construct circle

[[Jo...]]

Is there a way to construct a circle that is tangent to 3 lines ? The tangent function will let me pic 2 lines but forces diameter.

[[Da...]]

AFC (Away from Calypso) so I can't verify but, if you just recall all those points from all lines and do max inscribed circle? Or do you need OD?

[[Jo...]]

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Think of it as a 3 sided box. I need Tangent to two sides(roughly parallel to each other) and bottom(perpendicular to other two lines) , and report center location and diameter. I tried circle in contour but not having any success.

Your method would work if the were formed as a polygon.

[[Ry...]]

Dane - You nailed it - That's what I would do. Works great - check this out:

Just make sure to scan the lines with lots of points. Recall Feature Points into a circle and choose MAXIMUM INSCRIBED or INNER TANGENTIAL in the EVALUATION. Make sure to Update your nominals to the Acutals so you get a current visualization in Calypso.

[[Ry...]]

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Well then, the only thing controlling the Circle is the two sides. Do a Cartesian Distance between the two sides - that's the Diameter you are looking for. The Center would be the intersection of the CONSTRUCTION>SYMMETRY of the two sides and the CONSTRUCTION>PLANE WITH OFFSET from the Bottom Plane offset by the distance (input as a formula).

Good luck!

[[Ri...]]

I'm confused by this request. The circle would be fully constrained after being tangent to two lines. What is the third line doing?

[[Er...]]

A circle is not fully defined if tanget to two points. It has different tangent points depending on the diameter. And two points alone cant make a circle.

Updated, I expressed my self wierd 🙂

[[Jo...]]

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I want a best fit tangent to 3 surfaces. The diameter to be defined by 3 tangents, the location can translate two directions. So I need the largest diameter with its location..

______
|
| circle inside these three lines
______

[[Jo...]]

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A symmetry of the two parallel lines intersecting the perpendicular line gives me what would be a tangent point and constraint of translation in one axis. I haven't worked out how to intersect the other two tangents.. If I could a 3 point circle would work.

[[Ri...]]

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I should have clarified, if the two lines are parallel. If they are not parallel, then the circle would be fully constrained (location wise). The third line in this sense could only work in controlling the size.

I guess that's what they are going for?

[[Ri...]]

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Does the option above not work for you (the inner tangential of the three lines with recalled feature points)?

[[Jo...]]

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No it's not working, I think because the vector of data ?? This is a theoretical section of a part which has tapered surfaces. I'll work on the circle some more. It should work, dunno why it doesn't.

[[Jo...]]

Thanks everyone. Finally got the circle feature to work. Deleted it and started over, and corrected nominals like suggested for display.

[[Er...]]

Richard: Calculate a circle based on two parallell lines is impossible. You might have a diameter and one center coordinate. But you cant make a circle out of that.

Calculate a circle based on two lines that isn't parallell is also impossible. Then you dont have either radius or any coordinate for the center.

To calculate a circle you need atleast 3 points.
See example:
Find distance between p1 and p2.
Find distance between p2 and p3.
Find symmetry between p1 and p2.
Find symmetry between p2 and p3
Calculate projection from sym p1 p2 perp to distance line.
Same with p2 p3.
Intersection point from both projections is circle center.
Calculate distance from center point to either p1, p2 or p3. Thats radius.

p1=point(1.1368683772162e-13,1898.5,0)
p2=point(-16.567327643663,1898.4277109373,0)
p3=point(-33.133393621183,1898.2108492544,0)

x1=p1.x
y1=p1.y

x2=p2.x
y2=p2.y

x3=p3.x
y3=p3.y

temp = (x2 * x2) + (y2 * y2)
bc = (x1 * x1 + y1 * y1 - temp) / 2
cd = (temp - x3 * x3 - y3 * y3) / 2
det = (x1 - x2) * (y2 - y3) - (x2 - x3) * (y1 - y2)

if abs(det) < 0
    message("Ingen beräkning kan ske då punkterna bildar en linje"+cr()+"Allvarligt fel")
else
	cx = (bc*(y2 - y3) - cd*(y1 - y2)) / det
	cy = ((x1 - x2) * cd - (x2 - x3) * bc) / det
	display("Center: "+cx+" "+cy)
	display("Radius: "+sqrt((cx - x1)**2 + (cy - y1)**2))
endif

[[Ri...]]

Eric,

Not disagreeing about size. I was mainly talking about location. I was assuming originally that they were looking for location of a known size.

Now I see that they were looking for something different.