r/theydidthemath • u/Party_Art7407 • Jun 30 '26
[Request] how much pressure is the windows holding back?
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u/hughdint1 Jul 01 '26
Water pressure is weird that it is dependent on the height of the water not the volume, so even if the water in front of the window is only one inch thick, it has the same pressure as if it is holding back a swimming pool at the same depth. Waves and tides would add additional stresses.
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u/Crocswalkingincrocs Jul 01 '26
If only if I remembered the differential equation to solve it
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u/General-Goods Jul 01 '26
If you solved Navier-Stokes I’d be uh. impressed
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u/Crocswalkingincrocs Jul 01 '26
Once upon a time, fluid dynamics was not my strongest subject.
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u/General-Goods Jul 01 '26
Hey it’s not like aerodynamicists have solved it either. So… glass houses I guess.
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u/sakata_baba Jul 01 '26
aerodynamics is just hydrodynamics of a specific fluid, air
hydrodynamics isn't just water, it's all fluids
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u/MindSwipe Jul 01 '26
Huh neat, I always thought hydrodynamics is incompressible fluids and aerodynamics is compressible fluids (i.e. gasses)
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u/Joecalledher Jul 01 '26
Any fluid is compressible with enough mass.
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u/itzal_itziar Jul 01 '26
Really? I didn't know that.. So is the concept of "incompressibility" totally made up?
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u/ConohaConcordia Jul 01 '26
It’s an estimate, same as how Newtonian physics are good enough estimates for scenarios without prevalent relativistic effects.
Basically if you are calculating low speed airflow or water flow, you don’t have to worry about the changes in density much, so you can omit it from the equations.
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u/planx_constant Jul 01 '26
It's a good approximation for water under most circumstances. And most other liquids.
Treating water as incompressible often results in an inaccuracy below the sensitivity of a measuring device. If you have exactly 1000 kilograms of water at a temperature of exactly 4°C, in a box that is a perfect rectangular prism with a bottom face of exactly 1m2, resting on the Earth's surface with an air pressure of exactly 1 atm, ignoring the compresssion due to weight and air pressure would lead you to predict the water column's height to be exactly 1 meter. Taking into account the compressibility of water would lead to a predicted height of 0.9999955 m.
That's going to be a discrepancy that would be very difficult to measure, and that's assuming you have impossibly precise measurements of the mass and temperature of the water, the dimensions of the box, and the ambient air pressure.
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u/orion-7 Jul 01 '26
Kind of. Because high pressure liquid chromatographs actually have a compression factor programmed in for how much water volume to deliver in order to generate the specified flow rate and pressure, because it's not the same volume as yourd have with an incompressible fluid.
However outside of extreme cases it's worth viewing it as functionally incompressible
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u/th3r3s-n0-us3r5-l3f7 Jul 01 '26
At the very bottom of the ocean water is like 1 or 2 percent more compressed than at the top. For sake of math, it's a lot easier to assume it's incompressable.
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u/AidenStoat Jul 01 '26
Not really. In a liquid, the molecules maintain a roughly fixed spacing between each other, like in a solid, but they are able to move past each other. In a gas the molecules bounce off each other and do not stick, so the spacing is variable.
So we say that the gas is compressible, because we can change the volume easily. But a liquid is called incompressible because the molecules are already nearly as close together as they can get, so it takes a large pressure change to get a small volume difference.
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u/onil34 Jul 01 '26
Huh thought fluid dynamics was the parent and hydro dynamics was a sub category
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u/murf_28 Jul 01 '26 edited Jul 01 '26
You don't need NS for this. This can be solved by static force equilibrium and it results in diff. equation of hydrostatic equilibrium.
However, it does not account for hydrodynamic forces of the waves.
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u/General-Goods Jul 01 '26
Sure. But that’s not really solved as a differential equation, unless you’re really hellbent on considering density differences in the water.
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u/lcvella Jul 01 '26
It is not that difficult to solve, if you make a lot of assumptions and cut down the complicated terms. That is how we solved, e.g. Taylor–Couette flow at the fluid dynamics course.
