How to convert a 3D point into 2D perspective projection

Question

I am currently working with using Bezier curves and surfaces to draw the famous Utah teapot  Using Bezier patches of 16 control points  I have been able to draw the teapot and display it using a  world to camera  function which gives the ability to rotate the resulting teapot  and am currently using an orthographic projection   The result is that I have a  flat  teapot  which is expected as the purpose of an orthographic projection is to preserve parallel lines   However  I would like to use a perspective projection to give the teapot depth  My question is  how does one take the 3D xyz vertex returned from the  world to camera  function  and convert this into a 2D coordinate  I am wanting to use the projection plane at z 0  and allow the user to determine the focal length and image size using the arrow keys on the keyboard   I am programming this in java and have all of the input event handler set up  and have also written a matrix class which handles basic matrix multiplication  I ve been reading through wikipedia and other resources for a while  but I can t quite get a handle on how one performs this transformation

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All of the answers address the question posed in the title. However, I would like to add a caveat that is implicit in the text. Bézier patches are used to represent the surface, but you cannot just transform the points of the patch and tessellate the patch into polygons, because this will result in distorted geometry. You can, however, tessellate the patch first into polygons using a transformed screen tolerance and then transform the polygons, or you can convert the Bézier patches to rational Bézier patches, then tessellate those using a screen-space tolerance. The former is easier, but the latter is better for a production system.

I suspect that you want the easier way. For this, you would scale the screen tolerance by the norm of the Jacobian of the inverse perspective transformation and use that to determine the amount of tessellation that you need in model space (it might be easier to compute the forward Jacobian, invert that, then take the norm). Note that this norm is position-dependent, and you may want to evaluate this at several locations, depending on the perspective. Also remember that since the projective transformation is rational, you need to apply the quotient rule to compute the derivatives.

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You can project 3D point in 2D using  Commons Math  The Apache Commons Mathematics Library with just two classes   Example for Java Swing   import org apache commons math3 geometry euclidean threed Plane  import org apache commons math3 geometry euclidean threed Vector3D    Plane planeX   new Plane new Vector3D 1  0  0    Plane planeY   new Plane new Vector3D 0  1  0       Must be orthogonal plane of planeX  void drawPoint Graphics2D g2  Vector3D v        g2 drawLine 0  0               int   world unit   planeX getOffset v                 int   world unit   planeY getOffset v        protected void paintComponent Graphics g        super paintComponent g        drawPoint g2  new Vector3D 2  1  0        drawPoint g2  new Vector3D 0  2  0        drawPoint g2  new Vector3D 0  0  2        drawPoint g2  new Vector3D 1  1  1        Now you only needs update the planeX and planeY to change the perspective-projection  to get things  like this

