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arithma

Boyancy

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arithma    226
I have an object partly immersed in water. The result is a boyant force pushing upward. So far am cool with that. But in some cases the boyant force causes moment (analogical for force in the world of rotation)... In the act of calculating that moment I need to know the resultant POINT OF APPLICATION of the boyant forces. IS IT A SAFE ASSUMPTION TO MAKE THE CENTER OF GRAVITY OF THE "IMMERSED VOLUME" AS THE POINT OF APPLICATION? PLEASE HELP

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Motorherp    613
Quote:
Original post by arithma
IS IT A SAFE ASSUMPTION TO MAKE THE CENTER OF GRAVITY OF THE "IMMERSED VOLUME" AS THE POINT OF APPLICATION?


Unfortunately not really. The boyancy force is directly proportional to the volume of water displaced by the object. In reality the force is evenly distributed across the objects surface and is at any one point proportional to the volume displaced by the object directly above that point. To solve this exactly would require integrals, however this can be approximated by solving for evenly distributed points along the objects submerged surface. For each point cast a ray upward to the water surface to get the length of displacement and treat the segment as a cuboid with width and depth equal the spacing between points and height equal to the length of the ray above to give you the volume. You can then calculate the buyoncy force to apply at each point from here which should lead to the rotational behaviour you'd expect.

Hope this helps

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arithma    226
The forces are not evenly distributed over the surface since pressure increases as depth increases, so upthrust at the buttom is greater than closer into the surface... Did I get something mixed up here?

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Motorherp    613
But unless your dealing with objects which are extremely volumous its safe to disregard the pressure difference. And I've just realised that i shouldn't have said evenly distributed. Rather what I mean is that the force is distributed across the submerged surface but is not even because it is proportional to the submerged volume directly above it as I go on to say.

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Dmytry    1151
Indeed, force is applied on surface, as well as weight is applied to every particle. But in sum, force is the same as if it would be applied to single point.
As about disregarding pressure difference, damnit, as all buoyant force is result of pressure difference, don't you know that?

In hydrostatical case (that is, no turbulence/something), force is applied to center of mass of displaced volume of water, and if i understood arithma's post right, exactly as guessed.

It is very easy to prove ala Archimedes theorem. Water in water doesn't spin, and there's no turning force. All forces is compensated. If we have water in the bag, we can assume in computations that weight is applied to center of mass. And bag with water doesn't turn in water, therefore buoyant force neutralizes weight, and to do that it have to in sum be applied to center.

In summary:
1: Sum of forces applied acts exactly as one force applied to center of mass of displaced volume of water. (well, almost exactly, if you gonna to take curvature of Earth and non-linearity of gravitation field into account, you'll get some extremely small difference)
2: Motorherp's idea about disregarding pressure difference is completely incorrect.

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Motorherp    613
My bad. I had it in my head you were refering to applying the force at the objects centre of mass. Totally missed the submerged volume centre of mass thing. However, computationaly speaking, unless you are dealing with very simple shapes you'll find it difficult to approximate this 'submerged volume' and harder still to find its centre of mass. Hence for complex shapes you might be better sticking with an idea like I said where the buyancy force is approximated at points distributed accross the submerged surface.

Quote:
Original post by Dmytry
2: Motorherp's idea about disregarding pressure difference is completely incorrect.


Not exactly. What I was refering to is that in most cases the pressure difference between two points on the bottom of the submerged surface, where in my example the force is approximated, is negligable.

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Dmytry    1151
Quote:
Original post by Motorherp
My bad. I had it in my head you were refering to applying the force at the objects centre of mass. Totally missed the submerged volume centre of mass thing. However, computationaly speaking, unless you are dealing with very simple shapes you'll find it difficult to approximate this 'submerged volume' and harder still to find its centre of mass. Hence for complex shapes you might be better sticking with an idea like I said where the buyancy force is approximated at points distributed accross the submerged surface.

Quote:
Original post by Dmytry
2: Motorherp's idea about disregarding pressure difference is completely incorrect.


Not exactly. What I was refering to is that in most cases the pressure difference between two points on the bottom of the submerged surface, where in my example the force is approximated, is negligable.

not neligable. Buoyancy force is result of pressure difference.

