When falling down with air resistance, y-axis vertical (positive is up), x-axis is horizontal, we have that: $F_y=-mg-kv=ma_y$, where $k$ is the drag coefficient. Now if the man opens the parachute, then I can analize it in two ways:

  1. At time $t_0$ of the deployment, I only change $k$ with $k_2>k$ and the rest stays the same.
  2. I consider that the parachute will take a period of time $\delta t$ to fully open and therefore I need to consider $k(t)$.

Now If my aim is to study the force felt when opening the parachute, how should I do it?

My first idea was to work with (1), but I didn't come up with anything interesting. The second idea was to consider the derivative of the acceleration, the jerk. Hence to consider $m\frac{d^3y}{dt^3}=-k\frac{d^2y}{dt^2}$ which is the derivative of the expression above for the force. However I don't know how to minimize it. It can't be zero, otherwise, nothing would change, right? Should I then consider $k$ as in (2) and use the chain rule when differentiating? And what should I do after that if I wanted to minimize this force felt when opening the parachute?

  • $\begingroup$ Gravity and drag act in opposite directions here, so $F_y=-mg-kv=ma_y$ cannot be correct. During the opening of the chute: $ma_y=mg-k(t)v$, with boundary condition $v=v_{op}$, where the latter is the speed just before opening the chute and $k_{op}=k$, also at that moment. $\endgroup$
    – Gert
    Commented Feb 27, 2016 at 2:22
  • $\begingroup$ Calculate $v(t)$ from $m\frac{dv}{dt}=mg-kv$ by integration (before chute opening). $\endgroup$
    – Gert
    Commented Feb 27, 2016 at 2:28
  • $\begingroup$ The equation $F_y=-mg-kv=ma_y$ should be correct. Indeed the velocity will be used negative in that formula. The air drag is opposite to the velocity, which is poiting downwards (so negative). Also the force $-mg$ is pointing downwards, hence they both have minus sign. Of course when plugging in the values for $g$, $k$, $v$ etc, we will have that $-mg$ and $-kv$ have different signs. However do you know how I could calculate the force felt when opening the parachute? $\endgroup$ Commented Feb 27, 2016 at 2:34
  • $\begingroup$ It's easier to have $y$ pointing down here, IMO, but yours works too. I think I know how to solve your problem but it's very late here. I'll bookmark and see what happens tomorrow. Nite. $\endgroup$
    – Gert
    Commented Feb 27, 2016 at 2:40
  • 1
    $\begingroup$ related: physics.stackexchange.com/q/238894/26969 $\endgroup$
    – Floris
    Commented Feb 27, 2016 at 3:58

1 Answer 1


Your model for the force is reasonable - there is a constant acceleration due to gravity, and a force that depends on velocity and size of the parachute. It is also reasonable that the parachute should take some time to deploy, and that the rate at which it opens may affect the force. One can make simplifying assumptions about this - for example, we can say that it takes the parachute a time $\tau$ from deployment to fully open, and that during that time the coefficient $k$ grows linearly - that is, $k(t) = k_0 + k_p\frac{t}{\tau}$ for $t\lt \tau$, and $k(t)=k_0+k_p$ for $t\ge \tau$ - where $k_0$ is the drag due to the body (without the parachute deployed), and $k_p$ is the drag coefficient due to the parachute.

Now we can write down the equation of motion. Let's use "down" as the positive Y direction - so the sign of gravity is positive, the velocity is positive, and the sign of the drag is negative. Then

$$\begin{align} F &= mg - k(t) v\\ m \dot v&= mg - \left(k_0+k_p\frac{t}{\tau}\right)v\\ \end{align}$$

If you want to know the derivative of the acceleration, you just divide that expression by $m$ and differentiate with respect to $t$, giving

$$\dot a = -\frac{k_p}{\tau} \left(v + \dot vt\right)$$

Substituting for the expression for $\dot v$, we get

$$\dot a = -\frac{k_p}{\tau}\left(v + \left(g - \left(\frac{k_0}{m}+\frac{k_p}{m\tau}t\right)\right)t\right)$$

Which tells us that at the time that the parachute first opens, the rate of change of acceleration is negative (you are accelerating less quickly as the drag gets bigger), but that it can actually change sign (greater acceleration in the other direction) as the parachute deploys; but the larger $\tau$, the less the jerk.

Of course this is one reason why parachutes have a little "helper chute" - not only does it provide the force needed to pull the main chute from the pack, but it also starts to slow down the parachutist - in essence, increasing $\tau$ in the equation above.

In reality, equations are not nice and linear - but this is a start.

  • $\begingroup$ Plus, of course, aerodynamic drag is generally proportional to the square of the velocity, not linearly. $\endgroup$ Commented Feb 27, 2016 at 5:06

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