# Metaballs - Normals

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#21
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Posted 24 May 2000 - 09:37 AM

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Float like a butterfly, bite like a crocodile.

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#22
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Posted 25 May 2000 - 04:05 AM

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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#23
Moderators - Reputation: **4409**

Posted 25 May 2000 - 05:18 AM

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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#24
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Posted 25 May 2000 - 06:13 AM

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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#25
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Posted 25 May 2000 - 07:19 AM

- WitchLord

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#26
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Posted 25 May 2000 - 08:01 AM

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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#27
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Posted 25 May 2000 - 08:54 AM

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...

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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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#28
Members - Reputation: **123**

Posted 25 May 2000 - 10:05 AM

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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#29
Members - Reputation: **122**

Posted 25 May 2000 - 10:28 AM

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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#30
Members - Reputation: **123**

Posted 25 May 2000 - 10:42 AM

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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#31
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Posted 25 May 2000 - 11:50 AM

I was inspired by this discussion so I sat down to make my own Metaballs, 2 hours later it was working (I stripped an old program I had for marching cubes). After some tweaking with the numbers I actually managed to get them to look good as well. As of yet the code is totally unoptimized though so the performance is truly poor: 8.7 fps on a Celeron 500 with GeForce. (Impressive isn''t it )

- WitchLord

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#32
Members - Reputation: **123**

Posted 25 May 2000 - 02:22 PM

We are not done yet! At the moment my surface is smooth, and the phong is shining brightly, but it''s still not right!

This thread is finished when I say so

and only when an image of perfect metaballs is posted!

...

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#33
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Posted 25 May 2000 - 04:09 PM

Now that I have gotten my metaballs to run at a steady rate of 105fps with perfect smoothness (at least as perfect as it gets with low detail) I'm willing to give you a hand again.

You might wonder how I managed to optimize from 8.7 fps to 105 fps. Well the biggest change I did was that I no longer compute the field energy if there are no balls in close proximity. This got the fps up to a steady 32 fps (367% increase). The second optimization I made was that I switched to release build, and voila 105 fps (328% increase).

Here's a screenshot of my metaballs:

(Image no longer available)

For the sake of this screenshot I increased the density of the energy field to 64*64*64 from 32*32*32 (lowered fps to 14)

- WitchLord

Edited by - WitchLord on May 25, 2000 11:25:48 PM

Edited by - WitchLord on May 26, 2000 11:45:35 PM

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#40
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Posted 26 May 2000 - 06:25 AM

You may also check the source for my metaballs, it''s really well commented in my own humble opinion.

- WitchLord