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Perspective.ipynb
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Perspective.ipynb
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{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "5f5e1e09-487e-421d-ad34-94a8e3b42629",
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"metadata": {
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"tags": []
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},
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"source": [
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"# Deriving a perspective matrix given field of view and near & far z-planes\n",
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"\n",
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"I am looking for a function $f(\\theta, Z_n, Z_f) = \\begin{bmatrix}\\ddots\\end{bmatrix}$ that will return a 4x4 perspective matrix, where:\n",
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"\n",
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" * $\\theta$ is the vertical field of view,\n",
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" * $Z_n$ is the near z clipping plane, and\n",
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" * $Z_f$ is the far z clipping plane."
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]
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},
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{
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"cell_type": "markdown",
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"id": "575394e6-9924-4691-b212-e98fbb824bf8",
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"metadata": {
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"tags": []
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},
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"source": [
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"## Vertex Pipeline\n",
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"\n",
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"Suppose I start with a set of untransformed vertices, $V$, that make up my 3D scene, where $v_w = 1$ for all $v \\in V$.\n",
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"\n",
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"OpenGL will follow these steps when transforming vertices:\n",
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" \n",
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" 1. Allow me to **transform my vertices**, $V$, into transformed vertices, $V'$.\n",
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" 1. **Perspective divide** my transformed vertices, $V'$, into projected vertices, $V''$.\n",
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" 1. **Clip each point** $v''$ using a $2\\times2\\times2$ cube centered at the origin.\n",
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"\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "95b7bd1c-93a9-4504-a60b-08b57c2b0aab",
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"metadata": {},
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"source": [
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"### Transforming\n",
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"\n",
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"I want to get away with a single matrix multiply per vertex:\n",
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"$$\n",
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"v' = f(\\theta, Z_n, Z_f) \\cdot v\n",
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"$$\n",
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"so I need to carefully build the function $f$ while giving consideration to the steps that come after."
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]
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},
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{
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"cell_type": "markdown",
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"id": "7901a150-bd3a-48e0-8d11-a4b12a2c2ca3",
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"metadata": {},
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"source": [
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"### Perspective divide\n",
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"\n",
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"OpenGL will divide each $v'$'s $x, y, z$ by its $w$:\n",
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"$$\n",
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"v''= <\\frac{v'_x}{v'_w}, \\frac{v'_y}{v'_w}, \\frac{v'_z}{v'_w}, v'_{w}>\n",
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"$$"
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]
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},
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{
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"cell_type": "markdown",
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"id": "18f57f5a-c1a4-4bd1-b572-9965039ca2c6",
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"metadata": {},
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"source": [
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"### Clipping\n",
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"\n",
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"After the perspective divide, all values are clipped to be inside a $2\\times2\\times2$ cube centered at the origin so that. So that\n",
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"\n",
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"$$\n",
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"-1 \\le v''_x \\le 1\\\\\n",
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"-1 \\le v''_y \\le 1\\\\\n",
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"-1 \\le v''_z \\le 1\\\\\n",
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"$$\n",
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"\n",
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"*Side note*: Direct3D's clipping bounds are different than OpenGL's, so $f$ needs to be different for it.\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "1e7ac1fc-5dc1-4a40-95e4-2fb2ee5d2357",
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"metadata": {
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"tags": []
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},
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"source": [
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"## Finding $v'$\n",
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"\n",
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"I'll start by putting variables ($a$ through $p$) in each spot in the matrix which we can solve for below:\n",
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"\n",
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"$$\n",
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"f(\\theta, Z_n, Z_f) = \\begin{bmatrix}\n",
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"a & b & c & d\\\\\n",
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"e & f & g & h\\\\\n",
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"i & j & k & l\\\\\n",
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"m & n & o & p\n",
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"\\end{bmatrix}\n",
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"$$\n",
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"\n",
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"Since $v'$ = $f(\\ldots) \\cdot v$, I can multiply that out to get:\n",
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"\n",
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"$$\n",
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"\\begin{align}\n",
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"v'_x &= av_x + bv_y + cv_z + d\\\\\n",
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"v'_y &= ev_x + fv_y + gv_z + h\\\\\n",
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"v'_z &= iv_x + jv_y + kv_z + l\\\\\n",
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"v'_w &= mv_x + nv_y + ov_z + p\n",
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"\\end{align}\n",
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"$$\n",
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"\n",
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"(I'm using the fact that $v_w = 1$ here)\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "8efa1546-9022-4d3a-9f46-162ee45cb22f",
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"metadata": {},
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"source": [
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"## Now finding $v''$\n",
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"\n",
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"I said earlier that:\n",
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"\n",
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"$$\n",
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"v''= <\\frac{v'_x}{v'_w}, \\frac{v'_y}{v'_w}, \\frac{v'_z}{v'_w}, v'_{w}>\n",
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"$$\n",
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"\n",
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"so substituting $v'$ in from above gives me:\n",
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"\n",
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"$$\n",
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"\\begin{align}\n",
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"v''_x &= \\frac{av_x + bv_y + cv_z + d}{mv_x + nv_y + ov_z + p}\\\\\n",
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"v''_y &= \\frac{ev_x + fv_y + gv_z + h}{mv_x + nv_y + ov_z + p}\\\\\n",
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"v''_z &= \\frac{iv_x + jv_y + kv_z + l}{mv_x + nv_y + ov_z + p}\\\\\n",
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"\\end{align}\n",
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"$$\n",
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"\n",
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"I don't really care what happens to $v''_w$ so I left it out."
