distance travelled byparachute in terminal velocity

Dec 2019
4
0
india
Hi so I'm calculating the distance travelled by a payload attached to a parachute. I have calculated the terminal velocity and then drag force. I need to find the descent time so that I can use it to find distance travelled in the horizontal direction. so I used the formula:
t=sqrt(2H/g)
but changed it to this:
t=sqrt(2H/g−a)
where a is the acceleration due to drag which I found by the formula - a=D−W/m. But then I realised that at terminal velocity there is no acceleration. So how do I include the drag force to calculate the time and distance?
 

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
If there is no acceleration (such as when the parachute and its load reach terminal velocity), then the time to travel distance comes from the simple formula distance = rate times time, which in this case can be rearranged to time = distance/rate. So:

t = H/v_term

where v_term = terminal velocity. Have you calculated what the terminal velocity is? If you have a formula for drag, then terminal velocity is the velocity at which drag = weight.
 
Oct 2017
642
330
Glasgow
The origin of your formula is the SUVAT equation for motion under a constant acceleration:

\(\displaystyle s = ut + \frac{1}{2}at^2\)

if u = 0 and a =g, then we can rearrange this for t:

\(\displaystyle s = \frac{1}{2}gt^2\)
\(\displaystyle t = \sqrt{\frac{2s}{g}}\)

However, this assumes a constant acceleration under gravity, which is not a suitable assumption for your parachute problem. You can't take a formula that makes that assumption and then retrofit an acceleration onto it("acceleration due to drag"). Instead, we have to lift the assumption that there is no air resistance and create a new formula.

I think you have two possible approaches to solve your problem. The first approach is the simplest, but also the most crude. The second is more complex, but will better model the whole system
  • First approach: assume the parachute + payload just drop at constant velocity (terminal velocity). Then, \(\displaystyle s = vt\) will give you the solution, where v is the (constant) terminal velocity and t is duration of fall. It's a crude model, but it will have minimal errors in situations where the fall duration or distance is very large.
  • Second approach: assume an equation for air resistance, develop an equation of motion based on Newton's laws and then solve the corresponding second order ODE.
The second of these approaches is mindful of the fact that air resistance is a force that contributes to a Newtonian dynamics problem, so in order to determine the fate of your parachute + payload, the drag force needs to be known. This drag force will compete with the weight of the parachute + payload and the result is the direct consequence of Newton's first and second laws. Most modellers doing this second approach assume an air resistance of the form \(\displaystyle F_r \propto v^b\) where b is usually something between 1 and 2. This, unfortunately, makes the maths rather difficult and sometimes non-linear.

Example of second approach: assume \(\displaystyle F_r = k v\), where k is some constant that calibrates the problem (case b=1).

Newton's second law:

Resultant force = mass \(\displaystyle \times\) acceleration:

\(\displaystyle Mg - kv = Ma\)
\(\displaystyle Mg - k \frac{dx}{dt} = M \frac{d^2 x}{dt^2}\)
\(\displaystyle \frac{d^2 x}{dt^2} + \frac{k}{M} \frac{dx}{dt} - g = 0\)

Now to solve this second order ODE. Because the constant term, g, this equation is imhomogeneous. The homogeneous equivalent is of the form:

\(\displaystyle \frac{d^2 x}{dt^2} + B\frac{dx}{dt}= 0\)

where \(\displaystyle B = \frac{k}{M}\). The auxiliary equation is consequently:
\(\displaystyle r^2 + Br = 0\)
\(\displaystyle r(r + B) = 0\)
solution are r=0 and r=-B. Since these are real and distinct, the trial solution is therefore:

\(\displaystyle x = c_1 + c_2 e^{-Bt} \)

However, this will only work for the homogeneous form. For the imhomogenous form, it turns out that the particular equation is \(\displaystyle \frac{g}{B}t\). Consequently, the full solution has the form:

\(\displaystyle x = c_1 + c_2 e^{-Bt} + \frac{g}{B} t\)

Let's confirm that this solution works. Differentiate to find derivatives:

\(\displaystyle \frac{dx}{dt} = - c_2 B e^{-Bt} + \frac{g}{B}\)
\(\displaystyle \frac{d^2x}{dt^2} = c_2 B^2 e^{-Bt}\)

Substitute these into our ODE to confirm that it is a solution:

\(\displaystyle c_2 B^2 e^{-Bt} - c_2 B^2 e^{-Bt} + g - g = 0\)

LHS = RHS, so it works. The full solution is therefore, substituting for B,

\(\displaystyle x = c_1 + c_2 e^{-\frac{k}{M}t} + \frac{Mg}{k} t\)

In order to eliminate the unknowns \(\displaystyle c_1\) and \(\displaystyle c_2\), you need to introduce boundary conditions (e.g. when t=0, v=0, x=0).

Note: x here is just s from the SUVAT equations.
 
Last edited:
Apr 2017
538
140
Hi so I'm calculating the distance travelled by a payload attached to a parachute. I have calculated the terminal velocity and then drag force. I need to find the descent time so that I can use it to find distance travelled in the horizontal direction. so I used the formula..... :
You want to find HORIZONTAL distance traveled ???? this is completely dependent on wind speed ....

