Triangle calculation

[Opie]

I'm working on a routine to draw two lines at equal lengths. The two lines represent the midpoint between a line and an arc along a second line, which may or may not be perpendicular to the first line. The line coming from the arc is radial from the center point of the arc.

Does anyone have any ideas on how to determine the lengths of the two equal length lines?

[BIG-AL]

You have the point where the horizontal line hits the angle line so can work out a 90deg line, ok so do a loop stepping a distance for a new point then do a closestpointto the circle when the two values offset and new distance are within a fudge value that is the answer.

[LiDo]

The problem in fact consists in determining the position of the point P. Solution is simple. See below image.
New Picture.png

[Opie]

Thanks. I went with more of a fudge by traversing down the green line and testing distance from the center point less the radius. If the two distances were within a margin of error, it was close enough. I would like a more exact algorithm, but I am comfortable with my current solution.

[LiDo]

Thanks. I went with more of a fudge by traversing down the green line and testing distance from the center point less the radius. If the two distances were within a margin of error, it was close enough. I would like a more exact algorithm, but I am comfortable with my current solution.

Considering the notations in the previous message, try this:

(defun bepe (pointO pointA pointB AD / pointD pointQ)
  (setq pointD (polar pointO (angle pointO pointA) (+ (distance pointO pointA) AD))
;;pointQ = (d) intersected with BD
        pointQ (polar pointB (angle pointB pointD) (/ (distance pointB pointD) 2.))
  )
;;Distance between point B and point P
  (distance pointB (inters pointQ (polar pointQ (+ (angle pointB pointD) (/ pi 2)) 100.) pointO pointB nil))
)

[Opie]

Considering the notations in the previous message, try this:

(defun bepe (pointO pointA pointB AD / pointD pointQ)
  (setq pointD (polar pointO (angle pointO pointA) (+ (distance pointO pointA) AD))
;;pointQ = (d) intersected with BD
        pointQ (polar pointB (angle pointB pointD) (/ (distance pointB pointD) 2.))
  )
;;Distance between point B and point P
  (distance pointB (inters pointQ (polar pointQ (+ (angle pointB pointD) (/ pi 2)) 100.) pointO pointB nil))
)

Unfortunately, the location of Point B is not known. I know it falls on the arc ... somewhere. If I knew the location of Point B, I could calculate Point P's location.

[LiDo]

Unfortunately, the location of Point B is not known. I know it falls on the arc ... somewhere. If I knew the location of Point B, I could calculate Point P's location.

In this case the problem is undetermined. With a little effort could write the equation of the curve on which point P moves, considering that point B moves on the circle.
On the other hand, if the angle ODP is given (on your sketch the value is 26.593 deg), then the problem has one solution but my approach needs to be slightly changed.

[markmarkarian679925]

use the cosine formula to find value of 'c'
in the triangle OPD (a+c)^2 = c^2 + (a+b)^2 - 2c(a+ b)cosD
if u need angle at 'O' use the sine formula
sinO/c=sinD/(a+c)


(defun TC (a b D / c)
(setq D (/ (* D pi) 180.0)) ;; convert D to radians
(setq c (/ (+ (* b b)(* 2 a b))(* 2 (+ a (* (cos D)(+ a b)))))) ) )


;;; (setq D 26.593)
;;; (setq a 25.0)
;;; (setq b 5.0


(TC 25.0 5.0 26.593)


HTH

[Opie]

That is exactly what I was looking for. Thanks.