This is one possible simplification: http://grasshopper3d.googlegroups.com/web/Vector+Substraction+2.ghx
I don't know if i can simplify it further without using the scripting component. On Jan 7, 10:09 pm, basbasbas <[email protected]> wrote: > Done. See the file named Vector Substraction 1.XML (zipped) or try the > follwing > link:http://grasshopper3d.googlegroups.com/web/Vector+Substraction+1.zip?h... > > I hope you or someone else can simplify this. > > Cheers Bas > > On Jan 7, 8:56 pm, visose <[email protected]> wrote: > > > Click on the 'files' link at the right hand side of this site. Then > > click on 'upload file' (you must be logged in). Remember to include > > both the .3dm and .ghx files if you are referencing any rhino > > geometry. > > > On Jan 7, 8:48 pm, basbasbas <[email protected]> wrote: > > > > How can I post the definition? > > > > On Jan 7, 8:20 pm, visose <[email protected]> wrote: > > > > > 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 > >