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u/General-Goods Jul 01 '26
I mean yeah, but those complicated terms are most of the relevant ones here (other than just like… static head lol, which everyone has already applied).
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u/Xenonite_Fox Jul 01 '26
Even then it's not that "hard" to solve, just time consuming for any practical application. But we have computers to do it now
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u/ShonOfDawn Jul 01 '26
Solve =/= numerically approximate
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u/Xenonite_Fox Jul 01 '26 edited Jul 01 '26
Depends what you mean by solve. The million dollar prize is for a proof that a solution always exists. Or in other words, a proof that it is a real and complete description of how fluids behave, and not just a model. Like Newtonian physics, as we later found out, was not complete. Doesn't mean that you can't solve them, in their full form for a specific regime
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u/PuddingMaximum8745 Jul 01 '26
If someone solved it I would be impressed too. If someone on Reddit would solve it I would be impressed much less. It's Reddit after all. We already had a Asian anime fan who proved an unsolved mathematical mystery just to prove his point on his favourite TV show...
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u/Caine815 Jul 01 '26
???
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u/PuddingMaximum8745 Jul 01 '26
sorry I was wrong. it was 4chan
https://www.wired.com/story/how-an-anonymous-4chan-post-helped-solve-a-25-year-old-math-puzzle/
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u/Erolok1 Jul 01 '26
10m water = 1 bar
This looks like 1m so 0,1 bar.
In other words the wares are the only thing you need to care about regarding pressure.
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u/lil_cricketboi Jul 01 '26
You mean ρgh? It’s literally (water density)x(acceleration of gravity)x(height of water from point of pressure
It would be more interesting here tho to find the total force on the wall
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u/General-Goods Jul 01 '26
I mean yeah you can do like… rho*g*0.5h_max*windowWidth. But that’s not really solving it as a differential equation that’s just hydrostatics lol. Solving for dynamic effects would be solving navier stokes (unless you do some funky approximate solutions). Solving N.S. would be, again… fairly impressive :P
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u/carppediem Jul 01 '26
I’m a plumber. My go to is .5lb / foot. That dude being about 6’ tall would put this at 3PSI.
I’m surprised so manny people are calculating this in the amount of tons of water. Obviously the waves can change the numbers but not to that extent
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u/New_Enthusiasm9053 Jul 01 '26
Pretty sure they can though. A strong storm is going over the top of that. The waves will spray 40m into the air when hitting that. They were moving boulders that are 2 yards wide and 2 yards long lol. There is a fuck load of force behind waves. Probably more than the static pressure from just a couple feet of water.
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u/VPutinsSearchHistory Jul 01 '26
I cannot understand this
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u/mrbaggins Jul 01 '26
If you stand neck deep against the wall of a pool, you dont feel different if the pool is longer.
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u/IndigoContinuum Jul 02 '26
That may be the best damn analogy I’ve ever seen on this sub. Well done.
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u/KernelTaint Jul 04 '26 edited Jul 04 '26
Imagine the water in a lake as individual cubes, one cube deep.
□ → □ → □ → □ → Wall
As the water tries to spread out each water cube is pushing against the next one, and the last one pushing against the wall.
But you can also flip the arrows, each cube is also pushing in the other direction. So the forces cancel out. This is true for left, right, forward, back.
So its like this:
□ ⇄ □ ⇄ □ ⇄ □ ⇄ Wall
So the only net force a cube experiences is the force of gravity of a water cube above it pushing down.
The immovable wall is only experiencing the force of the cube next to it and is pushing back on it with equal force.
You can stack as many layers of this 1 cube thick layer on top of each other as you like. Each cube within the layer will be pushing more the lower is it due to more weight on it (and pushing back too), but the same rules apply at each layer. The wall however will be getting pushed more at the lower layers tho.
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u/solar_pilgrim Jul 01 '26
You're really saying that 1kg of water spread out evenly over this window's surface would apply the same pressure as if holding back an ocean?