And if you properly compute surface integral of force (for every submerged triangle of mesh), it will be the same as computing submerged volume and center of mass of that volume. It is relatively easy.

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Motorherp    613
Quote:
Original post by Dmytry
not neligable. Buoyancy force is result of pressure difference.

Yes....between a point on the top of the object and a point on the bottom it is very significant. However in my method where the force is approximated at points across the bottom, the pressure difference between any two points on the bottom when compared to the pressure difference between top and bottom is negligable. Because of this the pressure diff at force application points can be taken as constant, ie: the variation is negligable (at least enough of the time to make a convincing if not completely physicaly accurate simulation).

Out of interest, since we're talking about simulations for real time applications, just how practical is it to properly compute the surface integral of force for every submerged triangle of the mesh, every frame for every submerged object? I'm not splitting hairs.....I honestly don't know.

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Dmytry    1151
if you gonna to make any physics, you anyway will have to do many things on per-triangle basis, and will have to have low-poly models for physics anyway. So probably not that costly. Integral for every triangle is analitical, you need to check every triangle of mesh, and pseudocode:

Vector3 sum(0,0,0);
Float distlaced_mass;
for every triangle{
if triangle is completely underwater, compute effect of volume of water enclosed by this triangle and it's projection onto water surface, kinda prism-like. That is, find mass of water in that volume and add to displaced_mass, find center of mass of that volume, add coordinate of center of mass multiplied by mass to the sum
else
if triangle is partially underwater
cut triangle with water plane leaving unly underwater part, do the same as above with that cut surface.
}
center_of_mass_of_underwater_volume=sum*(1/total_mass)

Note: For clockwise or counterclockwise looking-from-top triangles, volume in calculations must be negative. That way, it should work even with concave mesh.

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arithma    226
Take this example and you will immediatly know where I want to take you regarding the resultant point of application (used for converting forces in angular uses):

A cylinder is immersed perpendicularly with respect to the surface of water. Since the forces on the (curved) sides of the cylinder cancel each other out, the resultant force is exclusively the pressure applied to the base surface. This in other words leads us to the conclusion that the point of application is also in that plane (containing the button surface). In conclusion, we cannot take the center of mass of displaced water as the point of application.

Any way I would like to thank Dmytry for his explanation on volumes and such (very insightful).
And thank Motorhep for doing mistakes so that Dmytry can point them out and warn us about them :)
Please Continue to help

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Dmytry    1151
Quote:
Original post by arithma
Take this example and you will immediatly know where I want to take you regarding the resultant point of application (used for converting forces in angular uses):

A cylinder is immersed perpendicularly with respect to the surface of water. Since the forces on the (curved) sides of the cylinder cancel each other out, the resultant force is exclusively the pressure applied to the base surface. This in other words leads us to the conclusion that the point of application is also in that plane (containing the button surface). In conclusion, we cannot take the center of mass of displaced water as the point of application.

Not really. All points on the direction of force (on the line that pass "application point" and is directed in direction of force) is equally valid as points of application (if body is rigid (that is, if you are not interested in deformation/etc.)). It seems somewhat counter-intuitive at first, but it is so. There's actually a line of application that pass some point, and we can use this point as point of application. But effect of forces placed anywhere on line of application is equal. In this case, force is direted upward and center of top surface finally gives same result as center of cylinder or center of bottom surface or center of immersed volume, because them all is on same line that contains force vector.

Torque is equal to cross product of point of application and force. If point is P and force is F, torque around zero point is F x P (or P x F, matter of convention) Note that
F x (P+a*F) = F x P + F x a*F = F x P
That is, addition of scaled vector F to P doesn't change torque.

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arithma    226
But it is just that, I know that the point of application is down there (different from where the center of mass is) but because it is displaced along gravity it has no rotational effect. But if you take another example (ie a tilted bowling ball where the more volumous part is downward) we observe rotational effects. It is like such since the point of application (which is in conclusion not the CM itself) causes that rotation (in conjuntion with the resultant force of course).

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Guest Anonymous Poster   
Guest Anonymous Poster
Buoyancy force is not the result of pressure difference it is the result of weight difference.
A body immense in fluid experiences the same upward buoyancy force at 1 meter depth than at 1000 meter depth

I think there is confusion with the buoyancy force and the fluid pressure, Archimedes principle says that the buoyancy force is equal to the weight of the displaced fluid, point in the direction opposite to the gravity, and is applied at the center of gravity of the displaced fluid.

