Find Parallel or Offset SVG path











up vote
0
down vote

favorite
1












I need a parallel SVG path.
I have path as 'M0 0 H50 A20 20 0 1 0 100 50 v25 C50 125 0 85 0 85 z'.
Let Offset be 2px.



Original Path O/P:
Path Image
Path Segments when further divided:



0:  ["M", 0, 0]
1: ["H", 50]
2: ["A", 20, 20, 0, 1, 0, 100, 50]
3: ["V", 75]
4: ["C", 50, 125, 0, 85, 0, 85]


Any algorithm or way to find the offset path for individual segments or on whole??



Updated Image:
Red lines represent the offset SVG image, I need to get.
Example Offset SVG Path










share|improve this question
























  • If you want a second, identical shape that is simply offset from the first, you can make a copy of the original image and just adjust the starting x, y coordinates in M. It's possible that I haven't understood your goal, however. So, M2 0 H50 A20 20 0 1 0 100 50 v25 C50 125 0 85 0 85 z would be identical to the original shape, but offset two units on the x-axis from the original shape.
    – Tex
    Nov 21 at 12:38












  • Have you had a look at scale and translate SVG transforms? developer.mozilla.org/en-US/docs/Web/SVG/Attribute/transform
    – Mehdi
    Nov 21 at 12:54










  • @Tex that is wrong it just offsets in the beginning, I need the whole path or individual segments for more clarity. Did you check an update on my question.The image will give you more clarity.
    – user10519853
    Nov 21 at 12:55










  • you could apply a scale transform, that should be pretty close. e.g. transform="scale(0.9)"
    – Robert Longson
    Nov 21 at 13:16












  • I mistakenly thought we were dealing with a collection of relative paths (i.e. each path starts off where the last one ends). It also was not clear to me from your original question that your goal was to draw a smaller version of the original image inside the original image.
    – Tex
    Nov 21 at 14:28

















up vote
0
down vote

favorite
1












I need a parallel SVG path.
I have path as 'M0 0 H50 A20 20 0 1 0 100 50 v25 C50 125 0 85 0 85 z'.
Let Offset be 2px.



Original Path O/P:
Path Image
Path Segments when further divided:



0:  ["M", 0, 0]
1: ["H", 50]
2: ["A", 20, 20, 0, 1, 0, 100, 50]
3: ["V", 75]
4: ["C", 50, 125, 0, 85, 0, 85]


Any algorithm or way to find the offset path for individual segments or on whole??



Updated Image:
Red lines represent the offset SVG image, I need to get.
Example Offset SVG Path










share|improve this question
























  • If you want a second, identical shape that is simply offset from the first, you can make a copy of the original image and just adjust the starting x, y coordinates in M. It's possible that I haven't understood your goal, however. So, M2 0 H50 A20 20 0 1 0 100 50 v25 C50 125 0 85 0 85 z would be identical to the original shape, but offset two units on the x-axis from the original shape.
    – Tex
    Nov 21 at 12:38












  • Have you had a look at scale and translate SVG transforms? developer.mozilla.org/en-US/docs/Web/SVG/Attribute/transform
    – Mehdi
    Nov 21 at 12:54










  • @Tex that is wrong it just offsets in the beginning, I need the whole path or individual segments for more clarity. Did you check an update on my question.The image will give you more clarity.
    – user10519853
    Nov 21 at 12:55










  • you could apply a scale transform, that should be pretty close. e.g. transform="scale(0.9)"
    – Robert Longson
    Nov 21 at 13:16












  • I mistakenly thought we were dealing with a collection of relative paths (i.e. each path starts off where the last one ends). It also was not clear to me from your original question that your goal was to draw a smaller version of the original image inside the original image.
    – Tex
    Nov 21 at 14:28















up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





I need a parallel SVG path.
I have path as 'M0 0 H50 A20 20 0 1 0 100 50 v25 C50 125 0 85 0 85 z'.
Let Offset be 2px.



Original Path O/P:
Path Image
Path Segments when further divided:



0:  ["M", 0, 0]
1: ["H", 50]
2: ["A", 20, 20, 0, 1, 0, 100, 50]
3: ["V", 75]
4: ["C", 50, 125, 0, 85, 0, 85]


Any algorithm or way to find the offset path for individual segments or on whole??



