Some crazy maths.

[EvAsion]

Given several points, and a 3 dimensional line (determined by a point and a rotation from that point), how can i determine the point closest to that line.

Note: (x,y,z not x,y)

[weaponx]

What does this have to do with AutoIt?

[EvAsion]

i don't know.

Edit: i want to calculate this with autoit..

Problem is i don't know how to calculate the closest distance betweeen the line and a certain point(so determine the closest point on the line to a certain point).. but i can easily calculate the distance between 2 points.

Edit2: Oh wait i could draw a line perpendicular to the line from the point.. and then determine the point of intersection.. calculate the distance between the point and the point of intersection.. do it for all the points.. yay... now i just need to convert it from my head into autoit.

Edited April 15, 2008 by EvAsion

[d4rk]

i don't know.

how about going somewhere ? funadvice.com ? Edited April 15, 2008 by d4rk

[EvAsion]

how about going somewhere ? funadvice.com ?

I would.. but i have to sign up!?!

[mrbig1479]

I would probebly do

1. start in the begining of that line . and jumping in small steps on the line direction..

2. every step check to see is the perpendicular line to the point is smaller then the last line. (mark the point also)

3. do that untill i finish the steps

[EvAsion]

I would probebly do

1. start in the begining of that line . and jumping in small steps on the line direction..

2. every step check to see is the perpendicular line to the point is smaller then the last line. (mark the point also)

3. do that untill i finish the steps

I could do that, but that would be incredibly inefficient. I think it would be best if i drew a line from the point, perpendicular to the original line, at a rotation where it would intersect the original line, mark the intercept, get the distance between the point and the intercept, and repeat for all the points.

Edit: am i making any sense?

Edited April 15, 2008 by EvAsion

[weaponx]

I converted the example found here:

http://mathforum.org/library/drmath/view/54731.html

I tested using a line from 0,0,0 to 1,0,0 with a point at 0,1,0 which should return 1 as the distance.

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#cs ----------------------------------------------------------------------------
Let,    A(x1,y1,z1), B(x2,y2,z2) = Two points defining the line 
        P(x,y,z) = Point whose distance from line AB is to be found

We can find vectors,
        AB =  = 
        AP = < x-x1,  y-y1, z-z1 > = 

I am using angle brackets < > to distinguish vectors from points.

Now, let's take the cross product AB X AP.  This is how you can calculate it:

    AB X AP = < y3*z4 - z3*y4, -(x3*z4 - x4*z3), x3*y4 - x4*y3 >

Let's let (a) be the length of AB X AP.  Then

    a = Sqrt{ (y3*z4 - z3*y4)^2 + (x3*z4 - x4*z3)^2 + (x3*y4 - x4*y3)^2 }

If you divide (a) by the distance AB you will get the distance of P 
from line AB. Distance AB can be found using the distance formula as,
    AB = square root of (x3^2 + y3^2 + z3^2)

Thus the distance we are looking for is a/AB.

Now, did you see why it works? Because the length of the cross product (a)
gives the area of the Parallelogram formed by the vectors AB and AP as
shown. If you rest the parallelogram on AB as the base, its area (a) is
AB*h. Where (h) is the height, which is nothing but the distance we are
looking for!

      P
      o------+------o
       \     |       \
        \    |h       \
         \   |         \
    ------o--+----------o-----------
          A             B

As for the last part of your question : When a line is given only in 
terms of its inclinations with the coordinate axes, then the line could 
'float' anywhere parallel to itself, because its not 'anchored' to (or 
through) any point. In such a case, its distance from point P is not 
fixed.
#ce ----------------------------------------------------------------------------

;A(x1,y1,z1), B(x2,y2,z2) = Two points defining the line 
Const $x1 = 0, $y1 = 0, $z1 = 0
Const $x2 = 1, $y2 = 0, $z2 = 0

;P(x,y,z) = Point whose distance from line AB is to be found
Const $x = 0, $y = 1, $z = 0

;Vectors
Dim $AB[3] = [$x2-$x1, $y2-$y1, $z2-$z1]
Dim $AP[3] = [$x-$x1, $y-$y1, $z-$z1]

;Now, let's take the cross product AB X AP.  This is how you can calculate it:
$ABAP = cross($AB,$AP)

;Let's let (a) be the length of AB X AP.  Then
$a = Sqrt($ABAP[0]^2 + $ABAP[1]^2 + $ABAP[2]^2 )

;Distance AB can be found using the distance formula as, AB = square root of (x3^2 + y3^2 + z3^2)
$dAB = Sqrt($AB[0]^2 + $AB[1]^2 + $AB[2]^2)

;Thus the distance we are looking for is a/AB.
$answer = $a/$dAB

MsgBox(0,"","Distance from point " & $x & "," & $y & "," & $z & " to line from " & $x1 & "," & $y1 & "," & $z1 & " to " & $x2 & "," & $y2 & "," & $z2 & @CRLF & "= " & $answer)

Func cross($V1, $V2)
    Local $x3 = $V1[0], $y3 = $V1[1], $z3 = $V1[2]
    Local $x4 = $V2[0], $y4 = $V2[1], $z4 = $V2[2]
    
    Local $CP[3]
    
    $CP[0] = $y3*$z4 - $z3*$y4
    $CP[1] = -($x3*$z4 - $x4*$z3)
    $CP[2] = $x3*$y4 - $x4*$y3
    
    Return $CP
EndFunc

[EvAsion]

You are amazing weaponx.

[weaponx]

You are amazing weaponx.

"Its nothing any American with balls of steel wouldn't do for his community." -The Mask

[PsaltyDS]

"Its nothing any American with balls of steel wouldn't do for his community." -The Mask

Those would be adamantium, wouldn't they...?

[weaponx]

Haha. That would be correct.

[d4rk]

"balls of steel"

yes, you're amazing but ball of steel , what is it ?

[Skrip]

yes, you're amazing but ball of steel , what is it ?

Nothing but of course this!

[d4rk]

Nothing but of course this!

yoh,