#8
(2019-10-30, 2:48)N. W. Perry Wrote: There's another way to do this, by finding the intersection of two circles. This might be handy if you need to do it in a program that doesn't have LDCad's triangle calculation feature (or if, as I did, you forgot that feature exists and spent an hour figuring it out for yourself and are feeling all clever about it). I setup the selection info stuff before the scripting existed and didn't want to force a certain way to apply the angles etc.

But you could script your method, and apply the angles in one go if wanted.

FYI: This is what the selection info does internally:

Code:
```        const double a=relPnt.length(relPnt); const double b=relPnt.length(relPnt); const double c=relPnt.length(relPnt); const double angAB=rad2Deg(acos((a*a+b*b-c*c)/(2*a*b))); const double angBC=rad2Deg(acos((b*b+c*c-a*a)/(2*b*c))); TGLVector3d BADir=relPnt-relPnt; BADir.normalize(); TGLVector3d BCDir=relPnt-relPnt; BCDir.normalize(); TGLVector3d CBDir=BCDir*-1.0; TGLVector3d CDDir=relPnt-relPnt; CDDir.normalize(); TGLVector3d planeNor(true); planeNor.comp[sCompIdx]=1.0; const double intAngAB=rad2Deg(planeNor.calcSignedAngleBetween(BCDir, BADir)); const double intAngBC=rad2Deg(planeNor.calcSignedAngleBetween(CBDir, CDDir)); const double angABCor=intAngAB+angAB; const double angBCCor=intAngBC-angBC; const double angABCorInv=intAngAB-angAB; const double angBCCorInv=intAngBC+angBC;```
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