Topic: Calculating a point defining a line

[Keith™]

I have a unique situation where I need to calculate programmatically the point as indicated in the picture below. The known dimensions, points, and angles are shown as "known" or a value. The shape is subject to change at any time with the exception of the relationship of the shortest known length line to the top line. The indicated angle will always be 90 degrees.

I need to be able to calculate the final point, in lisp, without resorting to drawing temporary objects, since the operation may take place on drawings that are not open in the editing window.

I can do it really easily by drawing a circle and then a line tangent to the circle, but that requires manually editing the drawings.
Since I am having a difficult time with calculus this morning, I thought some kind soul would help me with the mathematics so I can tranfer it to lisp.

[CAB]

Hum, I don't think you have enough information to calculate it. <I have been know to be mistaken though>
If you don't have the angle of the angled line, no can do.
Could you use vlax-curve-getClosestPointTo ?
Using inters might be doable too, but you need the line entity for both.

[T.Willey]

Hum, I don't think you have enough information to calculate it. <I have been know to be mistaken though>
If you don't have the angle of the angled line, no can do.
Could you use vlax-curve-getClosestPointTo ?
Using inters might be doable too, but you need the line entity for both.

You don't need the line, you only need the points of the line to pass to inters.  But I'm with you about the amount of information.  You have a complete 360 degrees the small line can be.  You need one of the unknow angles.

[Keith™]

Well, obviously that isn't the answer I was hoping for .... I have a few ideas I am playing with at the moment ... perhaps ...

how would this work ...

I know the length of the shortest line and the distance associated with 2 known points ... this distance (my hypotenuse) and the side (shortest line) can be mathematically calculated to determine the length of the unknown line ... now I only need to determine the angle so all of the points close ...

I'll be back

[Josh Nieman]

I think there is enough information to get it... the key is in the 90d relation of the short chord and the length of it... picturing the unconstrained lines moving around, I can't see them moving, while keeping that short chord at 90 AND at a given length at the same time.  I'll have to work this out on paper though.

I love problems like these

[Arizona]

Trig Functions: The Functions
  

sine(q) = opp/hyp
cosine(q) = adj/hyp
tangent(q) = opp/adj
cotangent(q) = adj/opp
cosecant(q) = hyp/opp
secant(q) = hyp/adj

[T.Willey]

I know the length of the shortest line and the distance associated with 2 known points ... this distance (my hypotenuse) and the side (shortest line) can be mathematically calculated to determine the length of the unknown line ... now I only need to determine the angle so all of the points close ...

I think you are on the right path with this.  Glad my thinking was wrong.  Must be too early here on the west coast for me to be thinking.

[uncoolperson]

ehhh...

To get the equation of the tangent to a circle from a point not on the circle:
You must use the perpindicular distance formula (the perpindicular distance from the tangent to the centre of the circle is equal to the radius). The tangent has the equation
y - y1 = m(x - x 1). First get this equation into the form: ax + by + c = 0 using the given point (x1,y1) as one point on the line. If you put this equation into the perpindicular distance formula, with the centre of the circle as the point and make it equal the radius, you should be able to calculate two values for m (the slope).
Putting these values back into the equation:
y - y1 = m(x - x 1), you get two line equations, one for each tangent.

http://www.netsoc.tcd.ie/~jgilbert/maths_site/applets/circles/tangents_to_circles.html

[CAB]

Maybe this?
I hate it when I'm wrong. Opened my mouth too soon.
<EDIT: 1st picture was not quite right>

[VVA]

Code: [Select]

(setq P2 (getpoint "\nPoint P2: "))
(setq P4 (getpoint "\nPoint P4: "))
(setq distance-P2P3 (getdist "\nDistance P2-P3:"))
(setq a distance-P2P3)
(setq c (distance P2 P4))
(setq b (sqrt (- (* c c)(* a a))))
(setq tan_alfa (/ b a))
(setq alfa (atan tan_alfa))
(setq ang (angle P2 P4))
(setq new_ang (+ ang alfa))
(setq P3 (polar P2 new_ang a))


[Arizona]

If you have one side (a) and the hypoteneus (c) then it should be simple to get the last line, thereby getting the angle.

a^2 + b^2 = c^2
Solve for b.

[Josh Nieman]

man... i made mine way harder on myself than I needed to.  after seeing how ya'll did it I'm not even showing my work lol...

that's why I like this stuff though... learning new ways of looking at things that can be approached a hundred ways and finding a simpler one.

[ElpanovEvgeniy]

Code: [Select]

(defun test (p1 p2 d1 / B D2 DD)
  (if (< (setq p2(mapcar '/ (mapcar '+ p1 p2)(list 2 2))
               d2 (distance p1 p2)) (+ d1 d2))
    (progn
      (setq
        dd (* d1 d1)
        b  (/ dd (* 2 d2))
      ) ;_  setq
      (polar p1 (+ (angle p1 p2) (angle '(0 0) (list b (sqrt (abs(- dd (* b b))))))) d1)
    ) ;_  progn
  ) ;_  if
)


[uncoolperson]

man... i made mine way harder on myself than I needed to.  after seeing how ya'll did it I'm not even showing my work lol...

that's my approach today...

[CAB]

Here is my code & test routine

Code: [Select]

(defun getPx (L0 L1 L2 L4)
  (defun asin (z /)
    (atan z (sqrt (- 1.0 (* z z))))
  )
  (setq L3 (- L2 L1)
        L5 (sqrt (+ (* L3 L3) (* L4 L4)))
        A2 (- (/ pi 2.) (asin (/ L0 L5)))
  )
)


(defun c:test ()
  (setq p1 (getpoint "\nPick point P1"))
  (setq L0 (getdist p1 "\nDistance L0 ?"))
  (setq L1 (getdist p1 "\nDistance L1 ?"))
  (setq L4 (getdist (getpoint "\nDistance L4 ?")))
  (setq L2 (getdist (getpoint "\nDistance L2 ?")))

  (command ".point"
           "_non"
           (polar P1 (+ (getPx L0 L1 L2 L4) (atan (/ L3 L4))) L0)
  )
  (princ)
)