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]
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},
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{
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"cell_type": "markdown",
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"id": "6795607c-23ab-43ea-ae1a-16ba655b015d",
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"metadata": {},
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"source": [
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"## Figuring out what I want $v''_x$ to actually be\n",
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"\n",
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"I'll skip the derivation for now, but knowing field of view, $\\theta$, and the size of the clipping box, i.e. $2\\times2\\times2$, I can use a pinhole camera model to say:\n",
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"\n",
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"$$\n",
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"v''_x = d\\frac{v_x}{v_z}\n",
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"$$\n",
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"\n",
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"where $d$ is the distance to the back of the pinhole camera. I can form a right triangle with an angle of $\\frac{\\theta}{2}$ and an opposite edge being the extent of the clipping cube, i.e. $1$, plus the definition of $\\tan(\\theta)$, to get:\n",
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"\n",
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"$$\n",
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"\\tan(\\frac{\\theta}{2}) = \\frac{1}{d}\n",
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"$$\n",
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"\n",
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"Solving for $d$ gives me:\n",
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"\n",
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"$$\n",
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"d = \\frac{1}{\\tan(\\frac{\\theta}{2})}\n",
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"$$\n",
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"\n",
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"(Insert cool diagram here)\n",
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"\n",
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"Putting $d$ back in, I can find what I want $v''_x$ to be:\n",
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"\n",
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"$$\n",
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"v''_x = \\frac{1}{\\tan(\\frac{\\theta}{2})}\\frac{v_x}{v_z}\n",
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"$$"
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]
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},
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{
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"cell_type": "markdown",
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"id": "98b168fe-b93e-481e-b747-3ea0517d5907",
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"metadata": {},
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"source": [
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"## Deriving the first row of $f(\\ldots)$\n",
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"\n",
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"I'm finally ready to try to figure out what $a$, $b$, $c$, & $d$ are to put in my matrix for function $f$.\n",
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"\n",
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"I can set my desired $v''_x$ equal to my multiplied-out version of $v''_x$ and solve to find the values in my matrix:\n",
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"\n",
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"$$\n",
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"\\frac{av_x + bv_y + cv_z + d}{mv_x + nv_y + ov_z + p} = \\frac{1}{\\tan(\\frac{\\theta}{2})}\\frac{v_x}{v_z}\n",
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"$$\n",
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"\n",
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"It's immediately obvious that $b$, $c$, $d$, $m$, $n$ & $p$ must all be zero. That leaves me with:\n",
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"\n",
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"$$\n",
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"\\frac{a}{o}\\frac{v_x}{v_z} = \\frac{1}{\\tan(\\frac{\\theta}{2})}\\frac{v_x}{v_z}\n",
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"$$\n",
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"\n",
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"So algebra tells me that:\n",
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"$$\n",
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"\\frac{a}{o} = \\frac{1}{\\tan(\\frac{\\theta}{2})}\n",
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"$$\n",
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"\n",
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"I'm going to use a bit of foresight here realize that the variable $o$ is used in 4 equations, where $a$ is only used in this one, so I'll save figuring out $o$ for later by pushing it to the other side of the equation:\n",
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"\n",
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"$$\n",
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"a = \\frac{1}{\\tan(\\frac{\\theta}{2})}o\n",
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"$$"
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]
