Anyone written a lisp to divide 4 sided shape into two equal areas with vertical line?

so take 4 points that make a 4 sided closed shape and find a vertical line that divides them into two equal areas.
Anyone done this?
thx
James Maeding
jmaeding at hunsaker dot com
Civil Engineer/Programmer

 

Can you post an image?

 

Anyway... James

Have a look at BDivide v1.02.... you may find it useful...

http://www.draftteam.com/app.php?id=13&lang=en

It does better job than... AutoCAD Land Development Desktop subdivisions tools...

Luis Esquivel
http://www.draftteam.com

 

Only for rectangles. Not for trapezoids or other polygons.

--
Bill DeShawn

http://my.sterling.net~bdeshawn

that divides them into two equal areas.

 

James,
Is the dividing line "always" a vertical line or is it in any direction?
Alan Henderson @ A'cad Solutions

that divides them into two equal areas.

 

Hi James,

I wish you hadn't asked this question - My mind has been reverting to it all
day.

Eventually I couldn't resist sketching on some graph paper.

For $25 I'd go straight to Luis' site and buy his program - get all the
extra functionality he offers and get on with my life.

However, an almost* general method of solution is to divide the rectangle
into three with vertical lines through the two inner vertices. (*If either
of the east pair of sides or the west pair are re-entrant this model needs
more thinking)

Check the sizes of each of the end triangles and see if either is more than
half the area of the original rectangle. If it is, then the solution is a
direct computation inside that triangle (by solving the quadratic equation
for the width to give the required area).

If it isn't then the solution is in the remaining rectangle with two
vertical sides and a direct solution is again available - picture warping
this rectangle till the bottom side is horizontal and you'll see what I
mean.

An easy way to get the area of a rectangle without any computation in VBA at
least, is:

Sub fred()

Dim op As AcadLWPolyline
Dim p(0 To 7) As Double
p(0) = 1
p(1) = 1
p(2) = 10
p(3) = 2
p(4) = 3
p(5) = 12
p(6) = 11
p(7) = 0
Set op = ThisDrawing.ModelSpace.AddLightWeightPolyline(p)
With op
p(0) = .Area
.Delete
End With
End Sub


Another interesting outcome of my sketching was that it appears (don't take
this as true) that if you create a series of inner rectangles - each joining
the midpoints of the sides of the previous one, you end up near the centroid
of the original rectangle.
--

Regards,


Laurie Comerford
www.cadapps.com.au

 

You must mean "quadrilateral" rather than "rectangle." If it's really a
rectangle, it's easy (a vertical line through the point halfway between
opposite corners). It's the NON-rectangular quadrilateral that's harder. I
guess the line through the centroid of a region isn't going to be the
answer, is it (too easy)?

 

Hi Kent,

You're right. Thanks.

--

Regards,


Laurie Comerford
www.cadapps.com.au

 

I spent about 3 hours last night and this morning working on the problem.
I could only get this to work using an iteration method.

Here is what I came up with -

;=========================================
; Subdivide 4-side Polygon to equal areas
; Copyright (C) 2005 A'cad Solutions
; Alan R. Henderson
; All Rights Reserved
; Website: www.acadsolutions.biz
;=========================================

;GETAREA
; 1. Pick 4 corners of polygon
; 2. Pick new dividing line at final angle for division
; Line must be through opposite sides in direction of 1st to 2nd corner
of polygon
;DIVAREA
; 1. Offset original dividing line by starting OFFSET amount (default set to
0.1)
; If AREAS are with (* OFFSET 10), then change OFFSET to (/ OFFSET 10.0)
; Keep offseting until areas are within AREAOK amount (defualt set to
0.00000001)
; The smaller the AREAOK amount the longer the program runs

(defun CALC_AREA (PP1 PP2 PP3 PP4 /)
(setq D13 (distance PP1 PP3))
(setq D24 (distance PP2 PP4))
(setq R13 (angle PP1 PP3))
(setq R24 (angle PP2 PP4))
(setq R31 (angle PP3 PP1))
(setq R42 (angle PP4 PP2))
(cond
((equal (- R42 R13) (- R24 R31) 0.00000001)
(setq AREA (* 0.5 D13 D24 (abs (sin (- R42 R13)))))
)
((equal (- R13 R24) (- R31 R42) 0.0000000001)
(setq AREA (* 0.5 D13 D24 (abs (sin (- R13 R24)))))
)
(T
(getstring "\nProblems with CALC_AREA...")
(exit)
)
)
)