Updated Image:
Red lines represent the offset SVG image, I need to get.
Example Offset SVG Path










share|improve this question















I need a parallel SVG path.
I have path as 'M0 0 H50 A20 20 0 1 0 100 50 v25 C50 125 0 85 0 85 z'.
Let Offset be 2px.



Original Path O/P:
Path Image
Path Segments when further divided:



0:  ["M", 0, 0]
1: ["H", 50]
2: ["A", 20, 20, 0, 1, 0, 100, 50]
3: ["V", 75]
4: ["C", 50, 125, 0, 85, 0, 85]


Any algorithm or way to find the offset path for individual segments or on whole??



Updated Image:
Red lines represent the offset SVG image, I need to get.
Example Offset SVG Path







javascript math svg polygon






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 21 at 12:50

























asked Nov 21 at 12:25









user10519853

105




105












  • If you want a second, identical shape that is simply offset from the first, you can make a copy of the original image and just adjust the starting x, y coordinates in M. It's possible that I haven't understood your goal, however. So, M2 0 H50 A20 20 0 1 0 100 50 v25 C50 125 0 85 0 85 z would be identical to the original shape, but offset two units on the x-axis from the original shape.
    – Tex
    Nov 21 at 12:38












  • Have you had a look at scale and translate SVG transforms? developer.mozilla.org/en-US/docs/Web/SVG/Attribute/transform
    – Mehdi
    Nov 21 at 12:54










  • @Tex that is wrong it just offsets in the beginning, I need the whole path or individual segments for more clarity. Did you check an update on my question.The image will give you more clarity.
    – user10519853
    Nov 21 at 12:55










  • you could apply a scale transform, that should be pretty close. e.g. transform="scale(0.9)"
    – Robert Longson
    Nov 21 at 13:16












  • I mistakenly thought we were dealing with a collection of relative paths (i.e. each path starts off where the last one ends). It also was not clear to me from your original question that your goal was to draw a smaller version of the original image inside the original image.
    – Tex
    Nov 21 at 14:28




















  • If you want a second, identical shape that is simply offset from the first, you can make a copy of the original image and just adjust the starting x, y coordinates in M. It's possible that I haven't understood your goal, however. So, M2 0 H50 A20 20 0 1 0 100 50 v25 C50 125 0 85 0 85 z would be identical to the original shape, but offset two units on the x-axis from the original shape.
    – Tex
    Nov 21 at 12:38












  • Have you had a look at scale and translate SVG transforms? developer.mozilla.org/en-US/docs/Web/SVG/Attribute/transform
    – Mehdi
    Nov 21 at 12:54










  • @Tex that is wrong it just offsets in the beginning, I need the whole path or individual segments for more clarity. Did you check an update on my question.The image will give you more clarity.
    – user10519853
    Nov 21 at 12:55










  • you could apply a scale transform, that should be pretty close. e.g. transform="scale(0.9)"
    – Robert Longson
    Nov 21 at 13:16












  • I mistakenly thought we were dealing with a collection of relative paths (i.e. each path starts off where the last one ends). It also was not clear to me from your original question that your goal was to draw a smaller version of the original image inside the original image.
    – Tex
    Nov 21 at 14:28


















If you want a second, identical shape that is simply offset from the first, you can make a copy of the original image and just adjust the starting x, y coordinates in M. It's possible that I haven't understood your goal, however. So, M2 0 H50 A20 20 0 1 0 100 50 v25 C50 125 0 85 0 85 z would be identical to the original shape, but offset two units on the x-axis from the original shape.
– Tex
Nov 21 at 12:38






If you want a second, identical shape that is simply offset from the first, you can make a copy of the original image and just adjust the starting x, y coordinates in M. It's possible that I haven't understood your goal, however. So, M2 0 H50 A20 20 0 1 0 100 50 v25 C50 125 0 85 0 85 z would be identical to the original shape, but offset two units on the x-axis from the original shape.
– Tex
Nov 21 at 12:38