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},
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{
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"cell_type": "markdown",
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"id": "e817fdab-92ef-4a7b-bd92-3dd1b2c260cb",
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"metadata": {},
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"source": [
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"## Checking in on $f$ after working on $v''_x$\n",
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"\n",
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"Let's see how $f$ is doing:\n",
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"\n",
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"$$\n",
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"f(\\theta, Z_n, Z_f) = \\begin{bmatrix}\n",
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"\\frac{1}{\\tan(\\frac{\\theta}{2})}o & 0 & 0 & 0\\\\\n",
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"e & f & g & h\\\\\n",
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"i & j & k & l\\\\\n",
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"0 & 0 & o & 0\n",
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"\\end{bmatrix}\n",
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"$$\n",
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"\n",
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"So, $v''_y$ is going to be nearly identical except $e$ will be zero instead of $f$, letting me skip straight to this:\n",
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"\n",
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"$$\n",
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"f(\\theta, Z_n, Z_f) = \\begin{bmatrix}\n",
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"\\frac{1}{\\tan(\\frac{\\theta}{2})}o & 0 & 0 & 0\\\\\n",
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"0 & \\frac{1}{\\tan(\\frac{\\theta}{2})}o & 0 & 0\\\\\n",
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"i & j & k & l\\\\\n",
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"0 & 0 & o & 0\n",
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"\\end{bmatrix}\n",
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"$$"
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]
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},
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{
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"cell_type": "markdown",
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"id": "54508209-1324-4d54-aa5b-eebcf9a31032",
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"metadata": {},
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"source": [
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"## Figuring out what I want $v''_z$ to actually be\n",
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"\n",
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"This is where it gets tricky. I want to remap any $v_z$ that's between $Z_n$ through $Z_f$ to $-1$ to $1$ (i.e. stuffing it into the $2\\times2\\times2$ clipping cube). That seems simple enough. First, I'll turn $v_z$ into a ratio like:\n",
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"\n",
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"$$\n",
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"\\frac{v_z - Z_n}{Z_f - Z_n}\n",
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"$$\n",
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"\n",
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"Now remaping it to $-1$ through $1$, I get what I want $v''_z$ be:\n",
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"\n",
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"$$\n",
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"v''_z = 2\\frac{v_z - Z_n}{Z_f - Z_n}-1\n",
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"$$\n",
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"\n",
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"I'm going to use a little bit of code to rearrange this for me:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"id": "821cac4d-cf5a-452e-b776-09536ddaec85",
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"metadata": {
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"tags": []
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},
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"outputs": [],
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"source": [
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"from sympy import *\n",
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"init_printing(use_unicode=True)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"id": "e463c065-f51b-468c-a90e-7d5da5aac035",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/latex": [
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"$\\displaystyle \\left\\{ q : \\frac{2}{Z_{f} - Z_{n}}, \\ r : - \\frac{Z_{f} + Z_{n}}{Z_{f} - Z_{n}}\\right\\}$"
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],
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"text/plain": [
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"⎧ 2 -(Z_f + Zₙ) ⎫\n",
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"⎨q: ────────, r: ────────────⎬\n",
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"⎩ Z_f - Zₙ Z_f - Zₙ ⎭"
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]
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},
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"execution_count": 2,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"q, r, v_z, Z_n, Z_f = symbols('q, r, v_z, Z_n, Z_f')\n",
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"solve(Eq(\n",
