If I drop a bucket of water from a height, what (if anything) happens to the air pressure on the water?

Factual Questions
41

The issue with drafting is not (much) the static pressure reduction. it’s in there, but only about 5 spots to the right of the decimal point.

The benefit to drafting is that lead vehicle entrains a blob of air moving along with it at nearly its own speed. And expends a bunch of energy creating and pushing that blob along at e.g. 30mph.

If you can slip in close enough behind them, you can travel 30mph over the ground inside a blob of air already moving 25mph the same way you are. So the air drag force your car/bike/bucket feels is only 5mph.

It is exactly the same effect as traveling with the wind vs no wind vs into the wind. It’s no more complicated than that.

Yes, you can describe the drag force as a pressure vector. But that really mixes up a bunch of meanings of the word “pressure” and IMO confuses more than it illuminates what’s going on.

42

Well complain to the physicists who break down drag into its two components: skin (or friction) drag and pressure drag.

Communicating it to us general public folk people at places like the National Air and Space Museum put it like this:

It makes for a pretty straightforward simple as that explanation to me, and with greater understanding of what is going on than thinking about an entrained blob of air. Which really is not accurate. The low pressure zone in the water behind a speed boat, for example, is not traveling forward all that quickly.

43

You’re once again comparing the pressure behind an object in motion to the pressure in front of it. There’s no question the pressure behind the object is lower than in front of it. That doesn’t mean the pressure behind the object is lower than the ambient air pressure at that altitude.

44

The object has pushed particles ahead of it, compressing them, increasing the pressure relative to atmospheric.

The object has left a relative void with fewer particles in the space of the wake, decompressing it, lowering the pressure relative to atmospheric. Which is why the fluid (air in this case) flows from atmospheric into that space.

Hence explanations that read like this:

How could a region of the same temperature not be at lower pressure when the same volume has fewer particles in it???

45

This. Note that if the lead vehicle is extremely aerodynamic, like this:

it doesn’t drag much air along with it, and any vehicle behind it will not have much to gain by drafting it.

OTOH, if you’re on a bicycle staying close behind a city bus, you can maintain 30 MPH without much pedal effort at all. You can even go a lot faster than that:

She rode her bike to 183 mph and broke a cycling record that stood for 2 decades

Muller-Korenek tethered herself to a pace car driven by drag racer Shea Holbrook until she reached about 90 miles per hour. Then, she unclipped and pedaled through the draft for several miles, reaching an average top speed of 183.9 mph between miles four and five.

Turbulence reduces form drag by reducing the pressure drop in a moving object’s wake. This is because turbulence helps delay separation of the boundary layer from the surface of the object. It is this flow separation that causes an aerodynamic stall condition on an airplane wings operated at a high angle of attack. Aircraft wings often have small turbulence-generating devices on their wings to delay flow separation, allowing safe operation at higher angles of attack and lower airspeeds. You can use the same sort of turbulence generators on other vehicles to reduce form drag by delaying or eliminating flow separation

Flow separation also increases the total area behind a moving object that experiences low wake pressure. So yes, for a reasonably aerodynamic object (like the faired bicycle in the above photo), some fine-scale turbulence may help decrease total drag. You can even achieve a good drag reduction with a simplistic “boattail” design, as is sometimes done on bullets and artillery shells:

But for an object as un-aerodynamic as a bucket of water falling straight down, no amount of turbulence will prevent flow separation as the air moving up the sides of the bucket reaches the bucket’s rim; it’s just not going to move smoothly around that sharp corner and cleanly fill the void at the surface of the water. There will be a low-pressure region above the water that contributes to the total aero drag experienced by the bucket of water.

46

Maybe the confusion dissipates if we distinguish between static pressure (due to the effect of gravity on air) and dynamic pressure (due to motion relative to the air).

You are correct that the benefit of drafting has principally to do with a reduction in dynamic pressure felt by the following vehicle; changes in static pressure are small and their effects insignificant.

47

Based on this and the strength of @DSeid’s arguments I’m leaning towards this as the correct explanation. I think flow separation occurs at the leading edge of the bucket and the entire bucket is creating a low pressure turbulent region over the bucket.