(defun FIND_AREAS ()
(setq PX nil PS nil AR1 nil AR2 nil)
(if (inters P1 P2 P5 P6 T) (setq PS (append PS (list "12")) PX (append PX
(list (inters P1 P2 P5 P6 T)))))
(if (inters P2 P3 P5 P6 T) (setq PS (append PS (list "23")) PX (append PX
(list (inters P2 P3 P5 P6 T)))))
(if (inters P3 P4 P5 P6 T) (setq PS (append PS (list "34")) PX (append PX
(list (inters P3 P4 P5 P6 T)))))
(if (inters P4 P1 P5 P6 T) (setq PS (append PS (list "41")) PX (append PX
(list (inters P4 P1 P5 P6 T)))))
(cond
((and (= (car PS) "12") (= (cadr PS) "34"))
(setq AR1 (CALC_AREA P1 (cadr PX) (car PX) P4))
(setq AR2 (CALC_AREA (cadr PX) P2 P3 (car PX)))
)
((and (= (car PS) "23") (= (cadr PS) "41"))
(setq AR1 (CALC_AREA P1 P2 (car PX) (cadr PX)))
(setq AR2 (CALC_AREA P3 P4 (cadr PX) (car PX)))
)
)
(princ)
)

(defun C:GETAREA (/ P1 P2 P3 P4 P5 P6 DONE AR1 AR2 LAR1 LAR2 ANG OFFSET
AREAOK DON PS PX)
(setvar "OSMODE" 1)
(setq P1 (getpoint "\nPick 1st Point ? "))
(setq P2 (getpoint P1 "\nPick 2nd Point ? "))
(setq P3 (getpoint P2 "\nPick 3rd Point ? "))
(setq P4 (getpoint P3 "\nPick 4th Point ? "))
(setvar "OSMODE" 0)
(setq P5 (getpoint "\nPick 1st Point of Dividing Line ? "))
(setq P6 (getpoint P5 "\nPick 2nd Point of Dividing Line ? "))
(grdraw P5 P6 1)
(setq DONE nil LAR1 nil LAR2 nil ANG (/ pi 2))
(setq OFFSET 0.1) ;starting offset amount
(setq AREAOK 0.00000001) ;final equal area amount
(CALC_AREA P1 P2 P3 P4)
(FIND_AREAS)
(while (not DONE)
(setq LAR1 AR1 LAR2 AR2)
(setq P5 (polar P5 (+ (angle P5 P6) ANG) OFFSET))
(setq P6 (polar P6 (+ (angle P5 P6) ANG) OFFSET))
(FIND_AREAS)
(if (and (= (type AR1) 'real) (= (type AR2) 'real))
(cond
((equal AR1 AR2 (* OFFSET 100.0))
(setq OFFSET (/ OFFSET 10.0))
)
((equal AR1 AR2 AREAOK)
(setq OFFSET (/ OFFSET 10.0))
(setq DONE T)
)
((and LAR1 LAR2 (< AR1 AR2) (< AR1 LAR1))
(setq ANG (+ ANG pi))
(setq DONE nil)
)
((and LAR1 LAR2 (< AR2 AR1) (< AR2 LAR2))
(setq ANG (+ ANG pi))
(setq DONE nil)
)
)
(progn
(princ "\nProblems with FIND_AREAS....")
(exit)
)
)
)
(setvar "OSMODE" 0)
(if PX (command "PLINE" (car PX) (cadr PX) "") (princ "\nProblem with PX
value.."))
(princ)
)

 

Gary, What do you mean by horizontal or vertical? Alan

 

Works horizontally but not vertically.

added (command "redraw")

Gary

 

I get different areas when picking the two dividing line points; picking
horizontally
versers picking vertically.

After picking the four points, I tried picking the next two points in
different directions.
The four points picked were for an unequal sided box.

 

wow, 13 replies! too cool.

For those that were asking for clarification, the four points can be any four.
I guess to simplify things a bit, the shape is really never like a V.
That is, a vertical line wil never split the shape into more than two other shapes.

Whether the solution works for a horizontal or vertical line does not matter, I can temporarily flip points around if
that was an issue, so I could solve either case.

I'll look at the code more, I just got back into my newsreader this am.
thanks

James Maeding <jmaeding at hunsaker dot com>
|>so take 4 points that make a 4 sided closed shape and find a vertical line that divides them into two equal areas.
|>Anyone done this?
|>thx
|>James Maeding
|>jmaeding at hunsaker dot com
|>Civil Engineer/Programmer

James Maeding
jmaeding at hunsaker dot com
Civil Engineer/Programmer

 

Now that I think of it, this is the same as asking for the center of gravity point for any 4 sided shape.
The x value would be the vertical line and y would be the horizontal.
combine those and you have a point about which you can calc any angle dividing line.

