4 ms·

This already does look like it has falloff... mutual attractors clearly seem more attracted to nearby particles than to larger groups further away.

wetmore 3y ago
I assume it falls off at inverse of r squared, which is why the person you're replying to is mentioning other functions besides that one.
- noduerme 3y ago
  I read them differently. >> I have a feeling that such a system would grind to a halt even with clever optimizations. I took this to mean that they thought there was no falloff calculated at all. If there is, I don't see why substituting a different function, e.g. cube vs square, would be significantly more CPU-intensive.
  - at_a_remove 3y ago
    You focused on just one clause. Consider the whole: 1) Three dimensions, not two. Therefore distance is not the square root of (delta-x * delta-x + delta-y * delta-y) but the cube root of (delta-x * delta-x + delta-y * delta-y + delta-z * delta-z). More operations. 2) Three dimensions, not two. As you start increasing the number of dimensions, the simulation feels more and more empty. One hundred particles on a line a thousand units long is crowded. One hundred particles on a grid of one thousand by one thousand feels like more "room" for that same number of particles. In a volume of a thousand by a thousand by a thousand, one hundred particles feels too few, and one would naturally increase the number of particles. Naively, which is to say without optimizations, the number of force computed grow as N-squared. More operations. 3) Cube falloff goes by r * r * r, rather than r * r. More operations, by fifty percent. And I did suggest some more exotic functions in there which certainly could be more daunting.
    - mattashii 3y ago
      > 1) Three dimensions, not two. Therefore distance is not the square root of (...) but the cube root of (...). No, it is still a square root. The term under the root is correct, but distance in N dimensions (assuming euclidian space) is just sqrt(sum(delta-n ^ 2))
      - at_a_remove 3y ago
        Hah, you're right. Serves me for typing late at night.

wetmore 3y ago

I assume it falls off at inverse of r squared, which is why the person you're replying to is mentioning other functions besides that one.

noduerme 3y ago
I read them differently. >> I have a feeling that such a system would grind to a halt even with clever optimizations. I took this to mean that they thought there was no falloff calculated at all. If there is, I don't see why substituting a different function, e.g. cube vs square, would be significantly more CPU-intensive.
- at_a_remove 3y ago
  You focused on just one clause. Consider the whole: 1) Three dimensions, not two. Therefore distance is not the square root of (delta-x * delta-x + delta-y * delta-y) but the cube root of (delta-x * delta-x + delta-y * delta-y + delta-z * delta-z). More operations. 2) Three dimensions, not two. As you start increasing the number of dimensions, the simulation feels more and more empty. One hundred particles on a line a thousand units long is crowded. One hundred particles on a grid of one thousand by one thousand feels like more "room" for that same number of particles. In a volume of a thousand by a thousand by a thousand, one hundred particles feels too few, and one would naturally increase the number of particles. Naively, which is to say without optimizations, the number of force computed grow as N-squared. More operations. 3) Cube falloff goes by r * r * r, rather than r * r. More operations, by fifty percent. And I did suggest some more exotic functions in there which certainly could be more daunting.
  - mattashii 3y ago
    > 1) Three dimensions, not two. Therefore distance is not the square root of (...) but the cube root of (...). No, it is still a square root. The term under the root is correct, but distance in N dimensions (assuming euclidian space) is just sqrt(sum(delta-n ^ 2))
    - at_a_remove 3y ago
      Hah, you're right. Serves me for typing late at night.