Oh yay I got it! I never thought that finding my left from my right was so bloody hard!
Yes, I know this could be better, but I also still have to do two other axis's and remaim sane while I put in my code that makes this worth-while.
For selected.selecteddata = Each selecteddata ;DO THIS NOW! YOU KNOW WHAT IT IS DYLAN! ;should be easy, you just have to choose the most suitable of the 3 vertex's and use it according to the sheet of note paper that is somewhere in here... If leftx# > VertexX(selected\surface,selected\vert0) Then LeftX# = VertexX(selected\surface,selected\vert0) If leftx# > VertexX(selected\surface,selected\vert1) Then LeftX# = VertexX(selected\surface,selected\vert1) If leftx# > VertexX(selected\surface,selected\vert2) Then LeftX# = VertexX(selected\surface,selected\vert2) If leftx# = 0.0 If leftx# < VertexX(selected\surface,selected\vert0) Then LeftX# = VertexX(selected\surface,selected\vert0) If leftx# < VertexX(selected\surface,selected\vert1) Then LeftX# = VertexX(selected\surface,selected\vert1) If leftx# < VertexX(selected\surface,selected\vert2) Then LeftX# = VertexX(selected\surface,selected\vert2) EndIf ;too many if's... we're all going to die! A jusgement is upon us!!! If rightx# < VertexX(selected\surface,selected\vert0) Then rightx# = VertexX(selected\surface,selected\vert0) If rightx# < VertexX(selected\surface,selected\vert1) Then rightx# = VertexX(selected\surface,selected\vert1) If rightx# < VertexX(selected\surface,selected\vert2) Then rightx# = VertexX(selected\surface,selected\vert2) If rightx# = 0.0 If rightx# > VertexX(selected\surface,selected\vert0) Then rightx# = VertexX(selected\surface,selected\vert0) If rightx# > VertexX(selected\surface,selected\vert1) Then rightx# = VertexX(selected\surface,selected\vert1) If rightx# > VertexX(selected\surface,selected\vert2) Then rightx# = VertexX(selected\surface,selected\vert2) EndIf ; box\width = Rightx# - Leftx# Next
Yes, I know this could be better, but I also still have to do two other axis's and remaim sane while I put in my code that makes this worth-while.