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Thanks to  Mads Elvenheim for a proper example code  I have fixed the minor syntax errors in the code  just a few const problems and obvious missing operators   Also  near and far have vastly different meanings in vs   For your pleasure  here is the compileable  MSVC2013  version  Have fun  Mind that I have made NEAR Z and FAR Z constant  You probably dont want it like that    include  lt vector gt   include  lt cmath gt   include  lt stdexcept gt   include  lt algorithm gt    define M PI 3 14159   define NEAR Z 0 5  define FAR Z 2 5  struct Vector       float x      float y      float z      float w       Vector     x  0    y  0    z  0    w  1          Vector  float a  float b  float c     x  a    y  b    z  c    w  1              Assume proper operator overloads here  with vectors and scalars        float Length   const               return std  sqrt  x x   y y   z z              Vector amp  operator   float fac  noexcept               x    fac          y    fac          z    fac          return  this            Vector  operator  float fac  const noexcept               return Vector  this   fac            Vector amp  operator   float div  noexcept               return operator   1 div        avoid divisions  they are much                                        more costly than multiplications            Vector Unit   const               const float epsilon   1e-6          float mag   Length            if  mag  lt  epsilon                std  out of range e                    throw e                    return Vector  this   mag            inline float Dot  const Vector amp  v1  const Vector amp  v2         return v1 x v2 x   v1 y v2 y   v1 z v2 z     class Matrix   public      Matrix     data  16                 Identity              void Identity                 std  fill  data begin    data end    float  0              data 0    data 5    data 10    data 15    1 0f            float amp  operator    size t index                 if  index  gt   16                std  out of range e                    throw e                    return data index             const float amp  operator    size t index   const               if  index  gt   16                std  out of range e                    throw e                    return data index             Matrix operator   const Matrix amp  m   const               Matrix dst          int col          for  int y   0  y lt 4    y                col   y   4              for  int x   0  x lt 4    x                    for  int i   0  i lt 4    i                        dst x   col     m i   col    data x   i   4                                                     return dst            Matrix amp  operator    const Matrix amp  m                  this     this    m          return  this                The interesting stuff        void SetupClipMatrix  float fov  float aspectRatio                 Identity            float f   1 0f   std  tan  fov   0 5f            data 0    f aspectRatio          data 5    f          data 10     FAR Z   NEAR Z     FAR Z- NEAR Z           data 11    1 0f     this  plugs  the old z into w            data 14     2 0f NEAR Z FAR Z     NEAR Z - FAR Z           data 15    0 0f             std  vector lt float gt  data       inline Vector operator   const Vector amp  v  Matrix amp  m         Vector dst      dst x   v x m 0    v y m 4    v z m 8    v w m 12       dst y   v x m 1    v y m 5    v z m 9    v w m 13       dst z   v x m 2    v y m 6    v z m 10    v w m 14       dst w   v x m 3    v y m 7    v z m 11    v w m 15       return dst     typedef std  vector lt Vector gt  VecArr  VecArr ProjectAndClip  int width  int height  const VecArr amp  vertex         float halfWidth    float width   0 5f      float halfHeight    float height   0 5f      float aspect    float width    float height      Vector v      Matrix clipMatrix      VecArr dst      clipMatrix SetupClipMatrix  60 0f    M PI   180 0f   aspect           Here  after the perspective divide  you perform Sutherland-Hodgeman clipping     by checking if the x  y and z components are inside the range of  -w  w       One checks each vector component seperately against each plane  Per-vertex     data like colours  normals and texture coordinates need to be linearly     interpolated for clipped edges to reflect the change  If the edge  v0 v1      is tested against the positive x plane  and v1 is outside  the interpolant     becomes   v1 x - w     v1 x - v0 x      I skip this stage all together to be brief             for  VecArr  const iterator i   vertex begin    i    vertex end      i            v     i    clipMatrix          v    v w     Don t get confused here  I assume the divide leaves v w alone            dst push back  v                  TODO  Clipping here         for  VecArr  iterator i   dst begin    i    dst end      i            i- gt x    i- gt x    float width     2 0f   i- gt w    halfWidth          i- gt y    i- gt y    float height     2 0f   i- gt w    halfHeight            return dst     pragma once

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I m not sure at what level you re asking this question   It sounds as if you ve found the formulas online  and are just trying to understand what it does   On that reading of your question I offer    Imagine a ray from the viewer  at point V  directly towards the center of the projection plane  call it C   Imagine a second ray from the viewer to a point in the image  P  which also intersects the projection plane at some point  Q  The viewer and the two points of intersection on the view plane form a triangle  VCQ   the sides are the two rays and the line between the points in the plane  The formulas are using this triangle to find the coordinates of Q  which is where the projected pixel will go

