In the OpenSCAD documentation for minkowski(), examples are given for 2
child objects, ref: https://en.wikibooks.org/wiki/OpenSCAD_User_Manual/Transformations#minkowski But what if you supply 3 or more, for example: $fn=50; minkowski() { cube([10,10,10]); cylinder(r=2,h=10); sphere(r=4); } This seems to be accepted, but it is not obvious what happens in this case. It looks like the same as below? $fn=50; minkowski() { minkowski() { cube([10,10,10]); cylinder(r=2,h=10); } sphere(r=4); } In other words, when more than 2 child objects are supplied the child objects are subject to a series of implicit minkowski operations until the list is exhausted. Is this a correct interpretation? Carsten Arnholm _______________________________________________ OpenSCAD mailing list [hidden email] http://lists.openscad.org/mailman/listinfo/discuss_lists.openscad.org 
I don't think the order matters does it? Basically the result is the set with all combinations of points added together. On 18 March 2018 at 11:41, Carsten Arnholm <[hidden email]> wrote: In the OpenSCAD documentation for minkowski(), examples are given for 2 child objects, ref: https://en.wikibooks.org/wiki/ _______________________________________________ OpenSCAD mailing list [hidden email] http://lists.openscad.org/mailman/listinfo/discuss_lists.openscad.org 
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On Mar 18, 2018, at 7:41 AM, Carsten Arnholm <[hidden email]> wrote:
> > In other words, when more than 2 child objects are supplied the child objects are subject to a series of implicit minkowski operations until the list is exhausted. Is this a correct interpretation? > Nophead is right: Minkowski sums are commutative, but they’re implemented by accumulating binary sums as you describe. The exception is if all operators are convex. In that case we can exploit the fact that the "convex hulls of a minkowski sum” equals the “minkowski sum of convex hulls”. ..and simply collect vertices and calculate a single convex hull: https://en.wikipedia.org/wiki/Minkowski_addition Marius _______________________________________________ OpenSCAD mailing list [hidden email] http://lists.openscad.org/mailman/listinfo/discuss_lists.openscad.org 
On 18. mars 2018 14:52, Marius Kintel wrote:
> Nophead is right: Minkowski sums are commutative, but they’re implemented by accumulating binary sums as you describe. > > The exception is if all operators are convex. In that case we can exploit the fact that the "convex hulls of a minkowski sum” equals the “minkowski sum of convex hulls”. ..and simply collect vertices and calculate a single convex hull: > https://en.wikipedia.org/wiki/Minkowski_addition Thanks, that makes it clearer. Perhaps the Wiki could mention minkowski with more than 2 operators and that the order is not significant (i.e. commutative). Ok, I see that if all operators are convex then the minkowski sum is convex as well, and the calculation becomes much simpler. Good point. Carsten Arnholm _______________________________________________ OpenSCAD mailing list [hidden email] http://lists.openscad.org/mailman/listinfo/discuss_lists.openscad.org 
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