Discussion forum
CAD discussion |
Please abide by the rules of this forum.
How to post questions: register or login, go to the specific forum and click the NEW TOPIC button.
modeling a mathematical shape |
Post Reply |
Author | |
Frank98
Newbie Joined: 15.Feb.2017 Location: Bahamas Using: autocad 2015 Status: Offline Points: 4 |
Topic: modeling a mathematical shape Posted: 15.Feb.2017 at 21:28 |
how can i model such a plot in autocad.
this plot what created using a mathematical formula ,demostrated in mathlab. thanks
|
|
Kent Cooper
Senior Member Joined: 12.Mar.2013 Location: United States Using: AutoCAD2020, 2023 Status: Offline Points: 627 |
Posted: 15.Feb.2017 at 22:13 |
The equations would be necessary....
That looks like some kind [two kinds, really] of formula in which the angle to a point on the curve is based on the inverse of a trigonometric function whose value swings between positive and negative without ever going to infinity [such as sine or cosine], applied to some multiplier of the local radius, and for the browner curve with a constant angle [looks like 45 degrees] added to the result. That shouldn't be very hard to work out -- for a related routine that draws polynomial functions [y = ax^n + bx^(n-1) + .... + px^2 + qx + r], see my PolynomialFunction.lsp routine, available here. But this one, being radial, would not involve X and Y coordinates, but instead (polar) functions measured from the origin as basepoint, in a (repeat) function stepping through small increments of radius as the distance arguments, and calculating the angle arguments for each based on the equation, and presumably used in a SPLINE, or perhaps a POLYLINE that you could spline-curve to smooth it out.
|
|
Vladimir Michl
Moderator Group Arkance Systems CZ Joined: 26.Jul.2007 Location: Czech Republic Using: Autodesk software Status: Offline Points: 2018 |
Posted: 17.Feb.2017 at 08:34 |
You can use our 2DPLOT utility (see http://www.cadstudio.cz/en/apps/2dplot/). You equation then should be expressed in cartesian space (X/Y):
; for range (0-1.5> (defun fPolar1 (u / r tau alfa R0) ;x = (* r (cos th)) ;y = (* r (sin th)) (setq r u) (setq tau 1.1) ; example (setq alfa 1.0) ; example (setq R0 0.1) ; example (setq th (* alfa (sin (/ (* pi (log (/ r R0))) (log tau))))) ; your expression (list (* r (cos th)) (* r (sin th))) ; return (X Y) ) Then call: (2Dplot fPolar1 0.001 1.5 0.001) meaning - draw my function "fPolar1" from 0.001 to 1.5 (the radius), with ministeps of 0.001 |
|
Vladimir Michl (moderator)
Arkance Systems - arkance-systems.cz - Autodesk reseller |
|
Kent Cooper
Senior Member Joined: 12.Mar.2013 Location: United States Using: AutoCAD2020, 2023 Status: Offline Points: 627 |
Posted: 17.Feb.2017 at 20:56 |
Another, (polar)-based, solution (along with Vladimir's solution above) is over here. See also other Posts in that same thread.
EDIT: That link doesn't seem to work. Try copy/pasting this: https://forums.autodesk.com/t5/autocad-2013-2014-2015-2016-2017/modeling-a-mathematical-shape/m-p/6886507#M163554 Edited by Kent Cooper - 17.Feb.2017 at 21:02 |
|
Post Reply | |
Tweet
|
Forum Jump | Forum Permissions You cannot post new topics in this forum You cannot reply to topics in this forum You cannot delete your posts in this forum You cannot edit your posts in this forum You cannot create polls in this forum You cannot vote in polls in this forum |
This page was generated in 0,352 seconds.