Increasig Atmospheric Pressure
 

Has anyone calculated the extent of the rise in average sea level atmospheric pressure caused by the combined weight of all aircraft flying?

I'm not sure aircraft weight in flight increases atmospheric pressure (weight) on the ground. But using the analogy of the weight of people in a building increasing the pressure the total building + people puts on the ground...

1) Only aircraft in flight at a given instant count. That would be on the order of 10000 aircraft at any one instant. Could range from 8000 to 12000 depending on time of day (and where "day" is at any instant - U.S.? EU/Africa? Asia?).

2) Weight of those aircraft - C150 to A380? Let's take the 739ER as a reasonable average: MTOW = 71350 kg

x 10000 = 713500000 kg or 7.135 x 10ˆ8 kg

3) Weight of the atmosphere = 5.1 x 10ˆ18 kg

4) aircraft weight adds a bit over one 10-billionth (or ten 1000-millionths) to the total.

5) In millibars, average atmospheric pressure with no planes flying = 1013.25 millibars

- with planes included: 1013.25000017418 millibars

A microscopic difference, in other words....

(Assumptions and math open to criticism).

I'll pass this one on to my sister-in-law who interviews physics candidates for a well-known university. :E

Thinks: surely the atmosphere just expands a bit, there is no lid on it, after all?

I'm pretty sure the atmosphere's already big enough to hold the aircraft when it's sitting at the gate.

Mean density rises, but I'm not sure that pressure does = I'll have to think about it!

Why would an aircraft add to the weight of the atmosphere as it's developing lift and and not actually resting or sliding along an area of air excerting pressure on it?

How much do mid to high altitude clouds add to the pressure on the ground?

It is, but it's not supporting it at that point.

edit: just thought about that again, I've written rubbish. I think.

As the aircraft develops lift, it exerts a force downward on the airmass equal to its weight. So it is "actually resting or sliding along an area of air excerting pressure on it."

But the atmosphere under the aircraft is not sealed on the side, so it escapes sideward and upward until it reaches equilibrium.

Some points:

- If the atmosphere expands in radius, the air column above any given square meter or centimeter of the earth's surface gets taller and thus weighs more = higher pressure at sea level.

- for aircraft not flying, their weight is supported directly by the ground - and would be, even in a vacuum. Not a factor in this thought-experiment about air pressure.

- clouds are equivalent to lighter-than-air balloons. Water vapor and humid air are less dense than dry air. So they float, most of the time.

http://images.slideplayer.com/14/441...s/slide_15.jpg

- don't confuse "volume" with "weight." An unpressurized aircraft is just a tube with equal atmospheric pressure/density inside and outside the thin skin. A pressurized aircraft has higher air density inside, and thus a slightly higher weight - but I left that out as being an even less significant effect than the total aircraft weight(s), which are already insignificant.

The atmosphere is really big - all the aircraft flying are just bacteria by comparison:

https://pix-media.priceonomics-media...ash_plane.jpeg

- I'd like to hear from overstress's sister-in-law, too. ;)

This is an extension of the question if a cargo load of birds take to the wing would a flying aircraft weigh less, or no change.

How Airplanes Fly - The Real Story (With Experimental Verification!) | Science 2.0

You could request a copy of the scientific paper here, probably cost.

https://www.nnv.nl/en/english-home/

What a fascinating question.

I like to think about it this way. The atmosphere exerts a certain force on the surface of the earth. That force is caused by gravity pulling on the mass of the atmosphere. This is known as the weight of the atmosphere. It is this weight that causes sea-level atmospheric pressure to be what it is.

If you picture an aircraft sitting on the ground, the total weight of the atmosphere plus the aircraft will be the same as the total weight of the atmosphere plus the aircraft if the aircraft is airborne. However, in the airborne case, the atmosphere must transfer the weight of the aircraft to the surface rather than the aircraft's wheels. The only way the atmosphere can do this is to increase its MSL pressure slightly.

