[Mathias.]

Here it is, the vector question of the day!

I got a position for example A (-1231.0, 123.0, 1337.0) and I want to make A move toward an other point B (-1231.0, 321.0, 7331.0) of X units.

so the question of the day is... well... you know!

Thx

[11530]

If you're wanting X to be always between 0 and 1: F(X) = A + X(B-A)

For example, in three steps. 0, 0.5, 1:
F(0) = A
F(0.5) = A/2 + B/2
F(1) = B

[Mathias.]

But I need to make it move toward B for only X units...

[KissLick]

For 2D (just modify for 3D..):

1) make vector v with just right length v(0,10) = length is 10 (points are 0,0 and 0,10)
2) rotate vector v (makes vector v2) (viz google: https://www.google.cz/search?=rotate...+around+origin)
3) move vector v2 to point A (or add vector v2 to vector origin -> A)
4) you have point A2 what is what you want

EDIT: vector rotation: http://www.siggraph.org/education/ma...ran/2drota.htm

[11530]

Assuming you mean the mathematical definition of "unit", that is a distance of 1, then you can do the following where A = (a₁, a₂, a₃), B = (b₁, b₂, b₃) and C = (c₁, c₂, c₃) is partway along the line.


c₁ = a₁ + X(b₁ - a₁)/(√((b₁ - a₁)² + (b₂ - a₂)² + (b₃ - a₃)²))
c₂ = a₂ + X(b₂ - a₂)/(√((b₁ - a₁)² + (b₂ - a₂)² + (b₃ - a₃)²))
c₃ = a₃ + X(b₃ - a₃)/(√((b₁ - a₁)² + (b₂ - a₂)² + (b₃ - a₃)²))


Just used my first answer, but now used a coefficient which has been divided by the length of A to B (the Pythagoras part of the equation) in order to get unit distances.

[Mathias.]

Originally Posted by 11530 Assuming you mean the mathematical definition of "unit", that is a distance of 1, then you can do the following where A = (a₁, a₂, a₃), B = (b₁, b₂, b₃). F(X) = (a₁, a₂, a₃) + X(b₁ - a₁ + b₂ - a₂ + b₃ - a₃)/(√((b₁ - a₁)² + (b₂ - a₂)² + (b₃ - a₃)²)) I'll clean that up a bit. Final point on the line between A and B is C: (c₁, c₂, c₃) c₁ = a₁ + X(b₁ - a₁)/(√((b₁ - a₁)² + (b₂ - a₂)² + (b₃ - a₃)²)) c₂ = a₂ + X(b₂ - a₂)/(√((b₁ - a₁)² + (b₂ - a₂)² + (b₃ - a₃)²)) c₁ = a₃ + X(b₃ - a₃)/(√((b₁ - a₁)² + (b₂ - a₂)² + (b₃ - a₃)²)) Just used my first answer, but now used a coefficient which has been divided by the length of A to B (the Pythagoras part of the equation) in order to get unit distances.

Thanks, that what I needed, will test it out.

[blodia]

use MakeVectorFromPoints on A and B, NormalizeVector the result then ScaleVector by X and add it to A

[Peace-Maker]

Let the code speak..

PHP Code:

stock MovePointTowardsOtherPoint(const Float:A[3], const Float:B[3], const Float:X, Float:result[3])
{
  MakeVectorFromPoints(A, B, result);
  NormalizeVector(result, result);
  ScaleVector(result, X);
  AddVectors(A, result, result);
} 


Edit: Oh, ^ what blodia said!

[GsiX]

Originally Posted by Peace-Maker Let the code speak.. PHP Code: stock MovePointTowardsOtherPoint(const Float:A[3], const Float:B[3], const Float:X, Float:result[3]) {   MakeVectorFromPoints(A, B, result);   NormalizeVector(result, result);   ScaleVector(result, X);   AddVectors(A, result, result); }  Edit: Oh, ^ what blodia said!

If i understand this, it should change your pos from A to B, yes?

[Marcus_Brown001]

c₁ = a₁ + X(b₁ - a₁)/(√((b₁ - a₁)² + (b₂ - a₂)² + (b₃ - a₃)²))
c₂ = a₂ + X(b₂ - a₂)/(√((b₁ - a₁)² + (b₂ - a₂)² + (b₃ - a₃)²))
c₃ = a₃ + X(b₃ - a₃)/(√((b₁ - a₁)² + (b₂ - a₂)² + (b₃ - a₃)²))


My brain hurts