Drawing arcs with ESTK ...

Adobe Community Professional ,
Dec 27, 2018 Dec 27, 2018

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Dear friends,

Next step after lines and polylines: I want to draw arcs (with a constant radius) and encounter a similar problem as with the lines in the different quadrants.

The blue lines are what I want to achieve. The outer three are correct, their starting angle is ≥ 0. The inner ones have starting angles -90, -60 and -30 degrees, which converted to 270, 300 and 330 respectively - because negative values are not allowed (they are interpreted as 0).

Arcs01.png

I do not see how to force the Arc object to draw what I want. Drawing with UI is not easier: To get the desired angles is quite simple, but to get the arc in correct position after rotation …

It seems that I need a comparable approach to the Line function (a multitude of transformations).

#target framemaker

main();

function main () {

var j, oDoc, oFrame, x0, y0, r, th0, thf, aArcs = [];

    CM = 1857713, PT = 65536, DEGREE = 65536, pi = Math.PI;

  oDoc = app.ActiveDoc;

  oFrame = oDoc.FirstSelectedGraphicInDoc;

  if (!oFrame.ObjectValid()) {

    Alert("Select an anchored frame and try again.", Constants.FF_ALERT_CONTINUE_WARN);

    return;

  }

  for (j= 0; j < 6; j++) {                       // FM-coordinate systemin cm

    r = 2 + j*0.5;                               //  2, 2.5, 3.0, ...

    th0 = -90 + j* 30;                           // -30, 0, 30, ...

    thf = th0 + 90;

    aArcs = DrawArc(oDoc, oFrame, 5, 5, r, th0, thf);

    aArcs.BorderWidth = 0.05*CM;

    aArcs.Color = oDoc.GetNamedColor("Magenta");

    aArcs.HeadArrow = 1;                      // default arrows

  }

} //--- end main

function DrawArc(oDoc, oFrame, x0, y0, r, th0, thf) { // === Draw an arc ==========================

// Arguments  x0/y0  coordinates of center point [CM] FM coord system

//            r      radius of circular arc

//            th0    Angle (degrees) of start-point

//            thf    Angle (degrees) of end-point

// Returns    object (e.g. for grouping)

var oArc, th1,

    CM = 1857713, PT = 65536, DEGREE = 65536, pi = Math.PI;

  oArc = oDoc.NewArc(oFrame);

  oArc.Width = r * CM;

  oArc.Height= r * CM;                            // let's start with a square 90° arc

  oArc.LocX = x0 * CM;

  oArc.LocY = (y0 - r) * CM;

  if (th0 < 0) {th0 = th0 + 360;}

  oArc.Theta  = th0 * DEGREE;

  oArc.DTheta = (th0 - thf)* DEGREE;

  oArc.Theta  = th0 * DEGREE;

  oArc.DTheta = (thf - th0)* DEGREE;

  return oArc;

} //--- end DrawArc

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correct answers 1 Correct answer

Adobe Community Professional , Jan 05, 2019 Jan 05, 2019
Well, after much fiddling around and expeiments with various angles etc. I have this insight: Angle is the rotation angle relative to the original object Theta and DTheta exist only for object Arc Trying to work in a mathematical/geometrical space for drawing requires some transformation. The solution I found is documented in https://daube.ch/zz_tests/FM-graphics.pdf see page 29 of current issueBut:There is a very strange thing with rotation. Any object rotated (by UI or by script) reflect the r...

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Adobe Community Professional ,
Dec 28, 2018 Dec 28, 2018

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Modifying line 42 to

  if (th0 < 0) {

    th0 = th0 + 360;

    thf = thf + 360;

  }

did the trick at least for this set of angles...

Arcs00.png

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Adobe Community Professional ,
Jan 05, 2019 Jan 05, 2019

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LATEST

Well, after much fiddling around and expeiments with various angles etc. I have this insight:

  • Angle is the rotation angle relative to the original object
  • Theta and DTheta exist only for object Arc
  • Trying to work in a mathematical/geometrical space for drawing requires some transformation.
  • The solution I found is documented in https://daube.ch/zz_tests/FM-graphics.pdf see page 29 of current issue

But:

There is a very strange thing with rotation. Any object rotated (by UI or by script) reflect the rotation in property Ange - but not so for Line. This object always displays 0 for the rotation angle!

See the experiment describe on pages 30-31 of the mentioned pdf.

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