Ronaldo wrote > .. although my very > basic knowledge of gear design stuffs can not grasp it. It is not so difficult. In following picture you will see that the right gear will travel a longer distance than the left gear. As they pass the same number of teeth they have equal angular speed. The secret is the profile shift. (+1 right half, -1 left half) <http://forum.openscad.org/file/t887/moeb.png> With a smooth transition the center gear will look like this <http://forum.openscad.org/file/t887/moeb1.png> How do you gain it? When cycling (and appropriately advancing) a toothed rack around a blank circle, the distance is modulated by means of sin(i). Try sin(3*i) instead. gear2D(z=20); Rack(z=20); module gear2D(m = 1, z = 10, x = 0, w = 20, clearance = 0.1) { iterations = 180; r_wk = m*z/2 + x; U = m*z*PI; dy = m; r_fkc = r_wk + dy *(1-clearance/2); s = 360/iterations; difference() { circle(r_fkc+2, $fn=300); // workpiece for(i=[0:s:360]) rotate([0, 0, -i]) translate([-i/360*U, sin(i), 0]) Rack(m, z, x, w, clearance); // Tool } } module Rack(m = 1, z = 10, x = 0, w = 20, clearance = 0) { p = m*PI; dy = 2*m; dx = dy * tan(w); ddx = dx/2 * clearance/2; ddy = dy/2 * clearance/2; r_wk = m*z/2 + x; y0 = r_wk+3*dy; y1 = r_wk+dy/2-ddy; y2 = r_wk+dy/2 - ddy; y3 = r_wk-dy/2 - ddy; x0 = p/4-dx/2 + ddx; x1 = p/4+dx/2 + ddx; x2 = 3*p/4-dx/2 - ddx; x3 = 3*p/4+dx/2 - ddx; polygon(points = tooth(z)); function tooth(z = 10) = concat([[-p, y0],[-p, y1]], [for(i=[-1:z], j=[0:3]) to(i*p)[j]], [[(z+1)*p, y1], [(z+1)*p, y0]]); function to(dx) = [[dx+x0, y2], [dx+x1, y3], [dx+x2, y3], [dx+x3, y2]]; } -- Sent from: http://forum.openscad.org/

That all seems a bit complicated. I think the scale of teeth required for this and the limits to printed resolution mean that precise tooth matching will not be required. I have made a ribbed mobius simply by modulating the size of the slices used to generate it. This should mesh reasonably well with gear teeth on the rollers: mp4 animation on imgur <https://imgur.com/gallery/KeBmY> -- Sent from: http://forum.openscad.org/

​Enlightening! Thank you. So that is the principle behind the elliptical gears? 2017-12-29 19:41 GMT-02:00 Parkinbot <rudolf@parkinbot.com>: > Ronaldo wrote > > .. although my very > > basic knowledge of gear design stuffs can not grasp it. > > > > It is not so difficult. In following picture you will see that the right > gear will travel a longer distance than the left gear. As they pass the > same > number of teeth they have equal angular speed. The secret is the profile > shift. (+1 right half, -1 left half) > <http://forum.openscad.org/file/t887/moeb.png> > > With a smooth transition the center gear will look like this > <http://forum.openscad.org/file/t887/moeb1.png> > > How do you gain it? When cycling (and appropriately advancing) a toothed > rack around a blank circle, the distance is modulated by means of sin(i). > Try sin(3*i) instead. > > gear2D(z=20); > Rack(z=20); > > module gear2D(m = 1, z = 10, x = 0, w = 20, clearance = 0.1) > { > iterations = 180; > r_wk = m*z/2 + x; > U = m*z*PI; > dy = m; > r_fkc = r_wk + dy *(1-clearance/2); > s = 360/iterations; > difference() > { > circle(r_fkc+2, $fn=300); // workpiece > for(i=[0:s:360]) > rotate([0, 0, -i]) > translate([-i/360*U, sin(i), 0]) > Rack(m, z, x, w, clearance); // Tool > } > } > > > module Rack(m = 1, z = 10, x = 0, w = 20, clearance = 0) > { > p = m*PI; dy = 2*m; dx = dy * tan(w); > ddx = dx/2 * clearance/2; ddy = dy/2 * clearance/2; > r_wk = m*z/2 + x; > y0 = r_wk+3*dy; y1 = r_wk+dy/2-ddy; y2 = r_wk+dy/2 - ddy; y3 = > r_wk-dy/2 > - ddy; > x0 = p/4-dx/2 + ddx; x1 = p/4+dx/2 + ddx; x2 = 3*p/4-dx/2 - ddx; x3 = > 3*p/4+dx/2 - ddx; > polygon(points = tooth(z)); > > function tooth(z = 10) = concat([[-p, y0],[-p, y1]], [for(i=[-1:z], > j=[0:3]) to(i*p)[j]], [[(z+1)*p, y1], [(z+1)*p, y0]]); > function to(dx) = [[dx+x0, y2], [dx+x1, y3], [dx+x2, y3], [dx+x3, y2]]; > } > > > > > -- > Sent from: http://forum.openscad.org/ > > _______________________________________________ > OpenSCAD mailing list > Discuss@lists.openscad.org > http://lists.openscad.org/mailman/listinfo/discuss_lists.openscad.org >

​Enlightening! Thank you. So that is the principle behind the elliptical gears?> Welcome! Well, that is more or less the principle of elliptical involute gears with *constant angular* speed for a rack ;-) You can use any other involute gear (made by a rack) as tool, to get a better radius adaption and so on ... Without the constant angular condition being very restrictive you obviouly can get much nicer looking solutions. For proper involute gearing (with instantanious radius limited on the downside) you would use constant speed and some constant profile shift. In this case you can use the involute function which is explicit, but lets you describe only the upper part of a tooth, where the engagement happens, not the foot part. Because of that most gears found e.g. in Thingiverse are +1-gears (+x-gears) which have a lot of friction, while 0-gears minimize relative movement and thus friction. -- Sent from: http://forum.openscad.org/