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Fit Circular Compound Curve 
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wtojp2013
Newbie Joined: 21.Oct.2013 Location: Hong Kong Using: autocad Status: Offline Points: 7 
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Posted: 29.Jun.2019 at 03:39 
As the coordinates of two tangent points are fixed by some constraints. I would like to fit a circular curves to joint them together. If begin radius say is 80m. How can I find the second radius which shown in attached file ?uploads/413380/ACADCCurveModel.dwg


philippe JOSEPH
Senior Member Joined: 14.Mar.2011 Location: France Using: AutoCAD Mechanical 2017 Status: Offline Points: 1069 
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Hello wtojp2013, I'm afraid some dimensions are missing to find ONE solution.
I can't open the AutoCAD file but if you don't fix the length of the R80000 radius there will be an infinity of "unknown radius". Eventually try PARAMETRIC CONSTRAINTS to do this but you must fix some other dimension( s ). By doing this you must use EXACT angle dimensions or exact length dimensions or intersection positions. Are the basic dimensions : 36470 , 21098 , 50952 and R80000 exact or aproximate ?????? I hope this will help you.


Kent Cooper
Senior Member Joined: 12.Mar.2013 Location: United States Using: AutoCAD2019 Status: Offline Points: 488 
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With different lengths of the 80mradius arc [blue], there would be different tangent continuations from it [green] that would end at the other fixed point, but I think only one that would be tangent to the short leg of the triangle at that point. That's the challenge  find the angle the blue Arc must swing through so that the the green continuation will end up at the other corner tangent to that short leg. Or, to put it another way, find the radius for the green Arc such that the blue Arc coming tangent to one side from one corner, and the green Arc coming tangent to another side from another corner, will meet tangent to each other. I haven't thought of a solution, but there may be one. [And no, the sizes shown, both lengths and angles, are not precise, but are all rounded, except for the 80m radius of the blue Arc and length of its two radial Lines.]
Edited by Kent Cooper  01.Jul.2019 at 15:00 

Kent Cooper
Senior Member Joined: 12.Mar.2013 Location: United States Using: AutoCAD2019 Status: Offline Points: 488 
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Having thought about it some more, I think this is a situation that would require iteration through small adjustments until a result falls within some tolerance.
Given the triangle defining points A/B/C, and a radius for the blue Arc, a routine could find D, and it could determine the dashed yellow from B on which the center of the green Arc must lie. It would then try swinging the DE angle (magenta variability), at each try calculating F, and comparing the distance FE with the distance FB. The solution would be when those are equal within whatever tolerance you need.
Edited by Kent Cooper  01.Jul.2019 at 16:13 

philippe JOSEPH
Senior Member Joined: 14.Mar.2011 Location: France Using: AutoCAD Mechanical 2017 Status: Offline Points: 1069 
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Hello wtojp2013 and Kent, so if the situation needs iteration to have a "informatically correct and not rounded solution" then it might be only with PARAMETRIC CONSTRAINTS the way to get it right.
That's what I tought at once ( and also because I can't have a this moment access to the AutoCAD file ). I have done a "manual CAD" try on a paper and I found that the solution(s) could be all within some values of the R80000 arc length. What we ( only ) need is to place the center of the unknown radius on a line perpendicular to the right fixed tangent point at the crossing to the right line coming from the center of the R80000 radius, in other words : on the line DE for a center F or on the line D(E) for a center (F) and this couls allow numberous values between reasonable ones.


Kent Cooper
Senior Member Joined: 12.Mar.2013 Location: United States Using: AutoCAD2019 Status: Offline Points: 488 
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Except that at any (E) & (F) location other than the one and only right one, the blue and green Arcs will not be tangent to each other. Maybe it's only my assumption that tangency between the two of them is required, but if not, you're right that it's a comparatively trivial problem. That tangency happens only when distance FE is equal to FB, which happens in only one location. It's pretty obvious that (F)(E) is not the same distance as (F)B. My parentheses around (E) and (F) were intended to indicate a trial direction, to be checked for that equality, and when unequal, the direction would be adjusted and a new trial checked.


philippe JOSEPH
Senior Member Joined: 14.Mar.2011 Location: France Using: AutoCAD Mechanical 2017 Status: Offline Points: 1069 
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You are right Master Kent, only iteration ( unprecise ) or parametric constraints ( informatically correct ) can give you FE = FB.
wtojp2013 are you still alive ?


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