I think you just need to use offset() to expand the shape in 2D before
extruding it to 3D if I understand the question right.

On 21 November 2016 at 12:05, nigeljohnson73 nigel@nigeljohnson.net wrote:

Nigel,

There are some parameters to offset() which can be used to control sharp corners, see:
https://en.wikibooks.org/wiki/OpenSCAD_User_Manual/Transformations#offset

-Marius

Check out http://www.thingiverse.com/thing:8483 (this tells you how to make
a heart using Bezier curves) and http://www.thingiverse.com/thing:186660
(tells you how to extrude along a path).

Combine the two to extrude a square or rectangle along a heart-shaped path
and you should have what you're after?

Frank

On Wed, Nov 23, 2016 at 5:37 AM, nigeljohnson73 nigel@nigeljohnson.net
wrote:

Another way to achieve this is with the 2-D minkowski which is
moderately efficient.

function heartRadius(t) = (1.5 - 1.5* sin(t) + (sin(t) *
(sqrt(abs(cos(t)))/(sin(t)+1.6))));
$fn=23;

module pHeart(xy,off=0) {

td = 1; // theta delta - the resolution of the curve

// Calculate x/y co-ords separetely so I can scale correctly later
xp = [ for (t = [0: td : 360-td]) heartRadius(t) * cos(t) ]; // xco-ordinates on the curve
yp = [ for (t = [0 : td : 360-td]) heartRadius(t) * sin(t) ]; // yco-ordinates on the curve

xsc = xy/(max(xp) - min(xp)); // x scaler to get to fit expected size
ysc = xy/(max(yp) - min(yp)); // y scaler to get to fit expected size

// Generate a vector of points for the polygon to work
p = [ for (i = [0 : 360/td-1]) [xsc*xp[i], ysc*yp[i]] ];

// translate polygon so it is centralised at +/- xy/2	
translate([0,ysc*(-max(yp)+(max(yp)-min(yp))/2)])
if(off)
{    
minkowski(){ polygon(p); circle(off); }
}
else polygon(p);

}

difference()
{
linear_extrude(5) pHeart(xy=78,off=2);
translate([0,0,-.5]) linear_extrude(6) pHeart(xy=78, off=.4);
}

On Mon, 21 Nov 2016 13:36:36 +0000
nop head nop.head@gmail.com wrote:

Note that 2D minkowski is equivalent to 2D offset with a radius, but 2D offset should be considerably faster.

-Marius

On Wed, 23 Nov 2016 20:38:02 -0500
Marius Kintel marius@kintel.net wrote:

Testing the code shows no difference in speed.  The test code is
below.  Make sure to clear the cache!  div should be a power of 2.
Incidentally, as div gets large both versions show they same failure
artifacts.  I think that the offset module is simply a wrapper around
the minkowski.  In this program they seem to behave
identically to the agree that they have the same identical timed speed
and with the same identical artifacts.  Is there a known algorithm that
can calculate the offset faster than a minkowski algorithm or
that doesn't use the minkowski or some equivalent?

Regards
Otto

function heartRadius(t) = (1.5 - 1.5* sin(t) + (sin(t) *
(sqrt(abs(cos(t)))/(sin(t)+1.6))));
$fn=23;
div=512;  //ONLY USE POWERS OF 2

module pHeart(xy,off=0) {

td = 1/div; // theta delta - the resolution of the curve

// Calculate x/y co-ords separetely so I can scale correctly later
xp = [ for (t = [0: td : 360-td]) heartRadius(t) * cos(t) ]; // xco-ordinates on the curve
yp = [ for (t = [0 : td : 360-td]) heartRadius(t) * sin(t) ]; // yco-ordinates on the curve

xsc = xy/(max(xp) - min(xp)); // x scaler to get to fit expected size
ysc = xy/(max(yp) - min(yp)); // y scaler to get to fit expected size

// Generate a vector of points for the polygon to work
p = [ for (i = [0 : 360/td-1]) [xsc*xp[i], ysc*yp[i]] ];

// translate polygon so it is centralised at +/- xy/2	
translate([0,ysc*(-max(yp)+(max(yp)-min(yp))/2)])

//// UNCOMMENT LINE TO TEST
//// CLEAR THE CACHE FIRST

///////////////////////  MINKOWSKY
//if(off){minkowski(){polygon(p);circle(off);}}else polygon(p);
///////////////////////  OFFSET
//offset(r=off) polygon(p);

}

difference()
{
linear_extrude(5) pHeart(xy=78,off=2);
translate([0,0,-.5]) linear_extrude(6) pHeart(xy=78, off=.4);
}

I might be wrong but I think OpenSCAD uses the clipper library for offset
and presumably CGAL for 2D Minkowski.

The artefacts from low div values come from the polygon generation, not the
offsetting, so it is no surprise they look identical. Provided clipper
obeys $fn they should produce the same results as they are mathematically
identical operations. There is no ambiguity where each vertex should be, so
any differences should only be down to the different number representations.

I am surprised there is little difference in speed but both are very fast
until you get to tens of thousands of vertices.

BTW rather than linear extruding twice and subtracting 3D shapes I think it
is more efficient to subtract the 2D shapes and extrude that.

linear_extrude(5, convexity = 2)
difference() {
pHeart(xy = 78, off = 2);
pHeart(xy = 78, off = 0.4);
}

On 24 November 2016 at 20:29, otto otto@123phase.com wrote:

We use the clipper library for both offset and 2D minkowski. Perhaps they implement offset using minkowski internally?

-Marius

I didn't realise clipper did Minkowski. A quick look at the code reveals
they don't use Minkowski for offset. Minkowski is just the union of a
polygon shifted by each point in a path. Offset is a lot more complex.

On 25 November 2016 at 15:07, Marius Kintel marius@kintel.net wrote:

I agree.

Otto

On Fri, 25 Nov 2016 10:53:27 +0000
nop head nop.head@gmail.com wrote: