Dropping a package with a parachute from a moving plane

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  • #1
lawliet
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Homework Statement:
so basically, i need to drop a package from a moving plane so that it lands at a given location. the package has a parachute attached to it which opens as soon as it is dropped from the plane. so i need to figure out how much distance before the drop location i need to release the package, taking into account air resistance (drag). im having trouble because of the air resistance part, which is usually neglected in most problems.
Relevant Equations:
drag force = Cd * .5 * r * V^2 * A
Homework Statement:: so basically, i need to drop a package from a moving plane so that it lands at a given location. the package has a parachute attached to it which opens as soon as it is dropped from the plane. so i need to figure out how much distance before the drop location i need to release the package, taking into account air resistance (drag). I am having trouble because of the air resistance part, which is usually neglected in most problems.
Homework Equations:: drag force = Cd * .5 * r * V^2 * A

so i have a fair idea on what factors i need to consider but i have problem putting them all together.

i know the drag force formula. and i can find the terminal velocity by equating this formula with the weight of the package.
so , v = sqrt((mg * 2)/r*cd*A))

i also know that we can calculate how far the package will land if we ignore the parachute part, which is, using projectile equations.
but i don't know how to proceed further. how do i combine the parachute part with the package part?
 
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  • #2
PeroK
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Did you make this problem up yourself?
 
  • #3
Abhishek11235
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Well,start writing equations of motion. Initially,the box has speed in the direction of plane. Now,as it comes out, the drag force acts on it on both direction. In x direction, air sees an effective area of parachute and in y direction,it sees effective area in y direction. Now, write equations of motion with final condition.
 
  • #4
PeroK
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Well,start writing equations of motion. Initially,the box has speed in the direction of plane. Now,as it comes out, the drag force acts on it on both direction. In x direction, air sees an effective area of parachute and in y direction,it sees effective area in y direction. Now, write equations of motion with final condition.
I wonder how relevant the vertical motion is?
 
  • #5
jim mcnamara
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I think the problem is not solvable as stated.
Some of several things missing:

Laminar air flow (wind velocities -direction and magnitude - change with AGL ) -
very short youtube on the Albuquerque box:

Air density (altitude change, base elevation the target)

Helpful --
Calculating chute drag: http://www.aeroconsystems.com/chutes/drag_calculator.htm

In the real world pilots really want to make drops from the lowest safe feasible Altitude Above Ground level in order to remove as many fudge factors as possible. They receive training in execution of the maneuver. Exact positioning of the drop point is HARD to attain reliably - all this is secondhand from (now) balloon pilots who have done this stuff in combat. Apparently modern computer assisted navigation control improves accuracy.

I claim no direct knowledge. Hopefully someone else has direct knowledge on the calculations.
 
  • #6
sysprog
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It partly depends on the shape of the chute, and on how long the cords are -- a more rectangular chute with shorter cords will at first produce more braking against further forward travel then a more semi-spherical one with long cords would. Once the forward movement has slowed so that there's no more significant tilt, the semi-spherical one will be more susceptible to horizontal influences, whether tailwind, headwind or sidewind, because of its larger vertical area. One the whole, there's not enough information in the problem as stated to make more than an educated guess, but that's generally also true in the real world when you're making parachute package drops.
 
  • #7
lawliet
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Did you make this problem up yourself?
hey, i actually have to write a code for dropping a package from a UAV( unmanned aerial vehicle) at a target location which is given to us. i will also be taking into account wind direction and speed, plane's altitude, plane's speed, to make the drop accurate. but like i said in my question, I am not sure how to proceed.
 
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  • #8
lawliet
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It partly depends on the shape of the chute, and on how long the cords are -- a more rectangular chute with shorter cords will at first produce more braking against further forward travel then a more semi-spherical one with long cords would. Once the forward movement has slowed so that there's no more significant tilt, the semi-spherical one will be more susceptible to horizontal influences, whether tailwind, headwind or sidewind, because of its larger vertical area. One the whole, there's not enough information in the problem as stated to make more than an educated guess, but that's generally also true in the real world when you're making parachute package drops.
hello, thanks for replying.
this is not just a problem question, ill actually be dropping something from a UAV plane. the chute that i have is circular with a diameter of 1.6m. let's assume i have calculated the drag coeff of the chute. what would i do next? And also, to make an accurate drop, i will be taking into account wind speed and direction,plane altitude and speed, just before the drop.
 
