Vector to Angle , and vice versa

Jarvis Swift · 2012-06-17T14:33:02

I'm currently trying to figure out how to convert XYZ to PYR. Here are my classes. Also i'm currently looking it up. I know of sin/cos. I'm learning what i can from online docs. Vector: [SPOILER] #region Properties public float X; public float Y; public float Z; public float Forward { get { return Y; } set { Y = value; } } public float Right { get { return X; } set { X = value; } } public float Up { get { return Z; } set { Z = value; } } #endregion #region Constructors public Vector( ) { X = 0; Y = 0; Z = 0; } public Vector( float x , float y ) { X = x; Y = y; Z = 0; } public Vector( float x , float y , float z ) { X = x; Y = y; Z = z; } #endregion #region Methods public Vector Normalize( ) { } #endregion #region Operators public static Vector operator +( Vector vector1 , Vector vector2 ) { return new Vector ( vector1.X + vector2.X , vector1.Y + vector2.Y , vector1.Z + vector2.Z ); } public static Vector operator -( Vector vector1 , Vector vector2 ) { return new Vector ( vector1.X - vector2.X , vector1.Y - vector2.Y , vector1.Z - vector2.Z ); } public static Vector operator /( Vector vector1 , Vector vector2 ) { return new Vector ( vector1.X / vector2.X , vector1.Y / vector2.Y , vector1.Z / vector2.Z ); } public static Vector operator *( Vector vector1 , Vector vector2 ) { return new Vector ( vector1.X * vector2.X , vector1.Y * vector2.Y , vector1.Z * vector2.Z ); } public static void operator ++( Vector vector ) { vector.Z++; } public static void operator --( Vector vector ) { vector.Z--; } #endregion [/SPOILER] Angle: [SPOILER] #region Properties public float P; public float Y; public float R; public float Pitch { get { return Pitch; } set { P = value; } } public float Yaw { get { return Yaw; } set { Y = value; } } public float Roll { get { return Roll; } set { R = value; } } #endregion #region Constructors public Angle( ) { P = 0; Y = 0; R = 0; } public Angle( float pitch , float yaw ) { P = pitch; Y = yaw; R = 0; } public Angle( float pitch , float yaw , float roll ) { P = pitch; Y = yaw; R = roll; } #endregion #region Methods public Vector Forward( ) { } public Vector Right( ) { } public Vector Up( ) { } #endregion [/SPOILER]

frob

46,221

June 11, 2012 07:37 PM

The goal is to convert Pitch , Yaw , Roll to X , Y , Z and then back.
So Angle( 0 , 0 , 0 ) would be in XYZ ( 0 , 1 , 0 ).
So Angle( 0 , 90 , 0 ) would be in XYZ ( 1 , 0 , 0 ).
So Angle( 0 , 180 , 0 ) would be in XYZ ( 0 , -1 , 0 ).
So Angle( 0 , 270, 0 ) would be in XYZ ( -1 , -1 , 0 ).
So Angle( 0 , 360, 0 ) would be in XYZ ( 0 , 1 , 0 ).

Your numbers don't make sense. How can (0,270,0) be (-1,-1,0) ?

Perhaps it is time for to go find a primer on 3D mathematics?

As for converting between them, it looks like you want to project one unit along the forward vector of the orientation and find that position.
As for reversing it, there are infinitely many orientations that can generate the forward vector.

DarkScience

36

Author

June 11, 2012 07:41 PM

[quote name='DarkScience' timestamp='1339442924' post='4948262']
The goal is to convert Pitch , Yaw , Roll to X , Y , Z and then back.
So Angle( 0 , 0 , 0 ) would be in XYZ ( 0 , 1 , 0 ).
So Angle( 0 , 90 , 0 ) would be in XYZ ( 1 , 0 , 0 ).
So Angle( 0 , 180 , 0 ) would be in XYZ ( 0 , -1 , 0 ).
So Angle( 0 , 270, 0 ) would be in XYZ ( -1 , -1 , 0 ).
So Angle( 0 , 360, 0 ) would be in XYZ ( 0 , 1 , 0 ).

Your numbers don't make sense. How can (0,270,0) be (-1,-1,0) ?

Perhaps it is time for to go find a primer on 3D mathematics?

As for converting between them, it looks like you want to project one unit along the forward vector of the orientation and find that position.
As for reversing it, there are infinitely many orientations that can generate the forward vector.
[/quote]

Sorry wrote that in a hurry to represent the directions in relation to x,y,z not the proper numbers.

DarkScience

36

Author

June 11, 2012 07:43 PM

Alvaro is correct: a triplet of pitch, yaw and roll (Euler angles) represents a 3D orientation. An euclidean 3D vector can at most represent a direction in 3D space (plus a magnitude). That direction can be e.g. an axis of rotation, or a look-at direction, but it does not capture a full 3D orientation.

If representing an axis of rotation with a 3D euclidean vector, one needs to add in a single scalar (the angle of rotation about that axis) to make it a complete representation of a 3D orientation. This is commonly called an axis-angle representation of an orientation. If you are looking to convert an axis-angle representation to a 3D rotation matrix (and then to an Euler triplet), see e.g. MathGeoLib's float3x3::RotateAxisAngle function (boils down to Quat::SetFromAxisAngle).

