Sengna Posted May 29, 2014 Share Posted May 29, 2014 Hi, just a simple question, i had a tough time drawing an arc or circle to make it tangent to the top horizaltal line, i used TTR for circle, it tangened with the vertical line but not the top line, when i use TTR for circle it will ask to specify Radius of the circle i just throw out the random R? what is the trick to calculate the radius? when used arc command it looks awkward. is there other tool that i can use? Quote Link to comment Share on other sites More sharing options...
rkent Posted May 29, 2014 Share Posted May 29, 2014 There are any number of R that would work, you have to know what you want as the result. Also, I would use Fillet command unless there is more to this than meets the eye. Quote Link to comment Share on other sites More sharing options...
Sengna Posted May 29, 2014 Author Share Posted May 29, 2014 There are any number of R that would work, you have to know what you want as the result. Also, I would use Fillet command unless there is more to this than meets the eye. the result i want is that the end of the horizontal and vertical lines tangent with the circle. Fillet would work as well but again how do i guess what would be the appropriate radius? Quote Link to comment Share on other sites More sharing options...
BIGAL Posted May 29, 2014 Share Posted May 29, 2014 Two solutions then rad1 = tan end of line 1, rad2 = tan end of line 2, only time rad1 = rad2 is when both ends are the same distance from intersection point, anglea does not come into it, drop line of desired end do fillet R 0, DI end to intersection is radius, fillet again with new R delete temp line. Someone will have a lisp to do this. Quote Link to comment Share on other sites More sharing options...
Dana W Posted May 29, 2014 Share Posted May 29, 2014 Two solutions then rad1 = tan end of line 1, rad2 = tan end of line 2, only time rad1 = rad2 is when both ends are the same distance from intersection point, anglea does not come into it, drop line of desired end do fillet R 0, DI end to intersection is radius, fillet again with new R delete temp line. Someone will have a lisp to do this.I don't understand what rad1 and rad2 are. The only time the distance from tan to line intersect equals the radius of the fillet is when the angle between the lines is exactly 90 degrees, but the distance from tan to point of line intersect on both lines will always be equal, no matter what the angle between the lines. If the angle between the lines is not equal to 90 degrees, the radius of the fillet will never be equal to the distance from tan to the point of intersect. Quote Link to comment Share on other sites More sharing options...
eldon Posted May 29, 2014 Share Posted May 29, 2014 I think that you should explain a bit further. Are the ends of the two lines fixed in place? You could use a polyline and define the end tangent directions. A circle with TTR and Fillet produce basically the same result, but you have to know the radius before you start. If you used TTR in the approved manner, it would be impossible to produce circles which were not tangential to the picked lines. Quote Link to comment Share on other sites More sharing options...
BIGAL Posted May 29, 2014 Share Posted May 29, 2014 Dana a bit more hopefully makes sense to what I posted before if the distance from the end points to the intersection is different then there will always be two answers if you want to use a fixed end point of 1 of the lines hopefully this image makes sense, I think the required answer is pick 1 line as the control pt for auto radius with other line. Quote Link to comment Share on other sites More sharing options...
JD Mather Posted May 29, 2014 Share Posted May 29, 2014 Hi, just a simple question, .... A simple request - attach your file here. The way I read your question (line endpoints fixed) there might not be a solution for a curve tangent to both lines and connected at both line endpoints. In the TTR image the circle IS tangent to the "top" line by the definition of tangent, it is not coincident at any point. Quote Link to comment Share on other sites More sharing options...
ReMark Posted May 29, 2014 Share Posted May 29, 2014 Someone obviously forgot the meaning of the word tangent. The next request will be for a lisp routine to fix the "problem". One should not be "guessing" what the radius should be. They should know what radius they want to use. Here's a solution. Use the Fillet command instead and specify the radius. Quote Link to comment Share on other sites More sharing options...