Last edited:
  • #9
lawliet
6
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I think the problem is not solvable as stated.
Some of several things missing:

Laminar air flow (wind velocities -direction and magnitude - change with AGL ) -
very short youtube on the Albuquerque box:

Air density (altitude change, base elevation the target)

Helpful --
Calculating chute drag: http://www.aeroconsystems.com/chutes/drag_calculator.htm

In the real world pilots really want to make drops from the lowest safe feasible Altitude Above Ground level in order to remove as many fudge factors as possible. They receive training in execution of the maneuver. Exact positioning of the drop point is HARD to attain reliably - all this is secondhand from (now) balloon pilots who have done this stuff in combat. Apparently modern computer assisted navigation control improves accuracy.

I claim no direct knowledge. Hopefully someone else has direct knowledge on the calculations.

Hi, thanks for the reply.
i will actually be dropping something from a UAV to a given target location. it is a package with a parachute attached to it which will open as soon as it is dropped. from the plane's sensors, i can get the altitude, speed, wind direction, wind speed. i will be taking into account all these factors to calculate the best time to make the drop so it lands as close to the target location as possible. any ideas on how to proceed further?
 
  • #10
PeroK
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hello, thanks for replying.
this is not just a problem question, ill actually be dropping something from a UAV plane. the chute that i have is circular with a diameter of 1.6m. let's assume i have calculated the drag coeff of the chute. what would i do next? And also, to make an accurate drop, i will be taking into account wind speed and direction just before the drop.

That is a specialist application. Not least you are going to need experimental data for your chute. That will involve a lot of real testing.
 
  • #11
lawliet
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That is a specialist application. Not least you are going to need experimental data for your chute. That will involve a lot of real testing.
thanks, i understand the part about calculating the drag coeff for my chute. what i don't understand it , like i mentioned in my question, what do i do after calculating the drag force acting on the package? so i have the distance that my package will drop at, because of the plane's initial velocity. and i have the drag force acting on the parachute . what do i do with these both?
 
  • #12
PeroK
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thanks, i understand the part about calculating the drag coeff for my chute. what i don't understand it , like i mentioned in my question, what do i do after calculating the drag force acting on the package? so i have the distance that my package will drop at, because of the plane's initial velocity. and i have the drag force acting on the parachute . what do i do with these both?

You search for a specialist site. For example:

http://www.rocketmime.com/rockets/descent.html

Whatever this is, it's not introductory physics homework.

For example, as an object moves through air, the drag force will be in a direction opposing the motion (relative to the air). Your first step would be to analyse what happens to an object moving in two dimensions. The overall drag coefficient will depend on the direction the chute is moving and, given you have a ##v^2## model, the drag does not decompose simply into horizontal and vertical drag.
 
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  • #13
lawliet
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You search for a specialist site. For example:

http://www.rocketmime.com/rockets/descent.html

Whatever this is, it's not introductory physics homework.

For example, as an object moves through air, the drag force will be in a direction opposing the motion (relative to the air). Your first step would be to analyse what happens to an object moving in two dimensions. The overall drag coefficient will depend on the direction the chute is moving and, given you have a ##v^2## model, the drag does not decompose simply into horizontal and vertical drag.


ill look further into it , thank you so much for your time!
ps - i did post this under advanced physics, but the moderator shifted it to introductory physics.
 
  • #14
PeroK
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ill look further into it , thank you so much for your time!
ps - i did post this under advanced physics, but the moderator shifted it to introductory physics.
Here's my idea from post #4. The box is dropped with the parachute open. So, it falls slowly. Meanwhile, it's moving in the horizontal direction - decelerating from the plane's airspeed to zero airspeed.

A simple approximation is to ignore the vertical motion. I.e. simply calculate how far the chute travels horizontally under the horizontal drag/air resistance.

In that case, the formula for vertical drag is irrelevant. What you need is the formula for horizontal drag of the chute.

I have no idea how accurate this model might be.

To include wind speed and direction, you would need to calculate the time to fall, hence you would need the vertical motion. Again a simple approximation is to model the chute falling at its terminal vertical speed.
 

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