Have you had a look at scale and translate SVG transforms? developer.mozilla.org/en-US/docs/Web/SVG/Attribute/transform
– Mehdi
Nov 21 at 12:54




Have you had a look at scale and translate SVG transforms? developer.mozilla.org/en-US/docs/Web/SVG/Attribute/transform
– Mehdi
Nov 21 at 12:54












@Tex that is wrong it just offsets in the beginning, I need the whole path or individual segments for more clarity. Did you check an update on my question.The image will give you more clarity.
– user10519853
Nov 21 at 12:55




@Tex that is wrong it just offsets in the beginning, I need the whole path or individual segments for more clarity. Did you check an update on my question.The image will give you more clarity.
– user10519853
Nov 21 at 12:55












you could apply a scale transform, that should be pretty close. e.g. transform="scale(0.9)"
– Robert Longson
Nov 21 at 13:16






you could apply a scale transform, that should be pretty close. e.g. transform="scale(0.9)"
– Robert Longson
Nov 21 at 13:16














I mistakenly thought we were dealing with a collection of relative paths (i.e. each path starts off where the last one ends). It also was not clear to me from your original question that your goal was to draw a smaller version of the original image inside the original image.
– Tex
Nov 21 at 14:28






I mistakenly thought we were dealing with a collection of relative paths (i.e. each path starts off where the last one ends). It also was not clear to me from your original question that your goal was to draw a smaller version of the original image inside the original image.
– Tex
Nov 21 at 14:28














2 Answers
2






active

oldest

votes

















up vote
4
down vote



accepted










Algorithmic solution



Offsetting lines is relatively trivial. For arcs, it is solvable by changing both radii by the same value. (You still have to find the end points.)



The problem are Bezier curves. There is a library bezier.js that solves this mathematically. For background, look at the accompaigning "Primer on Bézier Curves" by Pomax, especially the chapter on curve offsetting.



As is pointed out there, it is impossible to find a single Bezier curve that offsets another single curve; you need to divide it in "simpler" sub-parts. The library implements a function that combines these parts again and gives back the offset path: .offset(d).



Grafical solution



Vector grafic GUIs implement this sort of thing. The following describes it for Inkscape, but I am sure Adobe Illustrator (and maybe Sketch) can do more or less the same.




  • Draw your path. Inkscape has an "XML editor" where you can enter the path definition string directly.

  • Remove the fill and define a stroke with a width that is double the offset you want to achieve.

  • Choose "Stroke to Path" from the "Path" menu.

  • Choose "Break Apart" from the "Path" menu.

  • Now you have two (filled) paths, one offset to the outside, on offset to the inside; discard the one you don't need.


Note btw that your path definition is invalid. The arc between points [50, 0] and [100, 50] is given with a radius of 20, but the two points are 70.71 apart. In accordance with the spec, the path is instead drawn as



M 0,0 H 50 A 35.3553,35.3553 0 1 0 100,50 V 75 C 50,125 0,85 0,85 Z


If I find a path inset by 2 with Inkscape, I get



M 2,2 H 45.7988 C 34.2583,16.6514 35.0764,37.9045 48.5859,51.4141 62.0955,64.9236 83.3486,65.7417 98,54.2012
V 74.1094 C 73.6278,98.1373 49.7442,100.409 31.6426,96.7891 14.9635,93.4533 3.8673,85.3962 2,83.9785 Z


Note that Inkscape has computed a cubic Bezier approximation for the arc segment. You can revert to an arc simply by increasing the radii by 2, changing the large arc flag and retaining the end points:



M 2,2 H 45.7988 A 37.3533 37.3533 0 0 0 98,54.2012
V 74.1094 C 73.6278,98.1373 49.7442,100.409 31.6426,96.7891 14.9635,93.4533 3.8673,85.3962 2,83.9785 Z





share|improve this answer




























    up vote
    1
    down vote













    I understand you have this path and you need to draw a line inside following the border. My solution is using the <feMorphology> filter with the operator="erode". I hope this is what you need.