So for bodies fully immerse the center is the same (assuming equal density) and the force is just the weight difference. For bodies partially immerse gets more complicated because the volume of the immerse body have to be integrated, but you can do it by populating the volume with pointe evenly spaced and just counting how many are under the surface and determining the volume and the center for the submerged ones.

Or you can just integrate the face areas using the Green, and divergence theorem by applying the flow traveling along the gravity vector. But this is more complex.

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Dmytry    1151
Quote:

Buoyancy force is not the result of pressure difference

it's exactly result of pressure difference. If you have 1x1x1 cube aligned top side to top, you can easily see that buoyancy force is equal to pressure difference between top side and bottom side multiplied by area. Same for any other shape. (And, indeed, pressure difference * area is shown to be equal to weight of water in that volume). In incompressible fluid in constant gravitation field, indeed, pressure difference is same at depth 1 meter as at depth 1000 meters.

Heck, if you have a bottle with water, increased weight of bottle is a result of pressure difference :-) and pressure difference is result of weight of water.

Quote:

But it is just that, I know that the point of application is down there (different from where the center of mass is) but because it is displaced along gravity it has no rotational effect. But if you take another example (ie a tilted bowling ball where the more volumous part is downward) we observe rotational effects. It is like such since the point of application (which is in conclusion not the CM itself) causes that rotation (in conjuntion with the resultant force of course).

i'm unsure what you mean, but you can easily see that equal forces applied along line parallel to force produce exactly same torque. In case of ball staying on surface, for any orientation, line of application of surface force passes ball center. You can use ball center in calculations as point of application of force, instead of point of contact with surface, and will get exactly same torque in result.

(again, assuming objects we talk about is perfectly rigid)

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Guest Anonymous Poster   
Guest Anonymous Poster
Actually it is not result of a pressure difference. It is a result of weight difference.
If the force is the same a regardless of the depth then it means it is independent of the depth. Furthermore more in any demonstration of the principle the concept of pressure in not required at all what is used is the weigh difference of a column of water at depth h and a column of water of depth h1.

Why making more confusing when there is a principle that says how to do it?
Just calculate the volume by whatever mean you can, determine how much this water volume weights and the different between that weight and the solid weight it the buoyancy force. The center of application for the force is the geometry center of the submerge volume (which means that for fully submerged volumes the torque is zero since the center coincides with the rigid body), and the direction of the force a vector parallel to the gravity. The problem is completely well defined.

I do not now what the incompressibility, gravity field, objects are perfectly rigid, etc has to do with it since it was not stated as part of the problems.
I can easily see engineers never making anything if all these things were in the way

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Guest Anonymous Poster   
Guest Anonymous Poster
Your are right it is teh rsult of presure difference?
But presure diffrence is not part of the calculation in any way.


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Dmytry    1151
seems you don't understand what is pressure and what is difference.

Put 1mx1mx1m cube into water so top is at depth 1 m. You have pressure on top = 9.8 kN/M2 and pressure on bottom = 18.6 kN/M2 . Pressure difference is 9.8 kN/M2 and buoyancy is 9.8 kN
Put that cube at some other depth where pressure is A (say, at depth 1000 m where A=9.8*106 N/M2). On top you'll have A and on bottom you'll have A+9.8 kN/M2, and difference is again 9.8 kN/M2 and buoyancy is again 9.8 kN. Understand now? If pressure would be same on top and bottom, there would be no force.

Basically, buoyancy force (and in general, any force caused by liquid) is equal to integral of pressure*dA , (where "dA" stands for vector that is normal to surface and have length equal to area), and it could be nonzero only if on different parts of object you have different pressures. Also, that integral of pressure*dA is equal to [minus] weight of same water in volume of object, it can be proven in numerous ways(e.g. classically by observing that if there would be water, forces would be compensated, or using calculus)

(Speaking of gravity, and compressibility, in real water, there will be bit bigger buoyancy (for same volume of object) at 10 000 m depth than on surface, because water is densier and gravity is stronger. But it of course doesn't matter. Still, it is better to say that explicitly)

[Edited by - Dmytry on March 30, 2005 3:41:56 PM]

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Guest Anonymous Poster   
Guest Anonymous Poster
Oh I do, but it seems you do not have common sense. I said yes to keep you happy and get over with.