If representing a look-at direction with a 3D euclidean vector, one needs to add in the concept of a 'up' vector (also called a 'world up' or a 'scene up' vector, a fixed vector specifying the upwards direction convention in the scene), or a single scalar 'roll' (being equivalent to above axis-angle) to make it a complete representation of a 3D orientation. This orientation however cannot have any roll, if the up vector is kept static. If you are looking how to turn a look-at direction vector into a 3D orientation, see e.g. MathGeoLib's float3x3::LookAt.

To convert a rotation matrix to pitch-yaw-roll components (Euler angles), see e.g. MathGeoLib's float3x3::ToEuler*** conversion functions.

Alvaro is incorrect. I explained that im looking for the direction not the absolute full 3d orientation. He should of figured given the deffinition of Angle. Alvaro thanks for this stressful encounter. Go to hell.

clb thanks for an actual reply to my inquiry. Though go to hell for saying "Alvaro is correct".

Sorry i thought he would understand that i'm trying to get a Angle not a Position or Vector.

DarkScience

36

Author

June 11, 2012 07:46 PM

Oh and Alarvo http://en.wikipedia.org/wiki/Angle

frob

46,221

June 11, 2012 07:47 PM

Each of your posts say you want to go back again. Only one direction makes sense.

You can project one unit along your orientation to find a 'forward' normal. That is a matter of simple trig.

The other direction, finding an orientation that generates the same forward normal, does not really make sense. There are infinitely many solutions.

clb

2,152

June 11, 2012 07:49 PM

The goal is to convert Pitch , Yaw , Roll to X , Y , Z and then back.
So Angle( 0 , 0 , 0 ) would be in XYZ ( 0 , 1 , 0 ).
So Angle( 0 , 90 , 0 ) would be in XYZ ( 1 , 0 , 0 ).
So Angle( 0 , 180 , 0 ) would be in XYZ ( 0 , -1 , 0 ).
So Angle( 0 , 270, 0 ) would be in XYZ ( -1 , -1 , 0 ).
So Angle( 0 , 360, 0 ) would be in XYZ ( 0 , 1 , 0 ).

This sounds like you're trying to define a 3D orientation based on a LookAt direction of an object. See if the functions in my previous post can help you with that (float3x3::LookAt, and then float3x3::ToEuler***).

DarkScience

36

Author

June 11, 2012 07:51 PM

Each of your posts say you want to go back again. Only one direction makes sense.

What ???

You can project one unit along your orientation to find a 'forward' normal. That is a matter of simple trig.

Thats what i'm asking how to do. I've been getting an output which i posted here.

first set
[spoiler]

double A_P = math.DegreeToRadian ( P ); // 0 to 360
double A_Y = math.DegreeToRadian ( Y );// 0 to 360
double A_R = math.DegreeToRadian ( R );// 0 to 360

normal.X = ( Math.Sin ( A_Y ) );
normal.Y = ( Math.Cos ( A_Y ) );
normal.Z = ( Math.Tan ( A_P ) );

seems to give.

Angle: {P=0,Y=0,R=0}
Forward: {X=0,Y=1,Z=0}

Angle: {P=0,Y=90,R=0}
Forward: {X=1,Y=6.12303176911189E-17,Z=0}

Angle: {P=0,Y=180,R=0}
Forward: {X=1.22460635382238E-16,Y=-1,Z=0}

Angle: {P=0,Y=270,R=0}
Forward: {X=-1,Y=-1.83690953073357E-16,Z=0}

Angle: {P=0,Y=360,R=0}
Forward: {X=-2.44921270764475E-16,Y=1,Z=0}
[/quote]

Y seems to be correct at 0,180,360 but incorrect in between.
X seems to be correct at 0,90,270 but incorrect in between and oddly enough at 360.
[/spoiler]

second set
[spoiler]

double A_P = P; //math.DegreeToRadian ( P );
double A_Y = Y; //math.DegreeToRadian ( Y );
double A_R = R; //math.DegreeToRadian ( R );

normal.X = ( Math.Sin ( A_Y ) );
normal.Y = ( Math.Cos ( A_Y ) );
normal.Z = ( Math.Tan ( A_P ) );
[/quote]

seems to give

Angle: {P=0,Y=0,R=0}
Forward: {X=0,Y=1,Z=0}

Angle: {P=0,Y=90,R=0}
Forward: {X=0.893996663600558,Y=-0.44807361612917,Z=0}

Angle: {P=0,Y=180,R=0}
Forward: {X=-0.80115263573383,Y=-0.598460069057858,Z=0}

Angle: {P=0,Y=270,R=0}
Forward: {X=-0.176045946471211,Y=0.984381950632505,Z=0}

Angle: {P=0,Y=360,R=0}
Forward: {X=0.958915723414307,Y=-0.283691091486527,Z=0}
[/quote]

Y is correct at 0 and incorrect every where else.
X is correct at 0 and incorrect every where else.

though the values are -1 to 1 which is the correct format.

