- Thread starter lawliet
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First lets just consider the horizontal component of the velocity , and assume there is no wind ... leaving the plane the package will have the velocity of the plane , but will rapidly slow ... you need to figure out how the HORIZONTAL velocity changes , it will rapidly come close to zero , and maybe possible to ignore

Then need to find terminal (vertical) velocity of pack and parachute , how quickly it falls , this can easily be done by experiment ...

Then it's simple to make adjustments for wind speed and height of UAV . Wind speed will be the biggest factor and impossible to accurately know as it varies from moment to moment and for different altitudes

There is a very short period when the parachute is opening that pack will not be at terminal velocity , this can probably be ignored.

hey, i understand but what would i do next? i have the distance the package will travel considering only horizontal velocity. and lets assume i have found the terminal velocity of the package and chute. what do i with these both?

First lets just consider the horizontal component of the velocity , and assume there is no wind ... leaving the plane the package will have the velocity of the plane , but will rapidly slow ... you need to figure out how the HORIZONTAL velocity changes , it will rapidly come close to zero , and maybe possible to ignore

Then need to find terminal (vertical) velocity of pack and parachute , how quickly it falls , this can easily be done by experiment ...

Then it's simple to make adjustments for wind speed and height of UAV . Wind speed will be the biggest factor and impossible to accurately know as it varies from moment to moment and for different altitudes

There is a very short period when the parachute is opening that pack will not be at terminal velocity , this can probably be ignored.

\(\displaystyle V=U+At\)

\(\displaystyle D=O+Ut+1/2At^2\)

Where:

A is acceleration

t is the elapsed time

U is initial velocity (when t=0)

O is initial position (when t=0)

V is the velocity at time t

D is the displacement at time t

The acceleration is given by F=mA (or A=F/m)

Where F is the Drag created by the parachute m is the mass of the package.

That is horizontal velocity and displacement,

You also have a second set of these equations for vertical displacement, but here the acceleration is due to gravity,

Note that as the package falls you might need to consider how much drag is acting in the vertical direction and how much in the horizontal direction.

This will change as the angle of the trajectory of the package changes.

Watch out for your reference point, is it on the ground or in the aircraft?

Either, or even both may be used, (but be careful if you try using both, just use one at a time)

(It is even possible to use a reference point based in the package, but non-inertial reference points can get confusing).

Suppose the package and chute fall with a terminal velocity of 5meters/sec ... If they were dropped from a height of 500 meters this would mean they were in the air for 100secs ... all this time the wind is carrying the package off target , if wind is 5meters/sec (10MPH) to the North , this will land the package 500 meters to the North of drop point ... 20MPH wind means 1 km to the North ....lets assume i have found the terminal velocity of the package and chute. what do i with these both?

This maybe a good way to deliver packages in the future , just as some companies are using drones now , but chutes could be cheaper . human parachuters now use steerable chutes and can land where they want within about 5 miles ... so could parcels ...

A cargo plane could pass over a city , dump out thousands of parcels , each parcel with a chute and GPS . The parcel would know where it was and where it wanted to go , servos could pull strings on the chute changing it's profile and aim it towards the delivery address landing on the door step ..