Dana W Posted May 29, 2014 Share Posted May 29, 2014 (edited) Dana a bit more hopefully makes sense to what I posted before if the distance from the end points to the intersection is different then there will always be two answers if you want to use a fixed end point of 1 of the lines hopefully this image makes sense, I think the required answer is pick 1 line as the control pt for auto radius with other line. [ATTACH=CONFIG]49162[/ATTACH] I see what you are saying now. The picture shows two examples of what I was trying to explain as well. However, if the distance to the intersection point from the line ends is NOT equal, then the Cad Operator has made an error. One cannot fillet to two fixed end points with a circular arc in that manner. That solution would require an elliptical curve, which is only generically a FILLET, not a geometric one. Edited May 29, 2014 by Dana W Quote Link to comment Share on other sites More sharing options...
Dana W Posted May 29, 2014 Share Posted May 29, 2014 (edited) The way I read your question (line endpoints fixed) there might not be a solution for a curve tangent to both lines and connected at both line endpoints. The only condition that there would not be a solution for is if there were a fixed radius that would not fit the distance presented by the fixed end points of the lines. Any two lines, at any angle from each other, with endpoints fixed anywhere in space can have a perfect fillet tangent to both, with the fillet radius being the variable. Edited May 29, 2014 by Dana W Quote Link to comment Share on other sites More sharing options...
Dana W Posted May 29, 2014 Share Posted May 29, 2014 the result i want is that the end of the horizontal and vertical lines tangent with the circle. Fillet would work as well but again how do i guess what would be the appropriate radius?No, the result you want is an accurate drawing that is useful. Guess? That is just the point. You do not guess the radius. If one is drawing up a machine part for instance, one needs ALL the dimensions prior to starting the drawing. Now, if you are practicing drawing this condition, and you do not have a set of specifications to go by, you have put yourself in a bind by drawing two lines with fixed end points. Just draw the lines all the way to the intersecting point, then TTR a circle with ANY radius > 0 to both of them. Then trim away the unwanted parts of the lines and circle. Quote Link to comment Share on other sites More sharing options...
rkent Posted May 29, 2014 Share Posted May 29, 2014 what is the trick to calculate the radius? [ATTACH=CONFIG]49151[/ATTACH] To answer your question (which appears to involve two lines 90 degrees apart) when using Circle, TTR, pick the vertical line, then the horzontal line, for radius use apparent intersection osnap, pick both lines, finally pick the end point of one of the lines. If there is more to this question please provide details. Quote Link to comment Share on other sites More sharing options...
Dana W Posted May 30, 2014 Share Posted May 30, 2014 The only condition that there would not be a solution for is if there were a fixed radius that would not fit the distance presented by the fixed end points of the lines. Any two lines, at any angle from each other, with endpoints fixed anywhere in space can have a perfect fillet tangent to both, with the fillet radius being the variable.OK so my statement is only true if the end points of both lines is equidistant from the apparent intersection point of the lines. Yes, one would need at least two arcs or maybe an elliptical arc to fillet two lines if their fixed end points were different distances from the apparent intersection. I must have been half asleep after a looooong day. I had to draw up an example or two myself to see what a load I dropped. Quote Link to comment Share on other sites More sharing options...
JD Mather Posted May 30, 2014 Share Posted May 30, 2014 T.... Any two lines, at any angle from each other, with endpoints fixed anywhere in space can have a perfect fillet tangent to both, with the fillet radius being the variable. Hmmm, Geometry constraints in Inventor, SolidWorks, Creo and AutoCAD would seem to disagree. It appears that the two fixed points have to meet special condition. Quote Link to comment Share on other sites More sharing options...
BIGAL Posted May 30, 2014 Share Posted May 30, 2014 I think we need to slow down or give up, the responses are starting to out way the original question, so I think we should wait for Senga to reply again before we guess any more. Quote Link to comment Share on other sites More sharing options...
Dana W Posted May 30, 2014 Share Posted May 30, 2014 Hmmm, Geometry constraints in Inventor, SolidWorks, Creo and AutoCAD would seem to disagree. [ATTACH]49178[/ATTACH] It appears that the two fixed points have to meet special condition. Yeah, I realized later what I posted only applies to the special condition of both fixed endpoints being equidistant from the apparent intersection. My, this thread has developed legs. Quote Link to comment Share on other sites More sharing options...
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