48

The “correct explanation” for what? It’s a universal truth that the “right answer” to a question depends on what the question is! And ISTM that the reason we’re going around in circles on this is that the OP’s simple question has been misinterpreted and over-complicated, despite Herculean efforts by some to answer the original simple question. Let me try one last time.

The OP re-iterates the thread title in the third post thusly:

I have a bucket of water at sea level, and the water is at 1 atm at its surface. I then drop the bucket from some height. Does the air pressure on the water’s surface change?

Now, if we go back to the thread’s original post, the first sentence is (emphasis mine):

My initial thought would be that without the column of air pushing on the surface, it would decrease until it reached some new equilibrium with the terminal velocity of the water bucket.

Clearly (unless I am now the one misunderstanding the question) the OP is wondering whether the downward acceleration of a bucket of water affects the static air pressure exerted by the column of air above it. Not about dynamic pressure. Not about aerodynamic drag. Not about “drafting”.

And the answer, once again, is no – the downward acceleration of the bucket does not affect the static atmospheric air pressure at all – not one bit.

OK, end of story. Now, if we wanted to explore the effects of changing air pressure with altitude, or aerodynamic effects like air resistance, drag, draft, and turbulence, it’s a whole different question and the answer becomes a whole lot more complicated.

I might note, though, per my post #25 that if the bucket was dropped from only a modest height such that its maximum velocity was 30 mph, the aerodynamic pressure at the front (bottom) at that point would be just 0.016 PSI. Calculating the effect of draft on the top surface of the water becomes a complicated matter of fluid dynamics but it would likely be much less than the dynamic pressure at the flat front, so pretty insignificant. But that’s not the question that was asked!

You betcha! :grin:

49

For exactly what @Machine_Elf stated. That there will be a low pressure zone over the bucket as it falls.

50

To me you are answering a nonsensical question then.

There is no static pressure on an object in motion through air. The object is not static. The air relative to the object is not static.

There is measurable pressure in the air immediately behind the object in motion. That number is impacted by atmospheric pressure at any specific elevation and by the impacts of the motion on pressure at that specific point in space relative to the object. Which includes the impact of drag and resultant low pressure zones inside any wake.

51

There absolutely is! That’s the whole point. If there weren’t, the water in the bucket would boil away, as it would do in outer space.

If you dropped a barometer from some height, do you think its pressure reading would be any different while it was falling than if it was standing still?

52

That pressure, the air pressure measurable at the surface of the water, is the pressure we are both talking about. And that pressure is definitely lowered because of the fact that it is inside the wake of a rapidly moving object.

Gravity really is immaterial other than as the force acting upon the object.

You could do it upside down (not with water in a bucket of course) - accelerating an object upwards with 1g force net upwards. The air pressure on the rear surface of the object or otherwise inside whatever wake there is, is lower to the same degree as if the same object was dropping under g, other than by impacts of surrounding atmospheric pressure changing with elevation. Do it horizontally. Same.

Yes a barometer, a pressure transducer, would measure a different number inside the wake of a moving object dropping.

53

OK, it’s pointless for me to keep arguing about what I think is my interpretation of the OP’s question, as clear and obvious as I think that interpretation is. At this point we can only ask @Maserschmidt whether he feels his question has been adequately addressed. I saw nothing in it pertaining to anything about aerodynamic forces. AFAICT it was all about how downward acceleration in itself (and possibly also downward constant velocity) affected static atmospheric pressure acting on the object. The answer is that it does not.

Remember, too, that even if aerodynamic forces were of any interest, at modest speeds like 30 mph for an object dropped from modest heights, these forces would be negligible, as already discussed. In the context of the question as I understood it, it’s an irrelevant red herring.

54

I’m really not wanting to argue about interpretation of the OP question, and I am curious about whether you think a barometer attached to the rear of a falling object would measure “any different while it was falling than if it was standing still?”

Am I understanding correctly that you think it would measure the same in both conditions at the same elevation?