Within the first few seconds the parachute will have zero horizontal velocity (relative to wind).... the terminal velocity is vertical , you know this , and know the height so you can find the TIME of fall .... multiply this TIME by wind speed to get horizontal distance traveled ....
 
Oct 2017
642
330
Glasgow
Oh... the horizontal direction...

You can do the same thing as above for the x-direction, except that you don't have the g-term so you don't need the addition of the particular equation to the trial solution.
 
Dec 2019
4
0
india
The origin of your formula is the SUVAT equation for motion under a constant acceleration:

\(\displaystyle s = ut + \frac{1}{2}at^2\)

if u = 0 and a =g, then we can rearrange this for t:

\(\displaystyle s = \frac{1}{2}gt^2\)
\(\displaystyle t = \sqrt{\frac{2s}{g}}\)

However, this assumes a constant acceleration under gravity, which is not a suitable assumption for your parachute problem. You can't take a formula that makes that assumption and then retrofit an acceleration onto it("acceleration due to drag"). Instead, we have to lift the assumption that there is no air resistance and create a new formula.

I think you have two possible approaches to solve your problem. The first approach is the simplest, but also the most crude. The second is more complex, but will better model the whole system
  • First approach: assume the parachute + payload just drop at constant velocity (terminal velocity). Then, \(\displaystyle s = vt\) will give you the solution, where v is the (constant) terminal velocity and t is duration of fall. It's a crude model, but it will have minimal errors in situations where the fall duration or distance is very large.
  • Second approach: assume an equation for air resistance, develop an equation of motion based on Newton's laws and then solve the corresponding second order ODE.
The second of these approaches is mindful of the fact that air resistance is a force that contributes to a Newtonian dynamics problem, so in order to determine the fate of your parachute + payload, the drag force needs to be known. This drag force will compete with the weight of the parachute + payload and the result is the direct consequence of Newton's first and second laws. Most modellers doing this second approach assume an air resistance of the form \(\displaystyle F_r \propto v^b\) where b is usually something between 1 and 2. This, unfortunately, makes the maths rather difficult and sometimes non-linear.

Example of second approach: assume \(\displaystyle F_r = k v\), where k is some constant that calibrates the problem (case b=1).

Newton's second law:

Resultant force = mass \(\displaystyle \times\) acceleration:

\(\displaystyle Mg - kv = Ma\)
\(\displaystyle Mg - k \frac{dx}{dt} = M \frac{d^2 x}{dt^2}\)
\(\displaystyle \frac{d^2 x}{dt^2} + \frac{k}{M} \frac{dx}{dt} - g = 0\)

Now to solve this second order ODE. Because the constant term, g, this equation is imhomogeneous. The homogeneous equivalent is of the form:

\(\displaystyle \frac{d^2 x}{dt^2} + B\frac{dx}{dt}= 0\)

where \(\displaystyle B = \frac{k}{M}\). The auxiliary equation is consequently:
\(\displaystyle r^2 + Br = 0\)
\(\displaystyle r(r + B) = 0\)
solution are r=0 and r=-B. Since these are real and distinct, the trial solution is therefore:

\(\displaystyle x = c_1 + c_2 e^{-Bt} \)

However, this will only work for the homogeneous form. For the imhomogenous form, it turns out that the particular equation is \(\displaystyle \frac{g}{B}t\). Consequently, the full solution has the form:

\(\displaystyle x = c_1 + c_2 e^{-Bt} + \frac{g}{B} t\)

Let's confirm that this solution works. Differentiate to find derivatives:

\(\displaystyle \frac{dx}{dt} = - c_2 B e^{-Bt} + \frac{g}{B}\)
\(\displaystyle \frac{d^2x}{dt^2} = c_2 B^2 e^{-Bt}\)

Substitute these into our ODE to confirm that it is a solution:

\(\displaystyle c_2 B^2 e^{-Bt} - c_2 B^2 e^{-Bt} + g - g = 0\)

LHS = RHS, so it works. The full solution is therefore, substituting for B,

\(\displaystyle x = c_1 + c_2 e^{-\frac{k}{M}t} + \frac{Mg}{k} t\)

In order to eliminate the unknowns \(\displaystyle c_1\) and \(\displaystyle c_2\), you need to introduce boundary conditions (e.g. when t=0, v=0, x=0).

Note: x here is just s from the SUVAT equations.
hey, in the first approach, if the payload were to be dropped from a moving plane, how would we take into account the plane's speed?
 
Apr 2017
538
140
hey, in the first approach, if the payload were to be dropped from a moving plane, how would we take into account the plane's speed?
As soon as the parachute is open it very rapidly looses all horizontal speed relative to air , within about 5 secs the only movement is vertically downward at terminal velocity . But it will be moved horizontally by the wind ... so you need to know the wind speed and direction and the time of decent to know horizontal distance and direction traveled
 
Oct 2017
642
330
Glasgow
hey, in the first approach, if the payload were to be dropped from a moving plane, how would we take into account the plane's speed?
That's your boundary condition: when \(\displaystyle t=0, x=0\) and \(\displaystyle v = v_{plane}\)

Just be careful to make sure your calculation is in the horizontal direction (see my earlier post; #5)
 
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