MotorHerp answer completely covered the question, if it any clarification was needed; it was perhaps how to go about implementation details for calculating the buoyancy center.

A simple explanation of how to implement the Archimedes principle is more than sufficient. However you brought gratuitous confusion by adding the variables pressure, gravity, compressibility, deformable bodies, turbulence, and how knows what else (why not gravity variation due to elevation, oh wait you did bring it up in the end, what next quamtum variations).

The problem with you is that more often than not you go completely off the tangent answering the question that was not asked. What you like is to grandstand displaying how much you have know about math and physics and you have the tendency to think you invented these things.

Now since you brought it up saying I do not underneath what pressure is, I am going to say I think you are right. In my time pressure was not a quantity it was a convenient scalar function of force that just expresses force per unit area. To have pressure you need weight and in fact you can completely disregard the concept of pressure when calculating the buoyancy force. All it is needed to know is the weight difference between the fluid and the body.

It is very simple if at depth H1 the force, and buoyancy center is the same than at depth H2, However the pressure at H1 is very different than at H2 then I do not know how the pressure influence the calculation. In the old days when we run experiments with such outcome we use to conclude something like “Buoyancy force is independent of the surface pressure”

Even if the force per unit area is different at the body top than it is a the box bottom, the fact is that the bouyance force is same regarless of the fuild depth.

If we did the same experiment using two fluids with different density and we found that at depth H1 the buoyancy force in fluid A is different than the Buoyancy force at same depth H1 in fluid B, then we would conclude “buoyancy force is a function of the fluid density, (which is relate to the fluid specific weight)”

But that was in the old days maybe the science is different now.
Or perhaps you can show the variable pressure is applied in the Archimedes principle.

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Dmytry    1151
I even gave example HOW at depth of 1000M you have same buoyancy as on depth 1M, as you have same pressure difference. As buoyancy is a direct result of *difference* of pressure (to stay simple, difference between top and bottom of object, that's is!), as i have said really many times. And no, addition of constant pressure does not change difference (you don't understand that a-b = (a+c) - (b+c) ?) And, as I also said many times, buoyancy can be computed as weight of liquid in volume of object. My point is that you can not assume that pressure on surface is equal when computing buoyancy by integrating pressure over surface to find "application point". If you don't understand what does word "difference" mean.... well.

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Dmytry    1151
and, read that , BTW.

(note that it's the case when you can't assume that point of application of force is in center of submerged volme, as net force is not vertical.)

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Eelco    301
AP:

please shut up, because you clearly have little knowledge of what youre talking about. it annoys me like hell.

buoyancy is nothing but the net result of pressures acting over the suface of a body, pressures which happen to be equal to the amount of mass leaning on that portion of the fluid, thus linearly increasing with depth. archimedes says the same thing (easy to derive from the above), but less. archimedes doesnt tell you anything about point of application of the force, for instance. nor does it give you a clear insight as to HOW to compute the correct buoyancy for a given situation, such a partially submerged objects, whereas integration over the surface does.

also, dmytrys explaination of why it is valid to apply the force of buoyancy at the CM of the displaced water is correct: just look at any volume of water: the bouancy force on it doesnt cause any rotation there either.

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Guest Anonymous Poster   
Guest Anonymous Poster
Dude this was the question

“)... In the act of calculating that moment I need to know the resultant POINT OF APPLICATION of the boyant forces.
IS IT A SAFE ASSUMPTION TO MAKE THE CENTER OF GRAVITY OF THE "IMMERSED VOLUME" AS THE POINT OF APPLICATION?

PLEASE HELP”


This is part of Motrohelp answer:

“Out of interest, since we're talking about simulations for real time applications, just how practical is it to properly compute the surface integral of force for every submerged triangle of the mesh, every frame for every submerged object? I'm not splitting hairs.....I honestly don't know.”

Sound like he was concerned with the practicality of the solution.

This is part of my practical answer:

“Archimedes principle says that the buoyancy force is equal to the weight of the displaced fluid, point in the direction opposite to the gravity, and is applied at the center of gravity of the displaced fluid.