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" q*v_z + r,\n",
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" 2*(v_z - Z_n)/(Z_f - Z_n) - 1\n",
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"), q, r)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "8cec6c52-f3f5-48b7-98cf-c3351e3af7a0",
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"metadata": {},
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"source": [
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"That gives me a desired $v''_z$ like:\n",
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"\n",
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"$$\n",
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"v''_z = \\frac{2}{Z_f-Z_n}v_z - \\frac{Z_f+Z_n}{Z_f-Z_n}\n",
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"$$\n",
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"\n",
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"Neat.\n"
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]
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},
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{
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||||
"cell_type": "markdown",
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||||
"id": "80171350-d1ae-429d-a57f-a6d2cafe9970",
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||||
"metadata": {},
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||||
"source": [
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||||
"## Deriving the third row of $f(\\ldots)$\n",
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"\n",
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"I'll do the same thing I did earlier and set my multiplied-out $v''_z$ equal to my desired $v''_z$ and figure out what values the matrix should have for $f$:\n",
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||||
"\n",
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||||
"$$\n",
|
||||
"\\frac{iv_x + jv_y + kv_z + l}{mv_x + nv_y + ov_z + p} = \\frac{2}{Z_f-Z_n}v_z - \\frac{Z_f+Z_n}{Z_f-Z_n}\n",
|
||||
"$$\n",
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||||
"\n",
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"It's immediately obvious that $i$ & $j$ must both be zero. I already established that $m$ & $n$ & $p$ must be zero too. That leaves me with:\n",
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||||
"\n",
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||||
"$$\n",
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||||
"\\frac{kv_z + l}{ov_z} = \\frac{2}{Z_f-Z_n}v_z - \\frac{Z_f+Z_n}{Z_f-Z_n}\n",
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||||
"$$\n",
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||||
"\n",
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||||
"There are two $v_z$s on the left side, so I will rearrange it:\n",
|
||||
"\n",
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||||
"$$\n",
|
||||
"\\frac{l}{o}v_z^{-1} + k = \\frac{2}{Z_f-Z_n}v_z - \\frac{Z_f+Z_n}{Z_f-Z_n}\n",
|
||||
"$$\n",
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||||
"\n",
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||||
"Well that's not going to work! I could try $k = -\\frac{Z_f+Z_n}{Z_f-Z_n}$ and $\\frac{l}{o} = \\frac{2}{Z_f-Z_n}$, but I'm never going to escape the fact that $v_z^{-1} \\ne v_z$. 😞"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "7ab42c13-f71d-4381-b0be-cbaed3b63649",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Rethinking what I want $v''_z$ to be\n",
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||||
"\n",
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||||
"My goal is to remap $v_z$ from the range $Z_n$ through $Z_f$ to the range $-1$ through $1$. I'll call this remapping function $r(z)$.\n",
|
||||
"\n",
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||||
"The only other constraint is that if $s \\le t$ then $r(s) \\le r(t)$ as well.\n",
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||||
"\n",
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||||
"If that's the case then we can just as easily remap $v_z^{-1}$ from the range $Z_n^{-1}$ through $Z_f^{-1}$ to the range $-1$ through $1$ and the constraints will hold. A very important thing to note is that doing this means that z-depth is no longer a *linear* interpolation, which explains the warnings about not letting $Z_n$ get too close to zero for fear of losing bits of precision.\n",
|
||||
"\n",
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||||
"Working it out looks almost the same as before:\n",
|
||||
"\n",
|
||||
"$$\n",
|
||||
"v''_z = 2\\frac{v_z^{-1} - Z_n^{-1}}{Z_f^{-1} - Z_n^{-1}} - 1\n",
|
||||
"$$\n",
|
||||
"\n",
|
||||
"I'll have sympy rearrange it for me:"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 3,
|
||||
"id": "716e1825-f31e-4796-a6a9-aaaa7da3f42b",
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/latex": [
|
||||
"$\\displaystyle \\left\\{ q : - \\frac{2 Z_{f} Z_{n}}{Z_{f} - Z_{n}}, \\ r : \\frac{Z_{f} + Z_{n}}{Z_{f} - Z_{n}}\\right\\}$"
|
||||
],
|
||||
"text/plain": [
|
||||
"⎧ -2⋅Z_f⋅Zₙ Z_f + Zₙ⎫\n",
|
||||
"⎨q: ──────────, r: ────────⎬\n",
|
||||
"⎩ Z_f - Zₙ Z_f - Zₙ⎭"
|
||||
]
|
||||
},
|
||||
"execution_count": 3,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"solve(Eq(\n",
|
||||
" q*v_z**(-1) + r,\n",
|
||||
" 2*(v_z**(-1) - Z_n**(-1))/(Z_f**(-1) - Z_n**(-1)) - 1\n",
|
||||
"), q, r)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "14d13eae-9015-473c-a718-3e56a82d5ec7",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"That gives me a desired $v''_z$ like:\n",
|
||||
"\n",
|
||||
"$$\n",
|
||||
"v''_z = \\frac{2 Z_f Z_n}{Z_n-Z_f}v_z^{-1} + \\frac{Z_f+Z_n}{Z_f-Z_n}\n",
|
||||
"$$\n",
|
||||
"\n",
|
||||
"Neat. This should work."