[/spoiler]
[/quote]

PYR are in degrees in second set...

Seems to be a slight inaccuracy. I cant get it working properly even though from what i know second set should work fine.

Ignore the Z = nonsense. Its just a placeholder.

EDIT The issue is that when X or Y is suppose to be full (-1 ,0, 1) they are but the other number is 6.E~~~ (first set)

EDIT AGAIN:

Remember that 6.12303176911189E-17 is 0.00000000000000006 (I may have even missed a zero there!) so it is a very, very small deviation.[/quote]

That's the issue i was having, now that i know what it is i can fix it. So my calculations are correct. Just a issue that i must work around.

Also as to the -1 rating. Thanks. It means alot to me.

DarkScience

36

Author

June 11, 2012 08:42 PM

SOLUTION



private PointF DegreesToXY( float degrees , float radius , Point origin )

  {

   PointF xy = new PointF ( );

   double radians = degrees * Math.PI / 180.0;

   double x = Math.Cos ( radians ) * radius + origin.X;

   double y = Math.Sin ( -radians ) * radius + origin.Y;

			xy.X = (float) Math.Round ( x * 1000000 ) / 1000000;

			xy.Y = (float) Math.Round ( y * 1000000 ) / 1000000;

   return xy;

  }

or
[spoiler]



public static double DegreesToX( double degrees , double radius )

		{

			double radians = degrees * ( Math.PI / 180.0 );



			double x = Math.Cos ( radians ) * radius;



			if ( x < 0.1 )

				x = Math.Round ( x * 100000000000000 ) / 100000000000000;



			return x;

		}

		public static float DegreesToXF( float degrees , float radius )

		{

			float radians = degrees * (float) ( Math.PI / 180.0 );



			float x = (float) Math.Cos ( radians ) * radius;



			if ( x < 0.1 )

				x = (float) Math.Round ( x * 1000000 ) / 1000000;



			return x;

		}



		public static double DegreesToY( double degrees , double radius )

		{

			double radians = degrees * ( Math.PI / 180.0 );



			double y = Math.Sin ( -radians ) * radius;



			if ( y < 0.1 )

				y = Math.Round ( y * 100000000000000 ) / 100000000000000;



			return y;

		}

		public static float DegreesToYF( float degrees , float radius )

		{

			float radians = degrees * (float) ( Math.PI / 180.0 );



			float y = (float) Math.Sin ( -radians ) * radius;



			if ( y < 0.1 )

				y = (float) Math.Round ( y * 1000000 ) / 1000000;



			return y;

		}

[/spoiler]

alvaro

21,604

June 15, 2012 02:30 PM

That's a solution to what exactly? I don't see a conversion from XYZ to PYR anywhere.

The strange rounding you do at the end will give you the illusion that the result is more precise, but you are actually throwing precision away in general.

_WeirdCat_

1,512

June 17, 2012 02:33 PM



float __fastcall T3DPlanet::GetLongitude(t3dpoint cpos)   //poludniki (Dlugosc)

{

float	   angle;

angle = n2dGetPolarCoordAngleA(cpos.x-pos.x,cpos.z-pos.z) / 6.2731860f;



angle = -360.0f*angle;

angle = VALIDUJ(angle)+90.0f;

angle = VALIDUJ(angle);

	return angle;



}



float __fastcall T3DPlanet::GetLatitude(t3dpoint cpos)

{

t3dpoint cpn;

cpn = vectorAB(pos,cpos);

cpn = Normalize(cpn);

float t = cpn.x*cpn.x + cpn.z*cpn.z;

t = sqrt(t);

	   if (cpn.y == 0.0f) return 0.0f;

float tanges = t / cpn.y;

float angle;

   angle = atan(tanges);

	 angle = RadToDeg( angle );

	 if (angle < 0) angle = -1.0f*(90.0f + angle); else

	 angle = 90.0f - angle;

  return angle;

}

float __fastcall n2dGetPolarCoordAngleA(float x,float y)

{

if ( (x > 0) && (y >= 0) ) {  return ArcTan(y/x);		}

if ( (x > 0) && (y < 0) )  {  return ArcTan(y/x)+2.0*3.1415926535897932384626433832795;	 }

if  (x < 0)				{ return ArcTan(y/x)+3.1415926535897932384626433832795;	  }

if ( (x == 0) && (y > 0) )  {  return 3.1415926535897932384626433832795/2.0;				   }

if ( (x == 0) && (y < 0) )  {  return 3.0*3.1415926535897932384626433832795/2.0;				 }

//last versions were without .0 just simple 2 division

return - 1000.0;

}

pos is a point(0,0,0)

glop is acually heading angle! imopi =pi / 180



SUPERPOINT.x = 100000.0*(sin((glop)*imopi)*cos((heading)*imopi));

SUPERPOINT.z = 100000.0*(cos((glop)*imopi)*cos((heading)*imopi));

SUPERPOINT.y = 100000.0*(sin((heading)*imopi));

then the vector is +SUPERPOINT.x,
-SUPERPOINT.y,-SUPERPOINT.z

need more ?

Master & Mentor
https://sites.google.com/site/customprog/

Vector to Angle , and vice versa

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