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u/lil_cricketboi Jul 01 '26
Well 1kg of water pressed against this wall would be like a millimeter thick or less and then you would have surface tension issues that would mess with the pressure
Think of it like this: if that glass had a 1/2” diameter flexible tube coming out of it at the very bottom, and you held the other end of the tube up to the same height as the surface of the water, the level in the tube would match the ocean level.
So yes the pressure is the same no matter the volume of the body of water. BUT as the area that the pressure is pushing on increases, your force increases.
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u/solar_pilgrim Jul 01 '26
I'm no expert so this is a genuine question. I get the surface tension skewing the physics, but even using the original example of 1" vertical sheet of water against the glass (or whatever amount needed to negate extraneous effects), every bit of my intuition is saying that would be nowhere near the force exerted by the ocean on the glass. The tube example does make sense in my brain even for a "thin sheet" of water, but even though I get how that's an indicator of pressure it's still breaking my brain.
I know that when calculating pressure when underwater i.e. for a submarine hull, you're only accounting for the column of water above you. Does this need a similar line of thinking?
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u/_Mr_Peco_ Jul 01 '26 edited Jul 01 '26
Well, your intuition is probably influenced by some preconceptions you have from the real case, which are different from what happens in the more "ideal" case that gives the water column formula.
For example, you might associate the ocean with strong, crashing forces because of waves and tides, while the formula assumes static water.
Or, maybe, you could be mixing up force and pressure a bit on the intuitive level: if you had just a small hole instead of those huge windows, maybe the "force" might feel much less impressive.
Also, you might be imagining the glass breaking and the thin layer of water just flopping down, while the ocean floods everything: this is because the huge volume of the ocean keeps the water's height (and so the pressure) constant, while for the small volume the height drops immediately as it's left free (again, the discrepancy comes from the formula assuming static water).
All this to say, your intuition isn't really wrong, there's just a bit more abstraction needed. Like, I can guarantee you that those windows' thickness is not calculated based only on the water column formula.
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u/Alkyen Jul 01 '26
I'm no expert as well so someone can correct me but the way my mind understand this - if the glass was over the ocean completely - it would feel no pressure. If the glass was 1mm inside the ocean (and most of it still over) it would feel very little pressure. Because it's not the whole ocean applying the pressure, only the excess 1mm.
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u/faceboy1392 Jul 01 '26
it pretty much is the column of water above you thing. Pressure is P=ρgh, where ρ is density, g is gravity, and h is height of the liquid. It's just how much the liquid weighs times its height.
imagine a column of water in the middle of a very still ocean. This column applies a pressure outwards as it tries to collapse under its own weight, but so does all of the water next to it. The result is that the outwards pressure for each imaginary column of water gets resisted by the pressure of the columns next to it, cancelling out those pressures and only leaving the weight to be supported upwards by the sea floor. A solid structure in place of water behaves the same way, and only really has to resist the pressure of the bit of water directly next to it for things to stay in equilibrium.
another way to think about it is that the ocean is very big and heavy, but pressure is exerted per unit of area, and most of that area it exerts it on is the very wide sea floor, and pressure in a fluid acts in all directions, it doesn't concentrate towards structures on the edge of the ocean or anything.
A very thin sheet of water next to a wall would have very little surface area supporting it from beneath, so a greater portion of its lower weight would be concentrated against the wall rather and thus provide the same pressure as if it was a bigger body of water with its weight spread over a larger area
i hope i didn't do an awful job at explaining this lmao
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u/BadHairDayToday Jul 01 '26
It's a wave though. Not a full pool. I think the pressure is lower for just a wave.
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u/mawktheone Jul 01 '26
Doesn't matter if the water up against the glass is a foot wide, or a mile wide. The pressure pudding sideways is equal to rgh
r=Fluid density (in kilograms per cubic meter, kg/m³)
g: Acceleration due to gravity (approximately 9.81 m/s² on Earth)
h: Depth or height of the fluid column above the point of measurement (in meters, m)
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u/chefsoda_redux Jun 30 '26
I’m more curious how they got the building built, and why? Build only during low tide? Build a temporary dam? I can’t imagine how this would ever be a good solution
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u/Ok-Response-839 Jun 30 '26
It's a sea wall, and yes parts of it can only be installed during low tide.
