How can I consider a UAV thrust power for wind effects?

Question

Sorry that I can't write English well.

I was wondering how can I consider a UAV thrust power for wind effects.

Assuming a wind velocity of -7 m/s, uav velocity of 10m/s and the stall speed is 10 m/s.

Considering that the desired velocity of the uav must be close to the stall speed. And, assuming that the wind is opposite to the flight direction.

With this information the uav velocity should be 3m/s? And if that's the case, how can I calculate the thrust force.

I'll add a figure and clarifying question. first, I desire minimize thrust power of UAV for flight.

Like a figure, If the wind speed is 7m/s and the UAV speed is 10 m/s, ground speed must be 3m/s.

however, I thought relative velocity for a UAV is 17m/s. And To minimize thrust, The UAV closed to the stall speed(stall speed is 10m/s). I thought the UAV need only UAV speed over than 3m/s.

And if that's the case, how can I calculate the thrust force.

Answer 1

As i understand your image:

• $V_{uav}$ is the velocity of the plane relative to the air.
• $V_{wind}$ is the velocity of the air relative to the ground.
• $V_{ground}$ is the velocity of the plane relative to the ground.

If the UAV velocity is 10 m/s and the stall speed is 10 m/s the plane is at its limit of flying no matter what the wind speed is (neglecting turbulences). The relative speed of the plane and the air is crucial, this speed is mandatory for the physics beyond the lift.

There is simple equation how to calculate sum velocities. Note that they are vectors (speed and direction) not just numbers. $V_{wind}$ can be named as drift velocity.

$$\overrightarrow V_{ground}=\overrightarrow V_{uav}+\overrightarrow V_{wind}$$

If the wind direction is opposite to the plane direction its ground speed, at the stall limit, is smaller than the UAV limit. It can be zero (the plane seems stand still) or even negative (plane seems to reverse).

Side notes: $V_{uav}$ is The velocity a pilot cares the most, $V_{ground}$ is the velocity his manager cares. This wind effect is also the reason why flying wit tailwind is preferred by airlines - they spare fuel - and it is also the reason why takeoff and landing is preferred in headwind - you need shorter runway to start/land sucesfully. With tailwind the runway may not be long enough.

Answer 2

Thrust required is going to involve a lot more than airspeed, stallspeed and the speed of the headwind. It deals a lot with the coefficient of drag and the coefficient of lift as well as the weight.

$T=\frac{W}{(Cl/Cd)}$

where $T$ is thrust, $W$ is weight, $Cl$ is coefficient of lift and $Cd$ is coefficient of drag.

You can find lift and drag data for the wing you will be using on airfoiltools.com, but finding the overall lift and drag of the aircraft will require actual experimental testing.

You say that you want to minimize your thrust required to get to speed, so you will want to operate at $Cl/Cd$ max, which is the most efficient way of flying anyway.

here's a useful formula to figure Cl.

$L = Cl(dv^2/2)A$

where $L$ is lift (you can substitute your weight here) $d$ is air density, $v$ is velocity and $A$ is wing area