Math/Physics question

[Radagast]

Hi all,
I wonder if one of you smart folk could help me with this -

If my MP were to fall like a stone from an altitude of 200 feet, with a crosswind of 20 mph, how far would it would it have drifted on impact with the ground?

The reason I ask is I want film along side a stretch of road, and would like to know minimum safe distance in the event of any type of failure

thanks

 

[Lake_Flyer]

Great to see someone who actually thinks before flying!

It won't drift off more than a few meters, given it falls like a stone with engines off. Quads typically fall completely vertical. That's from hovering. Unless you are flying with a decent speed, in that case it will travel for a few seconds before falling straight down. Try to fly parallel as much as possible to avoid that it ever reaches the highway and fly slow so it will not travel more than a few meters after engine failure.
You could test it in the simulator, set the desired wind and do a CSC at 200 feet, straight above the take off point. It will be quite accurate. You can do that while hovering or flying at different speeds.

 

[Simmo]

So many variables. Are you flying forward? At what speed. At what direction to the wind are you flying? I'm sure @sar104 has some form of software to work this sort of problem out??

 

[Ken Nash]

⨊3∜3x20⨧= 120feet

 

[Radagast]

Great to see someone who actually thinks before flying!

It won't drift off more than a few meters, given it falls like a stone with engines off. Quads typically fall completely vertical. That's from hovering. Unless you are flying with a decent speed, in that case it will travel for a few seconds before falling straight down. Try to fly parallel as much as possible to avoid that it ever reaches the highway and fly slow so it will not travel more than a few meters after engine failure.
You could test it in the simulator, set the desired wind and do a CSC at 200 feet, straight above the take off point. It will be quite accurate. You can do that while hovering or flying at different speeds.

Great idea using the simulator, thanks.
After numerous tests the drift distance ranged between 60-75 feet, from a hover. That seems quite far to me.
I guess the safest option for me would be to fly downwind-side of the road

So many variables. Are you flying forward? At what speed. At what direction to the wind are you flying? I'm sure @sar104 has some form of software to work this sort of problem out??

I was thinking of planning a Litchi waypoint mission, flying forward at 15 mph. The wind direction would be a worst case scenario, i.e. at 90 degrees pushing me towards the road

 

[Radagast]

⨊3∜3x20⨧= 120feet

Thanks Ken, that definitely meets my 'smart' criteria

 

[sar104]

So many variables. Are you flying forward? At what speed. At what direction to the wind are you flying? I'm sure @sar104 has some form of software to work this sort of problem out??

Unfortunately there is no simple answer to that question. In the zero drag limit the answer is clearly distance = 0, but as soon as you add drag the problem becomes uncertain.

The net horizontal force on the aircraft, F₁, will be just the aerodynamic drag and approximately proportional to the square of the airspeed:

F₁ = k₁(w - v₁)² ,
​

while the vertical force, F₂, will be the weight minus a drag force that is proportional to the square of its vertical velocity:

F₂ = mg - k₂v₂²​

Those allow us to construct the horizontal and vertical equations of motion but, even with some simplifying assumptions such as isotropic drag (i.e. k₁ = k₂), we still don't know the values of those constants and so there is no numerical solution. If you do the thought experiment in the opposite limit to zero drag (which would k → ∞) then the aircraft obviously never descends, and drifts indefinitely. Clearly the problem doesn't lie at either of those extremes but bounding it to within a useful range is impossible without an aerodynamic model for the aircraft.

 

[JoostGT3]

The vertical movement of the drone can be described as follows:
st = 0.5 a * t^2 + v0 * t + s0

Where:

• st = position [m] at time t
• a = accelleration [m/s^2]
• t = time
• v0 = velocity [m/s] (in vertical direction) at time t = 0
• s0 = position [m] at time t = 0

Now, we know the following data:

• st = 0
• a = -9.81 [m/s^2]
• t = ?
• v0 = 0 [m/s]
• s0 = 200 [ft] = 60.96 [m]

So

0 = (0.5 * -9.81 * t^2) + (0 * t) + 60.96
0 = -4.905 * t^2 + 60.96 -> 4.905 * t^2 = 60.96
t^2 = 60.96/4.905
t = SQRT(60.96/4.905) = 3.5

So, the drone will be falling for 3.5 seconds.

