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baskuenen

Metaballs - Normals

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bas: what i meant is this.. if you remember your calculus, the first derivative of a function defines the vector tangent to the function at a given point. in other words, if a surface is defined as A(s,t) then the partial derivatives of A, A''s and A''t define the plane tangent to the surface at (s,t). taking the cross product of the partial derivatives results in the normal vector at (s,t), i.e. nA(s,t) = A''s x A''t. this is pretty standard in parametric surfaces such as bezier surfaces, NURBS, etc.



--
Float like a butterfly, bite like a crocodile.

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I'm not sure if any real metaball-angle-calculation-formula has been here yet.

I once made a program that calculates the real angle in a metaball system for single point. It's only 2d but it's very easy to implement to 3d also.

And this is the only _real_ way to calculate vertex normals (the original question).



If you're intrested in how it works, check out my mighty qbasic (!!) demonstratin program (& source of course)

http://www.kolumbus.fi/hannes.k/TESMI.BAS

I don't want to explain it here but it's quite basic vector calculus though. And the program is quite old so don't ask if it looks odd

-Hans


Edited by - Hans on May 25, 2000 11:13:16 AM

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Hi Hans!

How much computations does it require to compute the normal the way you do it? I could figure it out myself but if you already know the answer...

The way I do it above requires about four (educated guess) crossproducts per normal. Is it faster to do it your way?

After a quick glance at the source for your program it seems that you compute the normals like this:

V : The vertex who's normal is to be computed
C : The center of a metaball
Sq = SquareMagnitude(V-C)

N = (V-C0)/Sq0 + (V-C1)/Sq1 + (V-C2)/Sq2 + ...

That seems to be logical. Baskuenen you should probably try this. If it works it will improve your performance alot


- WitchLord

Edited by - WitchLord on May 25, 2000 12:26:51 PM

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Of course it works

But

N = (V-C0)/Sq0 + (V-C1)/Sq1 + (V-C2)/Sq2 + ...

needs to be

M = magnitude(V-C)
N = (V-C0)/M0^3 + (V-C1)/M1^3 + (V-C2)/M2^3 + ...

I don't know why it must be power 3 but I was trying different powers and noticed that power 3 is the correct one (whoops the BAS code has ^2.5.. well it should be ^3). So I haven't proven the formula mathematically, but if someone can, I'll be happy to hear.

-Hans


Edited by - Hans on May 25, 2000 1:17:34 PM

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Well, the first power is to normalize the vector to unit length. The two other powers is to compensate for the metaballs'' dropoff in energy with distance. As I have never made any metaballs I assumed that the dropoff was linear but it might be better if it is proportional to the square distance.

- WitchLord

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This is really starting to get interresting. I'll look up your source in a few minutes Hans, it already seems promising.
Here's a few lines I use to calculate the force on my balls (the meta ones):


ForceGrid[ index ] =
MB_SIZE*MB_SIZE/((x-bx1)*(x-bx1) + (y+by1)*(y+by1) + (z-bz1)*(z-bz1)) +
MB_SIZE*MB_SIZE/((x-bx2)*(x-bx2) + (y+by2)*(y+by2) + (z-bz2)*(z-bz2)) +
MB_SIZE*MB_SIZE/((x-bx3)*(x-bx3) + (y+by3)*(y+by3) + (z-bz3)*(z-bz3)) +
MB_SIZE*MB_SIZE/((x-bx4)*(x-bx4) + (y+by4)*(y+by4) + (z-bz4)*(z-bz4));


Edited by - baskuenen on May 25, 2000 3:03:17 PM

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So all you''re balls are the same size?
If you want different sized balls, you must do this:

C = center of ball
S = size of ball
V = The vertex who''s normal is to be computed
M = magnitude(V-C)

N = (V-C0)*S0/M0^3 + (V-C1)*S1/M1^3 + (V-C2)*S2/M2^3 + ...

And I calculated my metaballs like:

sqr((x-bx1)*(x-bx1) + (y+by1)*(y+by1) + (z-bz1)*(z-bz1))

but you have only:

(x-bx1)*(x-bx1) + (y+by1)*(y+by1) + (z-bz1)*(z-bz1)

So you''ll need different power that ^3.. I''m not sure what but ^2 felt good when I tested it
Wild quess: maybe now ^1 normalizes the vector and the other ^1 is like (1/2)*2. In the old formula ^1 was normalize and ^2 was (1)*2..
I originally came up with the algo before I even knew what vector is, and I just tested things...

--
MB_SIZE*MB_SIZE
--
But why do you square your balls'' sizes? If you''re going to use that formula above, you shouldn''t sqare the sizes or else you''ll need square root in the formula:

N = (V-C0)*sqr(S0)/M0^2 + (V-C1)*sqr(S1)/M1^2 + (V-C2)*sqr(S2)/M2^2 + ...


-Hans

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I''m still looking through the overload of recent info.
Meanwhile I noticed the folowing:

Can''t Sqrt(x)^3 be described as x^2 ?

So after some radical algo optimizations:
N = 1/(V-C0) + 1/(V-C1) + 1/(V-C2) + ...

Am I getting this right?

If you want to use different sizes you get:
N = MB_SIZE/(V-C0) + MB_SIZE/(V-C1) + MB_SIZE/(V-C2) + ...


Please correct me if I''m wrong...

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>Can''t Sqrt(x)^3 be described as x^2 ?

No, sqrt(x)=x^(1/2)
And so x^(1/2)^3=x^(3/2)=x^1.5, not x^2
Sqrt(x)^4=x^2.

But since you should use:

N = (V-C0)*S0/M0^2 + (V-C1)*S1/M1^2 + (V-C2)*S2/M2^2 + ...

And M = sqr((x-x0)^2+(y-y0)^2)
M^2=(x-x0)^2+(y-y0)^2 , which is quite fast to calculate and you don''t need sqrt() at all.

(and we''re back at the form that WitchLord suggested earlier)

Whether it is M^2 or M^3 depends on how you calculate the metaballs. (but you, baskuenen, should use M^2 because you don''t use sqr to calculate metaballs'' forces, read my earlier posts)

-Hans

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Me = little bit stupid sometimes. You''re right! Optimized my amount of braincells a bit too much...
Good thing there are some sober people around!

But we optimized it!

Optimization is real important in my case, because I''m writing a software rendered runtime 3D animation of the metaballs. It''s still too slow (~20 FPS on PIII-450).

But you made a good point with metaball normal calculation.
- There is a better way -
(Although the "smooth surface normal calculation" topic was getting real interresting!)

I''m still implementing...hope it''ll work...

Thanks Hans and all of you guys!

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