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I see this question is a bit old  but I decided to give an answer anyway for those who find this question by searching  The standard way to represent 2D 3D transformations nowadays is by using homogeneous coordinates   x y w  for 2D  and  x y z w  for 3D  Since you have three axes in 3D as well as translation  that information fits perfectly in a 4x4 transformation matrix  I will use column-major matrix notation in this explanation  All matrices are 4x4 unless noted otherwise  The stages from 3D points and to a rasterized point  line or polygon looks like this   Transform your 3D points with the inverse camera matrix  followed with whatever transformations they need  If you have surface normals  transform them as well but with w set to zero  as you don t want to translate normals  The matrix you transform normals with must be isotropic  scaling and shearing makes the normals malformed  Transform the point with a clip space matrix  This matrix scales x and y with the field-of-view and aspect ratio  scales z by the near and far clipping planes  and plugs the  old  z into w  After the transformation  you should divide x  y and z by w  This is called the perspective divide  Now your vertices are in clip space  and you want to perform clipping so you don t render any pixels outside the viewport bounds  Sutherland-Hodgeman clipping is the most widespread clipping algorithm in use  Transform x and y with respect to w and the half-width and half-height  Your x and y coordinates are now in viewport coordinates  w is discarded  but 1 w and z is usually saved because 1 w is required to do perspective-correct interpolation across the polygon surface  and z is stored in the z-buffer and used for depth testing   This stage is the actual projection  because z isn t used as a component in the position any more  The algorithms  Calculation of field-of-view This calculates the field-of view  Whether tan takes radians or degrees is irrelevant  but angle must match  Notice that the result reaches infinity as angle nears 180 degrees  This is a singularity  as it is impossible to have a focal point that wide  If you want numerical stability  keep angle less or equal to 179 degrees  fov   1 0   tan angle 2 0   Also notice that 1 0   tan 45    1  Someone else here suggested to just divide by z  The result here is clear  You would get a 90 degree FOV and an aspect ratio of 1 1  Using homogeneous coordinates like this has several other advantages as well  we can for example perform clipping against the near and far planes without treating it as a special case  Calculation of the clip matrix This is the layout of the clip matrix  aspectRatio is Width Height  So the FOV for the x component is scaled based on FOV for y  Far and near are coefficients which are the distances for the near and far clipping planes   fov   aspectRatio          0                  0                        0                  0                 fov                 0                        0                  0                  0           far near   far-near             1                  0                  0           2 near far   near-far           0          Screen Projection After clipping  this is the final transformation to get our screen coordinates  new x    x   Width      2 0   w    halfWidth  new y    y   Height     2 0   w    halfHeight   Trivial example implementation in C    include  lt vector gt   include  lt cmath gt   include  lt stdexcept gt   include  lt algorithm gt   struct Vector       Vector     x 0  y 0  z 0  w 1        Vector float a  float b  float c    x a  y b  z c  w 1            Assume proper operator overloads here  with vectors and scalars        float Length   const               return std  sqrt x x   y y   z z                  Vector Unit   const               const float epsilon   1e-6          float mag   Length            if mag  lt  epsilon               std  out of range e  quot  quot                throw e                    return  this   mag            inline float Dot const Vector amp  v1  const Vector amp  v2        return v1 x v2 x   v1 y v2 y   v1 z v2 z     class Matrix       public      Matrix     data 16                Identity              void Identity                 std  fill data begin    data end    float 0            data 0    data 5    data 10    data 15    1 0f            float amp  operator   size t index                if index  gt   16               std  out of range e  quot  quot                throw e                    return data index             Matrix operator  const Matrix amp  m  const               Matrix dst          int col          for int y 0  y lt 4    y               col   y 4              for int x 0  x lt 4    x                   for int i 0  i lt 4    i                       dst x col     m i col  data x i 4                                                     return dst            Matrix amp  operator   const Matrix amp  m                 this     this    m          return  this                The interesting stuff        void SetupClipMatrix float fov  float aspectRatio  float near  float far                Identity            float f   1 0f   std  tan fov   0 5f           data 0    f aspectRatio          data 5    f          data 10     far near     far-near           data 11    1 0f     this  plugs  the old z into w            data 14     2 0f near far     near-far           data 15    0 0f             std  vector lt float gt  data      inline Vector operator  const Vector amp  v  const Matrix amp  m        Vector dst      dst x   v x m 0    v y m 4    v z m 8     v w m 12       dst y   v x m 1    v y m 5    v z m 9     v w m 13       dst z   v x m 2    v y m 6    v z m 10    v w m 14       dst w   v x m 3    v y m 7    v z m 11    v w m 15       return dst     typedef std  vector lt Vector gt  VecArr  VecArr ProjectAndClip int width  int height  float near  float far  const VecArr amp  vertex        float halfWidth    float width   0 5f      float halfHeight    float height   0 5f      float aspect    float width    float height      Vector v      Matrix clipMatrix      VecArr dst      clipMatrix SetupClipMatrix 60 0f    M PI   180 0f   aspect  near  far           Here  after the perspective divide  you perform Sutherland-Hodgeman clipping          by checking if the x  y and z components are inside the range of  -w  w           One checks each vector component seperately against each plane  Per-vertex         data like colours  normals and texture coordinates need to be linearly         interpolated for clipped edges to reflect the change  If the edge  v0 v1          is tested against the positive x plane  and v1 is outside  the interpolant         becomes   v1 x - w     v1 x - v0 x          I skip this stage all together to be brief             for VecArr  iterator i vertex begin    i  vertex end      i           v     i    clipMatrix          v    v w     Don t get confused here  I assume the divide leaves v w alone            dst push back v                 TODO  Clipping here         for VecArr  iterator i dst begin    i  dst end      i           i- gt x    i- gt x    float width     2 0f   i- gt w    halfWidth          i- gt y    i- gt y    float height     2 0f   i- gt w    halfHeight            return dst     If you still ponder about this  the OpenGL specification is a really nice reference for the maths involved  The DevMaster forums at http   www devmaster net  have a lot of nice articles related to software rasterizers as well