    <svg viewBox="-10 -10 120 120" width="300">
    <defs>

    <filter id="erode">
    <feMorphology in="SourceAlpha" result="eroded"
    operator="erode" radius="2"/>
    <feFlood flood-color="white" result="white" />
    <feComposite in ="white" in2="eroded" operator="in" />
    </filter>


    <filter id="erode1">
    <feMorphology in="SourceAlpha" result="eroded1"
    operator="erode" radius="3"/>


    </filter>
    <path id="g" d="M0, 0
    H50
    A20, 20, 0, 1, 0, 100, 50
    V75
    C50, 125, 0, 85, 0, 85" />
    </defs>

    <use xlink:href="#g" />
    <use xlink:href="#g" filter="url(#erode)" />
    <use xlink:href="#g" filter="url(#erode1)" />
    </svg>








    share|improve this answer





















      Your Answer






      StackExchange.ifUsing("editor", function () {
      StackExchange.using("externalEditor", function () {
      StackExchange.using("snippets", function () {
      StackExchange.snippets.init();
      });
      });
      }, "code-snippets");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "1"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














       

      draft saved


      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53411999%2ffind-parallel-or-offset-svg-path%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      4
      down vote



      accepted










      Algorithmic solution



      Offsetting lines is relatively trivial. For arcs, it is solvable by changing both radii by the same value. (You still have to find the end points.)



      The problem are Bezier curves. There is a library bezier.js that solves this mathematically. For background, look at the accompaigning "Primer on Bézier Curves" by Pomax, especially the chapter on curve offsetting.



      As is pointed out there, it is impossible to find a single Bezier curve that offsets another single curve; you need to divide it in "simpler" sub-parts. The library implements a function that combines these parts again and gives back the offset path: .offset(d).



      Grafical solution



      Vector grafic GUIs implement this sort of thing. The following describes it for Inkscape, but I am sure Adobe Illustrator (and maybe Sketch) can do more or less the same.




      • Draw your path. Inkscape has an "XML editor" where you can enter the path definition string directly.

      • Remove the fill and define a stroke with a width that is double the offset you want to achieve.

      • Choose "Stroke to Path" from the "Path" menu.

      • Choose "Break Apart" from the "Path" menu.

      • Now you have two (filled) paths, one offset to the outside, on offset to the inside; discard the one you don't need.


      Note btw that your path definition is invalid. The arc between points [50, 0] and [100, 50] is given with a radius of 20, but the two points are 70.71 apart. In accordance with the spec, the path is instead drawn as



      M 0,0 H 50 A 35.3553,35.3553 0 1 0 100,50 V 75 C 50,125 0,85 0,85 Z


      If I find a path inset by 2 with Inkscape, I get



      M 2,2 H 45.7988 C 34.2583,16.6514 35.0764,37.9045 48.5859,51.4141 62.0955,64.9236 83.3486,65.7417 98,54.2012
      V 74.1094 C 73.6278,98.1373 49.7442,100.409 31.6426,96.7891 14.9635,93.4533 3.8673,85.3962 2,83.9785 Z


      Note that Inkscape has computed a cubic Bezier approximation for the arc segment. You can revert to an arc simply by increasing the radii by 2, changing the large arc flag and retaining the end points:



      M 2,2 H 45.7988 A 37.3533 37.3533 0 0 0 98,54.2012
      V 74.1094 C 73.6278,98.1373 49.7442,100.409 31.6426,96.7891 14.9635,93.4533 3.8673,85.3962 2,83.9785 Z





      share|improve this answer

























        up vote
        4
        down vote



        accepted










        Algorithmic solution



        Offsetting lines is relatively trivial. For arcs, it is solvable by changing both radii by the same value. (You still have to find the end points.)



        The problem are Bezier curves. There is a library bezier.js that solves this mathematically. For background, look at the accompaigning "Primer on Bézier Curves" by Pomax, especially the chapter on curve offsetting.



        As is pointed out there, it is impossible to find a single Bezier curve that offsets another single curve; you need to divide it in "simpler" sub-parts. The library implements a function that combines these parts again and gives back the offset path: .offset(d).



        Grafical solution



        Vector grafic GUIs implement this sort of thing. The following describes it for Inkscape, but I am sure Adobe Illustrator (and maybe Sketch) can do more or less the same.




        • Draw your path. Inkscape has an "XML editor" where you can enter the path definition string directly.

