The Terminal Velocity of a Falling Mavic 2

[AustinDrone]

If we are going to get that far into the details then this needs some clarification. Objects don't accelerate at 9.8 m/s² until they reach terminal velocity, and then discontinuously stop accelerating. The rate of acceleration follows from Newton's 2nd Law (F = ma), where F is the weight of the object minus the aerodynamic drag, which is approximately proportional to the square of the air speed v. So,

a = (mg - kv²)/m = g - kv²/m​

where k is the constant of proportionality in the form:

k = ρcA/2​

where ρ is the air density, A is the effective cross-sectional area and c is a geometric factor, which is approximately unity for a flat(ish) shape. So while the instantaneous initial acceleration of any object is g, it immediately starts to decrease for all v > 0.

If we take the cross-sectional area of a Mavic 2 to be about 25 cm x 25 cm (a conservative estimate including the propellers) and its mass to be 0.9 kg, and the and solve the resulting differential equation, we get the following acceleration, speed and distance as a function of time, which is actually quite consistent with observed falling Mavics.


View attachment 65532

Thank you! This is exactly what I was looking for.
From this I gather that a drone that experiences a critical failure and falls from 100ft would impact the ground at ~25 mph. From 150 or higher, it might impact at about 40 mph. I was looking for this so that I could estimate potential liability due to injury and property damages when flying over stuff (with any necessary waivers, of course).

 

[sar104]

Thank you! This is exactly what I was looking for.
From this I gather that a drone that experiences a critical failure and falls from 100ft would impact the ground at ~25 mph. From 150 or higher, it might impact at about 40 mph. I was looking for this so that I could estimate potential liability due to injury and property damages when flying over stuff (with any necessary waivers, of course).

The terminal velocity is aircraft dependent (specifically) mass and drag. From the actual data I plotted in post #40 you can see that the Mavic's terminal velocity is around 40 mph, in agreement with that drag model. Larger sUAS will typically have higher terminal velocities, but those can be estimated using geometric drag estimates.

 

[AustinDrone]

Very interesting. The three power-off events show similar initial evolution of pitch, roll and yaw, followed by varied, possibly chaotic, behavior:

View attachment 65882
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The vertical speed reflects that, with a transition to oscillatory behavior as the aircraft wobbles:

View attachment 65885

View attachment 65886

View attachment 65887

The terminal velocity is aircraft dependent (specifically) mass and drag. From the actual data I plotted in post #40 you can see that the Mavic's terminal velocity is around 40 mph, in agreement with that drag model. Larger sUAS will typically have higher terminal velocities, but those can be estimated using geometric drag estimates.

Right, and so this is for a Mavic 2 Pro at ~900 g, correct? What would is look like for a Mavic Air at ~430 g or a Mavic Mini at ~250 g? Or even the Anafi at ~375 g?

 

[The Editor]

Right, and so this is for a Mavic 2 Pro at ~900 g, correct? What would is look like for a Mavic Air at ~430 g or a Mavic Mini at ~250 g? Or even the Anafi at ~375 g?

The speed of impact is irrelevant - a feather could hit you at 40mph but do no damage.
A house brick at 40mph would be a different result!

It’s the kinetic energy you should be worried about (As speed doubles, kinetic energy quadruples).

KE = 0.5 x mv2

It is generally (and legally) accepted by most that an impact on the human body of <80 Joules is ‘acceptable’ and companies that manufacture arresting devices for UAV’s work to this figure.

 

[AustinDrone]

The speed of impact is irrelevant - a feather could hit you at 40mph but do no damage.
A house brick at 40mph would be a different result!

It’s the kinetic energy you should be worried about (As speed doubles, kinetic energy quadruples).

KE = 0.5 x mv2

Good point! I think that is actually 0.5 x mv^2, so would that be 0.5 x 900 g x 20 m/s ^2 = 180,000 KE? And what would I compare that to in order to estimate damage to a body or object?

Asking for a friend.

 

[The Editor]

Good point! I think that is actually 0.5 x mv^2, so would that be 0.5 x 900 g x 20 m/s ^2 = 180,000 KE? And what would I compare that to in order to estimate damage to a body or object?

Asking for a friend.

No - 900g would be 0.9 (formula is based on mass in kg)

See my edited post above re 80Joules impact for injury to a person etc.

 

[AustinDrone]

No - 900g would be 0.9 (formula is based on mass in kg)

See my edited post above re 80Joules impact for injury to a person etc.

Okay, so wouldn't 0.5 x 0.9 kg x 20 m/s ^2 be 180 J, not 80 J? And if so, how would one equate that to bodily or property damage.

