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Author Topic: Pendula  (Read 4266 times)

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Offline DANILIN

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Re: Pendula
« Reply #15 on: September 29, 2020, 04:07:06 pm »
more  of pendulums: long read text

http://rosettacode.org/wiki/Animate_a_pendulum
rosettacode.org/wiki/Animate_a_pendulum

and me learn another language graphic by this page
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Offline Ashish

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Re: Pendula
« Reply #16 on: September 30, 2020, 09:06:40 am »
My attempt to recreate this in 3D -

Code: QB64: [Select]
  1. 'Pendula in 3D : By Ashish in QB64 using OpenGL
  2. 'idea from here - https://www.qb64.org/forum/index.php?topic=3056.msg123288#msg123288
  3. _TITLE "Pendula in 3D"
  4. SCREEN _NEWIMAGE(800, 600, 32)
  5.  
  6. TYPE pendulum_type
  7.     ang AS SINGLE 'angle
  8.     ang_factor AS SINGLE
  9.     min_ang AS SINGLE
  10.     max_ang AS SINGLE
  11.     mass AS SINGLE
  12.     clr AS _UNSIGNED LONG 'color of the sphere
  13.     rad AS SINGLE 'radius of the sphere
  14.     rope_length AS SINGLE 'length of the rope on which the sphere is hanging
  16.  
  18.     SUB glutSolidSphere (BYVAL radius AS DOUBLE, BYVAL slices AS LONG, BYVAL stack AS LONG) 'use to draw a sphere
  19.     SUB gluLookAt (BYVAL eyeX#, BYVAL eyeY#, BYVAL eyeZ#, BYVAL centerX#, BYVAL centerY#, BYVAL centerZ#, BYVAL upX#, BYVAL upY#, BYVAL upZ#) 'for camera purpose
  21.  
  22. CONST MAX_PENDULUM = 15
  23.  
  24. DIM SHARED pendulum(1 TO MAX_PENDULUM) AS pendulum_type, init AS _BYTE
  25. DIM SHARED ix, iy, iz
  26.  
  27. ix = 0: iy = 1.5: iz = 2
  28.  
  29. FOR i = 1 TO MAX_PENDULUM
  30.     pendulum(i).rad = 0.25
  31.     pendulum(i).min_ang = _PI(1.25)
  32.     pendulum(i).max_ang = _PI(1.75)
  33.     pendulum(i).clr = _RGB(RND * 255, RND * 255, RND * 255)
  34.     pendulum(i).rope_length = 2
  35.     pendulum(i).mass = 3 + i * 0.5
  37.  
  38. init = 1
  40.     FOR i = 1 TO MAX_PENDULUM
  41.         pendulum(i).ang_factor = pendulum(i).ang_factor + 0.01 * pendulum(i).mass
  42.         pendulum(i).ang = map(SIN(pendulum(i).ang_factor), -1, 1, pendulum(i).min_ang, pendulum(i).max_ang)
  43.     NEXT
  44.     _LIMIT 60
  46.  
  47. SUB _GL ()
  48.     STATIC t
  49.     t = t + 0.01
  50.     IF init = 0 THEN EXIT SUB
  52.    
  53.     ' _glClearColor 0.3,0.3,0.3,1
  54.     _glClear _GL_COLOR_BUFFER_BIT OR _GL_DEPTH_BUFFER_BIT
  55.    
  56.     _glEnable _GL_LIGHTING
  57.     _glEnable _GL_LIGHT0