Bonus question! Does your answer change if it is a mercury barometer or a pressure transducer and based on the orientation of the pressure transducer?

55

I think it’s obvious that a barometer attached to the stable back end of a large falling object would register lower air pressure, but that wasn’t my question. Incidentally, I am living proof of that latter fact. During my foolish youth when I did a few skyldiving jumps, on one occasion the parachute seemed to be taking a disturbingly long time to deploy as I continued in free fall. As the watchers on plane informed me later, the pilot chute (drogue chute) deployed correctly but spent an unusual amount of time wobbling around in the partial vacuum behind my back before it finally caught the airstream and pulled out the main.

The question I asked is about dropping a barometer by itself. I’m no aeronautical engineer but IMO what it would measure in free fall would depend entirely on the design of the barometer and its size and shape and of course how fast it was going but in general wouldn’t differ much at all from static atmospheric pressure unless it was going super-fast.

Form drag (as opposed to skin friction drag) is a drag component that depends on the shape and size of the object and its aerodynamic stability (which would determine how much turbulence it creates around it). ISTM that the important thing from a basic theory perspective is that an object falling through air creates both higher pressure at the leading end and lower pressure at the trailing end, plus a whole lot of turbulence. It’s almost impossible to predict what the net effect of all this would be on what the falling barometer would measure, but it is possible to say (as has already been said upthread) that unless it was going very, very fast it would pretty much be indicating the same atmospheric pressure as if it was standing still.

56

I was depending on the concept that a turbulent region of air has the same average pressure as the ambient environment. I don’t know how and when that is applicable to anything after thinking about it. I’ve seen that stated various times, but looking up explanations for aerodynamic effects turns up a lot of simplifications. A turbulent region of air is full of movement in different directions so it doesn’t have a pressure measurable with a barometer but the question here is if it weighs less than than the ambient air at that altitude, and if it does then doesn’t that reduce the static air pressure on top of the bucket by reducing the weight of the column of air over it?

57

It’ll depend on the orientation of the port that’s actually sampling local pressure. A pitot tube is a good example of this:

If the port faces forward, the sensor will register the ram air pressure. That’s local atmospheric pressure, plus the local air density times the square of velocity divided by two. (this relationship is actually for incompressible fluids; it also works for air at modest speeds, but becomes less accurate as you get to speeds at which the air density starts to get affected by the pressure changes.)

If the port faces sideways, the sensor will measure something close to local atmospheric pressure.

If the port faces to the rear on a non-streamlined object (such as on the free liquid surface of a falling bucket of water), it’ll measure something less than local atmospheric pressure

If you’re talking about a mercury barometer with a free liquid surface, the whole thing can still be enclosed and connected to a hose the open end of which is oriented as described above.

58

Two questions.

One. The air on the sides would be moving quickly, yes? And therefore wouldn’t the pressure be lower there per Bernoulli’s principle?

Two. Would the average pressure on the object equal the atmospheric pressure at that elevation? I am thinking it would but would appreciate having my understanding confirmed.

Thanks.

59

Again, you’re correct, but somewhat obfuscating the point I’m trying to make. A pitot tube is expressly intended to measure dynamic pressure and ignore static pressure. Hook up a pitot tube to a calibrated dial and what you have is an airspeed indicator.

The point I’m making is best illustrated by a barometer inside a cardboard box that is thrown off a building – i.e.- one that is shielded from aerodynamic effects but not from static air pressure. Ignoring small and mostly unpredictable random effects, the barometer inside will register atmospheric pressure all the way down.

Yes. Hold a small piece of paper, like from a small notepad, directly in front of you mouth and blow. The paper will rise up into the lower pressure of the air stream.

My best guess is yes, as per my cardboard box example, though aerodynamic forces create various competing positive and negative pressures. What I’m certain of is that any measurable difference from atmospheric pressure would be exceedingly small.

60

Sorry about that. I think your point had been clearly made though and held up to my obfuscation. I just wanted to make sure I wasn’t missing or misunderstanding anything.

Thanks again.

What if the barometer is on a treadmill? :grin: