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Metalworking (rec.crafts.metalworking) Discuss various aspects of working with metal, such as machining, welding, metal joining, screwing, casting, hardening/tempering, blacksmithing/forging, spinning and hammer work, sheet metal work. |
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#1
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A Planar Geometry Problem
On Tuesday, May 3, 2011 5:32:27 PM UTC-5, Tim Wescott wrote:
The immediate problem is for making pseudo-ellipsis in a cheap CAD program (Qcad), so one could even restrict the two 'master' circles to being the same diameter. But there are times when I'd like to use different sized circles for this (imagine a triangle with rounded corners, and 'puffy cheeks'). I can see an algebraic approach for this involving over use of the Pythagorean theorem, but I'd rather know if there's a solution that can be done by construction, not by numbers. I assume your two circles are equidistant from your point, maybe like this if you are trying to construct an ellipse: http://i.imgur.com/na3nu.png Point A is what you want the arc to pass through, and Point B is the arc's center found by trial and error until the arc's size allows it to both pass through the point and be tangent to the circles to make the long side of your ellipse, like so: http://i.imgur.com/g7TUh.png Ken |
#2
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A Planar Geometry Problem
Ken Grunke wrote: On Tuesday, May 3, 2011 5:32:27 PM UTC-5, Tim Wescott wrote: The immediate problem is for making pseudo-ellipsis in a cheap CAD program (Qcad), so one could even restrict the two 'master' circles to being the same diameter. But there are times when I'd like to use different sized circles for this (imagine a triangle with rounded corners, and 'puffy cheeks'). I can see an algebraic approach for this involving over use of the Pythagorean theorem, but I'd rather know if there's a solution that can be done by construction, not by numbers. I assume your two circles are equidistant from your point, maybe like this if you are trying to construct an ellipse: http://i.imgur.com/na3nu.png With that particular set-up its fairly easy to do with compass and ruler. Let R equal radius of the 2 circles Find the distance R from A that is on centerline (call it point P) draw a line L1 from P to center of one of the circles Create a perpendicular line L2 from the mid point of L The point where the line L2 intersects the centerline is the center of the arc tangent to the two circles through point A -jim Point A is what you want the arc to pass through, and Point B is the arc's center found by trial and error until the arc's size allows it to both pass through the point and be tangent to the circles to make the long side of your ellipse, like so: http://i.imgur.com/g7TUh.png Ken |
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