        • Remove the fill and define a stroke with a width that is double the offset you want to achieve.

        • Choose "Stroke to Path" from the "Path" menu.

        • Choose "Break Apart" from the "Path" menu.

        • Now you have two (filled) paths, one offset to the outside, on offset to the inside; discard the one you don't need.


        Note btw that your path definition is invalid. The arc between points [50, 0] and [100, 50] is given with a radius of 20, but the two points are 70.71 apart. In accordance with the spec, the path is instead drawn as



        M 0,0 H 50 A 35.3553,35.3553 0 1 0 100,50 V 75 C 50,125 0,85 0,85 Z


        If I find a path inset by 2 with Inkscape, I get



        M 2,2 H 45.7988 C 34.2583,16.6514 35.0764,37.9045 48.5859,51.4141 62.0955,64.9236 83.3486,65.7417 98,54.2012
        V 74.1094 C 73.6278,98.1373 49.7442,100.409 31.6426,96.7891 14.9635,93.4533 3.8673,85.3962 2,83.9785 Z


        Note that Inkscape has computed a cubic Bezier approximation for the arc segment. You can revert to an arc simply by increasing the radii by 2, changing the large arc flag and retaining the end points:



        M 2,2 H 45.7988 A 37.3533 37.3533 0 0 0 98,54.2012
        V 74.1094 C 73.6278,98.1373 49.7442,100.409 31.6426,96.7891 14.9635,93.4533 3.8673,85.3962 2,83.9785 Z





        share|improve this answer























          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          Algorithmic solution



          Offsetting lines is relatively trivial. For arcs, it is solvable by changing both radii by the same value. (You still have to find the end points.)



          The problem are Bezier curves. There is a library bezier.js that solves this mathematically. For background, look at the accompaigning "Primer on Bézier Curves" by Pomax, especially the chapter on curve offsetting.



          As is pointed out there, it is impossible to find a single Bezier curve that offsets another single curve; you need to divide it in "simpler" sub-parts. The library implements a function that combines these parts again and gives back the offset path: .offset(d).



          Grafical solution



          Vector grafic GUIs implement this sort of thing. The following describes it for Inkscape, but I am sure Adobe Illustrator (and maybe Sketch) can do more or less the same.




          • Draw your path. Inkscape has an "XML editor" where you can enter the path definition string directly.

          • Remove the fill and define a stroke with a width that is double the offset you want to achieve.

          • Choose "Stroke to Path" from the "Path" menu.

          • Choose "Break Apart" from the "Path" menu.

          • Now you have two (filled) paths, one offset to the outside, on offset to the inside; discard the one you don't need.


          Note btw that your path definition is invalid. The arc between points [50, 0] and [100, 50] is given with a radius of 20, but the two points are 70.71 apart. In accordance with the spec, the path is instead drawn as



          M 0,0 H 50 A 35.3553,35.3553 0 1 0 100,50 V 75 C 50,125 0,85 0,85 Z


          If I find a path inset by 2 with Inkscape, I get



          M 2,2 H 45.7988 C 34.2583,16.6514 35.0764,37.9045 48.5859,51.4141 62.0955,64.9236 83.3486,65.7417 98,54.2012
          V 74.1094 C 73.6278,98.1373 49.7442,100.409 31.6426,96.7891 14.9635,93.4533 3.8673,85.3962 2,83.9785 Z


          Note that Inkscape has computed a cubic Bezier approximation for the arc segment. You can revert to an arc simply by increasing the radii by 2, changing the large arc flag and retaining the end points:



          M 2,2 H 45.7988 A 37.3533 37.3533 0 0 0 98,54.2012
          V 74.1094 C 73.6278,98.1373 49.7442,100.409 31.6426,96.7891 14.9635,93.4533 3.8673,85.3962 2,83.9785 Z





          share|improve this answer












          Algorithmic solution



          Offsetting lines is relatively trivial. For arcs, it is solvable by changing both radii by the same value. (You still have to find the end points.)



          The problem are Bezier curves. There is a library bezier.js that solves this mathematically. For background, look at the accompaigning "Primer on Bézier Curves" by Pomax, especially the chapter on curve offsetting.



