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angles between two vectors

apanjocko · 2003-10-26T19:15:29

if you have two vectors (in world space), how do you calculate the x, y (you could skip the z angle) between them? i have done like this, but it seems to sometimes give slightly bad results (looks right most of the time though!): // first vector, the surface vector of the terrain _terrainNormal = _terrain.GetTerrainNormal( _x, _y ); // second vector Vector3 tankNormal = new Vector3( 0, 0, -1 ); // calculate angles _angleX = (float) Math.Asin( ( _terrainNormal.X - tankNormal.X ) / ( Math.Sqrt( _terrainNormal.X * _terrainNormal.X + _terrainNormal.Z * _terrainNormal.Z )) ); _angleY = (float) Math.Asin( ( _terrainNormal.Y - tankNormal.Y ) / ( Math.Sqrt( _terrainNormal.Y * _terrainNormal.Y + _terrainNormal.Z * _terrainNormal.Z )) ); that is, im doing the regular sin(angle) = a/b thingy here, and sort of "projecting" the vectors on a 2d picture. like, when calculating angleX i totally forgett about the difference between the vectors in Y. can you do it like this? if you can do like that, maybe there is something wrong with the code that get the terrain normal? this code takes three points and makes it a normal: public Vector3 GetTerrainNormal( float x, float y ) { Vector3 p1 = Vector3.Empty; Vector3 p2 = Vector3.Empty; Vector3 p3 = Vector3.Empty; float fPosX = (x / _spacing); float fPosY = (_terrainHeight-y)/_spacing; int posX = (int) fPosX; int posY = (int) fPosY; float percentX = fPosX - (int) posX; float percentY = fPosY - (int) posY; float balance = percentX + percentY; int triangle; if( balance < 1 ) triangle = 0; else triangle = 1; if( triangle == 0 ) // top left triangle { p1.X = GridXToTerrainX( posX ); p1.Y = GridYToTerrainY( posY ); p1.Z = GridZToTerrainZ( GetGridHeight( posX, posY ) ); p2.X = GridXToTerrainX( posX + 1 ); p2.Y = GridYToTerrainY( posY ); p2.Z = GridZToTerrainZ( GetGridHeight( posX + 1, posY ) ); p3.X = GridXToTerrainX( posX ); p3.Y = GridYToTerrainY( posY + 1 ); p3.Z = GridZToTerrainZ( GetGridHeight( posX, posY + 1 ) ); } else // bottom right triangle { p1.X = GridXToTerrainX( posX ); p1.Y = GridYToTerrainY( posY + 1 ); p1.Z = GridZToTerrainZ( GetGridHeight( posX, posY + 1 ) ); p2.X = GridXToTerrainX( posX + 1 ); p2.Y = GridYToTerrainY( posY ); p2.Z = GridZToTerrainZ( GetGridHeight( posX + 1, posY ) ); p3.X = GridXToTerrainX( posX + 1 ); p3.Y = GridYToTerrainY( posY + 1 ); p3.Z = GridZToTerrainZ( GetGridHeight( posX + 1, posY + 1 ) ); } // TextWriter.AddText( " triangle: " + triangle ); // calculate the x and y angles to the terrain plane normal from the tank normal Vector3 v12 = p1 - p2; Vector3 v13 = p1 - p3; Vector3 normalTerrain = Vector3.Cross( v12, v13 ); // TextWriter.AddText( " normal: (" + normalTerrain.X + ", " + normalTerrain.Y + ", " + normalTerrain.Z + ")" ); normalTerrain.Normalize(); return normalTerrain; } thanks for any help! [edited by - apanjocko on October 23, 2003 9:21:32 PM]

General and Gameplay Programming Programming

Started by apanjocko October 23, 2003 01:24 PM

11 comments, last by apanjocko 20 years, 6 months ago

Dredge-Master

175

October 25, 2003 12:39 PM

it works on all of them, just depends on how many dimensions you use. The code that I use has 4 dimensions (yes, that''s x,y,z and that other one w).

but the descrepency that you are getting will still be there, but less unless you use atan2

_angleY = (float) Math.Atan( ( -_terrainNormal.Y + tankNormal.Y ) / tankNormal.Z );

to
_angleY = (float) Math.Atan2( ( -_terrainNormal.Y + tankNormal.Y ) , tankNormal.Z );

should do it.

bascially, the atan2 versions are alot more accurate than atan. It will provide the best accuracy.

when using floating point, atan/acos/asin is accurate to 8 sig figures (when about 1 or 2 degrees off) and atan2 is accurate to its full 20 sig figures.

Fiddle with it

Beer - the love catalystgood ol' homepage

apanjocko

122

Author

October 26, 2003 10:59 AM

thanks,
yeah i got it working i remember but it wasn''t that easy, just converting it strictly like that over atan2 yielded really strange results... had to change some signs here and there.

but what i meant was Optikal''s solution is less than adequate in this case. I will get an angle that is of no use to me whatsoever.

thanks for all help

Dredge-Master

175

October 26, 2003 07:15 PM

no problem

Beer - the love catalystgood ol' homepage

angles between two vectors

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