## Main Question or Discussion Point

I have a few questions that share a similar theme.

1) Is wind just fast-moving air particles?

2) When you're in a stationary car and it begins to move, does the back of the car push the air particles inside the car forward (I know they're always moving but is there some sort of net movement so that relative to the car, it's no longer moving?)

3) If you have a balloon floating in a car (let's say it's not touching anything) and the car accelerates, if the acceleration is great enough, would it be possible that the air particles being pushed to the front be enough to move the balloon a little?

4) So assuming that this is how wind works and if we can treat the air particles this way, if a really long train with a really long, narrow opening rolled past and you stuck your arm inside the opening, would you feel the the force of the air particles inside the train moving against your arm like a wind (assuming very little seeping)? Would it roughly the same speed as the train?

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jbriggs444
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I have a few questions that share a similar theme.

1) Is wind just fast-moving air particles?
Yes. Average movement in a particular direction by the air molecules in an area.
2) When you're in a stationary car and it begins to move, does the back of the car push the air particles inside the car forward (I know they're always moving but is there some sort of net movement so that relative to the car, it's no longer moving?)
Yes.
3) If you have a balloon floating in a car (let's say it's not touching anything) and the car accelerates, if the acceleration is great enough, would it be possible that the air particles being pushed to the front be enough to move the balloon a little?
Quite dramatically. If you have ever driven in a car with helium balloons floating in it, you will notice that when you accelerate, the balloons float "up" toward the front of the car and when you put on the brakes they float "up" toward the back of the car. The acceleration of the air under the force of the car has set up a pressure gradient.

For balloons filled with air there is no net effect of buoyancy. There is a some movement of air associated with the changing pressure gradient when the car speeds up or slows down, but it is very small. To a first approximation, air is uncompressible under these kinds of tiny pressure changes.

4) So assuming that this is how wind works and if we can treat the air particles this way, if a really long train with a really long, narrow opening rolled past and you stuck your arm inside the opening, would you feel the the force of the air particles inside the train moving against your arm like a wind (assuming very little seeping)? Would it roughly the same speed as the train?
Yes.

With the first question, since temperature has to do with the kinetic energy of particles, shouldn't it mean that fast wind would be of higher temperature?

For the third question, I'm not 100% sure what you mean by the pressure gradient and why it's different with an air-filled balloon

haruspex
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With the first question, since temperature has to do with the kinetic energy of particles, shouldn't it mean that fast wind would be of higher temperature?
This is quite an interesting question. If I move at 50kph through air at 0C I get very cold; if I were to move at 2000kph through the same air I would get very hot.
Think about the speeds involved. The typical speed of air molecules at room temperature is about 1800kph. That speed varies as the square root of the absolute temperature, so not much less at 0C. Adding 50kph doesn't warm it up much either, so the the main consequence will be the wind chill, i.e. encountering a lot more of the cold air in a given time period.
Moving at 2000kph, the relative speed is 3800kph. (Just adding them like that is probably wrong because one motion is in random directions and the other is uniform, but you get the idea.) So the temperature will feel like about 273 x (3800/1800)2 K = 950C.

jbriggs444
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For the third question, I'm not 100% sure what you mean by the pressure gradient and why it's different with an air-filled balloon
You understand that the lower down you go in the atmosphere, the more dense the air is and the higher the pressure, right?

A fancy word that is used to describe this is "pressure gradient". There are some technical mathematical underpinnings having to do with scalar fields, vector fields and partial derivitives that give the term a precise technical meaning, but basically it's just saying that the pressure is higher in one place than it is in another.

In an accelerating car, the air will tend to "pile up" in the back. The air at the back of the car is at a very slightly higher pressure than the air in the front of the car.

If you think about it, it has to be this way. Every little parcel of air in the accelerating car must be accelerating at the same rate as the car. That means that the pressure behind this little parcel of air must be just a little bit higher than the pressure in front of it... Just enough higher to make that parcel accelerate at the right rate.

This is a stable equilibrium. If the pressure differential is too small, that parcel will add to the pile in the back of the car. This will increase the pressure there, thereby increasing the pressure differential. If the pressure differential is too large, that parcel will be pushed toward the front of the car, reducing the pile-up in the back and reducing the pressure differential.