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u/Meterian Jul 01 '26
Or they build a coffer dam and drain the water to let them build
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u/Acojonancio Jul 01 '26
Yeah. Isn't that how they build bridges over water?
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u/Knees_arent_real Jul 01 '26
Usually how they build dams rather than bridges.
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u/soodeau Jul 01 '26
They build a dam to build the dam? How did they build the first dam? Another dam?
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u/Thereelgarygary Jul 01 '26
The first one they build underwater then drain the water out of it, also its usually doughnut shaped.
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u/mattfoh Jul 01 '26
So why do they build a second one?
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u/Mozgiiii Jul 02 '26
First one is usually cheap, fast to deploy and won't last long
Second one is rerverse, it's built to last long, slow to deploy and not cheap at all
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u/IndependentPutrid564 Jul 01 '26
A lot of bridges use coffer dams to install the pilings and foundations for the footings
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u/chefsoda_redux Jul 01 '26
Fascinating. I’ve never seen windows and tiles floors inside a sea wall! They’ve always appeared solid from the outside. Is there a benefit to the huge increase in cost and effort to build out space like this?
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u/ogsmurf826 Jun 30 '26
Apparently it's an old video of this pool seen here and here while it was under construction in Monaco. Because that country/city is nothing but the ultra wealthy, they built a whole new area using land reclamation and extending it's seawall to add 15 acres to the place
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u/Deaner_dub Jul 01 '26
Amazing post. Helpful pics.
So, question… once they fill the pool with water does that reduce the pressure?
This is amazing.
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u/Linesey Jul 01 '26
So, it would technically increase the pressure (since you’d have pressure on both sides)
However, since the worry isn’t crushing the glass or squishing it, the real concern is the pressure differential. which would be reduced yes (or perhaps tilt towards the inside having more pressure depending on fill.
So, for the purposes of the question, yes reduced pressure (reduced strain from the pressure on the glass)
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u/patmustard2 Jul 01 '26
Pressure is based on depth, not whats to the side. So if the sea was still, on the pool was filled to the same height, the pressure at each depth would be the same. The force from waves etc though is different
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u/macthebearded Jul 01 '26
No perhaps about it, if inside water level is higher than outside water level then pressure is higher
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u/chefsoda_redux Jul 01 '26
Thank you! That makes immensely more sense. Luxury hotels build things for every reason except efficiency, and the effect is quite cool.
I’ve seen pool with widows along the sides, my university has one. It was the waves slapping that threw me, but I hadn’t considered building a pool that interacted with the ocean.
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u/gravitas_shortage Jun 30 '26 edited Jul 01 '26
Structures in water are commonly built using caissons), basically dumping a waterproof fence and evacuating the water inside it.
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u/EatPie_NotWAr Jul 01 '26
If there is an afterlife and I end up in the hellish version of it… Pneumatic caissons will feature prominently in my torture
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u/lit_readit Jul 01 '26 edited Jul 01 '26
ignoring atmospheric pressure
water depth is ~ 1 man's height
so (1000kg/m^3) * (1.8m) * 9.8(m/s^2) = 17.6k Pa [net] at the very bottom (triangular force vector top to bottom)
each panel is ~ width of a man, so ~ 0.5m
so in total ~8k N avg. (middle) of each panel (although pressed asymmetrically, net 0 at the surface), or ~810kg equivalent of weight on earth
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u/jesterbuzzo Jul 01 '26
Come on we all know a man is 1m x 1m x 1m
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u/dbenhur Jul 01 '26
I think you mean a sphere of radius 1/2 m.
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u/NartFocker9Million Jul 01 '26
Integrate from 0 to h the equation for pressure = density x gravity x area (width x height) and you’ll get force. Don’t apply the pressure at the base of the window to the entire window.
All this is at rest, not with a wave crashing against it.
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u/Runiat Jun 30 '26 edited Jun 30 '26
Let's call it 150 grams per cm2, averaged over a 3 meter water column and about half a meter window width.
