cannon shell projectile

exercise 2.9

---

摘要

We consider a projectile such as a shell shot by a cannon.This problem can be viewed as the motion in two spatial dimensions.If we ignore air resistance,the Newton's second law in two spatial dimensions is given as

$$\frac{d^2x}{dt^2}=0$$ $$\frac{d^2y}{dt^2}=-g$$
where $x$ and $y$ are the horizontal and vertical coordinates of the projectile,and $g$ is the acceleration due to gravity.There are second-order differential equations,but we can deal with them by shifting them to first-order differentials.The algorithm is written as follows:
$$\frac{dx}{dt}=v_x$$ $$\frac{dv_x}{dt}=0$$ $$\frac{dy}{dt}=v_y$$ $$\frac{dv_y}{dt}=-g$$
where $v_x$ and $v_y$ are the $x$ and $y$ components of the velocity.While we multiple the equations which we have to deal with,we can use our standard Euler approach to solve each one.We write each derivative in finite form through Euler method,which leads to
$$x_{i+1}=x_i+v_{x,i}\Delta{}t$$ $$v_{x,i+1}=v_{x,i}$$ $$y_{i+1}=y_i+v_{y,i}\Delta{}t$$ $$v_{y,i+1}=v_{y,i}-g\Delta{}t$$
Given the initial values of $x$,$y$,$v_x$ and $v_y$,we can use the former equations to estimate their values at later times.Similarily,we ignore the $(\Delta{}t)^2$ and higher,but there exists another error term,the air resistance.
The air reistance is very important,so we add that to the model.We assume that the magnitude of the drag force on our cannon shell is given by
$$F_{drag}=-B_2v^2$$
where $v=\sqrt{v^2_x+v^2_y}$ is the speed of the shell.We must consider the vector components as following:
$$F_{drag,x}=F_{drag}cos\theta=F_{drag}(v_x/v)$$
with a similar expression for $F_{drag,y}$.
The componentsof the drag force are thus
$$F_{drag,x}=-B_2vv_x$$ $$F_{drag,y}=-B_2vv_y$$
Adding this force to the equations of motion leads to
$$x_{i+1}=x_i+v_{x,i}\Delta{}t$$ $$v_{x,i+1}=v_{x,i}-\frac{B_2vv_{x,i}}{m}\Delta{}t$$ $$y_{i+1}=y_i+v_{y,i}\Delta{}t$$ $$v_{y,i+1}=v_{y,i}-g\Delta{}t-\frac{B_2vv_{y,i}}{m}\Delta{}t$$
The time evolutio is continued until $y_{i+1}$ becomes negative,which means the shell struck the ground somewhere during the previous time step.We should estimate where the shell struck the ground.
The last point above the ground $(n)$ and the first point below the ground $(n+1)$ give $$x=\frac{x_n+rx_{n+1}}{r+1},y_l=0$$ $$r=-\frac{y_n}{y_{n+1}}$$
we can use it to complish figure.

背景介绍

In exercise 2.9,we should consider that the air density vary as a function of the altitude.A approach is to assume that the air is a poor conductor of heat.This leads to the so-called adiabatic approximation.It performes :

$$\rho=\rho_0(1-\frac{ay}{T_0})^\alpha$$
where $\alpha\approx6.5\times10^{10}k/m$ fits measured data fairly well.Here $T_0$ is the sea level temperature,and the exponent $\alpha\approx2.5$ for air.The drag force due to air resistance is proportional to the density,so
$$F^{\ast}_{drag}=\frac{\rho}{\rho_0}F_{drag}(y=0)$$
where $F_{drag}(y=0)$ is the force at the sea level,and $F^{\ast}_{drag}$ is the drag force at altitude.

正文

code
First,the equations don't involve the air resistance,show the figure for different firing angles:

Figure1 without air force

From this figure,we can find that angle which is close to $45^{\circ}$ has the maximum range.
Second,consider air resistance but no density variations:

Figure2 air resistance without density variations

Compared with Figure1,we find that the height and the range both are shortten a lot.But the $\theta$ for maximum range remains the same.
Third,we calculate the trajectory of cannon shell including both air drag and the reduced air density at high altitudes.The figure3 is taken on:

Figure3 air resistance with density variations

The $\theta$ for maximum range approaches $45^{\circ}$,the similar as the former conclusion.
Last,draw a figure that contains the case of no air resistance and air resistance with density variations for the $\theta=30^{\circ}$ and $45^{\circ}$:

Figure4 comparation,red and cambridge blue lines represents the lines of air resistance with density variations

结论

The air will weaken the cannon's range and height.For the case in which air density is involved,the value of the angle that gives the maximum range is close to $45^{\circ}$.

致谢

这次作业参考了13级学长的画图代码。