The bit where I get stuck is that the atmosphere doesn't have a "lid". If you kept pumping air into the atmosphere (from a hypothetical worm-hole), I don't think it would increase the atmospheric sea-level pressure. The excess air just escapes out the top into space.

It that is true, does getting an aircraft airborne actually increase MSL pressure, or does it just displace some air out the top into space?

True. But the atmosphere remains attached to the planet due to gravity. If your "worm hole" supplied sufficient air to double the mass of atmosphere here already, then the sea level pressure would double.

In old money, sea level pressure of 15 lbs per square inch is due to a column of air weighing 15 lbs over every square inch of the earth.

GR, you are correct. I was mistakenly under the belief that Earths atmospheric volume was limited by the escape velocity of atmospheric molecules. In fact, on Earth, this only happens to Helium and Hydrogen. Earth is capable of supporting a much larger and denser atmosphere. I have just learned something.

So yes, I believe MSL pressure is increased proportionally by the weight of all airborne aircraft (and birds) :)

As an aside, I believe the water pressure at the bottom of the ocean also increases for every floating vessel, but in this case it is not proportional to the weight of the vessel, because the ocean walls are not vertical.

Doesn't it depend whether the conveyor belt is in service?

Doesn't it also depend whether a lorry full of budgerigars startles the birds into flight and off their load-bearing perches?

We need to know these things. We also need to let the kiddies have the wanted answer in the ATPL question bank of their preferred crammer for slow learners.

As an extension of the nautical analogy, imagine a toy submarine at neutral buoyancy circling, submerged, in a bucket of water on a scale. Now replace some of the air inside the submarine with lead, but set the dive planes so it continues to circle at the same level.

The reading on the scale has to increase by the amount of lead added, but the water level in the bucket won't change. The answer seems to be that the hydrodynamic effects of the dive planes increase the dynamic pressure at the bottom of the bucket. I'd guess that the same is true for an aircraft flying in the atmosphere, even though the effect would be too small to measure.

I assume the effect would be localized as well -- after all, atmospheric pressure isn't the same everywhere to start with. Imagine a helicopter hovering above a set of truck scales.

Rain
 

A storm drops 100cm of rain over 1000 square km .
No NET water vapor is added or subtracted anywhere else, for arguments sake.:
Pressure must drop, due to mass of atmosphere ( rain ) resting on ground.

Not exactly.
Consider a water bottle with broad bottom and tall neck, which is full of water.
The water pressure on the bottom is high, and over the wide bottom sums out to a large force. Yet the weight of the water column in the tall but narrow neck does not sum up to so much.
The matter is that the water also exerts pressure on the upper part of the bottle, below the neck. That force is applied down to the water, and transferred to the bottom.
Likewise, if there is something pressing on air from above, such as a landed plane taking up volume from atmosphere but resting on wheels, that´s extra air pressure which need not be weight of air.

I don't know what you are saying with your water bottle, but if you are suggesting that the landed aircraft adds to atmospheric pressure as does an airborne aircraft, then no, I disagree.

Adds, but not equally.
Take a cylindrical vessel of water. Drop a stone in it.
Water level will rise due to the volume displaced by stone.
The pressure of the stone on the tiny contact points to the bottom will be the weight of the stone minus the weight of water displaced by the stone.
The pressure of the water on the bottom is the weight of the water actually in the vessel plus the weight of the water displaced by the stone.

Ditto about Earth atmosphere.
Reenter a shuttle into atmosphere so that it is no longer supported by centrifugal force but by aerodynamic force, and the pressure of air will be weight of atmosphere plus weight of shuttle.
Land the shuttle, and the pressure of air will be weight of atmosphere plus weight of air displaced by shuttle.

But no aircraft except spacecraft "enter" the atmosphere - they start out already in the atmosphere, from the moment they start to be assembled in the factory.

...or are you saying that the moment Boeing or Airbus finally seal every new fuselage with the last rivet, the world's air pressure jumps a tiny notch?

And spacecraft of earthly origin must reduce atmospheric pressure slightly when the leave the atmosphere to begine with, so when they reneter, they are simply restoring the original conditions.