There absolutely must be a formula out there for figuring this given a list of points forming a closed shape.
I bet this works out to be simple once we find that algorythm

James Maeding <jmaeding at hunsaker dot com>
|>so take 4 points that make a 4 sided closed shape and find a vertical line that divides them into two equal areas.
|>Anyone done this?
|>thx
|>James Maeding
|>jmaeding at hunsaker dot com
|>Civil Engineer/Programmer

James Maeding
jmaeding at hunsaker dot com
Civil Engineer/Programmer

 

nope, center of gravity does not work.
I made a solid and listed the center of mass, then drew a line and made two plines.
the areas are not equal.
this is because the areas are weighted by the dist from the CG point, no good

James Maeding <jmaeding at hunsaker dot com>
|>Now that I think of it, this is the same as asking for the center of gravity point for any 4 sided shape.
|>The x value would be the vertical line and y would be the horizontal.
|>combine those and you have a point about which you can calc any angle dividing line.
|>
|>There absolutely must be a formula out there for figuring this given a list of points forming a closed shape.
|>I bet this works out to be simple once we find that algorythm
|>
|>James Maeding <jmaeding at hunsaker dot com>
|>|>so take 4 points that make a 4 sided closed shape and find a vertical line that divides them into two equal areas.
|>|>Anyone done this?
|>|>thx
|>|>James Maeding
|>|>jmaeding at hunsaker dot com
|>|>Civil Engineer/Programmer
|>
|>James Maeding
|>jmaeding at hunsaker dot com
|>Civil Engineer/Programmer

James Maeding
jmaeding at hunsaker dot com
Civil Engineer/Programmer

 

I'd be surprised if there weren't a formula.
Have you looked at the math sites already?

gravity point for any 4 sided shape.

list of points forming a closed shape.

line that divides them into two equal areas.

 

a few but not real deep yet.
I keep thinking it has to be some variation of the center of gravity formula.
I'll see

"MP" <>
|>I'd be surprised if there weren't a formula.
|>Have you looked at the math sites already?
|>
|>"James Maeding" <jmaeding at hunsaker dot com> wrote in message
|>|>> nope, center of gravity does not work.
|>> I made a solid and listed the center of mass, then drew a line and made
|>two plines.
|>> the areas are not equal.
|>> this is because the areas are weighted by the dist from the CG point, no
|>good
|>>
|>> James Maeding <jmaeding at hunsaker dot com>
|>> |>Now that I think of it, this is the same as asking for the center of
|>gravity point for any 4 sided shape.
|>> |>The x value would be the vertical line and y would be the horizontal.
|>> |>combine those and you have a point about which you can calc any angle
|>dividing line.
|>> |>
|>> |>There absolutely must be a formula out there for figuring this given a
|>list of points forming a closed shape.
|>> |>I bet this works out to be simple once we find that algorythm
|>> |>
|>> |>James Maeding <jmaeding at hunsaker dot com>
|>> |>|>so take 4 points that make a 4 sided closed shape and find a vertical
|>line that divides them into two equal areas.
|>> |>|>Anyone done this?
|>> |>|>thx
|>> |>|>James Maeding
|>> |>|>jmaeding at hunsaker dot com
|>> |>|>Civil Engineer/Programmer
|>> |>
|>> |>James Maeding
|>> |>jmaeding at hunsaker dot com
|>> |>Civil Engineer/Programmer
|>>
|>> James Maeding
|>> jmaeding at hunsaker dot com
|>> Civil Engineer/Programmer
|>

James Maeding
jmaeding at hunsaker dot com
Civil Engineer/Programmer

 

Maybe I'm way off..but,
Couldn't you calculate the 'centroid' point by merely
drawing a line from (opposite) points, forming a X.
The 'intersection' of that X should be the centroid ?

Bob

 

nice! I'll get to the converting and post the code so everyone gets it.
thanks

"Kent Cooper, AIA" <>
|>By the way, here's an illustration of how far the intersection of the
|>diagonals can be from what you want. And this drawing is no
|>approximation -- it's QUITE accurate. It has nice round number
|>characteristics (ABCD is exactly 3, w is exactly 2, tanP is exactly 2, tanR
|>is exactly 6). The y offset ends up being a nice clean (in lisp terms)
|>
|>(/ 3.0 (+ 2.0 (sqrt 3)))
|>
|>I calculated that to seven decimal places, and used that for offsetting AB.
|>Boundary made two polylines whose areas were also equal to seven decimal
|>places (1.5000000).

James Maeding
jmaeding at hunsaker dot com
Civil Engineer/Programmer

 

I have a surveyor that has a legal description saying "east half of said lot".
This happens farily often apparently.
To figure out what is the east, wes, north or south half, you have to split the area with a vertcal or horiz line.
So I want to make a routine where you grab a pline and it will draw the vertical or horiz line splitting the area of the
shape.
thats it.

"Luis Esquivel" <>
|>Now... what exactly is what you are trying to do or come up?????????

James Maeding
jmaeding at hunsaker dot com
Civil Engineer/Programmer