Now, the accelleration in horizontal direction due to the wind can not be measured, but let's assume that worst case, it will be moving with 20 [MPH] for the entire time of falling, then it will have drifted the following distance:

20 [MPH] = 8.9408 [m/s], so falling for 3.5 seconds,
it will have drifted 3.5 * 8.9408 = 26.8 [m] ( = 88 [ft])

Now, this is assuming the following:

• Drone is falling without drag (in reality, it will be slowed down in vertical direction by it's drag, so it will be falling longer, resulting in more drift than calculated)
• When it starts falling, the sideways speed is instantaneously equal to the windspeed, which will not be happening in reality. So, in reality the drift will be less than calculated

All in all, I think you could account for a maximum drift of about 100 feet...

 

[Radagast]

Thank you sar104 and JoostGT3 for your very detailed replies.
I must admit I guessed there would be less drift, very useful info, thanks again

 

[LasVegasPilot]

E=mc hawking

 

[sar104]

The vertical movement of the drone can be described as follows:
st = 0.5 a * t^2 + v0 * t + s0

Where:

• st = position [m] at time t
• a = accelleration [m/s^2]
• t = time
• v0 = velocity [m/s] (in vertical direction) at time t = 0
• s0 = position [m] at time t = 0

Now, we know the following data:

• st = 0
• a = -9.81 [m/s^2]
• t = ?
• v0 = 0 [m/s]
• s0 = 200 [ft] = 60.96 [m]

So

0 = (0.5 * -9.81 * t^2) + (0 * t) + 60.96
0 = -4.905 * t^2 + 60.96 -> 4.905 * t^2 = 60.96
t^2 = 60.96/4.905
t = SQRT(60.96/4.905) = 3.5

So, the drone will be falling for 3.5 seconds.

Now, the accelleration in horizontal direction due to the wind can not be measured, but let's assume that worst case, it will be moving with 20 [MPH] for the entire time of falling, then it will have drifted the following distance:

20 [MPH] = 8.9408 [m/s], so falling for 3.5 seconds,
it will have drifted 3.5 * 8.9408 = 26.8 [m] ( = 88 [ft])

Now, this is assuming the following:

• Drone is falling without drag (in reality, it will be slowed down in vertical direction by it's drag, so it will be falling longer, resulting in more drift than calculated)
• When it starts falling, the sideways speed is instantaneously equal to the windspeed, which will not be happening in reality. So, in reality the drift will be less than calculated

All in all, I think you could account for a maximum drift of about 100 feet...

I agree with your assessment of that method and it's limitation, but at those velocities (up to 35 m/s) drag is significant - if I recall correctly, previous free fall events had a terminal velocity of much less than that. However, that does also provide an approach to assessing the drag coefficient, k, since at the terminal velocity, mg = kv². We can therefore vary terminal velocity over an appropriate range, compute the resulting drag factors, and numerically solve the vertical and horizontal equations of motion including drag.

I've seen the terminal velocities for these aircraft in the 40 - 60 mph range (indicating significant drag), so converting to SI and using values from 15 - 35 m/s ( 34 - 78 mph), we get the following relationship between distance fallen and horizontal distance from a stationary hover for a 20 mph (9 m/s) wind.



It's a non-linear relationship, as expected. For a 200 ft (60 m) fall, we get horizontal drift ranging from just 4 m (13 ft) for the lowest drag factor (terminal velocity 35 m/s or 78 mph) to 21 m (69 ft) at the highest drag factor (terminal velocity 15 m/s or 34 mph).

 

[Ken Nash]

We are so clever on this forum that we should get together and build our own drone and call it "Cleverdrone" - this time next year Rodney we could be Billionaire's.

 

[sar104]

Based on previous events and reports, it looks like the Mavic terminal velocity at sea level is no more than 45 mph, which would suggest around 12 m (40 ft) of drift in your scenario.

 

[Radagast]

Thanks again sar104

 

[MikeJ]

I can offer you to visit website of Math Tutors. They will help you to better understand the topic to avoid any mistakes in future. Because you can't come here and ask for the solution all the time. Plus at the same time you will learn some other topics in parallel

 

[Radagast]

Strange comment to make MikeJ, 2 months after my OP. Also, I am not sure you have any authority to tell me what I can or cannot post here

 

[colinfowler]

Strange comment to make MikeJ, 2 months after my OP. Also, I am not sure you have any authority to tell me what I can or cannot post here

With the smiley face I would give MikeJ the beniffit of the doubt that it may have been a shot at humour

 

[MavicCF]

With the smiley face I would give MikeJ the beniffit of the doubt that it may have been a shot at humour

Congratulations! I just noticed you latest edition to the family!

 

[colinfowler]

Congratulations! I just noticed you latest edition to the family!

thanks, now in yellow to match

 

[colinfowler]

thanks, now in yellow to match