Add z example tables
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@ -283,11 +283,11 @@
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{
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{
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"data": {
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"data": {
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"text/latex": [
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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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"$\\displaystyle \\left\\{ s : \\frac{2}{Z_{f} - Z_{n}}, \\ t : - \\frac{Z_{f} + Z_{n}}{Z_{f} - Z_{n}}\\right\\}$"
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],
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"text/plain": [
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"text/plain": [
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"⎧ 2 -(Z_f + Zₙ) ⎫\n",
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"⎧ 2 -(Z_f + Zₙ) ⎫\n",
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"⎨q: ────────, r: ────────────⎬\n",
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"⎨s: ────────, t: ────────────⎬\n",
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"⎩ Z_f - Zₙ Z_f - Zₙ ⎭"
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"⎩ Z_f - Zₙ Z_f - Zₙ ⎭"
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]
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]
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@ -297,11 +297,11 @@
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}
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}
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],
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],
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"source": [
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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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"s, t, v_z, Z_n, Z_f = symbols('s, t, v_z, Z_n, Z_f')\n",
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"solve(Eq(\n",
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"solve(Eq(\n",
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" q*v_z + r,\n",
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" s*v_z + t,\n",
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" 2*(v_z - Z_n)/(Z_f - Z_n) - 1\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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"), s, t)"
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]
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]
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@ -315,7 +315,76 @@
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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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"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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"\n",
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"\n",
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"Neat.\n"
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"Neat. I'm going to dump a table of data to show that it works like I expect:"
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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": 3,
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"id": "394187c9-7e95-4ffc-8148-0008c7190de5",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/markdown": [
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"Let $Z_n = 0.1$ and $Z_f = 10.0$"
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],
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"text/plain": [
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"<IPython.core.display.Markdown object>"
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]
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},
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"metadata": {},
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"output_type": "display_data"
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},
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{
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"data": {
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"text/html": [
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"<table>\n",
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"<thead>\n",
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"<tr><th style=\"text-align: right;\"> $z$</th><th style=\"text-align: right;\"> $r$($z$)</th></tr>\n",
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"</thead>\n",
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"<tbody>\n",
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"<tr><td style=\"text-align: right;\"> 0.100</td><td style=\"text-align: right;\"> -1.000</td></tr>\n",
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"<tr><td style=\"text-align: right;\"> 2.000</td><td style=\"text-align: right;\"> -0.616</td></tr>\n",
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"<tr><td style=\"text-align: right;\"> 4.000</td><td style=\"text-align: right;\"> -0.212</td></tr>\n",
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"<tr><td style=\"text-align: right;\"> 6.000</td><td style=\"text-align: right;\"> 0.192</td></tr>\n",
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"<tr><td style=\"text-align: right;\"> 8.000</td><td style=\"text-align: right;\"> 0.596</td></tr>\n",
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"<tr><td style=\"text-align: right;\">10.000</td><td style=\"text-align: right;\"> 1.000</td></tr>\n",
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"</tbody>\n",
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"</table>"
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],
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"text/plain": [
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"'<table>\\n<thead>\\n<tr><th style=\"text-align: right;\"> $z$</th><th style=\"text-align: right;\"> $r$($z$)</th></tr>\\n</thead>\\n<tbody>\\n<tr><td style=\"text-align: right;\"> 0.100</td><td style=\"text-align: right;\"> -1.000</td></tr>\\n<tr><td style=\"text-align: right;\"> 2.000</td><td style=\"text-align: right;\"> -0.616</td></tr>\\n<tr><td style=\"text-align: right;\"> 4.000</td><td style=\"text-align: right;\"> -0.212</td></tr>\\n<tr><td style=\"text-align: right;\"> 6.000</td><td style=\"text-align: right;\"> 0.192</td></tr>\\n<tr><td style=\"text-align: right;\"> 8.000</td><td style=\"text-align: right;\"> 0.596</td></tr>\\n<tr><td style=\"text-align: right;\">10.000</td><td style=\"text-align: right;\"> 1.000</td></tr>\\n</tbody>\\n</table>'"
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]
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},
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"metadata": {},
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"output_type": "display_data"
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}
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],
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"source": [
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"from tabulate import tabulate\n",
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"from IPython.display import display, Markdown\n",
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"\n",
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"def ztable(fn):\n",
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" Z_n, Z_f = 0.1, 10.0\n",
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" zs = [0.1, 2, 4, 6, 8, 10]\n",
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" display(\n",
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" Markdown(f\"Let $Z_n = {Z_n}$ and $Z_f = {Z_f}$\"),\n",
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" tabulate([[z, fn(z, Z_n, Z_f)] for z in zs], headers=('$z$', '$r$($z$)'), tablefmt='html', floatfmt=\".3f\"),\n",
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" )\n",
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"\n",
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"def r(z, Z_n, Z_f):\n",
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" return 2/(Z_f-Z_n)*z - (Z_f + Z_n)/(Z_f - Z_n)\n",
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"\n",
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"ztable(r)\n"
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]
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{
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"cell_type": "markdown",
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"id": "140bd95b-1614-45d9-a2f1-dab8f84af5ee",
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"metadata": {},
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"source": [
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"Yeah, that looks pretty good."
