Skip to main content

graphics - Moving a locator based on the movement of another



My problem is:


I want two Locators to simulate a vector in the following sense:




  • The first Locator is the base and the second the tip of the vector.




  • When I move the tip, the base does not move (hence the vector changes).





  • When I move the base, the vector is unchanged, and therefore the tip (and the Locator) moves.




How can I achieve that?


I have tried storing the previous value of the base and then test if the current value is different. And if so updated the position of the tip. I can however not make that work, when using 'Module'. I suspect that there is a more elegant solution.



Answer



This seems to be a duplicate but I can't find it :). Meanwhile, you can use second argument of Dynamic.


x = {0, 0}; y = {1, 1}; w = y - x;



Deploy@Graphics[{
                 Locator@Dynamic[x, (x = #; y = x + w;) &],
                 Locator@Dynamic[y, (y = #; w = y - x;) &],
                 Dynamic@Arrow[{x, y}]
                }
                , PlotRange -> 2]

enter image description here


In case of multiple vectors one may want to save space and make code more transparent so we can use extended version of Dynamic second argument to achieve this:


Deploy@Graphics[{

                 Locator@Dynamic[x, {(w = y - x;) &, (x = #; y = x + w) &, None}],
                 Locator@Dynamic[y],
                 Dynamic@Arrow[{x, y}]
                }
               , PlotRange -> 2]

Now we are working only with base, moreover w can be scoped to particular Locator.


There is huge advantage of the second method, well, not exactly the method but the usage of f_start and f_end. You can calculate w once, not all the time you are dragging the Locator.


Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more