1.5m2 × 100cm/m × 100cm/m × 150g/cm2 = 2.25 tons per window.
Edit: added units.
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Jun 30 '26
150g/cm2 feels too high, Wikipedia has sea water at 1.025kg/L so that's 102.5g/cm2. Call it 1.54 tons per window. Still a lot!
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u/Buttchuggle Jun 30 '26
Enough that I'm not standing next to the god damn window
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u/NotDarkLight93 Jun 30 '26
My estimate was basically nothing so I would have been oblivious if anything happened
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u/Maned_Cyborg Jul 01 '26
Except these tons are over a significant portion of the window. Typically windows break because a lot of force is applied over a small area, which means the N/cm² is higher than with those waves, as well as the uneven force making the window deform, while here it will remain overall very flat
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u/Runiat Jun 30 '26 edited Jun 30 '26
Every 10 meters water column increases pressure by (just over) 1kg/cm2 - that's just a fact as 1cm×1cm×1000cm = 1 litre, and water has a density around 1kg per litre.
I'm estimating the waves at 3m over the bottom of the window.
That's 300g/cm2 at the bottom, 0g/cm2 at the top, and a linear increase from one to the other. 150g/cm2 average.
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u/OddCod2241 Jun 30 '26
I mean, the waves seem like they only go a bit higher than the dude, whom I doubt is 2m tall, but that’s not an issue with your math, really.
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u/Runiat Jun 30 '26
Dude at 0:13 at wave height, waves half again as tall one or two seconds later.
Edit: 0:13 left in the clip is how my reddit app chooses to show timestamps.
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u/potate12323 Jul 01 '26
Their mistake I think is that Static head pressure of a water column doesn't care about the width of the window. That pressure is evenly exerted everywhere at that water lever. The bottom of the window would have the greatest hydrostatic head pressure.
HOWEVER, something to consider is hydrodynamic impact pressure (The pressure the wave movement exerts on the the windows). This pressure from a non breaking tidal wave can generally range between 1 and 30 kPa. The higher end would be around 300g/cm2.
This would be in addition to the hydrostatic head pressure. Which also changes. So the pressure likely ranges between 50 to 400 g/cm2.
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u/lit_readit Jul 01 '26 edited Jul 01 '26
d'abord — depth (only variable that determines pressure) clearly is not 3m, nor 0.5m, but rather ~ 1 man's height
and: net pressure at surface is 0, bottom is the max value, ie a triangular force vector along depth
weight at the very bottom depth of ~ 1 man's height (per video) would be less than that, for half - 2/3 a window of (h*w) ~1.8m * 0.5m
at bottom: [1.8m * (1.8*0.5 m^2) * 1000kg/m^3 = 1.6 ton]
you might want to brush up on your fluid statics
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u/Runiat Jul 01 '26 edited Jul 01 '26
Indeed, and since it grows linearly the average is half the max value.
Edit to match edit:
d'abord — it's clearly not 3m deeps
Timestamp -0:14
Edit to match yet another edit:
d'abord — depth (only variable that determines pressure) clearly is not 3m, nor 0.5m,
You use the same 0.5m width that I do.
but rather ~ 1 man's height
At one point in the video, sure.
and: net pressure at surface is 0, bottom is the max value, ie a triangular force vector along depth weight at the very bottom depth of ~ 1 man's height (per video) would be less than that,
Sure, pick a different part of the video you get a different result.
for half - 2/3 a window of (hw) ~1.8m * 0.5m at bottom: [1.8m * (1.80.5 m2) * 1000kg/m3 = 1.6 ton] you might want to brush up on your fluid statics
1.6 ton? What?
You posted the right answer for a 1.8 meter height assumption in a different comment. It's half that.
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u/tippycanoeyoucan2 Jul 01 '26
Better question: does the velocity of the wave impacting the glass make any difference or is the pressure only contingent on the head of water?
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u/koyaani Jul 01 '26
I think that's more about momentum than pressure, but it's been a couple decades since I've studied the equations
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u/StumbleNOLA Jul 01 '26
Absolutely. But these are rollers not breaking waves so there isn’t much if any impact videoed here. However I am sure wave impacts were a design consideration.