          As is pointed out there, it is impossible to find a single Bezier curve that offsets another single curve; you need to divide it in "simpler" sub-parts. The library implements a function that combines these parts again and gives back the offset path: .offset(d).



          Grafical solution



          Vector grafic GUIs implement this sort of thing. The following describes it for Inkscape, but I am sure Adobe Illustrator (and maybe Sketch) can do more or less the same.




          • Draw your path. Inkscape has an "XML editor" where you can enter the path definition string directly.

          • Remove the fill and define a stroke with a width that is double the offset you want to achieve.

          • Choose "Stroke to Path" from the "Path" menu.

          • Choose "Break Apart" from the "Path" menu.

          • Now you have two (filled) paths, one offset to the outside, on offset to the inside; discard the one you don't need.


          Note btw that your path definition is invalid. The arc between points [50, 0] and [100, 50] is given with a radius of 20, but the two points are 70.71 apart. In accordance with the spec, the path is instead drawn as



          M 0,0 H 50 A 35.3553,35.3553 0 1 0 100,50 V 75 C 50,125 0,85 0,85 Z


          If I find a path inset by 2 with Inkscape, I get



          M 2,2 H 45.7988 C 34.2583,16.6514 35.0764,37.9045 48.5859,51.4141 62.0955,64.9236 83.3486,65.7417 98,54.2012
          V 74.1094 C 73.6278,98.1373 49.7442,100.409 31.6426,96.7891 14.9635,93.4533 3.8673,85.3962 2,83.9785 Z


          Note that Inkscape has computed a cubic Bezier approximation for the arc segment. You can revert to an arc simply by increasing the radii by 2, changing the large arc flag and retaining the end points:



          M 2,2 H 45.7988 A 37.3533 37.3533 0 0 0 98,54.2012
          V 74.1094 C 73.6278,98.1373 49.7442,100.409 31.6426,96.7891 14.9635,93.4533 3.8673,85.3962 2,83.9785 Z






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 21 at 16:47









          ccprog

          8,2891825




          8,2891825
























              up vote
              1
              down vote













              I understand you have this path and you need to draw a line inside following the border. My solution is using the <feMorphology> filter with the operator="erode". I hope this is what you need.






              <svg viewBox="-10 -10 120 120" width="300">
              <defs>

              <filter id="erode">
              <feMorphology in="SourceAlpha" result="eroded"
              operator="erode" radius="2"/>
              <feFlood flood-color="white" result="white" />
              <feComposite in ="white" in2="eroded" operator="in" />
              </filter>


              <filter id="erode1">
              <feMorphology in="SourceAlpha" result="eroded1"
              operator="erode" radius="3"/>


              </filter>
              <path id="g" d="M0, 0
              H50
              A20, 20, 0, 1, 0, 100, 50
              V75
              C50, 125, 0, 85, 0, 85" />
              </defs>

              <use xlink:href="#g" />
              <use xlink:href="#g" filter="url(#erode)" />
              <use xlink:href="#g" filter="url(#erode1)" />
              </svg>








              share|improve this answer

























                up vote
                1
                down vote













                I understand you have this path and you need to draw a line inside following the border. My solution is using the <feMorphology> filter with the operator="erode". I hope this is what you need.






                <svg viewBox="-10 -10 120 120" width="300">
                <defs>

                <filter id="erode">
                <feMorphology in="SourceAlpha" result="eroded"
                operator="erode" radius="2"/>
                <feFlood flood-color="white" result="white" />
                <feComposite in ="white" in2="eroded" operator="in" />
                </filter>


                <filter id="erode1">
                <feMorphology in="SourceAlpha" result="eroded1"
                operator="erode" radius="3"/>


                </filter>
                <path id="g" d="M0, 0
                H50
                A20, 20, 0, 1, 0, 100, 50
                V75
                C50, 125, 0, 85, 0, 85" />
                </defs>

                <use xlink:href="#g" />
                <use xlink:href="#g" filter="url(#erode)" />
                <use xlink:href="#g" filter="url(#erode1)" />
                </svg>








                share|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  I understand you have this path and you need to draw a line inside following the border. My solution is using the <feMorphology> filter with the operator="erode". I hope this is what you need.






