Tangent, Radius and Point Arc
I'm afraid that in your example drawing [in which the Arc is not, in fact, tangent to the Circle], that's a geometric impossibility, unless the Arc curves in the opposite direction from the Circle. The Arc's radius is smaller than the Circle's, which means that if it is to be tangent to the Circle at any point, and curving in the same direction, then all of it except the tangent point must be inside the Circle.
If the curvature is allowed to be in the opposite direction [as long as the radius of the Arc is no less than half the distance from the point to the Circle], or if in real-life situations the radius of the Arc will always be greater than that of the Circle, something could probably be worked out.
Or, if the example is real, maybe you mean to say that the Arc should be tangent to the Line rather than to the Circle [it looks like it is, though I didn't verify that]. Certainly an Arc that is tangent to a Line at a given point, and of a given radius, could be constructed to the point(s) where it meets the Circle.
@marko_ribar wrote:...Apollonius problem Line, Point, Circle to get construction...
Marko, the OP's request is "drawing one arc tangnet to circle, with Radius input and any point on the line."
This is my attempt.
(defun c:TesT ( / *error* cir p lin pc pL rad_min rad pA pc dA dB dC pFc pF1 pF2)
(vl-load-com)
(defun *error* ( msg )
(vla-endundomark (vla-get-activedocument (vlax-get-acad-object)))
(if cir (redraw cir 4))
(if lin (redraw lin 4))
(if (not (wcmatch (strcase msg t) "*break,*cancel*,*exit*"))
(princ (strcat "\nError: " msg))
)
(princ)
)
(vla-startundomark (vla-get-activedocument (vlax-get-acad-object)))
(if (and
(setq cir (car (setq p (entsel "\nSelect Circle "))))
(setq p (cadr p))
(not (redraw cir 3))
(setq lin (car (entsel "\nSelect Line ")))
(not (redraw lin 3))
(setq pc (cdr (assoc 10 (entget cir))))
(setq pL (vlax-curve-getClosestPointTo (vlax-ename->vla-object lin) pc))
(setq rad_min (distance pL pc))
(not
(while
(progn
(setq rad (getreal (strcat "\nFillet Radius (min. = " (rtos rad_min) "): ")))
(not (<= rad_min rad))
)
)
)
)
(progn
(redraw cir 4)
(redraw lin 4)
(setq pA (polar pL (angle pL pc) rad))
(setq dA (- rad (cdr (assoc 40 (entget cir)))))
(setq dB (distance pc pA))
(setq dC (sqrt (- (* dA dA) (* dB dB))))
(if (minusp (sin (- (angle pc p) (angle pc pA))))
(setq pFc (polar pA (- (angle pA pc) (* pi 0.5)) dC))
(setq pFc (polar pA (+ (angle pA pc) (* pi 0.5)) dC))
)
(setq pF1 (polar pFc (angle pFc pc) rad))
(setq pF2 (polar pFc (angle pc pL) rad))
(if (minusp (sin (- (angle pc p) (angle pc pA))))
(vl-cmdf "_arc" "_non" pF2 "_c" "_non" pFc "_non" pF1)
(vl-cmdf "_arc" "_non" pF1 "_c" "_non" pFc "_non" pF2)
)
)
)
(vla-endundomark (vla-get-activedocument (vlax-get-acad-object)))
)
See notes in the attached drawing. All the stuff I added is yellow, plus I copied the original white Lines, Circle and Point for the right side.
Whether the Arcs curving the other direction, as in the left side solution, or in real situations the desired Arc radius will be greater than the Circle radius, as in the right side, I'd have to think about how to automate that in a routine, but it can probably be done.
@mehsan wrote:
... i have attached the dwg for further informations.
...
Here's the drawn solution to that. If that's what you're looking for, putting it into a routine is probably doable, though I'd have to think about whether it would be possible [or easy] to calculate the geometry of the Arc and build it directly, or easier to have it actually draw those Circles and the ones that include the solution Arcs, and do the Trimming.
[I assumed that the wording "Radius for this circle is 55 + 2 = 57" refers to the desired radius for the Arc that goes through Pt1 and is tangent to the original Circle. The red Arc in the drawing is not of that radius, which is why I mention it.]
@mehsan wrote:
Hi, The code send to me looks great but how can i use to get the required result, i am trying but the result is not achived, i have attached the dwg for further informations.
thanks
meshan,
I know that people typically don't like to deal with semantics, but using the wrong words quite often make it hard for people to understand what you're after. For something to be tangent to a circle, that object will only pass through ONE POINT on the circle -- even if you were to extend the object to infinity.
What I suspect you really want is to find the point of compound curvature. Or possibly Point of Reverse Curvature. I can't think of anything else that you could possibly be looking for. But knowing WHICH you want would be important.
Don Ireland
Engineering Design Technician
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@doni49 wrote:
@mehsan wrote:
Hi, The code send to me looks great but how can i use to get the required result, i am trying but the result is not achived, i have attached the dwg for further informations.
thanks
meshan,
I know that people typically don't like to deal with semantics, but using the wrong words quite often make it hard for people to understand what you're after. For something to be tangent to a circle, that object will only pass through ONE POINT on the circle -- even if you were to extend the object to infinity.
