post the definition. maybe there's a way to optimize it. Maybe using
the math approach and a scripting component for example.

On Jan 7, 8:13 pm, basbasbas <[email protected]> wrote:
> I hope that David is going to implement a function that creates the
> graphical approach of dividing a vector in two vectors with known
> directions. I used quit a lot of functions to get it. (orient, divide,
> item, scale, CCX, Line, and that four times for each line of the
> paralellogram)
>
> Cheers
> Bas
>
> On Jan 4, 12:17 am, visose <[email protected]> wrote:
>
> > Nice one. I used maths, you used the graphical approach. Using the
> > mathematical approach you get to use less components once you have the
> > proper equations (it's also more useful if it starts to get too
> > complex). Still, for simple things i prefer the graphical approach
> > (since my math are not very good). You understand better what you are
> > doing, it's more 'visual'.
>
> > On Jan 3, 11:43 pm,basbasbas<[email protected]> wrote:
>
> > > Thanks for this solution. In the mean while I tried another solution,
> > > which seems more simple. My main vector is indeed in the XY plane.
> > > I have drawn lines (using Orient) in the right amplitude at both the
> > > start point of my vector as on the end point. I have made these lines
> > > extremely long in both directions.
> > > Than I cut these lines at the intersections. The length of my new
> > > lines are my substracted vectors. simple!  Just like the old school
> > > days.  :)
>
> > > Cheers, Bas
>
> > > On Dec 31 2008, 4:55 am, visose <[email protected]> wrote:
>
> > > > Let's see, i did it using some basic math, but maybe there's a simpler
> > > > way.
> > > > For simplification, let's say you are working on 2D and the XY plane,
> > > > so you have the following equation:
> > > > a = X*b + Y*c
> > > > where 'a' is the initial vector, and 'b' and 'c' are the two unit
> > > > vectors you want to find the magnitude (right now they are unit
> > > > vectors).You want to find X and Y.
>
> > > > Since we are in 2d we have two numbers for every vector (x and y
> > > > coordinates), so we actually have two formulas:
> > > > a.x = X*b.x + Y*c.x
> > > > and
> > > > a.y = X*b.y + Y*c.y
> > > > Using substitution you get the following equations:
> > > > (a.y -a.x*c.y/c.x)/(-b.x*c.y/c.x + b.y) for X
> > > > (a.y -a.x*b.y/b.x)/(-c.x*b.y/b.x + c.y) for Y
> > > > Place this formulas into expression components, then use the magnitude
> > > > or multiply vector components to create the two final components.
>
> > > > The 3 vectors must be on the same plane, but maybe they are not in the
> > > > XY plane. For this i would use the 'orient' component to bring them to
> > > > the x,y plane and then use it again to bring them back to its initial
> > > > position.
>
> > > > There's probably a way of doing this using the vector components
> > > > without so much equation, I'm not sure.
>
> > > > On Dec 30, 9:58 pm,basbasbas<[email protected]> wrote:
>
> > > > > Dear Visose,
>
> > > > > I want to decompose it to two vectors not oriented to world
> > > > > coordinates and generaly not perpendicular to my vector.
>
> > > > > Cheers,
> > > > > Bas
>
> > > > > On 30 dec, 20:27, visose <[email protected]> wrote:
>
> > > > > > What exactly are you trying to achieve?
> > > > > > You want to decompose the vector into two other perpendicular 
> > > > > > vectors?
> > > > > > like when you want to calculate a diagonal force you decompose it 
> > > > > > in X
> > > > > > and Y forces. You want to decompose it to vectors not oriented to
> > > > > > world coordinates?
> > > > > > You want to get the magnitude of one vector, divide it by 2, and 
> > > > > > apply
> > > > > > it to two other vectors no matter what the direction of the first
> > > > > > vector is?
> > > > > > or something else?
>
> > > > > > If it's the first case it's easy, since vectors in grasshopper (and
> > > > > > rhino) are defined by 3 perpendicular components (x,y,z) and not by
> > > > > > magnitude and angle, you just need to use the 'decompose vector'
> > > > > > component, no need to do any trigonometry.
>
> > > > > > On Dec 30, 6:10 pm,basbasbas<[email protected]> wrote:
>
> > > > > > > Hi,
>
> > > > > > > I have a vector that I want to divide into two new vectors with 
> > > > > > > known
> > > > > > > directions. What is the best way to proceed?
>
> > > > > > > Cheers,
> > > > > > > Bas Goris
>
>

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