                  <svg viewBox="-10 -10 120 120" width="300">
                  <defs>

                  <filter id="erode">
                  <feMorphology in="SourceAlpha" result="eroded"
                  operator="erode" radius="2"/>
                  <feFlood flood-color="white" result="white" />
                  <feComposite in ="white" in2="eroded" operator="in" />
                  </filter>


                  <filter id="erode1">
                  <feMorphology in="SourceAlpha" result="eroded1"
                  operator="erode" radius="3"/>


                  </filter>
                  <path id="g" d="M0, 0
                  H50
                  A20, 20, 0, 1, 0, 100, 50
                  V75
                  C50, 125, 0, 85, 0, 85" />
                  </defs>

                  <use xlink:href="#g" />
                  <use xlink:href="#g" filter="url(#erode)" />
                  <use xlink:href="#g" filter="url(#erode1)" />
                  </svg>








                  share|improve this answer












                  I understand you have this path and you need to draw a line inside following the border. My solution is using the <feMorphology> filter with the operator="erode". I hope this is what you need.






                  <svg viewBox="-10 -10 120 120" width="300">
                  <defs>

                  <filter id="erode">
                  <feMorphology in="SourceAlpha" result="eroded"
                  operator="erode" radius="2"/>
                  <feFlood flood-color="white" result="white" />
                  <feComposite in ="white" in2="eroded" operator="in" />
                  </filter>


                  <filter id="erode1">
                  <feMorphology in="SourceAlpha" result="eroded1"
                  operator="erode" radius="3"/>


                  </filter>
                  <path id="g" d="M0, 0
                  H50
                  A20, 20, 0, 1, 0, 100, 50
                  V75
                  C50, 125, 0, 85, 0, 85" />
                  </defs>

                  <use xlink:href="#g" />
                  <use xlink:href="#g" filter="url(#erode)" />
                  <use xlink:href="#g" filter="url(#erode1)" />
                  </svg>








                  <svg viewBox="-10 -10 120 120" width="300">
                  <defs>

                  <filter id="erode">
                  <feMorphology in="SourceAlpha" result="eroded"
                  operator="erode" radius="2"/>
                  <feFlood flood-color="white" result="white" />
                  <feComposite in ="white" in2="eroded" operator="in" />
                  </filter>


                  <filter id="erode1">
                  <feMorphology in="SourceAlpha" result="eroded1"
                  operator="erode" radius="3"/>


                  </filter>
                  <path id="g" d="M0, 0
                  H50
                  A20, 20, 0, 1, 0, 100, 50
                  V75
                  C50, 125, 0, 85, 0, 85" />
                  </defs>

                  <use xlink:href="#g" />
                  <use xlink:href="#g" filter="url(#erode)" />
                  <use xlink:href="#g" filter="url(#erode1)" />
                  </svg>





                  <svg viewBox="-10 -10 120 120" width="300">
                  <defs>

                  <filter id="erode">
                  <feMorphology in="SourceAlpha" result="eroded"
                  operator="erode" radius="2"/>
                  <feFlood flood-color="white" result="white" />
                  <feComposite in ="white" in2="eroded" operator="in" />
                  </filter>


                  <filter id="erode1">
                  <feMorphology in="SourceAlpha" result="eroded1"
                  operator="erode" radius="3"/>


                  </filter>
                  <path id="g" d="M0, 0
                  H50
                  A20, 20, 0, 1, 0, 100, 50
                  V75
                  C50, 125, 0, 85, 0, 85" />
                  </defs>

                  <use xlink:href="#g" />
                  <use xlink:href="#g" filter="url(#erode)" />
                  <use xlink:href="#g" filter="url(#erode1)" />
                  </svg>






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 21 at 14:08









                  enxaneta

                  3,9271412




                  3,9271412






























                       

                      draft saved


                      draft discarded



















































                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53411999%2ffind-parallel-or-offset-svg-path%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Contact image not getting when fetch all contact list from iPhone by CNContact

                      count number of partitions of a set with n elements into k subsets

                      A CLEAN and SIMPLE way to add appendices to Table of Contents and bookmarks