So for bodies fully immerse the center is the same (assuming equal density) and the force is just the weight difference. For bodies partially immerse gets more complicated because the volume of the immerse body have to be integrated, but you can do it by populating the volume with points evenly spaced and just counting how many are under the surface and determining the volume and the center for the submerged ones.

…..”


However you decided on your own as usual that everyone here is ignorant and therefore you decided to lecture us in how to demonstrate the Archimedes principle but you omitted to mention the origin of the idea. You jump all over the place mentioning bug words just to make sure that every one know the you really know a lot. But you forgot to answer the question.

Here are some clues that the buoyancy is a force:

-The name Buoyancy Force, it say “Force”
-It has a direction, it point opposite to the gravity.
-The definition say the weight of the displaced fluid (not the pressure)

Here is a reflection for you, say you have two arbitrary point at difference height under water, I believe that even you will agree that there will be a difference of pressure between the to points. However the water does not jump around just because there is differential of water pressure. Now add a body and there will be a force parallel to the gravity. Some how this force is a function of how much the body weighs and how big the body is. But if you move the body up and down the force does not seems to change, so the fact that there is a difference of water pressure can be used to demonstrate that there will be force acting on the body and nothing else, it does no help on anything at all in how to calculate that force or the point of buoyancy.

What you two do not seems to understand is the question asked for and application of the principle not a demonstration of the principle.

Here is the work of some people that have done some works in that area.
http://www.cambrianlabs.com/Mattias/DelphiODE/BuoyancyParticles.asp

If you read the text you will notice that even thought they start with the premise of pressure, during the implementation the pressure is just use to explain why there is a force, but only the volume of the shape is used to calculate the force and the point of force.

I am going to go in a ledge here but I think that when Arithma asked the question he already new about the composed of buoyancy and the Archimedes principle he was asking how to calculate the force in a practical way, not for a reconfirmation of the principle.

As for Elco until you invent you own principle. Maybe you should search on the interned for Center of buoyancy. It is a concept been used for hundreds of years by ship makers and I think they all agree it was Archimedes how first explained, but perhaps you can come up with a better explanation.
Of course Archimedes principles states very clear how to calculate the force for bodies practically immensely in a fluid and how to calculate the center of the application of the force. It says the weight of the displaced fluid by the immersed volume.

I am going to tell you a secret here. All forces have a center of application and all volumes have centers, since the fluid have a uniform density, simple volume integration techniques is all that is required to solve the problem.

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Guest Anonymous Poster   
Guest Anonymous Poster
Quote:
Original post by Dmytry
I even gave example HOW at depth of 1000M you have same buoyancy as on depth 1M, as you have same pressure difference. As buoyancy is a direct result of *difference* of pressure (to stay simple, difference between top and bottom of object, that's is!), as i have said really many times. And no, addition of constant pressure does not change difference (you don't understand that a-b = (a+c) - (b+c) ?) And, as I also said many times, buoyancy can be computed as weight of liquid in volume of object. My point is that you can not assume that pressure on surface is equal when computing buoyancy by integrating pressure over surface to find "application point". If you don't understand what does word "difference" mean.... well.


Oh but I do understand, the problem is that, that was not the question. I prefer to have a solution even if it is approximated than not solution at all. You on the other hand seem to get conform in having a reason why the problem is not solvable and answer the quetion with more complex problems.

Of course it is completely 100% possible to calculate the center of a volume of water with a simple integral. The reason for that is that water and the majority of fluids are incompressible (for all practical purpose), therefore the density does not change with the depth.

Now let us say the fluid is compressible it is still possible to calculate the center of the volume with a simple volume integral. All it is required is and expression for the fluid density and used it with the divergence theorem so in any all cases the solution of this problem is not related to fluid pressure. And I do not think that the solution asked was for compressible fluids.
So it seem that it is you onfusing the question and is given and answer to a different problem or ratther not given any answer at all.

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Dmytry    1151
well. You kind of finally pissed me off. Read this thread, damnit. First, this surface integral is necessary to compute submerged volume, damnit.

Quote:

Motorherp
Quote:

Original post by arithma
IS IT A SAFE ASSUMPTION TO MAKE THE CENTER OF GRAVITY OF THE "IMMERSED VOLUME" AS THE POINT OF APPLICATION?

