Dropping a box from a quadcopter

Question

I am trying to estimate when and where to release a box carried by a delivery UAV such that it falls on a given target. I need to take the wind speed/direction into account. (See the visual below for reference.)

Through the pose estimator, I have access to the position and velocity of the UAV in $X, Y, Z$ axes, so as the position of the target. The box needs to be released from approximately $2.5m$. Also, there is a wind speed/direction sensor located by the target location, in average the wind speed is around $6 mph$ in the test area.

Note that the grabber that releases the box has a latency of $~3.5 s$.

There are a number of examples online that solve this problem using SUVAT equation in two dimensions without taking the wind into account. But I couldn't find any information regarding a robot operating in three dimensions.

Answer 1

I will give you a solution that contains air resistance etc and I hope someone here can solve this then because it’s less mathematical but more engineering like:

Let’s look at the object and all the forces applied in free fall:

Gravity: $$F_2 = m \cdot g$$ Horizontal wind resistance $$F_2 = \dfrac{1}{2} \rho v_h^2 A C$$

• A is the crosssectional area
• C is the drag coefficient (0,8 for an angled cube)
• v_h is the relative speed (in horizontal direction)
• $\rho$ is the density of air (I don’t know this value)

Vertical wind resistance:

$$F_3 = \dfrac{1}{2} \rho v_v^2 A C$$

• $v_h$ Is the vertical speed

Now let’s first do some assumptions:

1. The gravity constant is actually constant (which it isn’t for changing heights but the change is basically zero for your example
2. The wind speed is constant at any height
3. The box doesn’t start to rotate like crazy (because the drag formulae wouldn’t work anymore)

Let’s first see how long it’s actually in the air. Therefor we will look at the vertical forces applied: $$F_v = F_1 - F_3$$ Note that the drag force is opposite the speed (so its going upwards) whereas the gravity pulls the object downwards. With $F= ma$ we can solve for the vertical acceleration:

$$a_{ver} = \dfrac{F_1 - F_3}{m} = \dfrac{my - \dfrac{1}{2} \rho v_v^2 A C}{m}$$

This is an ODE 2nd order which needs to be solved in order to calculate the time the object is falling. Let’s assume that this ODE has been solved:

We can do the same with the horizontal acceleration:

$$a_{hor} = \dfrac{ \dfrac{1}{2} \rho v_h^2 A C}{m}$$

Again this is an ODE which needs to be solved. Because we got the time, we can calculate the displacement of the object while falling.

Now you have information about the displacement of the object depending on height and velocity of the wind. If you would drop the object with an initial velocity of 0, displaced above the target in wind direction, you should hit the target quiet accurately.

Hope this helps

Edit: when I am home, I can try to solve these ODEs