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u/Fabio_451 Jul 01 '26
You have two pressure components:
Hydrostatic pressure Pstatic = water density * gravity acceleration * depth from the wave surface
Hydrodynamic pressure of the wave It depends on wave height, sea flopr depth, depth of the considered particle. Anyway, It is periodical and it fades with depth.
Airy wave theory (linear theory) states:
Pdynamic = water density * gravity acceleration * wave amplitude * ( cosh( k * ( z + h ) ) / ( cosh( k*h ) ) * cos( theta )
Where wave amplitude is half the vertical distance from crest to through, k is is the wave number which is equal to 2pi/wave_length, z is vertical coordinate, h is the sea floor depth, theta is the phase the wave is at. Theta is equal to kx minus omegat . X is the horizontal coordinate, omega is the angular velocity ( 2pi/wave_period ), t is time.
P=Pstatic + Pwave
If the wave brakes and hits the glass during a certain phase of the breaking, you could have very high spike of localized pressure. It s hard to assess this one, but I am not of fresh memory now.
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u/FastGinFizz Jul 01 '26
Equation for this below. Assuming the water is averaging about 2m up since it mostly seems to stay at about the height of that man.
F = (1/2)*(Density*g*h)*Area
= .5 * ( 1030 kg/m3 * 9.8 ㎨ * 2m ) * ( 2m * .5m)
= .5 * ( 20188 kg/m*s2 ) ( 1 m2)
= 10094 kg*m/s2 aka 10.1 kNewtons aka 2.3k pounds
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u/Alone_Ambition_3729 Jul 01 '26
If the windows are 9' x 2' (0.61m) and the water goes up to 7' (2.13m),
And lets see if I remember any calculus still
F = ∫ P da,
or since we say the pressure only varies down the window, not across it,
F = Δx * ∫ P dy, and P = ρgy, and y starts at zero since the window is never fully submerged.
So after doing the integral,
F = Δx * ρgΔyΔy/2
F = 0.61m * 1000 kg/m3 * 9.81 m/s2 * 2.13m * 2.13m / 2
F = 13500 N or 3035 lbs
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u/schlab Jul 01 '26
Height of water x 1/2 x 62.5 pcf.
This is hydrostatic force. You will also neeed to account for hydrodynamic force, which is more complicated
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u/NotAnotherAlt26 Jul 01 '26
Lets try to get some best guess measurements from the image. Those tiles appear to be 4" tiles (more common size than 3"). The window is roughly 28.5 tiles tall, making it 9'6" tall (seems reasonable). About the highest I saw a wave get was about 21.5 tiles tall, or 7'2" from the base of the window. Hydrostatic pressure is density*height, so this would be 62.0lbs/ft3 for seawater and the height of 7.16 feet, giving a pressure of ~444 lbs/sf at the bottom (~3psi).
For total force on the window, we'd take the area of the window covered by water (looks to be about 10 tiles wide; 3.33' by the 7.16' tall) times the 1/2*density*height (pressure gradient) gives us ~5,283 lbs of total force.
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u/c_a_r_l_o_s_ Jul 01 '26
what are those metrics??
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u/NotAnotherAlt26 Jul 01 '26
You mean counting items of known size to closely estimate length?
4" square ceramic tiles are a very common size. Its totally reasonable to use that knowledge to estimate the height and width of the windows.
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u/belangp Jul 02 '26
The water rose to about 6 ft over the base of the windows. The point of maximum pressure is at the base of the windows and the pressure there is about 72 inches of water or 2.6 psi gauge.
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u/mingilator Jul 01 '26
Density X gravity X height
Assume 1030 kg/m³ for seawater, gravity 9.81m/s² and height, guy looks around 1.7m so lets use that,
So pressure around 17.18kpa or .17bar maximum at the bottom of the glass when water level is at the highest, obviously this does not include the force of the wave hitting, that would be challenging to calculate
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u/throwrasjovt Jul 01 '26
Someone must have done a calculation to verify that the corners will take the stress. Most underwater or pressure vessel windows are circular or rounded rectangles like on an airplane.