Accounting for the weight of the air displaced by an aircraft sitting on the runway has me stumped for a minute. But it made a lot more sense if I imagined that the aircraft was a slightly overloaded (or under-inflated) blimp.

If you're keeping track of volume, what about the oxygen from the alumina released into the air (as CO2) at the aluminum smelter?:\

Chu Chu, the oxygen would add to the atmospheric volume, but it's release would also reduce the gravitational mass of the Earth, which would reduce atmospheric pressure.

I don't know what the net effect would be, but since you have raised it, the same would apply to the aircraft which is made from material sourced from the Earth.

That's interesting (though we're getting a little far afield). In general, the gravitational mass of a sphere acts as if it's concentrated at the center of a sphere. And the atmosphere is part of of the Earth's mass. So smelting the alumina should have no effect on the orbit of the Moon, for example.

But it's different if you're looking at the the effect of gravity at a point within a sphere. My recollection is the it's as if only material at your radius or closer to the center counts (so, for example gravity at the center of the Earth would be zero). Trying to figure out the net effect of adding gas to the atmosphere on the amount of gravity affecting all the other gas in the atmosphere is hurting my head.

So I think you're right that the smelting would tend to reduce the net gravity affecting the atmosphere (because it would replace some mass that fully affects the atmosphere with mass that only partially affects it). But the effect would be less than that from simply reducing the mass of the Earth by the mass of oxygen liberated.

Going back a bit... 'Hydrogen and Helium are lost into space'.... I don't think so.... In a vacuum both hydrogen and helium will weigh something, it's only when weighed in air that the buoyancy makes them float.


So I would think they find a gravitational balance, on top of the air layer, much the same as the Ozone layer... We have invented the Hydrogen and Helium layers..! In practice, turbulence and heat energy keeps all the gasses mixed, or else the Nitrogen would separate from the Oxygen.
.

Well, you would need to calculate the volume of air now filling the holes in the ground created in mining of the aluminium used to construct the aircraft. LOL

"The average speed of hydrogen molecules and helium atoms is greater than the escape velocity from the Earth, and these light gases were lost and swept away through photoevaporation by the solar wind early in the Hadean Eon due to Earth's weak gravity."
Evolution of the Earth's Atmosphere

The atmosphere has too little helium to be able to extract it commercially. "Helium is one of these very limited resources that once you’ve used it, you can never recycle it,"
The best sources are found trapped in rocks after the natural decay of Uranium and Thorium.
https://www.chemistryworld.com/news/...010122.article

Pretty much wot Goldenrivett said...

It's been a while so I can't remember the joys of the statistical mechanics and probabilities of all of this but as I recall it whilst the average speed of the lighter gases such as Hydrogen and Helium may not be above escape velocity the distribution of velocities of the H/He atoms in the atmosphere means that at any one time some of those will be travelling above escape velocity. If they can exit the atmosphere unhindered by impacting other gas molecules they are indeed lost in space...then at some time later some more H/He atoms end up being at the top of the velocity curve, and they leak away, etc etc.....there will always be a few around in the atmosphere, but they haven't been around long enough to have (by chance) to reach the magic velocity in the right place...when they do, they're gone..It's an ongoing process.

In fact all lighter gases are subject to the same risk, it's just that H and He, being lightweight, have the highest average speeds, and therefore are at most risk of being lost....

Of course school texts should be amended to record the percentage mass of aluminum in the standard atmosphere.

After an excellent landing etc...

Mass of Earth's atmosphere dwarfs mass of aircraft
 

Earth's atmosphere has a total mass of about 5 x 10^18 kg. That so far dwarfs the mass of all aircraft - flying, parked, or long past their days of operation - that the impact on atmospheric pressure is well below anything that could ever be measured.

The big sky theory is alive and well! The only place that falls apart is when we have our aircraft tightly following common prescribed paths and heading to and from a limited number of terrestrial points. Unfortunately for airplane separation that is exactly what we do all day (and night) long every single day. I guess we can't do away with ATC.