  58.     _glEnable _GL_DEPTH_TEST
  59.     _glEnable _GL_COLOR_MATERIAL
  60.    
  61.     DIM v(3) AS SINGLE
  62.     v(0) = 0: v(1) = 0: v(2) = 1: v(3) = 0
  63.    
  64.     _glLightfv _GL_LIGHT0, _GL_POSITION, _OFFSET(v())
  65.    
  66.     _glMatrixMode _GL_PROJECTION
  68.     _gluPerspective 50, _WIDTH / _HEIGHT, 0.1, 25
  69.    
  70.     _glMatrixMode _GL_MODELVIEW
  72.    
  73.     gluLookAt 0, 2, 5, 0, 0, 0, 0, 1, 0
  74.    
  75.     drawXZPlane 0, -2, 0, 8, 16, -1
  76.    
  77.     DIM z
  78.     z = iz
  79.     _glLineWidth 5.0
  80.     FOR i = 1 TO MAX_PENDULUM
  81.         _glColor3ub 255, 255, 0
  82.         _glBegin _GL_LINES
  83.         _glVertex3f ix, iy, z
  84.         _glVertex3f ix + pendulum(i).rope_length * COS(pendulum(i).ang), iy + pendulum(i).rope_length * SIN(pendulum(i).ang), z
  85.         _glEnd
  86.        
  87.         _glPushMatrix
  88.         _glTranslatef ix + (pendulum(i).rope_length + pendulum(i).rad) * COS(pendulum(i).ang), iy + (pendulum(i).rope_length + pendulum(i).rad) * SIN(pendulum(i).ang), z
  89.         _glColor3ub _RED(pendulum(i).clr), _GREEN(pendulum(i).clr), _BLUE(pendulum(i).clr)
  90.         glutSolidSphere pendulum(i).rad, 50, 50
  91.         _glPopMatrix
  92.        
  93.         z = z - 2.3 * pendulum(i).rad
  94.     NEXT
  95.     _glColor3ub 255, 255, 0
  96.     _glBegin _GL_LINES
  97.     _glVertex3f ix, iy, iz
  98.     _glVertex3f ix, iy, z
  99.     _glEnd
  100.    
  101.     _glFlush
  103.  
  104. SUB drawXZPlane (x, y, z, w, d, n) '(x,z)->location to draw, w->width, d->depth of the plane, n->normal direction in Y direction
  105.     _glBegin _GL_QUADS
  106.     _glTexCoord2f 1, 0
  107.     _glNormal3f 0, n, 0
  108.     _glVertex3f x - (w / 2), y, z - d / 2
  109.     _glTexCoord2f 1, 1
  110.     _glNormal3f 0, n, 0
  111.     _glVertex3f x + (w / 2), y, z - d / 2
  112.     _glTexCoord2f 0, 1
  113.     _glNormal3f 0, n, 0
  114.     _glVertex3f x + (w / 2), y, z + d / 2
  115.     _glTexCoord2f 0, 0
  116.     _glNormal3f 0, n, 0
  117.     _glVertex3f x - (w / 2), y, z + d / 2
  118.     _glEnd
  120.  
  121. 'from p5js.bas
  122. FUNCTION map! (value!, minRange!, maxRange!, newMinRange!, newMaxRange!)
  123.     map! = ((value! - minRange!) / (maxRange! - minRange!)) * (newMaxRange! - newMinRange!) + newMinRange!
  125.  
  126.  
Screenshot_1.png
* Screenshot_1.png (Filesize: 74.48 KB, Dimensions: 801x625, Views: 184)
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My Projects - https://github.com/AshishKingdom?tab=repositories
OpenGL tutorials - https://ashishkingdom.github.io/OpenGL-Tutorials