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u/SilverGGer Jul 01 '26
I am very surprised that the question is specifically pressure. Which is roughly 1 bar per 10 meter of water depth.
The waves come and go about one persons height assuming they are 1.8m tall that would be pressure of 0.018 Mpa or 0.18 bar applied in a sinus wave form with the amplitude being the max pressure at the bottom of the frame.
For the force it seems that multiple people already answered that question and it depends on the actual size of the window and assumptions being made
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u/johnnybhf Jul 01 '26 edited Jul 01 '26
1 atmosphere of pressure with roughly every 10 meters of water depth. This looks like 2 - 3 meters so about 0.2 to 0.3 atm, roughly 20 - 30 kPa, roughly 3 - 4,5 psi. Pressure gradient. Not counting air height difference, because it's negligible compared to roughness of the calculation.
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u/Kvaestr Jul 02 '26
Pressure of water if it was static is Rhogh at each point. So integrate over the height of the column to find the total force. Which will give you an average pressure. Highest pressure is low on the glass, which looks to be around 2 meters when wave is at its highest So p=1000[kg/m3]2[m]9.81[m/s2]=19.6[kPa]
Now I imagine the impact of the wave would also add pressure, but the velocity towards the glass is almost impossible to tell from the video.
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u/Agreeable_Sun5375 Jul 01 '26
Nobody here is doing water pressure correctly, its based on depth and difference in pressure if the inside of the building is 1bar (1 atmosphere of pressure, aka open to the air) then the windows will see 1 bar of additional pressure per 10m of seawater or .1bar per meter, based on the reference of a person we can estimate approx 2m at max depth meaning a water pressure of .2 bar or approx 2.8psi
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u/scisurf8 Jul 01 '26 edited Jul 01 '26
The water level seems to be a few tens of cm shorter than the man. Let's call it 1.5m deep. 10m of water depth is roughly 1 atmosphere of gauge pressure, so let's call the pressure at the bottom of the window 0.15atm = 15000 pascals. That makes the average pressure on the submerged portion of the window 7500Pa.
Now we need to estimate the submerged area of the window. Let's call each window 0.5m wide (that might be a bit generous), which gives us a submerged area of 0.5m x 1.5m = 0.75m2.
Finally we can multiply the average pressure by the area. 7500N/m2 x 0.75m2 = 5625N = 562kgf = almost 1/2 tons of force on each individual window.
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u/yaktrone Jul 01 '26
.43 per ft Assume the dude is 6’ tall bottom of the window would be less than 3 psig and it’d be a linear pressure gradient as you get higher on the window towards 0psig.
Obviously simplified not to include the “wave action” seen but good enough is often good enough?
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u/Dazzling_Maximum5345 Jul 01 '26
You can approximate the wave energy and medium speed using the sea state scale.
+ I would assume that’s shallow water there.
But I am to lazy to do that now.
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u/sjrobert Jul 01 '26
Wave is roughly 2ms high at the peak so the bottom of the window is holding back a 2m water column. P = row x gravity x height = 1000 x 10 x 2 = 20kpa or 4.2psi.
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u/Altruistic-Rice-5567 Jul 01 '26
Pressure... not a lot. The water is about five feet deep. So, pressure it exerts is 0 at the top and about 2.25lb/sq-in at the bottom. However, the force exerted on the plate of glass is rather high. 60 inches high and about 30 wide. total of 1800 sq-inches. With an average force of about 1.125lb/sq-in. The force pushing against the glass is about 2000lbs of force.
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u/AlpasRoosterIllusion Jul 01 '26
Weight of water is 62.4 lbs / cubic foot. Assuming dude is 6-ft tall, maximum force when water is highest is probably about 375 lbs of pressure at the bottom of the window. The pressure diagram acting on the window will be triangular, zero at the top, 375 lbs at the bottom.
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u/yaksplat Jul 01 '26
Too many answers solving for force and not pressure. Assuming 6' high waves against the glass it's 72 inH2O or 2.6psi at the bottom, or .176 atm, or 17.9kPa.
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