Unfortunately not really. The boyancy force is directly proportional to the volume of water displaced by the object. In reality the force is evenly distributed across the objects surface and is at any one point proportional to the volume displaced by the object directly above that point. To solve this exactly would require integrals, however this can be approximated by solving for evenly distributed points along the objects submerged surface. For each point cast a ray upward to the water surface to get the length of displacement and treat the segment as a cuboid with width and depth equal the spacing between points and height equal to the length of the ray above to give you the volume. You can then calculate the buyoncy force to apply at each point from here which should lead to the rotational behaviour you'd expect.

if you do not understand that this answer essentially says that arithma's idea will not work and explains how to approximately compute integral(actually, what he describles is numerical integration) (and compute it incorrectly, BTW), you need to replace your head.

Following
Quote:

Dmytry
[b]In summary:
1: Sum of forces applied acts exactly as one force applied to center of mass of displaced volume of water. (well, almost exactly, if you gonna to take curvature of Earth and non-linearity of gravitation field into account, you'll get some extremely small difference) {note: by "difference" there i mean difference between what i suggest and accurate result}
2: Motorherp's idea about disregarding pressure difference is completely incorrect.

Now re-read arithma's post.

(Part in brakets is in there just for case somebody will start saying that accurate integral taking everything into account is better.)

Other than that
Quote:

Dmytry
And if you properly compute surface integral of force (for every submerged triangle of mesh), it will be the same as computing submerged volume and center of mass of that volume. It is relatively easy.


and this:
Quote:

Dmytry
Vector3 sum(0,0,0);
Float displaced_mass;
for every triangle{
if triangle is completely underwater, compute effect of volume of water enclosed by this triangle and it's projection onto water surface, kinda prism-like. That is, find mass of water in that volume and add to displaced_mass, find center of mass of that volume, add coordinate of center of mass multiplied by mass to the sum
else
if triangle is partially underwater
cut triangle with water plane leaving only underwater part, do the same as above with that cut surface.
}
center_of_mass_of_underwater_volume=sum*(1/total_mass)

Note: For clockwise or counterclockwise looking-from-top triangles, volume in calculations must be negative. That way, it should work even with concave mesh.

This integral essentially computes submerged volume and center of mass. It is clear enough from comments. I wasn't getting into details because as i understand from arithma's post, [s]he already know how to compute volume and find center of mass.(2arithma: if don't, ask me, I can provide more detailed pseudocode)
If you don't know that to "just compute volume of arbitrary mesh" you NEED to essentially compute surface integral on every triangle and add it (and that's using divergence theorem to find volume of object from surface integral), your knowledge about subject is essentially 0.
btw, clickster

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Eelco    301
Quote:
Original post by Anonymous Poster
Quote:
Original post by Dmytry
I even gave example HOW at depth of 1000M you have same buoyancy as on depth 1M, as you have same pressure difference. As buoyancy is a direct result of *difference* of pressure (to stay simple, difference between top and bottom of object, that's is!), as i have said really many times. And no, addition of constant pressure does not change difference (you don't understand that a-b = (a+c) - (b+c) ?) And, as I also said many times, buoyancy can be computed as weight of liquid in volume of object. My point is that you can not assume that pressure on surface is equal when computing buoyancy by integrating pressure over surface to find "application point". If you don't understand what does word "difference" mean.... well.


Oh but I do understand, the problem is that, that was not the question. I prefer to have a solution even if it is approximated than not solution at all. You on the other hand seem to get conform in having a reason why the problem is not solvable and answer the quetion with more complex problems.

Of course it is completely 100% possible to calculate the center of a volume of water with a simple integral. The reason for that is that water and the majority of fluids are incompressible (for all practical purpose), therefore the density does not change with the depth.

Now let us say the fluid is compressible it is still possible to calculate the center of the volume with a simple volume integral. All it is required is and expression for the fluid density and used it with the divergence theorem so in any all cases the solution of this problem is not related to fluid pressure. And I do not think that the solution asked was for compressible fluids.
So it seem that it is you onfusing the question and is given and answer to a different problem or ratther not given any answer at all.

it doesnt have jack to do with compressibility.

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