FellippeHeitor

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Re: Pendula
« Reply #17 on: September 30, 2020, 09:20:35 am »
Quote from: Ashish on September 30, 2020, 09:06:40 am
My attempt to recreate this in 3D -

Whoooooooooa!
🤩
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Offline DANILIN

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Re: Pendula
« Reply #18 on: January 06, 2022, 12:38:38 am »
improvement:

for colorize:
Randomize Timer

thik lines:
_glLineWidth 0.0

or symbol ':
'_glBegin _GL_LINES 
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Offline _vince

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Re: Pendula
« Reply #19 on: January 07, 2022, 05:13:58 am »
We've all seen the chaotic double pendulum.  I'll leave my simple Runge-Kutta one done with canned formulas from wikipedia at the end.

Challenge:  Generalize it to n-pendulum!  Triple, quadruple, whatever the user wants.  The next ball is to be attached to the previous.  It's going to be a fairly tedious system of differential equations based on the lagrangian of the system (https://en.wikipedia.org/wiki/Double_pendulum#Analysis_and_interpretation).  Probably so tedious you're going to need a large matrix and some linear algebra.  But goddamn I'll be so impressed if you do it, you'll come out of it totally competent to program everything DEs.

Here's the trivial case:
Code: QB64: [Select]
  1. defsng a-z
  2. pi = 3.141593
  3. r = 100
  4. a1 = -pi/2
  5. a2 = -pi/2
  6. h=0.001
  7. m=1000
  8. g=1000
  9. sw = 640
  10. sh = 480
  11.  
  13. p1 = _newimage(sw,sh,12)
  14. screen _newimage(sw,sh,12)
  15.  
  17.         a1p = (6/(m*r^2))*(2*m*v1-3*cos(a1-a2)*m*v2)/(16-9*(cos(a1-a2))^2)
  18.         a1k1 = a1p
  19.         a1k2 = a1p + h*a1k1/2
  20.         a1k3 = a1p + h*a1k2/2
  21.         a1k4 = a1p + h*a1k3
  22.         a2p = (6/(m*r^2))*(8*m*v2-3*cos(a1-a2)*m*v1)/(16-9*(cos(a1-a2))^2)
  23.         a2k1 = a2p
  24.         a2k2 = a2p + h*a2k1/2
  25.         a2k3 = a2p + h*a2k2/2
  26.         a2k4 = a2p + h*a2k3
  27.         na1=a1+h*(a1k1+2*a1k2+2*a1k3+a1k4)/6
  28.         na2=a2+h*(a2k1+2*a2k2+2*a2k3+a2k4)/6
  29.         v1p = -0.5*r^2*(a1p*a2p*sin(a1-a2)+3*g*sin(a1)/r)
  30.         v1k1 = v1p
  31.         v1k2 = v1p + h*v1k1/2
  32.         v1k3 = v1p + h*v1k2/2
  33.         v1k4 = v1p + h*v1k3
  34.         v2p = -0.5*r^2*(-a1p*a2p*sin(a1-a2)+g*sin(a2)/r)
  35.         v2k1 = v2p
  36.         v2k2 = v2p + h*v2k1/2
  37.         v2k3 = v2p + h*v2k2/2
  38.         v2k4 = v2p + h*v2k3
  39.         nv1=v1+h*(v1k1+2*v1k2+2*v1k3+v1k4)/6
  40.         nv2=v2+h*(v2k1+2*v2k2+2*v2k3+v2k4)/6
  41.         a1=na1
  42.         a2=na2
  43.         v1=nv1
  44.         v2=nv2
  45.  
  46.         _dest p1
  47.         pset (sw/2+r*cos(a1-pi/2)+r*cos(a2-pi/2), sh/2-r*sin(a1-pi/2)-r*sin(a2-pi/2)),12
  48.         _dest 0
  49.         _putimage, p1
  50.  
  51.         line (sw/2, sh/2)-step(r*cos(a1-pi/2), -r*sin(a1-pi/2))
  52.         circle step(0,0),5
  53.         line -step(r*cos(a2-pi/2), -r*sin(a2-pi/2))
  54.         circle step(0,0),5
  55.  
  56.         _limit 200
  57.         _display
  60.  
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Offline STxAxTIC

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Re: Pendula
« Reply #20 on: January 07, 2022, 10:14:12 am »
That's an interesting challenge. Bonus points if all pendula are constrained to never self-intersect, so that the resulting object should be like a hanging chain and exhibit the same physics. I wonder if this is even *that* terrible to do, becau - ya know what? I'll shut up.... Nice challenge!
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