Draw; Arc; Start, End, Length
Mike Poynton
(points) of a building between which I want to fit a curved wall, with 24"
masonry openings and 36" of solid masonry wall separating them. I want 10
openings altogether. This gives me a total arc length of 53"-0"
[(24x10)+(36x11)]. How can I draw an arc using start (1st corner), end (2nd
corner), and arch length (53'-0")?! In other words, locate the center of
the arc's radius based on 2 points in space and the arc length.
Along the same veign, it would be nice to have a way of dimensioning arch
lengths in AutoCAD |-(
411: Last time I went through this exercise, I had to call my wife and ask
her to calculate the radius for me - with MicroStation. I went back into
AutoCAD and drew the arc using start, end, radius.
--
Mike Poynton
poyn...@teng.com
Lila
in here somewhere. If we someone could figure out the formula to figure
this out, a lisp routine could be written.
I vaguely remember a custom routine somwhere for dimensioning arch
lengths. Might be worth looking at http://www.cadalog.com or
http://www.cadshack.com/
--
***Please remove "XXX" from email address when responding directly***
---
Mike Poynton wrote in message <7g5cl1$gc...@adesknews2.autodesk.com>...
>I've run into this a few times on projects with curves:
> ...
David Doane
Some time ago I wrote a program to draw arcs by length, and I have another
that will measure an arc. I also have written a routine as part of a
mapping
program to measure arcs with an output of arc radius, arc length and
bearing
and distance between points, placing it on the drawing at the end of an
arrow either inside or out side of the arc. The biggest problem occurs
when the arc is over 180 degrees.
However, in looking at this particular problem, I don't have a ready
answer. While I have formulas for catenary sag and catenary tension,
neither quite fit the bill.
Other than the problem at hand, if someone has a need for something that
could be modified from above, let me know.
Dave Doane (the ol' man in the mountains)
lectr...@cyberportal.net
(remove the dash for email)
Stephen Tate <no_...@stcad.co.uk> wrote in article
<7gam4v$rs...@adesknews2.autodesk.com>...
> Mike
>
> >How can I draw an arc using start (1st corner), end (2nd
> >corner), and arch length (53'-0")?! In other words, locate the center
of
> >the arc's radius based on 2 points in space and the arc length.
>
> This is a lot harder than it sounds. Needless to say there is no solution
> 'built into' AutoCAD. I have been unable to come up simple algebraic
formula
> that will calculate the radius of an arc given its length and the chord
> length (distance between end points). If this is a frequent problem for
you
> I may be able to come up with an iterative solution using LISP.
>
> Obviously, if it's a one off (and solved) problem it's not worth my time
> solving it, but if it's something you need regularly let me know and I'll
> give it a go.
>
> Cheers
>
> --
> Steve
> Stephen Tate CAD, UK
> E-mail: stephe...@stcad.co.uk
>
>
Stephen Tate
Stephen Tate
An excellent piece of work!
--
Steve
Stephen Tate CAD, UK
E-mail: stephe...@stcad.co.uk
Jon Fleming wrote in message ...
>That's an interesting one. Not trivial.
>
>I couldn't figure out an analytic solution (although it seems that there
>should be one). An iterative solution works fine unless the arc is _really_
>_really_ close to a straight line.
>
>Load the attached file and type:
>
> arcl
>
>jrf
Stephen Tate
What makes this problem so difficult is that you only have the two lengths
to go on. I got through about four sheets on my note pad before I concluded
an algebraic solution was beyond my abilities. I then asked the assistance
of some mathematicians to see if they could come up with a formula, as Jon
says, it seems like there should be one. They both concluded that an
algebraic method of solving the problem was unlikely/impossible (one said
unlikely, the other said impossible <g>). The problem is that whatever
formulae you try to apply to the problem you keep finding you need to know
the angle between the endpoints and the centre of the arc, which of course
is unknown until you know the radius, catch-22. They each came up with
solutions but, like the formula Jon used, they had to be solved using
iteration.
>However, in looking at this particular problem, I don't have a ready
>answer. While I have formulas for catenary sag and catenary tension,
>neither quite fit the bill.
