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" ![]()
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"
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"\n",
"\n",
"\n"
],
"text/plain": [
"\n",
" \n",
" \n",
" \n",
"
\n",
"![\"Drawing\"](\"Jupyter1_Beispiel1.png\")
\n",
"\n",
"**Gegeben**:\n",
"\n",
"* Anfangsgeschwindigkeit $v_0 = 20 \\frac{\\mathrm{m}}{\\mathrm{s}}$\n",
"* Abschusswinkel $\\theta = 15, 30, 45, 60, 75 ^\\circ$\n",
"* Erdbeschleunigung $g = 9.81 \\frac{\\mathrm{m}}{\\mathrm{s}^2}$ (entlang $x_2$)\n",
"\n",
"**Gesucht**:\n",
"\n",
"* Wertetabelle der Höhe $h(t)=x_2(t)$ über der Zeit für $\\theta=45$° mit 10 Einträgen.\n",
"* Flugbahnen für den jeweiligen Abschusswinkel in Form eines xy-Plots (Zeitdauer 3s).\n",
"* Animierte Flugbahn für $\\theta=45$°."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Theoretischer Hintergrund\n",
"\n",
"Die Flugbahn eine Punktes wird mittels eines Postionsvektors $\\vec x(t)$ beschrieben. \n",
"\n",
"![\"Drawing\"](\"Jupyter1_Beispiel1a.png\")
\n",
"\n",
"Der Geschwindigkeitsvektor $\\vec v(t)$ berechnet sich aus der zeitlichen Ableitung der Position (beides Funktionen der Zeit $t$) d.h.:\n",
"\n",
"$$ \\vec v(t) = \\frac{\\partial \\vec x(t)}{\\partial t} $$\n",
"\n",
"und der Beschleunigungsvektor $\\vec a(t)$ wir berechnet aus der Geschwindigkeitsänderung:\n",
"\n",
"$$ \\vec a(t) = \\frac{\\partial \\vec v(t)}{\\partial t} = \\frac{\\partial^2 \\vec x(t)}{\\partial t^2}.$$\n",
"\n",
"Im vorliegend Beispiel ist die Beschleunigung konstant (Erdbeschleunigung in negativer $x_2$ Richtung) und somit zeitunabhängig: \n",
"\n",
"$$ \\vec a(t) = \\vec a = \\left( \\begin{array}{c}\n",
"0 \\\\\n",
"-g \\\\\n",
"\\end{array}\\right) = \\mathrm{const}. $$\n",
"\n",
"Im Falle konstanter Beschleunigung folgt das Weg-Zeit Gesetz d.h. die Positions des Projektils\n",
"durch zweifache Integration dieser Differentialgleichung:\n",
"\n",
"\n",
"$$ \\frac{\\partial^2 \\vec x(t)}{\\partial t^2} = \\vec a \\qquad \\Rightarrow \\qquad \\vec x(t) = \\vec x_0 + \\vec v_0 + \\vec a \\frac{t^2}{2}$$\n",
"\n",
"\n",
"mit der Anfangsposition \n",
"\n",
"$$\\vec x_0 = \\left( \\begin{array}{c}\n",
"0 \\\\\n",
"0 \\\\\n",
"\\end{array}\\right) $$ \n",
"\n",
"und der Anfangsgeschwindigkeit $\\vec v_0$\n",
"\n",
"$$\\vec v_0 = \\left( \\begin{array}{c}\n",
"v_0 \\cos(\\theta) \\\\\n",
"v_0 \\sin(\\theta) \\\\\n",
"\\end{array}\\right).$$ \n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Numerische Umsetzung\n",
"\n",
"### Gegeben:"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"import numpy as np\n",
"\n",
"# Anfangsgeschwindigkeit [v0]=m/s\n",
"v0 = 20\n",
"\n",
"# Erdbeschleunigung [g]=m/s2\n",
"g = 9.81\n",
"\n",
"# Abschusswinkel [theta]=rad \n",
"np_theta = np.array([15, 30, 45, 60, 75])*np.pi/180.\n",
"\n",
"# Anfangsverschiebungsvektor [x0]=m\n",
"np_x0 = np.array([0, 0])\n",
"\n",
"# Beschleunigungsvektor [a]=m/s2\n",
"np_a = np.array([0, -g])"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Hilfsfunktionen"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
"outputs": [],
"source": [
"### Anfangsgeschwindigkeitsvektor [v0]=m/s\n",
"def np_v0(v0, theta): \n",
" return np.array([v0*np.cos(theta), v0*np.sin(theta)])"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {},
"outputs": [],
"source": [
"### Positionsvector [x]=m\n",
"def np_x(t, theta, v0, g):\n",
" return np_x0 + np_v0(v0, theta)*t + np_a*(t**2)/2."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Wertetabelle $h(t)$ für 𝜃=45 °"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Bodenkontakt nach 2.883 s\n"
]
}
],
"source": [
"# Vorgegebener Winkel\n",
"theta45=45*np.pi/180.\n",
"\n",
"# Hilfsfunktion zur Berechnung der Hoehe\n",
"def h(t) : \n",
" return np_x(t, theta45, v0, g)[1]\n",
"\n",
"# Zeit wenn Bodenkontakt erreicht wird d.h. h(t)=0 bei t>0!\n",
"from scipy import optimize\n",
"tmax45 = optimize.newton(h, 10) # Nullstellensuche, Startwert 10s d.h. rechter Nulldurchgang\n",
"\n",
"print(f\"Bodenkontakt nach {tmax45:.3f} s\")"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[0. 0.32035645 0.64071291 0.96106936 1.28142581 1.60178227\n",
" 1.92213872 2.24249517 2.56285163 2.88320808]\n"
]
}
],
"source": [
"# Zeit von 0...tmax \n",
"np_t45 = np.linspace(0, tmax45, 10) \n",
"\n",
"print(np_t45)"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Zeit s | Weg m\n",
"-----------------\n",
"0.000 | 0.000\n",
"0.320 | 4.027\n",
"0.641 | 7.047\n",
"0.961 | 9.061\n",
"1.281 | 10.068\n",
"1.602 | 10.068\n",
"1.922 | 9.061\n",
"2.242 | 7.047\n",
"2.563 | 4.027\n",
"2.883 | -0.000\n"
]
}
],
"source": [
"# Wertetabelle\n",
"print(\"Zeit s | Weg m\")\n",
"print(\"-----------------\")\n",
"for t in np_t45: \n",
" print(f\"{t:.3f} | {h(t):7.3f}\")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Plot der Flugbahnen"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[0. 0.15789474 0.31578947 0.47368421 0.63157895 0.78947368\n",
" 0.94736842 1.10526316 1.26315789 1.42105263 1.57894737 1.73684211\n",
" 1.89473684 2.05263158 2.21052632 2.36842105 2.52631579 2.68421053\n",
" 2.84210526 3. ]\n"
]
}
],
"source": [
"### betrachtete Zeitspanne 0-4sec \n",
"np_t = np.linspace(0, 3, 20)\n",
"print(np_t)"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"