The closest example I was able to find from Google searches is Red Wing Footwear for men and women must withstand an impact energy of 125 joules (the equivalent of a 50 lb. object dropped at a height of 22 inches) with no cracking through the thickness of the cap wall.

That was sounding pretty hard!

And then I found this study: https://www.casa.gov.au/sites/defau...nned-aircraft-impacts.pdf?acsf_files_redirect

In this study, they compare 120J as being the equivalent to a mid-size HAMMER (0.5kg) dropped from the EIGHTH FLOOR. That sounds REALLY hard! So, I think my math must be off in some way. Where am I getting off track?

By the way, thanks SO much for taking the time to explain this to an old man! I'm finding this fascinating!

 

[The Editor]

Okay, so wouldn't 0.5 x 0.9 kg x 20 m/s ^2 be 180 J, not 80 J? And if so, how would one equate that to bodily or property damage.

The closest example I was able to find from Google searches is Red Wing Footwear for men and women must withstand an impact energy of 125 joules (the equivalent of a 50 lb. object dropped at a height of 22 inches) with no cracking through the thickness of the cap wall.

That was sounding pretty hard!

And then I found this study: https://www.casa.gov.au/sites/defau...nned-aircraft-impacts.pdf?acsf_files_redirect

In this study, they compare 120J as being the equivalent to a mid-size HAMMER (0.5kg) dropped from the EIGHTH FLOOR. That sounds REALLY hard! So, I think my math must be off in some way. Where am I getting off track?

By the way, thanks SO much for taking the time to explain this to an old man! I'm finding this fascinating!

Yes - my 80J figure was not based on your calculation.

80 Joules is an industry figure that below which is generally accepted as little to no injury to a human being (property damage is not important).

The upcoming EASA regulations effective 1st July work on a maximum figure of 79 Joules.

 

[sar104]

Right, and so this is for a Mavic 2 Pro at ~900 g, correct? What would is look like for a Mavic Air at ~430 g or a Mavic Mini at ~250 g? Or even the Anafi at ~375 g?

I'll need to run those models to answer that question. I'll try to get to that this evening.

 

[ajlholt]

If we are going to get that far into the details then this needs some clarification. Objects don't accelerate at 9.8 m/s² until they reach terminal velocity, and then discontinuously stop accelerating. The rate of acceleration follows from Newton's 2nd Law (F = ma), where F is the weight of the object minus the aerodynamic drag, which is approximately proportional to the square of the air speed v. So,

a = (mg - kv²)/m = g - kv²/m​

where k is the constant of proportionality in the form:

k = ρcA/2​

where ρ is the air density, A is the effective cross-sectional area and c is a geometric factor, which is approximately unity for a flat(ish) shape. So while the instantaneous initial acceleration of any object is g, it immediately starts to decrease for all v > 0.

If we take the cross-sectional area of a Mavic 2 to be about 25 cm x 25 cm (a conservative estimate including the propellers) and its mass to be 0.9 kg, and the and solve the resulting differential equation, we get the following acceleration, speed and distance as a function of time, which is actually quite consistent with observed falling Mavics.


View attachment 65532

Hello.
That is a very interesting graph but it does not take account of the auto-rotation effect which will probably be caused by the propellors. in other words, as they start to rotate during the fall they will generate lift in the same way that a helicopter does when it descends by autorotation (and in the same way that a sycamore seed flutters on its descent).
Is it possible for you to factor that effect into your calculations?
Can you contact me separately on [email protected]?

 

[ajlholt]

Very interesting. The three power-off events show similar initial evolution of pitch, roll and yaw, followed by varied, possibly chaotic, behavior:

View attachment 65882
View attachment 65883

View attachment 65884

The vertical speed reflects that, with a transition to oscillatory behavior as the aircraft wobbles:

View attachment 65885

View attachment 65886

View attachment 65887

The blue dotted lines are the model that I posted previously, with a slightly lower drag coefficient than my initial guess (0.025 rather than 0.031), and predict the motion pretty well. The terminal velocity of around 20 m/s (45 mph) is in line with previous data and reports.

Very interesting indeed. Do you have any idea why the vertical speed starts to reduce for each falling drone? Surely they should each get to a fluctuating terminal velocity and stay there until impact when vertical speed should be zero?

 

[sar104]

Hello.
That is a very interesting graph but it does not take account of the auto-rotation effect which will probably be caused by the propellors. in other words, as they start to rotate during the fall they will generate lift in the same way that a helicopter does when it descends by autorotation (and in the same way that a sycamore seed flutters on its descent).
Is it possible for you to factor that effect into your calculations?
Can you contact me separately on [email protected]?

