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    <p>I see this question is a bit old, but I decided to give an answer anyway for those who find this question by searching.<br> The standard way to represent 2D/3D transformations nowadays is by using <strong>homogeneous coordinates</strong>. <em>[x,y,w]</em> for 2D, and <em>[x,y,z,w]</em> 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.<br> The stages from 3D points and to a rasterized point, line or polygon looks like this: </p> <ol> <li>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 <em>isotropic</em>; scaling and shearing makes the normals malformed.</li> <li>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 <em>perspective divide</em>.</li> <li>Now your vertices are in clip space, and you want to perform clipping so you don't render any pixels outside the viewport bounds. <strong>Sutherland-Hodgeman clipping</strong> is the most widespread clipping algorithm in use.</li> <li>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.</li> </ol> <p>This stage is the actual projection, because z isn't used as a component in the position any more.</p> <h2>The algorithms:</h2> <h3>Calculation of field-of-view</h3> <p>This calculates the field-of view. Whether tan takes radians or degrees is irrelevant, but <em>angle</em> must match. Notice that the result reaches infinity as <em>angle</em> nears 180 degrees. This is a singularity, as it is impossible to have a focal point that wide. If you want numerical stability, keep <em>angle</em> less or equal to 179 degrees.</p> <pre><code>fov = 1.0 / tan(angle/2.0) </code></pre> <p>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.</p> <h3>Calculation of the clip matrix</h3> <p>This is the layout of the clip matrix. <em>aspectRatio</em> 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. </p> <pre><code>[fov * aspectRatio][ 0 ][ 0 ][ 0 ] [ 0 ][ fov ][ 0 ][ 0 ] [ 0 ][ 0 ][(far+near)/(far-near) ][ 1 ] [ 0 ][ 0 ][(2*near*far)/(near-far)][ 0 ] </code></pre> <h3>Screen Projection</h3> <p>After clipping, this is the final transformation to get our screen coordinates.</p> <pre><code>new_x = (x * Width ) / (2.0 * w) + halfWidth; new_y = (y * Height) / (2.0 * w) + halfHeight; </code></pre> <h2>Trivial example implementation in C++</h2> <pre class="lang-cpp prettyprint-override"><code>#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(""); 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(""); 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; } </code></pre> <p>If you still ponder about this, the OpenGL specification is a really nice reference for the maths involved. The DevMaster forums at <a href="http://www.devmaster.net/" rel="noreferrer">http://www.devmaster.net/</a> have a lot of nice articles related to software rasterizers as well.</p>
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    1. POHow to convert a 3D point into 2D perspective projection?
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    1. CO@madas +1, Thanks for explanation.
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    2. COWhat does this mean? "you perform Sutherland-Hodgeman clipping by checking if the x, y and z components are inside the range of [-w, w]." Namely the -w and w. Where do those values come from?
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