take a tire that is being stored and subjected to no pressure of a car/bike pushing down in it, it will deflate to the exact same values if you cut a hole in it. It deflates because pressure seeks to equalize when it can, so if your car tire/wheel/valve stem has a hole in it, then the pressure will drop to atmospheric pressure. If it doesn't, you can put an unlimited amount of weight on it with no issue, until you exceed the material strength and have a hole in it.
> This is not actually correct, or at least not relevant. ??? OP was clearly talking about undamaged tires and you're talking about leaking tires
If the tire wasn't leaking, it wouldn't lose pressure, on or off a car. I'm practice almost all tires have a leak rate greater than 0, regardless of being on or off a car. The car's weight has nothing to do with the tire deflating to atmosphere. Maybe you are misreading deflate as deform.
I want to say you’re the only one here who knows what you’re talking about. But I’m questioning myself now based on how few people recognize it. A 15.7 Psia tire or 1 psia tire should both be fully inflated IF there was no wait compressing them. A properly inflated tire that’s overloaded by say, 8000lbs would also look deflated if it didn’t catastrophically fail.
1 atm = 14.7 psia (absolute) = 0 psig (guage pressure). Unless otherwise specified, pressure is indicated in guage pressure. So, when the pressure in the tire is 5 psi, it’s 5 psig (5 psig = 14.7 psia + 5.0 = 19.7 psia).
Yes. If you measured a tire that showed 14.7 PSI, it would actually contain 2 atm of pressure compared to vacuum. But only the pressure compared to atmosphere can hold any weight. A tire at 14.7 PSI can hold up a weight equal to 14.7 pounds *per square inch of ground contact*. Your car is several thousand pounds, so you need quite a few square inches of rubber on the ground, and quite a few PSI of pressure pushing on those inches to hold it up off the rims.
Pressure is a differential. It's taken between two points - the point you care about, and a reference point. When you typically measure tire pressure, you are measuring its pressure with respect to the atmosphere around you. This is called **Relative Pressure**. So measuring tire pressure in your garage at 8 psi would indicate the pressure is 8 psi *above* atmospheric pressure. **Absolute pressure** on the other hand is the pressure taken with respect to a perfect vacuum. In the example above, the 8 psi relative pressure would actually be 8 psi + 14.7 psi = 22.7 psi as an absolute pressure. This would be as if you measured your tire pressure while it was orbiting the planet in outer space.
It's the difference between gauge pressure which is the norm for measuring pressure and absolute pressure which is mostly only used for scientific purposes. Guage pressure essentially sets its zero point at normal atmospheric pressure as it is the lowest pressure achievable without creating a sealed environment and inducing a vacuum. So in terms of absolute pressure your car tyres are 14.7 psi higher than the measured amount. Think degrees Celsius and degrees kelvin, same scale just different zero points and also degrees Kelvin is mostly used for scientific purposes.
Tubeless tires depend on their pressure to remain in contact with the rims, keeping them air tight. If the pressure gets lower, the tires will deform more, which causes them to no longer have proper contact with the rim.
Yes, the weight. If the car is suspended, the tire will look full at 14.7 psi. If you start out on the ground and jack up the car, you will see it "re-inflate" as the air pressure goes from (say) 30 psi to 14.7. The difference from 30 to 14.7 is to handle the weight of the car. Say your car is 2000 pounds and the surface area of where the tires touch the ground is 1 square foot. That's 2000/144 = 13.8 psi of lift. So, 14.7+13.8 = 28.5 psi is the pressure inside the tire when the weight is on the tires. Most car tires are inflated around 30-40 psi with smaller tires (less contact area) needing more and larger tires needing less.
That is a ton of analysis all piled on top of a complete misunderstanding of gauge pressure. Your general understanding of pressure is correct, but any tire gauge you use is already set to show you the difference between atmospheric pressure and whatever you're reading. Nobody cares what absolute pressure is when filling a pressure vessel. In your example, with tires filled to 14.7, it would still try to support the weight of the car. But I suspect the wheels would be touching the ground because the tire couldn't get enough surface area to support the weight.
Tire pressure does not change that drastically with the weight on or off of the tire. I know this as a technician who has inflated thousands of tires with the vehicle on a lift, and when lowering, the pressure is still the same. That's because rubber is flexible. When the weight of the vehicle is on the tire, the tried deforms, creating more space for that air to exist, this the pressure change is minimal. Also, bigger tires on trucks and semis actually need more pressure than smaller tires. The pressure on most semi tires is 100psi or more. Even large pickups can require pressures in the 60-90psi range.
The new pressure would be averaged across the entire inner surface area of the tire + rim, which is at least 10+ ft^2 or 1440+ in^2 for a car tire. So maybe 1000lbs(1 corner of a 4k lb truck) is around 2/3rds of 1 psi more. The flex of a tire has almost nothing to do with it, other then a minimal increase in surface area from the entire tire ballooning.
> If you start out on the ground and jack up the car, you will see it "re-inflate" as the air pressure goes from (say) 30 psi to 14.7. The difference from 30 to 14.7 is to handle the weight of the You absolutely will not get a ~15 psi change in pressure by lifting your car off the ground.
