Skip to main content

Using graphics primitives to draw a cross as inset in graphics


I would like to draw a cross as an inset in a Graphics environment. I tried the following:


Graphics[{Inset[Style["\[Cross]", 100], {0, 0},{Center,Center}], Circle[]}, 
     Frame -> True]

Inset of cross in circle


Unfortunately, the {Center,Center} command for specifying the exact point of the cross that is drawn at the {0,0} position does not use the intercept of the two parts of the cross. Is there an easy way to produce a cross using graphics primitives rather than special characters? I have not done much with graphics yet and would really appreciate your help.




Answer



Nice & easy:-


a = 0.2; t = 6;
Graphics[{Circle[],
  Inset[Graphics[{AbsoluteThickness[t],
     Line[{{-a, -a}, {a, a}}]}], {0, 0}, Center, 2 a],
  Inset[Graphics[{AbsoluteThickness[t],
     Line[{{-a, a}, {a, -a}}]}], {0, 0}, Center, 2 a]},
 Frame -> True, ImageSize -> 250]


enter image description here


Using Inset ensures the resulting X stays within the 2 a bounds specified. The simpler red X shown, spills over slightly due to the thickness. The cross is also shown in green.


a = 0.2; t = 6;
Show[Graphics[{Circle[],
   AbsoluteThickness[t], Red,
   Line[{{-a, -a}, {a, a}}],
   Line[{{-a, a}, {a, -a}}]},
  Frame -> True, ImageSize -> 350],

 Graphics[{Circle[],

   Line[{{0, -1}, {0, 1}}],
   Line[{{-1, 0}, {1, 0}}],
   Line[{{a, -1}, {a, 1}}],
   Line[{{-1, a}, {1, a}}],
   Line[{{-a, -1}, {-a, 1}}],
   Line[{{-1, -a}, {1, -a}}],
   Inset[Style["\[Cross]", 100, Green], {0, 0}, {Center, Center}],
   Inset[Graphics[{AbsoluteThickness[t],
      Line[{{-a, -a}, {a, a}}]}], {0, 0}, Center, 2 a],
   Inset[Graphics[{AbsoluteThickness[t],

      Line[{{-a, a}, {a, -a}}]}], {0, 0}, Center, 2 a]},
  Frame -> True, ImageSize -> 350]]

enter image description here


Addendum


Inset X's can be made to scale. But first, using non-inset lines scales straight away:


a = 0.2; t = 6;
 Graphics[{Circle[], AbsoluteThickness[t], Red,
  Line[{{-a, -a}, {a, a}}], Line[{{-a, a}, {a, -a}}]},
 Frame -> True, ImageSize -> 250, AspectRatio -> 0.4]


enter image description here


Inset lines' sizes are not affected by the enclosing graphic's AspectRatio:


a = 0.2; t = 6; r = 0.4;
Graphics[{Circle[], Inset[Graphics[{AbsoluteThickness[t],
     Line[{{-a, -a}, {a, a}}]}], {0, 0}, Center, 2 a], 
  Inset[Graphics[{AbsoluteThickness[t],
     Line[{{-a, a}, {a, -a}}]}], {0, 0}, Center, 2 a]},
 Frame -> True, ImageSize -> 250, AspectRatio -> r]


enter image description here


Adding a factor to the insets can apply scaling:


a = 0.2; t = 6; r = 0.4;
Graphics[{Circle[], Inset[Graphics[{AbsoluteThickness[t],
     Line[{{-a, -a r}, {a, a r}}]}], {0, 0}, Center, 2 a],
  Inset[Graphics[{AbsoluteThickness[t],
     Line[{{-a, a r}, {a, -a r}}]}], {0, 0}, Center, 2 a]},
 Frame -> True, ImageSize -> 250, AspectRatio -> r]

enter image description here



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