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@ -357,7 +426,7 @@
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"\n",
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"\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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"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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"\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",
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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.\n",
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"\n",
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"\n",
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"Working it out looks almost the same as before:\n",
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"Working it out looks almost the same as before:\n",
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"\n",
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"\n",
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@ -370,31 +439,31 @@
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"cell_type": "code",
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"cell_type": "code",
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"execution_count": 3,
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"execution_count": 4,
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"id": "716e1825-f31e-4796-a6a9-aaaa7da3f42b",
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"id": "716e1825-f31e-4796-a6a9-aaaa7da3f42b",
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"metadata": {},
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"metadata": {},
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"outputs": [
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"outputs": [
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{
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{
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"data": {
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"data": {
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"text/latex": [
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"text/latex": [
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"$\\displaystyle \\left\\{ q : - \\frac{2 Z_{f} Z_{n}}{Z_{f} - Z_{n}}, \\ r : \\frac{Z_{f} + Z_{n}}{Z_{f} - Z_{n}}\\right\\}$"
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"$\\displaystyle \\left\\{ s : - \\frac{2 Z_{f} Z_{n}}{Z_{f} - Z_{n}}, \\ t : \\frac{Z_{f} + Z_{n}}{Z_{f} - Z_{n}}\\right\\}$"
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],
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],
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"text/plain": [
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"text/plain": [
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"⎧ -2⋅Z_f⋅Zₙ Z_f + Zₙ⎫\n",
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"⎧ -2⋅Z_f⋅Zₙ Z_f + Zₙ⎫\n",
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"⎨q: ──────────, r: ────────⎬\n",
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"⎨s: ──────────, t: ────────⎬\n",
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"⎩ Z_f - Zₙ Z_f - Zₙ⎭"
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"⎩ Z_f - Zₙ Z_f - Zₙ⎭"
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]
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]
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},
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},
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"execution_count": 3,
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"execution_count": 4,
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"metadata": {},
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"metadata": {},
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"output_type": "execute_result"
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"output_type": "execute_result"
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}
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}
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],
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],
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"source": [
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"source": [
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"solve(Eq(\n",
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"solve(Eq(\n",
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" q*v_z**(-1) + r,\n",
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" s*v_z**(-1) + t,\n",
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" 2*(v_z**(-1) - Z_n**(-1))/(Z_f**(-1) - Z_n**(-1)) - 1\n",
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" 2*(v_z**(-1) - Z_n**(-1))/(Z_f**(-1) - Z_n**(-1)) - 1\n",
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"), q, r)"
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"), s, t)"
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]
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]
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},
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},
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{
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{
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@ -408,7 +477,65 @@
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"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",
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"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",
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"$$\n",
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"$$\n",
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"\n",
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"\n",
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"Neat. This should work."
