Topic: change direction to start and arc LWPOLYLINE

[TopoWAR]

ronjonp , god, that simple, thank you for your wisdom.

[ronjonp]

Glad to help 

[LE3]

Hi Ron,
I think that will return the area of the sector no? - sorry no idea if that's what the OP wants... got the impression he is looking for the area of a segment... if not ignore my comment.

[ronjonp]

You are correct Luis ... let's see if the OP responds

[TopoWAR]

ronjonp , hello, LE is right, the area that I occupy is that of a circular segment!
I know it's more complicated, but there if I can help with this, thanks

[LE3]

^ then that will be easy, Ron gave to you the solution for the sector, simple you know the cord length and radius, create a triangle from those sides, calculate the area and subtract it from the area of the sector. HTH.

[TopoWAR]

LE , I fail to understand how, maybe with an example you can do

[Kerry]

LE , I fail to understand how, maybe with an example you can do

Area of Segment = ½ × (θ - sin θ) × r2   (when θ is in radians)

http://www.mathsisfun.com/geometry/circle-sector-segment.html

Have fun.

[TopoWAR]

Kerry , hello, thank you, I've seen several formulas out there, it's just that not be interpreted, I searched the meaning of the signs but I did not understand, maybe I translate the formula simple mathematics to pass a lisp I thank .

[TopoWAR]

LE, you're right, it is a matter of creating a triangle and go.
The area of ​​the circle segment is equal to the circular sector area minus the area of ​​the triangular portion.

[Lee Mac]

Area of triangle = 1/2 x base x height

For your sector, you have:

base = r
height = r sin(a) [Where 'a' is the included angle]

Therfore, the triangle area is:

1/2 x r x r sin(a) = r²/2(sin(a))

The area of the entire sector is the proportion that the sector constitutes of the area of the circle, i.e.:

a/2π x πr² = ar²/2

a/2π gives the proportion of the circle occupied by the sector
πr² obviously gives the area of the entire circle

Therefore, the area of the sector minus the area of the triangle is:

(ar²/2) - (r²/2(sin(a)))

We can take out a factor of r²/2 giving:

r²/2(a - sin(a))

Now, the only quantities we require from the polyline arc is the included angle & the arc radius.

The polyarc bulge is the tangent of 1/4 of the included angle, therefore, the inverse of this operation gives the included angle from the bulge value:

a = 4 x atan(b)

As for the bulge radius, there are various ways to acquire this value - here is one example:

Code - Auto/Visual Lisp: [Select]

1. ;; Bulge Radius  -  Lee Mac
2. ;; p1 - start vertex
3. ;; p2 - end vertex
4. ;; b  - bulge
5. ;; Returns the radius of the arc described by the given bulge and vertices
6.  
7. (defun LM:BulgeRadius ( p1 p2 b )
8.     (/ (* (distance p1 p2) (1+ (* b b))) 4 (abs b))
9. )

Now you have all the tools & information you require to calculate the required area 

[Kerry]

LE, you're right, it is a matter of creating a triangle and go.
The area of ​​the circle segment is equal to the circular sector area minus the area of ​​the triangular portion.

TopoWAR,
I'd like to see the complete code you design to resolve this problem.
It will be educational for anyone following along.