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modeling a mathematical shape

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Frank98 View Drop Down
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Frank98 Quote  Post ReplyReply Direct Link To This Post 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

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Kent Cooper View Drop Down
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Kent Cooper Quote  Post ReplyReply Direct Link To This Post 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.
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Vladimir Michl View Drop Down
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Vladimir Michl Quote  Post ReplyReply Direct Link To This Post 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)
CAD Studio - www.cadstudio.cz - Autodesk reseller
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Kent Cooper View Drop Down
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Kent Cooper Quote  Post ReplyReply Direct Link To This Post 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
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