Calculating normal of a quad
Blitz3D Forums/Blitz3D Programming/Calculating normal of a quad
| ||
| Say I have a xz quad with vertexes of differing y values how do I calculate the normal of the quad? eg. vertex 0 = (0,5,0) vertex 1 = (10,1,0) vertex 2 = (10,7,10) vertex 3 = (0,2,10) Cheers.. |
| ||
| Although the definition of a quad is just 4 vertices usually these vertices are coplanar i.e. the quad is flat. To get the normal of a quad is the same as getting the normal for either of its 2 triangles. Your quad however is not coplanar so it will have two normals, one for each triangle. |
| ||
| Get you. Is there a method that I can use to get the average normal of the quad? I know it has something to do with the cross products. I can't use the inbuilt update normals command for what I'm doing unfortunately. |
| ||
| To find the average normal of the quad I'd guess its just a matter of adding the normals of the two triangles together,divide by 2 then normalise the result. Here's some code which should show you the way... Type tri Field vx#[3],vy#[3],vz#[3] Field tnx#,tny#,tnz# End Type Function createtri(vx1#,vy1#,vz1#,vx2#,vy2#,vz2#,vx3#,vy3#,vz3#) ;make our triangle t.tri=New tri t\vx[1]=vx1 t\vy[1]=vy1 t\vz[1]=vz1 t\vx[2]=vx2 t\vy[2]=vy2 t\vz[2]=vz2 t\vx[3]=vx3 t\vy[3]=vy3 t\vz[3]=vz3 ;subtract vectors ax#=t\vx[2]-t\vx[1] ay#=t\vy[2]-t\vy[1] az#=t\vz[2]-t\vz[1] bx#=t\vx[3]-t\vx[2] by#=t\vy[3]-t\vy[2] bz#=t\vz[3]-t\vz[2] ;calculate face normal using cross product t\tnx=ay*bz-az*by t\tny=az*bx-ax*bz t\tnz=ax*by-ay*bx ;normalize it l#=Sqr(t\tnx*t\tnx+t\tny*t\tny+t\tnz*t\tnz) t\tnx=t\tnx/l t\tny=t\tny/l t\tnz=t\tnz/l End Function |
| ||
| EDIT: Ah, somebody was faster than me :P (And I forgot to normalize the vector :P) Code to calculate the normal of a triangle: vx0#=0: vy0#=5: vz0#=0 vx1#=10: vy1#=1: vz1#=0 vx2#=10: vy2#=7: vz2#=10 px#=vx1-vx0 py#=vy1-vy0 pz#=vz1-vz0 qx#=vx2-vx0 qy#=vy2-vy0 qz#=vz2-vz0 nx#=(py*qz)-(pz*qy) ny#=(pz*qx)-(px*qz) nz#=(px*qy)-(py*qx) Btw, this code is not tested, as I'm not using Blitz3D atm :) |
| ||
| Great - thanks guys!! |
   |