>
>Other than the problem at hand, if someone has a need for something that
>could be modified from above, let me know.
I'd be interested to see your routines for catenaries, especially if you
have one that draws a catenary (or close approximation).
Cheers
--
Steve
Stephen Tate CAD, UK
E-mail: stephe...@stcad.co.uk
David Doane wrote in message <01be9297$f7d37520$c756d2d0@lms-1>...
Stephen Tate
I thought this solution may interest you, the formula was provided by a
mathematician - the code is my own. It doesn't work for arcs greater than a
semi-circle but, maybe you can see why.
(defun C:ARCL ()
(initget 1)
(setq START_POINT (getpoint "\nPick arc start point :"))
(initget 1)
(setq END_POINT (getpoint START_POINT "\nPick arc end point :")
CHORD (distance START_POINT END_POINT)
) ;_ end setq
(initget 1)
(setq ARC (getdist "\nArc length : "))
(if (< ARC (* pi CHORD 0.5))
(command "._ARC"
START_POINT
"_E"
END_POINT
"_R"
(CAL-RADIUS CHORD ARC)
) ;_ end command
(princ "\nArc too large for this routine - sorry")
) ;_ end if
(princ)
) ;_ end defun
;;; Written by Stephen Tate
;;; Mathematics courtesy of:
;;; - Doctor Rob, The Math Forum
;;; With additional input from:
;;; Doctor Wilkinson and Doctor Jaffe
;;; <http://forum.swarthmore.edu/dr.math/>
;;; ...... and Sir Isaac Newton :)
(defun CAL-RADIUS (CHORD ARC / Y-N Y-LIST)
(setq Y-LIST '(1.0)
CHORD (float CHORD)
ARC (float ARC)
) ;_ end setq
(while
(not (equal (car Y-LIST) (cadr Y-LIST) 0.00000000000001))
(setq Y-N (car Y-LIST)
Y-LIST (cons (- Y-N
(/ (- (sin Y-N) (/ (* CHORD Y-N) ARC))
(- (cos Y-N) (/ CHORD ARC))
) ;_ end /
) ;_ end -
Y-LIST
) ;_ end cons
) ;_ end setq
) ;_ end while
(/ ARC (* 2 (car Y-LIST)))
) ;_ end defun
mike ford
If I understand Mike's original premise, it seems to me that this is a
curved corner of a building that would presumably be joining two walls
that are perp. to each other.
So, I thought...hmmm...then I could also take the easy way out and
assume that the arc is also tangent to both walls. This makes the
solution quite simple, and the resultant arc would move the endpoints of
the walls as required to permit the arc to be of the specified length.
However, I couldn't avoid thinking that tangency was the "perfect world"
condition, so I set about another approach. The quick solution is the
iterative approach, as mentioned by previous members, so now I'm faced
with the long approach.....a real live formula.
After digging through the archives and trying to remember why or why not
to use Huygen's approximation (and also wondering why I even remember
the name), I surrender. I would have to beg Mike to the arc be tangent,
or permit an iterative approach.
(but I haven't totally given up....I too have scribbled several sheets
of now incomprehensible formulas)
I'll be keeping my eye on this thread!
Mike
Jon Fleming
method. Newton's method requires differentiating the formula, which I
didn't want to bother to do. Reguli falsa is often better, and seems (in
the posting I put up this morning) to be as fast or faster than Newton's
method.
I'd be glad to discuss the benefits and drawbacks of various methods, but
we might get a little esoteric for this group; I did a lot of numerical
methods at a small trade school in Cambridge (Massachusetts, that is)
<grin>.
jrf
In article <7gf97p$6j...@adesknews2.autodesk.com>, Stephen Tate wrote:
> just thought I'd point out a better iterative solution (Newton's)
Mike Poynton
routine, so I took the liberty of editing it myself to do so. Cheers!
Jon Fleming wrote in message ...
>The idea I posted yesterday doesn't work, but the enclosed is much faster
>even for arcs that are essentially indistinguishable from a straight line.
>
>jrf