What I suspect you really want is to find the point of compound curvature. Or possibly Point of Reverse Curvature. I can't think of anything else that you could possibly be looking for. But knowing WHICH you want would be important.
Here's what I'm getting at.
In a Reverse Curve situation, if you were to draw a line between the two center points, the point where they intersect the line would be the point of
Reverse curvature.
In a Compound Curve situation, the centers of the two arcs/circles and the Point of Compound Curvature will all be along the same straight line.
Don Ireland
Engineering Design Technician
If a reply solves your issue, please remember to click on "Accept as Solution". This will help other users looking to solve a similar issue. Thank you.
Please do not send a PM asking for assistance. That's what the forums are for. This allows everyone to benefit from the question asked and the answers given.
@mehsan wrote:Hi, The code send to me looks great but how can i use to get the required result, i am trying but the result is not achived, i have attached the dwg for further informations.
thanks
Updated code
; FILLET Circle-Line
; 06.Apr.2015
; Gian Paolo Cattaneo
(defun c:FCL ( / *error* cir p lin pc pL rad_min rad pA pc dA dB dC pFc pF1 pF2)
(defun *error* ( msg )
(vla-endundomark (vla-get-activedocument (vlax-get-acad-object)))
(if cir (redraw cir 4))
(if lin (redraw lin 4))
(if (not (wcmatch (strcase msg t) "*break,*cancel*,*exit*"))
(princ (strcat "\nError: " msg))
)
(princ)
)
(vla-startundomark (vla-get-activedocument (vlax-get-acad-object)))
(if (and
(setq cir (car (setq p (entsel "\nSelect Circle "))))
(setq p (cadr p))
(not (redraw cir 3))
(setq lin (car (entsel "\nSelect Line ")))
(not (redraw lin 3))
(setq pc (cdr (assoc 10 (entget cir))))
(setq pL (vlax-curve-getClosestPointTo (vlax-ename->vla-object lin) pc))
(setq rad_min (distance pL pc))
(not
(while
(progn
(setq rad (getreal (strcat "\nEnter Fillet Radius (min. = " (rtos rad_min) "): " )))
(not (>= rad rad_min))
)
)
)
)
(progn
(redraw cir 4)
(redraw lin 4)
(setq dA (- rad (cdr (assoc 40 (entget cir)))))
(if (equal rad rad_min 1e-8)
(progn
(if (minusp (sin (- (angle pL p) (angle pL pc))))
(setq pFc (polar pc (- (angle pc pL) (* pi 0.5)) dA))
(setq pFc (polar pc (+ (angle pc pL) (* pi 0.5)) dA))
)
(setq pF1 (polar pFc (angle pFc pc) rad))
(setq pF2 (polar pFc (angle pc pL) rad))
)
(progn
(setq pA (polar pL (angle pL pc) rad))
(setq dB (distance pc pA))
(setq dC (sqrt (- (* dA dA) (* dB dB))))
(if (minusp (sin (- (angle pL p) (angle pL pc))))
(setq pFc (polar pA (- (angle pA pc) (* pi 0.5)) dC))
(setq pFc (polar pA (+ (angle pA pc) (* pi 0.5)) dC))
)
(setq pF1 (polar pFc (angle pFc pc) rad))
(setq pF2 (polar pFc (angle pc pL) rad))
)
)
(if (minusp (sin (- (angle pL p) (angle pL pc))))
(vl-cmdf "_arc" "_non" pF2 "_c" "_non" pFc "_non" pF1)
(vl-cmdf "_arc" "_non" pF1 "_c" "_non" pFc "_non" pF2)
)
)
)
(vla-endundomark (vla-get-activedocument (vlax-get-acad-object)))
)
(vl-load-com)
(princ "\nFILLET Circle-Line ... by Gian Paolo Cattaneo")
(princ "\nType \"FCL\" to invoke")
(princ)
That routine apparently assumes that the intent is for the resultant Arc to be tangent to the Line as well as to the Circle, determining for itself the point of tangency on the Line. I took the original question, supported [I think] by the wording in the posted drawings, to be about a specific User-specified point along the Line [their Pt1] from which an Arc of a User-specified radius would be drawn that would end tangentially on the Circle. But some clarification from mehsan is in order, because:
a) the example drawings have the Arc tangent to neither the Circle nor the Line [it looks pretty close to being tangent to the Line, but isn't];
b) the use of the wording "any point on the line" in the first Post leaves room for doubt about whether that is really to be a particular User-specified point or just a result of other factors;
c) the distance from the Circle to the Line happens to also be the same as the second example drawing's described difference in radius between the Circle and the desired Arc, but it's not clear to me whether that would always be the case, and the radius of the Arc in the drawing is not as described.
@Kent1Cooper wrote:
.... I took the original question, supported [I think] by the wording in the posted drawings, to be about a specific User-specified point along the Line [their Pt1] from which an Arc of a User-specified radius would be drawn that would end tangentially on the Circle. ....
... supported as well, come to think of it, by the very Subject of the thread. I really don't think tangency to the Line is part of it, but the OP will need to clarify.