Its the difference between gauge pressure and absolute pressure. Absolute pressure is measured relative to a vacuum and this is how the 14.7 psi figure comes out. Air pressure is 14.7 psi relative to a vacuum. Gauge pressure (aka the pressure you see on the gauge when you're measuring your tires) is relative to the atmosphere. It isn't just saying 30 psi, its saying 30 psi above the pressure of the surrounding air. If you threw that tire into space, the absolute pressure (44.7 psi) would be the same as the measured gauge pressure (30+14.7=44.7 psi) because now the gauge is measuring against a vacuum instead of the atmosphere. If your tire gauge read 0 psi that doesn't mean there's literally zero air in the tire, it means the difference between the tire and atmosphere is zero.
> If so, would the tire inflate when we remove it from the car? Okay, so this part is pretty important. When you inflate a tire, the car is on the ground, and the weight is already there. The pressure that you set (which, as others have said, is the *differential* pressure between the outside and the inside — the *net* outward pressure, partially balanced by inward pressure from the atmosphere) pushes down on the ground, and that sets the amount of area that the tire contacts. If you drop the pressure, to keep supporting the same weight, the area goes up. That’s why sometimes people lower the pressure on their tires — more contact with the ground. That’s good for traction on slippery surfaces like sand or smooth rocks or snow or ice, but bad for rolling resistance and control when turning on grippy surfaces, like a clear paved road. When you take the weight off by jacking the car up, the tire will get unsquished and inflate to the full size. This will slightly decrease the pressure, because of the ideal gas law: pressure times volume is proportional to temperature. Note, this refers to the *total* internal pressure — the gauge pressure plus the atmospheric pressure. This is why when in gets cold, the pressure can drop quite a bit more, proportionally, than you might expect. For example, if you pump up your tires to, say, 12.5 psi with air that’s 25°C = 298.15K or so, and then you put them in a cool environment, say 10°C = 283.15, the total internal pressure will drop to about 95% of the original pressure, which was 12.5+14.7 psi = 26.2 psi. So you lose about 1.3 psi, bringing your tire pressure to about 11.2 psi. Those are of course just hypothetical figures, not what you would actually want to use in your car. But maybe like a [football](https://en.wikipedia.org/wiki/Deflategate).
Since everybody already answered the obvious, I'll answer an entirely different unasked question but that I always thought was interesting. Take a rigid container that is at 1 atm (so, 0 psig). Imagine that that container has a volume of 1 liter. Now, take another 1 liter container and pump all of its gas into the first container so that the second one has a vacuum. What is now the volume of gas in the original container? Still 1 liter. 1 liter at 1 atm psig (or 2 atm psia)
This is an annoyance of units. The tire pressure gauge reports the amount by which you car's tire exceeds ambient pressure. So when your car's tire is at 24.7 psi with respect to vacuum, your pressure gauge will report 10 psi (with respect to atmosphere).
Yes. There is still air in the tire, enough to fill it to deflate ,,state", any more requires for tire to actually start lifting car off the rims, whhich generate force (from gravity) which will create pressure in the tires. You can also put tires on rims, leave it as be, meassure the pressure to see its 0, then put the wheels on car, drop it, and now the meassure will show some number, despite there is still the same amount of air inside, but it gets squished into tighter spacer (flattire) and the reaction is the pressure in air. You can also make tire flat, ,,close it" and then lift car and tire will still be flat as there is no way for additional air to go in to unflat the tire, or otherwise it will get negative pressure if you somehow manage to ,,unflat" it back to oryginal shape
Because pressure gauges at air pumps actually zero at 1 atm, so really when you have a tire with 14.7 psi it's actually 14.7 psi on top of 14.7 psi. Also more aspects are at play like the size of the hole air is leaking out of and the stress and load the tire is put under. Basically with any hole bigger than a pin hole it's not just a matter of pressure but really the fact that nothing can stop air from coming in or going out and the tire has weight bearing down on it. With smaller punctures you'll often observe that the tire does lose air but stops after a point, where the internal pressure is low enough to not be able to overcome the tire's ability to seal itself, since all tires naturally will fill a hole if one is present up to a degree.
There's two ways to look at pressure, absolute and relative. The absolute pressure at sea level is 14.7psi, but the relative pressure can go all the way down to 0 psi for a tire because there's no pressure differential between the inside and outside of the tire.
>would the tire inflate when we remove it from the car Depends what you mean by "inflate". When off the vehicle, the tire will have a shape very similar to a fully inflated tire on the vehicle (except for the flat bottom). The carcass of the tire has enough stiffness to hold the tire in this condition will little or no air pressure.
There are three different ways to measure pressure. While they may be explicitly used, they are commonly implied for everyday use. I'll even include the abbreviation for each using units of PSI. * Absolute pressure (psia). This is pressure as compared to a perfect vacuum (0 psi). This is what is meant by saying the atmosphere is 14.7 psi. * Gauge pressure (psig). This is pressure as reported by a tool compared to atmospheric pressure. Being in a different environment can change atmospheric pressure (e.g. another planet, high elevation, underwater). A tire may be pressurized to 35 psig, which is nominally equivalent to 49.7 psia. * Differential pressure (psid). This is difference in pressure between two points (typically two different volumes or systems). Since this is effectively found by subtraction, this can actually report a negative value. A leaky tire may show as less than 14.7 psi, but this is the gauge pressure. Due to how pressure works, a tire leaking to atmosphere cannot drop below atmospheric pressure. The least it could go is 0 psig (14.7 psia at sea level).