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "1a7691cb-14b7-40d0-bd7e-b8ca170518d4",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## Return to the third row of $f(\\ldots)$\n",
|
||||
"\n",
|
||||
"I'll do the same thing I did earlier and set my multiplied-out $v''_z$ equal to my new-and-improved $v''_z$ and figure out what values the matrix should have for $f$:\n",
|
||||
"\n",
|
||||
"$$\n",
|
||||
"\\frac{l}{o}v_z^{-1} + k = \\frac{2 Z_f Z_n}{Z_n-Z_f}v_z^{-1} + \\frac{Z_f+Z_n}{Z_f-Z_n}\n",
|
||||
"$$\n",
|
||||
"\n",
|
||||
"I'll throw $o$ over to the other side:\n",
|
||||
"\n",
|
||||
"$$\n",
|
||||
"lv_z^{-1} + k = \\frac{2 Z_f Z_n}{Z_n-Z_f}v_z^{-1}o + \\frac{Z_f+Z_n}{Z_f-Z_n}o\n",
|
||||
"$$\n",
|
||||
"\n",
|
||||
"Now I can set $l = \\frac{2 Z_f Z_n}{Z_n-Z_f}o$ and $k = \\frac{Z_f+Z_n}{Z_f-Z_n}o$.\n",
|
||||
"\n",
|
||||
"That should be all of them except $o$. This is how it looks:\n",
|
||||
"\n",
|
||||
"$$\n",
|
||||
"f(\\theta, Z_n, Z_f) = \\begin{bmatrix}\n",
|
||||
"\\frac{1}{\\tan(\\frac{\\theta}{2})}o & 0 & 0 & 0\\\\\n",
|
||||
"0 & \\frac{1}{\\tan(\\frac{\\theta}{2})}o & 0 & 0\\\\\n",
|
||||
"0 & 0 & \\frac{Z_f+Z_n}{Z_f-Z_n}o & \\frac{2 Z_f Z_n}{Z_n-Z_f}o\\\\\n",
|
||||
"0 & 0 & o & 0\n",
|
||||
"\\end{bmatrix}\n",
|
||||
"$$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "12219eba-6a44-49ef-a987-cfb395e1d6c5",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"# Completing $f$\n",
|
||||
"\n",
|
||||
"What to do $o$? I put factors of $o$ in all of my other values, so no matter what $o$ actually is, so long as it's not $0$, it can be anything. I will choose $1$.\n",
|
||||
"\n",
|
||||
"Here's the completed function:\n",
|
||||
"\n",
|
||||
"$$\n",
|
||||
"f(\\theta, Z_n, Z_f) = \\begin{bmatrix}\n",
|
||||
"\\frac{1}{\\tan(\\frac{\\theta}{2})} & 0 & 0 & 0\\\\\n",
|
||||
"0 & \\frac{1}{\\tan(\\frac{\\theta}{2})} & 0 & 0\\\\\n",
|
||||
"0 & 0 & \\frac{Z_f+Z_n}{Z_f-Z_n} & \\frac{2 Z_f Z_n}{Z_n-Z_f}\\\\\n",
|
||||
"0 & 0 & 1 & 0\n",
|
||||
"\\end{bmatrix}\n",
|
||||
"$$\n",
|
||||
"\n",
|
||||
"What does <code>gluPerspective</code> do? Check it out: https://www.khronos.org/registry/OpenGL-Refpages/gl2.1/xhtml/gluPerspective.xml"
|
||||
]
|
||||
}
|
||||
],
|
||||
"metadata": {
|
||||
"kernelspec": {
|
||||
"display_name": "Python 3 (ipykernel)",
|
||||
"language": "python",
|
||||
"name": "python3"
|
||||
},
|
||||
"language_info": {
|
||||
"codemirror_mode": {
|
||||
"name": "ipython",
|
||||
"version": 3
|
||||
},
|
||||
"file_extension": ".py",
|
||||
"mimetype": "text/x-python",
|
||||
"name": "python",
|
||||
"nbconvert_exporter": "python",
|
||||
"pygments_lexer": "ipython3",
|
||||
"version": "3.8.10"
|
||||
}
|
||||
},
|
||||
"nbformat": 4,
|
||||
"nbformat_minor": 5
|
||||
}
|
||||
Loading…
x
Reference in New Issue
Block a user