Autorotation is only going to be significant if the aircraft attitude remains approximately constant. With the observed tumbling rotation the effect of the props is simply captured in the overall drag coefficient. Also note that the props are much smaller than on helicopter, which is why they have to rotate much faster. Unpowered, their productive lift is going to be negligible.

 

[sar104]

Very interesting indeed. Do you have any idea why the vertical speed starts to reduce for each falling drone? Surely they should each get to a fluctuating terminal velocity and stay there until impact when vertical speed should be zero?

Yes - you misunderstood the maneuver. These were power-off descents followed by motor restart, bringing the vertical velocity back to zero. There is no impact with the ground.

 

[Cookedinlh]

Thanks for all that detailed analysis sar104 . . Here's a simple kinetic energy chart that shows the rough areas of weight and speed that can become lethal. Look at MASS down the left side and Velocity to the RIGHT and the chart gives "Joules of energy. The Colours are a bit arbitrary but approximate the human damage of that energy . . of course considering the cross section of how and where it strikes you . . but it shows that a well hit baseball of tennis ball can exceed an average speed Mavic 2 that's out of control.

 

[sar104]

Thanks for all that detailed analysis sar104 . . Here's a simple kinetic energy chart that shows the rough areas of weight and speed that can become lethal. Look at MASS down the left side and Velocity to the RIGHT and the chart gives "Joules of energy. The Colours are a bit arbitrary but approximate the human damage of that energy . . of course considering the cross section of how and where it strikes you . . but it shows that a well hit baseball of tennis ball can exceed an average speed Mavic 2 that's out of control.
View attachment 124516

That spreadsheet appears to be an over-constrained calculation. Mass and cross-sectional area should be independent variables. For example, a Mavic 2 impacting end on at 20 m/s will have a kinetic energy per unit area (in J/m²) of 90 kJ (9 x 10⁴ J). That's completely off the chart in the lethal direction.

In other words the assumed cross-sectional areas as a simple function of mass are completely unrealistic. In this case, the aircraft travelling at the observed terminal velocity of 20 m/s has a kinetic energy of 180 J, and the cross-sectional area of the rear of the aircraft is approximately 4 x 5 cm, = 2 x 10⁻³ m² (0.002 m²).

180/(2 x 10⁻³) = 9 x 10⁴ J

 

[Cookedinlh]

Ys thanks for adding that . . I did say it was a "Simple" approach . . it's just a rough guide of mass and velocity energy calc. You additional calculations are better

 

[zeusfl]

Jesus Christ. I just read this whole post and you guys are genius. I am glad you are resources in this forum. My Respects !!!

 

[Zbip57]

If you want to see how the FAA initially arrived at 250 grams or greater as being the threshold at which drone registration is required, here's a link to the November 21, 2015 Final Report of the "Unmanned Aircraft Systems (UAS) Registration Task Force (RTF) Aviation Rulemaking Committee".

See: www.hsdl.org/?view&did=788722

They assumed a brick-shaped object with projected area of 0.1m x 0.2m (approx 4" x 8"), but with a coefficient of drag of 0.3 (like a baseball), dropped in freefall from 500 feet height. Then they applied the 80 Joule kinetic energy limit, which is said to have a 30% probability of being lethal when striking a person in the head.

"Solving for mass and velocity, this equates to an object weighing 250 grams travelling at a terminal velocity of 25 meters/second or approximately 57 miles per hour."

They further assumed the Mean Time Between Failure of the average drone would be after every 100 hours of use, and then calculated the probability of it actual hitting anyone on the head when this object drops from 500 feet into a "densely packed urban area" of 10,000 people per sq.mile.

The calculated probability of fatality is: "4.7x10-8, or less than 1 ground fatality for every 20,000,000 flight hours of an sUAS"

Note that's a thousand times safer than the risk of being killed in general aviation.

"Some members of the task force questioned why sUAS risk level would ever be required to exceed the current general aviation risk level of 5x10-5."

Anyhow, that's where the 250 gram number originated from. Read the Task Force report. It's fascinating.
See: www.hsdl.org/?view&did=788722

 

[Meta4]

"Solving for mass and velocity, this equates to an object weighing 250 grams travelling at a terminal velocity of 25 meters/second or approximately 57 miles per hour."

They grossly overestimated the terminal velocity

 

[sar104]

They grossly overestimated the terminal velocity

Agreed, and the source of that overestimation is pretty clear - they assumed a brick-shaped object with dimensions that ignored the arms and props. The props, in particular, make a significant contribution to the drag, and so the assumed cross-sectional area was far too small, even though the drag coefficient was reasonable. A typical 250 g quad, such as the Mavic Mini, has an actual terminal velocity of closer to 15 m/s.

Of course they might argue that it was simply a very conservative calculation to arrive at a worst-case value.