Topic: Arc-length from known radius & chord

[DMSS1]

I have a radius of 1326.0
And a chord of 72.0

How can I find the arch length using autolisp?

Thanks,
DMSS1

[LE]

Welcome;

I think there are a lot of geometry formulas out there, here is what I had used"

Code: [Select]

(setq arcAngle
       (* (atan (/ (/ (sqrt (- (expt radius 2) (expt (/ chord 2.0) 2))) 2.0) chord))
  2.0))

(setq arcLength (* (* (/ (/ (* arcAngle 180.0) pi) 360.0) (* pi 2)) radius))

I did a quick translation to my formulas I have in C++, into autolisp, hope they are right (not tested)

Update: Removal of the C++ code - not needed here.

[DMSS1]

Thanks Esquivel!
I will try-it-out.

The only part I don't understand is "pow" (is this POINT-ON-WORLD coords?).

[LE]

The only part I don't understand is "pow" (is this POINT-ON-WORLD coords?).

No no.... the commented code is in C++.... I just posted as reference.

[DMSS1]

I tried it out, but got the wrong answer.

Radius=1326.0 of a circle.
Chord=72.0

Find the length of arc with a 72.0 chord on the circle.

My dwg says the angle bettween the intersections=3.11146537°
And the arc length=72.00884796

I drew a circle, r=1326.0
Then drew a line 72.0
Moved the middle of the line, to the top quadrant of the circle.
Then drew a line verticaly to the inside of circle on each side.
Drew a line from inersection to intersection (this should produce a 72.0 straight line across the to of the circle).
Trim the big circle with the new horizontal line.
Listing the arc = length above.

Did I do smething wrong?

Thanks,
DMSS1

[Jeff_M]

I didn't look at what Luis posted, but here's what I have in my toolbox:

Code: [Select]

(defun arclen_rc (r c / halfdelta)
  (setq halfdelta (atan (/ (/ c 2.0) (sqrt (- (expt r 2) (expt (/ c 2.0) 2))))))
  (* (* halfdelta 2.0) r)
  )


[DMSS1]

Thanks Jeff_m,

Works great!

Swamp Rat...  That use to be one of my CB-Handles when the CB-Radio was so popular.


Again,
Thanks!
DMSS1

[Dinosaur]

I have not found the chord length to be a very precise part of a curve definition, especially when there are only two parameters given to solve a curve.  I have tried various combinations of parameters in a curve solver program and the various methods that accept the chord as part will yield significantly different solutions.

[LE]

I didn't look at what Luis posted, but here's what I have in my toolbox:

Code: [Select]

(defun arclen_rc (r c / halfdelta)
  (setq halfdelta (atan (/ (/ c 2.0) (sqrt (- (expt r 2) (expt (/ c 2.0) 2))))))
  (* (* halfdelta 2.0) r)
  )


Thank you Jeff;

I am stuck into these new languages... that I simple lost my abilities on lisp  - I must need to be out of this forum - start to get dangerous by posting lisp code here.... 

Sorry about that.
Luis