Geometry - how to find points on line perpendicular/parallel to given line?

Geometry - how to find points on line perpendicular/parallel to given line?

spacepionier

Member #11,917

May 2010

I'm having trouble sorting a simple concept for some reason.
Basically I have a line element AB (as indicated on the attached picture) and I want to write a code which will calculate points L,M,N (Coordinates) so I could draw local axes of the element. Axis Y always will be perpendicular to element AB and axis X will be always parallel to element AB.

known:
point A(x1,y1)
point B(x2,y2)
point c(xm,ym) midpoint of the line AB
distance d (distance from the line AB to point M)
we can set as well the lengts of the axes so we know the distance from point M to N and from M to L

I was trying first to calculate point M from the distance equation unfortunately with no luck. Could you please help me and post a code in any programming language ( I will try to convert it to C# or VB.net then) or give me some hints or links where I could read about the soultion and refresh my highschool math

Thank you in Advance!

Jonatan Hedborg

Member #4,886

July 2004

I'm sure there easier ways to do this, but I would probably use sin,cos and atan2;
(pseudo code, obviously untested. Might need to change the orders in atan2... Also requires some vector class with overloaded operators )

ab_angle = atan2(a.y-b.y, a.x-b.x);
m.x = cos(ab_angle-M_PI/2);
m.y = sin(ab_angle-M_PI/2);
l = m * lm_length;
n.x = cos(ab_angle);
n.y = sin(ab_angle);
n *= nm_length;
m *= d;
m += c;
l += m;
n += m;

EDIT: Obviously there are better solutions below

Evert

Member #794

November 2000

Probably the easiest way to do this is to follow the Gram-Schmidt process.

EDIT: oh, looks like this is 2D? In that case it's much easier to find a vector that is perpendicular to the direction vector you already have. Those two vectors then span the space and if you express your coordinates with respect to those vectors you should be good.

EDIT2: Basically, what SiegeLord said below.

verthex

Member #11,340

September 2009

#SelectExpand
  1
  2float Ax, Ay, Bx, By,cXm,cYm,Lx, Ly, Mx, My, Nx, Ny, d;
  3float greenx, greeny;// I think these two have to be given theres no other way to find L or N
  4
  5cXm = (AX - Bx)/2;
  6cym = given;
  7
  8Mx = cXm;
  9My = cym + d;
 10
 11Lx = Mx;
 12Ly = My + greeny;
 13
 14Nx = cXm - x;
 15Ny = d;

SiegeLord

Member #7,827

October 2006

First, you need to calculate the unit vector perpendicular to the segment AB. You do that by simply rotating the AB vector by 90 degrees (look up 2D rotation matrices on Wikipedia). In short, the vector that points in the direction cL (and cM) is (y2 - y1, x1 - x2). To get the point M, then, you just normalize that vector, multiply it by d and add it to the point c.

See how that works out for you... if you can't follow that I suggest brushing up on your basic linear algebra .

"For in much wisdom is much grief: and he that increases knowledge increases sorrow."-Ecclesiastes 1:18
[SiegeLord's Abode][Codes]:[DAllegro5]:[RustAllegro]

gnolam

Member #2,030

March 2002

Pfft. Trig functions are completely unnecessary.

$<math>\vec u = \vec B - \vec A</math>$
$<math>\vec v = \{ -u_y, u_x \}</math>$
$<math>\vec M = \vec c + d \cdot \hat v</math>$
$<math>\vec L = \vec M + \hat v</math>$
$<math>\vec N = \vec M + \hat u</math>$

[EDIT]
Thoroughly beaten. But I took my time trying to beat some sense into a.cc's LaTeX.

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SiegeLord

Member #7,827

October 2006

Way to go, everyone, for making the OP work for his answer .

"For in much wisdom is much grief: and he that increases knowledge increases sorrow."-Ecclesiastes 1:18
[SiegeLord's Abode][Codes]:[DAllegro5]:[RustAllegro]

Jonatan Hedborg

Member #4,886

July 2004

Woo, I actually learned something Maybe I should dust of my old algebra book...

Evert

Member #794

November 2000

SiegeLord said:

Way to go, everyone, for making the OP work for his answer .

Well, he did ask for code or hints. Guess which one of those is easier to give.

spacepionier

Member #11,917

May 2010

Thank you very much for quick reply..

Yes, it would be good to get a code sample but looking on all replies I will have to find my old Linear Algebra book

gnolam: Would you mind to give some descripion to attached equations?

SiegeLord: will I be able to find coordinates of those M,L,N points using method you suggested?

I found a similar post on the forum
http://www.allegro.cc/forums/thread/589720 which I think will be helpful.

Could you please guys point the links from where I could try to sort out this issue ? some theory.

Thanks very much again for all replies!

gnolam

Member #2,030

March 2002

u: the vector from A to B.
v: counter-clockwise perpendicular vector to u. A 2D vector (x0, y0) rotated 90° counter-clockwise becomes (-y0, x0). You can derive this through rotation matrices, or just by drawing it on a piece of graph paper.

Hats denote unit vectors.

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spacepionier

Member #11,917

May 2010

Finally got it I have just modified slightly equations for L and N vectors from gnolam equations to get a specified length for x and y axes..

Thank you all for the input!

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EDIT: Just one quick question..I have all equations in place and my local axes are always parpendicular/parallel to the element, but depending on how I draw element - from left to right or from right to left I am getting axes above or below element. would it be possible to ammend those equations so that my local axes would be always above element and y axis would be always pointing upwards and x axis would be pointing in the same direction as element ie if element is drawn from right to left my X axis would go from 0,0 (local) to the left. I have attached a screen shot from my excel calculation with graph showing local axes below element because of the direction of element (from right to left)

I am sorry for edit but culd not reply..

Thank you in advance.

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EDIT 2: Never mind..this was a stupid question..I sorted out the issue.

Thomas Fjellstrom

Member #476

June 2000

Good to hear you got it fixed. And edits are ok here so long as you have a good excuse.

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