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"Neat. This should work. I'm going to dump a table of data to show that it works like I expect:"
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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": 5,
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"id": "f87f833a-5813-4b19-9adf-2e3d82bb2c75",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/markdown": [
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"Let $Z_n = 0.1$ and $Z_f = 10.0$"
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],
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"text/plain": [
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"<IPython.core.display.Markdown object>"
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]
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},
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"metadata": {},
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"output_type": "display_data"
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},
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{
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"data": {
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"text/html": [
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"<table>\n",
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"<thead>\n",
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"<tr><th style=\"text-align: right;\"> $z$</th><th style=\"text-align: right;\"> $r$($z$)</th></tr>\n",
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"</thead>\n",
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"<tbody>\n",
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"<tr><td style=\"text-align: right;\"> 0.100</td><td style=\"text-align: right;\"> -1.000</td></tr>\n",
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"<tr><td style=\"text-align: right;\"> 2.000</td><td style=\"text-align: right;\"> 0.919</td></tr>\n",
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"<tr><td style=\"text-align: right;\"> 4.000</td><td style=\"text-align: right;\"> 0.970</td></tr>\n",
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"<tr><td style=\"text-align: right;\"> 6.000</td><td style=\"text-align: right;\"> 0.987</td></tr>\n",
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"<tr><td style=\"text-align: right;\"> 8.000</td><td style=\"text-align: right;\"> 0.995</td></tr>\n",
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"<tr><td style=\"text-align: right;\">10.000</td><td style=\"text-align: right;\"> 1.000</td></tr>\n",
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"</tbody>\n",
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"</table>"
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],
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"text/plain": [
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"'<table>\\n<thead>\\n<tr><th style=\"text-align: right;\"> $z$</th><th style=\"text-align: right;\"> $r$($z$)</th></tr>\\n</thead>\\n<tbody>\\n<tr><td style=\"text-align: right;\"> 0.100</td><td style=\"text-align: right;\"> -1.000</td></tr>\\n<tr><td style=\"text-align: right;\"> 2.000</td><td style=\"text-align: right;\"> 0.919</td></tr>\\n<tr><td style=\"text-align: right;\"> 4.000</td><td style=\"text-align: right;\"> 0.970</td></tr>\\n<tr><td style=\"text-align: right;\"> 6.000</td><td style=\"text-align: right;\"> 0.987</td></tr>\\n<tr><td style=\"text-align: right;\"> 8.000</td><td style=\"text-align: right;\"> 0.995</td></tr>\\n<tr><td style=\"text-align: right;\">10.000</td><td style=\"text-align: right;\"> 1.000</td></tr>\\n</tbody>\\n</table>'"
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]
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},
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"metadata": {},
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"output_type": "display_data"
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}
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],
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"source": [
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"def r2(z, Z_n, Z_f):\n",
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" return (2*Z_f*Z_n)/(Z_n-Z_f)*z**(-1) + (Z_f + Z_n)/(Z_f - Z_n)\n",
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"\n",
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"ztable(r2)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "7202e8ff-ca9f-44c8-9676-8c2973cd05eb",
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"metadata": {},
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"source": [
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"It works, but it's crazy how quickly the values approach $1$ as $z$ gets away from $Z_n$."
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]
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]
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},
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},
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{
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@ -451,7 +578,7 @@
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"source": [
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"source": [
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"# Completing $f$\n",
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"# Completing $f$\n",
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"\n",
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"\n",
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"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",
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"What to do with $o$? 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",
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"\n",
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"\n",
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"Here's the completed function:\n",
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"Here's the completed function:\n",
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"\n",
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"\n",
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"\\end{bmatrix}\n",
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"\\end{bmatrix}\n",
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"$$\n",
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"$$\n",
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"\n",
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"\n",
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"What does <code>gluPerspective</code> do? Check it out: https://www.khronos.org/registry/OpenGL-Refpages/gl2.1/xhtml/gluPerspective.xml"
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"What does <code>gluPerspective</code> do? Check it out: https://www.khronos.org/registry/OpenGL-Refpages/gl2.1/xhtml/gluPerspective.xml\n",
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"\n",
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"Two things to note in comparison: their $k$ is the negated version of mine, and their $o$ is -1. TODO: figure out exactly why that is (probably coordinate system handedness)"
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]
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]
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}
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}
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],
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],
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