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.

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.

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.

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
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, he already know how to compute volume and find center of mass.(2arithma: if don't, ask me, I can provide more detailed pseudocode) <br>If you don't know that to "just compute volume of arbitrary mesh" you NEED to essentially compute surface integral &#111;n 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.<br>btw, <a href="http://www.crd.ge.com/~lorensen/munc/munc.html">clickster</a>

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.

Quote:Original post by Anonymous Poster
...stuff...

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.

yes, you can deduce such things from archimedes. but can your computer? any technique for actually calculating these forces and their point of application, are most easily done by integrating pressure over surface, except in the special case of completely submerged objects, for which you can precalculate buoyancy.

Quote:
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.

volume integration eh? indeed possible, but last time i checked 99.99% of all applications were working with surface models. seems more practical to work with those, no?

the surface pressure phenomena is what gives rise to buoyancy. indeed with the divergence theory you can reformulate it in the less inituative archimedes' way. but it doesnt make calculations easier, only harder. how would you go about actually calculating the desired quantities using a volume integral for a partially submerged object? seems quite complex to me. its all equally easy when doing a surface integral.

also, if you do this buoyancy calculation on a low-poly model, and exploit coherance, it doesnt have to be costly at all. were not 'over complicating' things. were giving a complete, correct and useable answer.

Quote:Original post by arithma
I see why Dmytry got pissed off. It is as if the anonymous poster didn't understand what am going for neither for what dmytry was..
Now, thanks to Dmytry, I know how that i need to find volume using an algorithm on each traingle in the mesh (brilliant indeed). However one more ambigouty still arises.
I will not argue whether or not a moment will arise (i will leave it to the comuter numerically to decide that).

What I actually need is the:
integration of (the cross product of the force F and the position of the application of that force on each triangle)
This will help me (alot!) in finding the total moment resulting from the submerged mesh (whether the volume was totally submerged or not)

Thanks again
Actually am putting a rate increase for Dmytry
And am a He

I actually could prove that integral of cross products of force with position will be equal to cross product of total buoyancy force and position of center of mass of submerged volume.
Simple proof: Let submerged part of object is replaced with "solid water" (Archimedes-style). Net force and torque is zero. Note that for it to be zero, torque caused by mass of "solid water" mush be exactly opposite. And this torque is equal to cross product of weight force(vertical, directed down) and center of mass of submerged volume. And it perfectly compensates buoyancy.
(More complex proof involves calculus and is lenghty to write.)
And that integral of cross product is not easier to accurately compute than submerged volume and center of mass of submerged volume. Actually both computations is kind of equivalent. It is like Integral(const x A) = const x Integral(A) (x is cross product).

(btw, speaking of volume integrals, them have n^3 complexity and surface integral have n^2, that's why volume integral is replaced with surface one. In this case, with polygonal mesh, surface integral can be computed precisely per-triangle)

for example of using surface integral, volume of closed triangle mesh = (1/6)* [sum for every triangle ABC]( ((B-A) x (C-A)).x * (A.x+B.x+C.x))
(where instead of x you can have y or z, doesn't matter. Note that it is essentially sum of volumes of that kinda prisms i told about) Mesh needs to have consistant winding order. Resulting colume could be negative or positive depending if it's clockwise or counterclockwise...

Center of mass of mesh (assuming mesh is filled with constant density thing) can be approximated as
(1/12)* [sum for every triangle ABC]( ((B-A) x (C-A)) per_component_vector_product (A+B+C)) /total_mass
where per_component_vector_product stands for operation like
C.x=A.x*B.x
C.y=A.y*B.y
C.z=A.z*B.z
(and maybe it's not an approximation but perfect result...)
You essentially need to compute that for submerged portion of volume...

[Edited by - Dmytry on April 1, 2005 5:37:59 AM]

Quote:Original post by arithma
Your proof for zero torque at first sight seems alot logical
Yet if you replace torque for acceleration, it is completly analogous and has the same peculiar logic.
But knowing that acceleration is not zero (as ships may protest) it shows your proof is false for torque being zero
NOTE: If you have lost interest I can understand that

seems I don't really understand what you mean regarding acceleration. In hydrostatic case, torque caused by buoyancy is perfectly compensated by torque caused by "solid water" weight. If object is moving or accelerating in the water, buoyancy models and stuff is pretty much invalid anyway, and you need to consider complete hydrodinamics model.

Some time ago there was a big thread about effect of water on moving and accelerating objects, i even wrote small demo with box falling into water (it simulates fast deceleration at impact and other stuff).
executable 1
executable 2
edit: unsure about locations
edit: fixed links
edit: requirs SDL dll to be placed in same folder to work.
edit: and i'm getting some weird bugs with newer SDL ....