Parachute Model

Question

Can you tell me if this model is correct or not? [assuming drag is linearly proportional to the velocity] Considering y-axis vertical and x-axis horizontal,with y positive upwards:

1) $F_y=-mg-kv_y=ma_y$ and 2) $F_x=-kv_x=ma_x$. Then considering only 1) we find the equation for the velocity ($t=0$,$v=o$ and $y=h$ are the initial conditions): \begin{equation} \frac{dy}{dt}+\frac{k}{m}y=-gt+\frac{k}{m}h \end{equation} then we can integrate again (initial conditions ,$t=0$ and $y=h$ again): \begin{equation} y=-g\frac{m}{k}t+g\frac{m^2}{k^2}h-g\frac{m^2}{k^2}e^{-\frac{k}{m}t} \end{equation} Is this correct as long as we don't open the parachute? and when we open the parachute and considering it takes a time $\tau$ to open and the drag coefficient increases linearly (i.e. $k(t)=k+k_p\frac{t}{\tau}$ ) what happens?

Answer 1

As regards modelling the interval during which the parachute deploys using a simple model for the drag coefficient as you (and Floris) suggest:

$$k(t)=k+k_p\frac{t}{\tau}$$

In my coordinate system the equation of motion becomes:

$$m\frac{dv}{dt}+(k+k_p\frac{t}{\tau})v=mg$$

This becomes difficult to intergrate. By setting $a=k_p/\tau$, Wolfram alpha provides the following solution.

Bearing in mind this still needs to be integrated to obtain displacement, that becomes unworkable.

I would suggest another way to model the chute deployment, see diagram below:

Split the expected deployment time into a number of intervals (5 shown), each with its average $k_i$ value. To each value $k_i$ and time interval can then be applied the derivation in the previous answer. Adding all obtained $y_i$ gives an approximated total displacement during deployment. This is of course a popular method of numerical integration.

Answer 2

Using the same axis and sign convention Floris used in the answer to your other parachute model question:

$$ma=mg-kv$$

$$m\frac{dv}{dt}=mg-kv$$

$$m\frac{dv}{mg-kv}=dt$$

$$-\frac{m}{k}\frac{d(mg-kv)}{mg-kv}=dt$$

Integrate between $t=0,v=0$ and $t,v(t)$:

$$\ln\frac{mg-kv(t)}{mg}=-\frac{k}{m}t$$

$$v(t)=\frac{mg}{k}\big(1-e^{-\frac{kt}{m}}\big)$$

Note that for $t \to \inf$, $v \to \frac{mg}{k}$, the terminal velocity.

Now $dy=v(t)dt$, so:

$$y(t)=\int_0^t v(t)dt$$

$$y(t)=\frac{mg}{k} \int_0^t dt \big(1-e^{-\frac{kt}{m}}\big)$$

$$y(t)=\frac{mg}{k}t-\frac{m^2g}{k^2}\big(1-e^{-\frac{kt}{m}}\big)$$

Here $y(t)$ is the distance travelled from the drop point. To convert back to your definition of $y$ as height above the ground:

$$\text{Height}=h-y(t)$$

---

I could be wrong of course but I do not see how you arrive at the following DE (even using your convention) and how an initial condition pops up in the DE, prior to integration:

$$\frac{dy}{dt}+\frac{k}{m}y=-gt+\frac{k}{m}h$$

If you want to go 'straight to $y$', your DE should be:

$$m\frac{d^2y}{dt^2}=-mg-k\frac{dy}{dt}$$

From:

$$ma_y=-mg-kv_y$$