Now consider a helium balloon. It is lighter than air. Under a pressure differential that would give a parcel of air just the right acceleration to keep up with the accelerating car, the helium balloon will feel enough force to have an larger acceleration (f = ma -- reduce the mass and you increase the acceleration). So as you accelerate the car, the helium balloon will float forward toward the windshield.

[Try the experiment. The effect is dramatic]

By contrast, with a balloon filled with air the density of the balloon is just about the same as the density of any other parcel of air. There is still a pressure differential, but it does not affect the balloon differently than the rest of the air. There is no net tendency to float forward under acceleration.

The effect where light objects in a fluid move under the influence of a pressure gradient has a name:

"Buoyancy"

In an accelerating car, the acceleration is just like gravity. If a balloon floats up in gravity, it will float forward in a car.

This is quite an interesting question. If I move at 50kph through air at 0C I get very cold; if I were to move at 2000kph through the same air I would get very hot.
Think about the speeds involved. The typical speed of air molecules at room temperature is about 1800kph. That speed varies as the square root of the absolute temperature, so not much less at 0C. Adding 50kph doesn't warm it up much either, so the the main consequence will be the wind chill, i.e. encountering a lot more of the cold air in a given time period.
Moving at 2000kph, the relative speed is 3800kph. (Just adding them like that is probably wrong because one motion is in random directions and the other is uniform, but you get the idea.) So the temperature will feel like about 273 x (3800/1800)2 K = 950C.
So if I understand you correctly, you're saying that the relatively small changes in speed isn't that big a factor when it comes to the temperature? Then what is the difference between cold air and hot air, like the one that comes out of an air conditioner?

What exactly is windchill?

@jbriggs444: saved the page and will read your post when I get home cause i'm out of time, will respond tomorrow!

jbriggs444
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What exactly is windchill?
http://en.wikipedia.org/wiki/Wind_chill

Calm air is an insulator. It does not conduct heat well. Moving air can carry heat away in process known as convection. So in cold temperatures, moving air feels colder than calm air.

haruspex
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So if I understand you correctly, you're saying that the relatively small changes in speed isn't that big a factor when it comes to the temperature?
Right. Let me correct something... I thought a bit more about how to add up the random velocities and the relative motion velocity. In terms of the total energy (temperature), one should add the squares. So for random velocities of around 1800kph plus a relative motion of 50kph, the effective temperature is the same as for random velocities of √(18002+502) = 1800.7 kph. That would make 0C feel like 0.21C (ignoring wind chill).
Making the relative motion 650kph raises the effective temperature to 35C.
Then what is the difference between cold air and hot air, like the one that comes out of an air conditioner?
As I said, the absolute temperature varies as the mean square of the velocity. So ambient air at 35C has typical molecules moving at 1900kph, compared with 1800kph at 0C.
What exactly is windchill?
As jbriggs444 says, it's an effective increase in the conductivity of air, achieved by constantly replacing the air that you've warmed from your body with fresh cold air. Same principle lies behind wetsuits; keeping the same water next to your skin avoids constantly having to heat fresh cold water.
Note that this means wind chill only happens when the air is colder than the chilled body. A thermometer left out in a cold wind won't measure any wind chill. If the air is hotter than you then you'll get a wind roast, not a wind chill.

Note that this means wind chill only happens when the air is colder than the chilled body. A thermometer left out in a cold wind won't measure any wind chill. If the air is hotter than you then you'll get a wind roast, not a wind chill.
Unless you're sweating...

You understand that the lower down you go in the atmosphere, the more dense the air is and the higher the pressure, right?
That's due to gravity?

A fancy word that is used to describe this is "pressure gradient". There are some technical mathematical underpinnings having to do with scalar fields, vector fields and partial derivitives that give the term a precise technical meaning, but basically it's just saying that the pressure is higher in one place than it is in another.

In an accelerating car, the air will tend to "pile up" in the back. The air at the back of the car is at a very slightly higher pressure than the air in the front of the car.

If you think about it, it has to be this way. Every little parcel of air in the accelerating car must be accelerating at the same rate as the car. That means that the pressure behind this little parcel of air must be just a little bit higher than the pressure in front of it... Just enough higher to make that parcel accelerate at the right rate.
So when a car accelerates, the air molecules kind of build up at the back because some molecules in the centre/front aren't bumped forward by air molecules at the back that are being pushed by the car...then there's higher pressure at the back because since all the parcels of air (so just large groups of molecules?) are accelerating at the same rate as the car, they can't accelerate forward to 'readjust' so that there's equal pressure….is that right?

