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please keep in mind that the following is homework, I do not want answers, only help. If you are confused as to what I refer to in this question, or I don't make any sense, please refer to the link below.

This is my assignment

I'm trying to plot the path of a firework that weighs $.0008 kg$ and has a density of $1900 kg/m^3$, and has a burn rate of $0.0027 kg/s$ in 2d using java. The firework has a launch velocity of $22 m/s$. The only force acting on the x-axis is the drag force, and wind velocity which I can change.

The forces acting on the x axis are: -the initial velocity $22 m/s$

-the drag force

-gravity $9.807 m/s^2$

-the wind force: anywhere from $-15$ to $15$

The drag force is computed by $F_D=(pvAC_D)/2$

$p$=fluid density$=1.2 kg/m^3$

$v$=velocity magnitude

$A$=cross sectional area$=0.00046140621$

$C_D$=drag coefficient$=.4$

Without any wind acting on the x axis, the velocity magnitude is merely: $\sqrt(22^2+0^2)=484$ Now using the formula: $F_D=(pvAC_D)/2$ we get: $F_D=(1.2*484*0.0004614*.4)/2=.05359$

Now we also add the mass times the gravity(9.8m/s^2)to get our total force on the y axis.

$F_y=-mg-F_D(v_y/v)$ ($v_y$, the velocity of y, is the same as v, the velocity magnitude, since the x axis velocity is 0, so $v_y/v=1$)

Which translates to $F_y=-0.0008*9.8-.5359*(1)=-.54374$

So there is a total of .54374 downwards force on the y axis.

My professor then says to use the Runge-Kutta ODE solver to find the velocity of $x_k+1$ and $y_k+1$ but I'm not sure how to interpret the Runge-Kutta solver into the formulas that I've already been given.

To conclude, I know the initial downward force on the object, I know that it has an initial velocity of 22 meters/s, and I know that the mass will be reducing as my path increases. I don't want to know the wind force at this time, as I want to add that in afterwards.

Could someone please help me understand how the Runge-Kutta ODE solver applies to this assignment?

To answer Kalvotom:

The total force being placed on the sphere for each axis is as such: $F_x=-F_D(v_x/v)$ and $F_y=-mg-F_D(v_y/v)$

quoting the assignment: "Since force equals mass times acceleration and acceleration is the first differential of velocity with time, these two equations can be expressed as two differential equations:"

$dv_x/dt=(F_Dv_x/mv)$ and $dv_y/dt=-g(F_Dv_y/mv)$

"These two ordinary differential equations cannot be solved using algebra, so we are going to have to resort to a numeric technique." AKA Runge-Kutta

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In order to see where the need for an ODE solver comes from, you have to be aware of the fact that you are trying to solve Newtonian equations of motion. I do not see those equations explicitly in your analysis.

Edit: Hmm, let me try it this way (I do not want to solve your HW :-)): you want to solve $$ z' = f(z), $$ where $z$ is a vector of unknown functions, let there be $n$ of them, and $f:\mathbb{R}^n\to\mathbb{R}^n$. The corresponding RK4 scheme (one of many RK methods available) works in the following way. Let $h$ be a small parameter (time step size) and let $z_0\in\mathbb{R}^n$ and $t\in\mathbb{R}$ be given. Then compute $$ z_1 = z_0 + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4), \quad t_{1} = t_0 + h, $$ where $$ k_1 = h f(z_0), \ k_2 = hf(z_0 + \frac{1}{2}k_1), \ k_3 = f(z_0 + \frac{1}{2}k_2), \ z_4 = h f(z_0 + k_3). $$ This gives you a mapping $z_0 \mapsto z_1$. $z_0$ are the initial data at $t$ and $z_1$ is an approximation of the solution at time $t+h$. Now you just have to iterate this map over and over again. That is, you take $z_1$ instead of $z_0$ and compute $z_2$, etc.

In your case the unknowns are $z=(x,y,v_x,v_y)$. You have $x' = v_x$, $y' = v_y$, $v_x' = \cdots$, $v_y' = \cdots$. So if we denote $f=(f_1,f_2,f_3,f_4)$ then for example $f_1(x,y,v_x,v_y) = v_x$. You have to find the other componentes, write down this iterative scheme (essentially a simple loop where you record values at each iteration) and you are done.

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  • $\begingroup$ I've answered Kolvotom's question at the end. $\endgroup$
    – Unknown
    Sep 29, 2012 at 22:53
  • $\begingroup$ Ok. So now you have to set up the RK numeric scheme. It is written in your assignment. You just have to find what the $f$ functions are in this case. $\endgroup$
    – kalvotom
    Sep 29, 2012 at 23:16
  • $\begingroup$ Wouldn't the f functions be the force on x and y? $\endgroup$
    – Unknown
    Sep 30, 2012 at 0:33
  • $\begingroup$ Sorry if I'm not understanding, I never took physics or calc and this is for a computer science class, perhaps I'm overthinking something that should be much simpler. $\endgroup$
    – Unknown
    Sep 30, 2012 at 0:34
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    $\begingroup$ Hmm, let me try it this way: you want to solve $$ x' = f(x), $$ where $x$ is a vector of unknown functions, let there be $n$ of them, and $f:\mathbb{R}^n\to\mathbb{R}^n$. The corresponding RK4 scheme (one of many RK methods available) works in the following way. Let $h$ be a small parameter (time step size) and let $x_0\in\mathbb{R}^n$ and $t\in\mathbb{R}$ be given. Then compute $\endgroup$
    – kalvotom
    Sep 30, 2012 at 8:24

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