• 作业L1 2.9题
• 作业L2 2.10题强化版（引入风阻）————“辅助精确打击系统”
• 作业L3 发展“超级辅助精确打击系统”（考虑炮弹初始发射的时候发射角度误差1%，速度有5%的误差，风阻误差10%，可以考虑引入Coriolis force等，以炮弹落点与打击目标距离差平方均值最小为优胜)

Homework 06 - The Trajectory of a Cannon Shell

ComputationalPhysics_HW 岳绍圣2013301020033

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• Homework 06 - The Trajectory of a Cannon Shell
  • Abstract
  • Background
  • Level 0 - No drag v.s. With air drag
    • • Conclusion 0
  • Level 1 - With v.s. Without Density Correction
    • 1.1 Isothermal v.s. Adiabatic
      • Conclusion 1.1
    • 1.2 Accurate Max Angle and Maxi Range
  • Reference

Abstract

In this homework several problems involving the motion of objects through the atmosphere are considered. Still, the problems are all described by ordinary differential equations in which initial values are given and can be solved using Euler method. This report solved cannon trajectory problem, in which air drag and reduced air density at high altitudes are considered.

Key Words: Projectile Motion; Euler Method; Air Drag; Air Density; Cannon Shell

Background

In realistic projectile motion[1], we will fail to to obtain a realistic description if we do not include Air resistance must be included if we are to obtain a realistic description of the problem.

In general, air resistance can be written in the fairly innocent form[2]:

$$F_{drag}\approx -B_1v-B_2v^2 \tag{1}$$

and the $v^2$ term dominates for most objects at any reasonable velocity.
To make an approximate estimate of $B_2$, we can use the fact that an object must push air in front it out of the way to move through the atmosphere. So $F_{drag}vdt=E_{air}$, i.e.

$$F_{drag}\approx -\frac{1}{2}C\rho Av^2 \tag{2}$$

$C$ is the drag coefficient, $\rho$ is air density, $A$ is the front area of the object.

Another important piece of physics we should consider is air density at high altitude. To investigate this effect, we need to know how the air density depends on the altitude.

The simplest model is isothermal ideal gas model. And it leads to

$$\rho=\rho_0exp(-y/y_0) \tag{3}$$

where $y_0=k_BT/mg\approx1.0\times10^4m$, and $\rho$ is density at sea level ($y=0$).
However, this isothermal model of atmosphere is perhaps not very realistic, since we know that air temperature can vary quite a bit over altitude changes of a few kilometers. A more accurate model is to assume thei air i a poor conductor of heat. This leads to the so-called adiabatic approximation:

$$\rho=\rho_0(1-ay/T_0)^\alpha \tag{4}$$

where $\alpha\approx6.5\times10^{-3}K/m$, $T_0$ is the sea level temperature, and the exponent $\alpha\approx2.5$ for air.

Level 0 - No drag v.s. With air drag

To be specific, we consider a projectile such as a shell shot by a cannon. According to Newton's law and using the Euler method, we can write the equations of motion with air drag in finite difference form:

$$\begin{matrix} x_{i+1}=x_i+v_{x,i}\Delta t\\ v_{x,i+1}=v_{x,i}-\frac{B_2vv_{x,i}}{m}\Delta t\\ y_{i+1}=y_i+v_{y,i}\Delta t\\ v_{y,i+1}=v_{y,i}-g\Delta t-\frac{B_2vv_{y,i}}{m}\Delta t \tag{5} \end{matrix}$$

The way to estimate the landing point of the shell used here[3] is to interpolate between the last point above geound ($n$) and the point that would have been below ground ($n+1$). If we let $r=-y_n/y_{n+1}$ then a linear interpolation gives:

$$x_l=\frac{x_n+rx_{n+1}}{r+1} \tag{6}$$

and $y_l=0$.

Initial State:

• initial speed: 700m/s
• $B_2/m=4\times10^{-5}m^{-1}$

(点击图片查看源代码)

Conclusion 0

The result shows that air resistance plays an important role in our problem. After considering air drag, both the altitude and the maximum range at identical initial speed and firing angle decrease.

Also the firing angle of maximum range also changes from $45^{\circ}$ to about $40^{\circ}$.

Level 1 - With v.s. Without Density Correction

作业L1 2.9题

1.1 Isothermal v.s. Adiabatic

Another important piece of physics we have not yet accounted for is that how the effect of different air density at different altitudes effect air drag.

We have two simple model to simulate the air density at different altitude. The simplest one is isothermal model, which treat the atmosphere as an isothermal ideal gas. To give the result, it simply leads to:

$$\rho=\rho_0exp(-y/y_0) \tag{7}$$

where $y_0=k_BT/mg\approx1.0\times10^4m$, and $\rho_0=1.225Kg/m^3$ is the air density at sea level.

This isothermal model of atmosphere is perhaps not realistic, since we know that the air temperature can vary quitea bit over altitude changes of a few kilometers. This leads to the second simplest model adiabatic model :

$$\rho=\rho_0(1-ay/T_0)^\alpha \tag{4}$$

where $\alpha\approx6.5\times10^{-3}K/m$, $T_0=288.15K$ is the sea level temperature, and the exponent $\alpha\approx2.5$ for air.

Putting this into calculation is straightforward, we replace $B_2$ in (5) with $B_2\rho/\rho_0$.

Initial State:

• initial speed: 700m/s
• $B_2/m=4\times10^{-5}m^{-1}$
• firing angle: $45^{\circ}$

(点击图片查看源代码)

Conclusion 1.1

We can see from the result picture, after considering air density change effect, the maximum firing range increased. That's because the air density at high altitude is thin, and the air drag is smaller than that in sea level, so the cannon can fly farer.

And the maximun firing range of isothermal model is larger than that of adiabatic model, which is for the fact that the air temperature can vary quitea bit over altitude changes of a few kilometers.

1.2 Accurate Max Angle and Maxi Range

(点击图片查看源代码)

Condition	Max Angle	Max Range
no drag	45.0$^{\circ}$	50004.950$m$
uniform air drag	38.8$^{\circ}$	22070.050$m$
isothermal air drag	45.9$^{\circ}$	26621.572$m$
adiabatic air drag	43.8$^{\circ}$	24641.195$m$

Reference