This is a stable equilibrium. If the pressure differential is too small, that parcel will add to the pile in the back of the car. This will increase the pressure there, thereby increasing the pressure differential. If the pressure differential is too large, that parcel will be pushed toward the front of the car, reducing the pile-up in the back and reducing the pressure differential.
So when there's a slightly higher pressure at the back when the car is accelerating, that's in stable equilibrium? The pressure differential - that's just the difference in pressure right? If so, why would the parcel of air move to the back of the car?

http://en.wikipedia.org/wiki/Wind_chill

Calm air is an insulator. It does not conduct heat well. Moving air can carry heat away in process known as convection. So in cold temperatures, moving air feels colder than calm air.

With regards to the wind chill stuff - would you consider the moving air to be wind?

Right. Let me correct something... I thought a bit more about how to add up the random velocities and the relative motion velocity. In terms of the total energy (temperature), one should add the squares. So for random velocities of around 1800kph plus a relative motion of 50kph, the effective temperature is the same as for random velocities of √(18002+502) = 1800.7 kph. That would make 0C feel like 0.21C (ignoring wind chill).
Making the relative motion 650kph raises the effective temperature to 35C.
So if we ignore the wind chill stuff, then if you were to stand in front of a 50kph wind when the temperature is 0°C, you'd feel as though it was 0.21°C (hypothetically of course)? Would a thermometer detect this? I'm not referring to any specific kind cause I don't know how they all work but I'm guessing it would feel all the windchill stuff too

As jbriggs444 says, it's an effective increase in the conductivity of air, achieved by constantly replacing the air that you've warmed from your body with fresh cold air. Same principle lies behind wetsuits; keeping the same water next to your skin avoids constantly having to heat fresh cold water.
Note that this means wind chill only happens when the air is colder than the chilled body. A thermometer left out in a cold wind won't measure any wind chill. If the air is hotter than you then you'll get a wind roast, not a wind chill.
So as I asked above, would the constantly replaced air being considered wind (even if a very weak wind, so a net movement somewhere rather than the random particle movements)? And if a thermometer doesn't measure wind chill, would it measure the small increase in temperature from strong winds? How exactly do they work anyway?

By the way, how are you figuring out how to work out all the velocities, squares and stuff?

Also, why is it harder to breathe in strong winds?

russ_watters
Mentor
Your lungs really aren't very strong and can't hold a lot of pressure. Still, it would take a pretty fast wind to make it hard to breathe due to pressure alone (remember, skydivers breathe pretty well at 120mph). Most of what you would probably notice during a stiff breeze would probably be due to difficulty standing and the resulting tensing-up of your core muscles.

With the first question, since temperature has to do with the kinetic energy of particles, shouldn't it mean that fast wind would be of higher temperature?

For the third question, I'm not 100% sure what you mean by the pressure gradient and why it's different with an air-filled balloon
Wind is directed movement and heat is undirected movement. More precisely, get the average momentum vector of the air molecules. That's wind. Subtract that from the momentum vector of each air molecule, then get the average of the magnitude of the resultant vector. That's heat.

That's the idea, but in real life it is a bit more complicated. There are other sorts of organized movement. Sound is also a form of organized movement, so are vortices, and maybe some other things I haven't thought of. In the end you measure temperature of air with with a dry thermometer and the organized movement cancels out.

This sunk in for me when I read about a relativistic jet that was at 1K. ??? The expansion of the jet cooled it faster than the microwave background of 2.7K could heat it. The high speed of the molecules is wind, not heat.

haruspex
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Also, why is it harder to breathe in strong winds?
It's your 'diving reaction'. It feels like being in water, so you automatically suppress breathing.

haruspex
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And if a thermometer doesn't measure wind chill, would it measure the small increase in temperature from strong winds?
Yes.
By the way, how are you figuring out how to work out all the velocities, squares and stuff?
Velocities add vectorially. Wind speed w + random speed r; if angle between is a and net speed is v then v2 = w2 + r2 - 2wr cos(a).
By symmetry, the average energy is therefore proportional to w2 + r2.

Thanks a lot everyone. If anyone else has any thoughts on this, please do post...this is super interesting stuff in my opinion