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I think this will probably answer your question   Here s what I wrote there   Here s a very general answer  Say the camera s at  Xc  Yc  Zc  and the point you want to project is P    X  Y  Z   The distance from the camera to the 2D plane onto which you are projecting is F  so the equation of the plane is Z-Zc F   The 2D coordinates of P projected onto the plane are  X   Y    Then  very simply  X      X - Xc     F Z     Xc Y      Y - Yc     F Z     Yc If your camera is the origin  then this simplifies to  X    X    F Z  Y    Y    F Z

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I know it s an old topic but your illustration is not correct  the source code sets up the clip matrix correct    fov   aspectRatio          0                  0                        0                  0                 fov                 0                        0                  0                  0           far near   far-near      2 near far   near-far            0                  0                  1                        0           some addition to your things   This clip matrix works only if you are projecting on static 2D plane if you want to add camera movement and rotation   viewMatrix   clipMatrix   cameraTranslationMatrix4x4   cameraRotationMatrix4x4    this lets you rotate the 2D plane and move it around  -

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To obtain the perspective-corrected co-ordinates  just divide by the z co-ordinate   xc   x   z yc   y   z   The above works assuming that the camera is at  0  0  0  and you are projecting onto the plane at z   1 -- you need to translate the co-ords relative to the camera otherwise   There are some complications for curves  insofar as projecting the points of a 3D Bezier curve will not in general give you the same points as drawing a 2D Bezier curve through the projected points

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You might want to debug your system with spheres to determine whether or not you have a good field of view   If you have it too wide  the spheres with deform at the edges of the screen into more oval forms pointed toward the center of the frame   The solution to this problem is to zoom in on the frame  by multiplying the x and y coordinates for the 3 dimensional point by a scalar and then shrinking your object or world down by a similar factor   Then you get the nice even round sphere across the entire frame   I m almost embarrassed that it took me all day to figure this one out and I was almost convinced that there was some spooky mysterious geometric phenomenon going on here that demanded a different approach   Yet  the importance of calibrating the zoom-frame-of-view coefficient by rendering spheres cannot be overstated   If you do not know where the  habitable zone  of your universe is  you will end up walking on the sun and scrapping the project   You want to be able to render a sphere anywhere in your frame of view an have it appear round   In my project  the unit sphere is massive compared to the region that I m describing   Also  the obligatory wikipedia entry  Spherical Coordinate System

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Looking at the screen from the top  you get x and z axis  Looking at the screen from the side  you get y and z axis   Calculate the focal lengths of the top and side views  using trigonometry  which is the distance between the eye and the middle of the screen  which is determined by the field of view of the screen  This makes the shape of two right triangles back to back   hw   screen width   2  hh   screen height   2  fl top   hw   tan   2   fl side   hh   tan   2    Then take the average focal length   fl average    fl top   fl side    2   Now calculate the new x and new y with basic arithmetic  since the larger right triangle made from the 3d point and the eye point is congruent with the smaller triangle made by the 2d point and the eye point   x     x   fl top     z   fl top   y     y   fl top     z   fl top    Or you can simply set  x    x    z   1   and  y    y    z   1

[java] How to convert a 3D point into 2D perspective projection?

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