Topic: return a 3dvector(xyz)?

[sybold]

i have 2 angles, horz. and vert.
how do i return the 3dvector(xyz)

base is 0,0,0

i'm breaking my head on it with .net

[Kerry]

Have a play with this.
You could use it as is by passing 0,0,0 as point an 1.0 as distance,
   and convert the point to a Vector3D passing the X Y Z of the returned point
   using  Vector3d(double x, double y, double z);

... or
fiddle with the math to only accept the 2 angles and return the Vector3D.

Code - C#: [Select]

1.  
2.         /// <summary>
3.         /// Return the Polar3D point calculated from angles in each plane
4.         /// Kdub Services
5.         /// </summary>
6.         /// <param name="point"></param>
7.         /// <param name="xyAngle"></param>
8.         /// <param name="xzAngle"></param>
9.         /// <param name="distance"></param>
10.         /// <returns>Point3d</returns>
11.         public static Point3d Polar3D(
12.             Point3d point,
13.             double xyAngle,
14.             double xzAngle,
15.             double distance)
16.         {
17.             double sinXy = Math.Sin(xyAngle),
18.                    cosXy = Math.Cos(xyAngle),
19.                    sinXz = Math.Sin(xzAngle),
20.                    cosXz = Math.Cos(xzAngle);
21.  
22.             return new Point3d(
23.                 point.X + distance*cosXy*cosXz,
24.                 point.Y + distance*sinXy*cosXz,
25.                 point.Z + distance*sinXz);
26.         }
27.  

[Kerry]

Something to play with :

Code - C#: [Select]

1. // (C) Copyright 2014 by KdubServices  
2. // Proof of concept ; theSwamp KdubServices 2015/09/05
3. // http://www.theswamp.org/index.php?topic=47778.msg527920#msg527920
4.  
5. using System;
6. using Autodesk.AutoCAD.Runtime;
7. using Autodesk.AutoCAD.ApplicationServices;
8. using Autodesk.AutoCAD.Geometry;
9.  
10. [assembly: CommandClass(typeof(Polar3DTest.MyCommands))]
11.  
12. namespace Polar3DTest
13. {
14.     public class MyCommands
15.     {
16.         public static double Deg2Rad(double degrees) { return degrees *  (Math.PI / 180.0); }
17.         public static double Rad2Deg(double radians) { return radians * (180.0 / Math.PI); }
18.  
19.         [CommandMethod("Test02", CommandFlags.Modal)]
20.         public void Test02()
21.         {
22.             var ed = Application.DocumentManager.MdiActiveDocument.Editor;
23.  
24.             Vector3d v3D = GetVector3D(Deg2Rad(15.0), Deg2Rad(45.0));
25.             ed.WriteMessage(
26.                             string.Format(
27.                                           "\nVector3D 15:45 :\nX:{0}\nY:{1}\nZ:{2}",
28.                                           v3D.X,
29.                                           v3D.Y,
30.                                           v3D.Z)
31.                 );
32.         }
33.         /// <summary>
34.         /// Return the Vector3d from angles in XY and XZ planes
35.         ///  
36.         /// </summary>
37.         /// <param name="xyAngle"></param>
38.         /// <param name="xzAngle"></param>
39.         /// <returns></returns>
40.         public static Vector3d GetVector3D(
41.             double xyAngle,
42.             double xzAngle
43.             )
44.         {
45.             double sinXy = Math.Sin(xyAngle),
46.                    cosXy = Math.Cos(xyAngle),
47.                    sinXz = Math.Sin(xzAngle),
48.                    cosXz = Math.Cos(xzAngle);
49.             return new Vector3d(
50.                 cosXy * cosXz,
51.                 sinXy * cosXz,
52.                 sinXz);
53.         }
54.     }
55. }
56.  

Code - Text: [Select]

1. Command:
2. Command: NETLOAD
3. Command: TEST02
4. Vector3D 15:45 :
5. X:0.683012701892219
6. Y:0.183012701892219
7. Z:0.707106781186547
8.  

edit kdub :> fixed error in Deg2Rad()

[Kerry]

I was testing the result with my calculator (HP35S) and had a moment until I realised that I was testing manually with decimal degrees instead of radians.
Easy to test because the vector length is expected to be 1.

Can be shown by :
note the angles differ from original sample

Code - C#: [Select]

1.         [CommandMethod("Test03", CommandFlags.Modal)]
2.         public void Test03()
3.         {
4.             var ed = Application.DocumentManager.MdiActiveDocument.Editor;
5.  
6.             Vector3d v3D = GetVector3D(Deg2Rad(15.0), Deg2Rad(30.0));
7.             ed.WriteMessage(
8.                             string.Format(
9.                                           "\nVector3D 15:30 :\nX:{0}\nY:{1}\nZ:{2}\nLength:{3}",
10.                                           v3D.X,
11.                                           v3D.Y,
12.                                           v3D.Z,
13.                                           v3D.Length)
14.                 );
15.         }
16.  
17.  

Code: [Select]

Command: TEST03
Vector3D 15:30 :
X:0.836516303737808
Y:0.224143868042013
Z:0.5
Length:1




[sybold]

that's great!

was totally missing the distance factor, had most of the rest.
thanks for helping out

[Kerry]

You're welcome.

[sybold]

just a note, deg2rad should be

Code: [Select]

public static double Deg2Rad(double degrees)
{
return degrees * Math.PI / 180.0;
}


[Kerry]

Sorry, I thought I'd repaired that.
... will fix now.

You can also use an extension method

Code - C#: [Select]

1. /// <summary>
2. /// Convert to Radians.
3. /// </summary>
4. /// <param name="deg">The value to convert to radians</param>
5. /// <returns>The value in radians</returns>
6. public static class NumericExtensions
7. {
8.     public static double ToRadians(this double deg)
9.     {
10.         return (Math.PI / 180.0) * deg;
11.     }
12. }
13.  
14.  

Code - C#: [Select]

1.     Vector3d v3D = GetVector3D(15.0.ToRadians(), 30.0.ToRadians());

[Bryco]

return degrees * Math.PI / 180.0; vs return degrees * Math.PI / 180;
I have been caught badly before by the lack of a .0
but a quick test shows int/int=int
double/int=double  int/double=double so 180 will be fine
Util.Debug.Print(3.0 / 5); =0.6
            Util.Debug.Print(3 / 5);=0
            Util.Debug.Print((double) 3 / 5);=0.6
            Util.Debug.Print((double)(3 / 5));=0
            Util.Debug.Print(3 / 5.0);=0.6