Drone dragging a hose behind it

[ethanesh]

Homework Statement

Find forces and moments acting on drone.

Relevant Equations

basic physics equations, moments, etc

This is not technically a homework question, but it seems trivial enough that it probably belongs in this forum (and last time I posted something similar it got moved here, apologies if I have now reversed the problem and posted in the wrong forum again.) I am trying to find the forces and moments experienced by a drone. It seems very straightforward, but it's been a while since I have done any physics problems like this and I would appreciate a second set of eyes letting me know if I have overlooked something. This drone is flying at around 30 feet in the air and dragging behind it a hose.

The hose will be at some angle θ; as far as I can tell I can't calculate this without more information about the forces like thrust from the drone, but it will probably be around 45 degrees. There are two main forces acting on the drone: the weight from the hose and the friction from dragging the hose.
The weight from the hose is only the weight of the hose that is in the air. If we know the weight per unit length λ, we can multiply that by the length 30/cos(θ) to get the downward force. This force acts at the center of the hose that is in the air, so it will create a moment. The moment will be equal to the force multiplied by the horizontal distance between the CG of the drone ad the point where the force acts: in this case, that will be 30*tan(θ) (plus and distance from the hose being mounted away from the drone CG but I will assume that is 0 for now.) So the moment caused by the weight of the hose is M = λ*(30/cos(θ))*30*tan(θ) (clockwise.)
The force of friction is the weight of the hose still on the ground multiplied by a friction coefficient μ. If the length of the hose on the ground is L and the weight per length is still λ, the friction force is μ*λ*L. The moment caused by this force is 30* μ*λ*L, and this time it is counterclockwise.

Hopefully a question this trivial doesn't bore anyone to death; I just want to be absolutely certain I haven't overlooked anything since this is for a real life project. Thank you!

 

[jbriggs444]

Do we know the total length of the hose, counting both the portion on the ground and the portion in the air? I assume that this length is fixed. So we are not unreeling hose from a spool. I assume that we are looking for an equilibrium situation, dragging the hose over the ground at a fixed speed.

You do realize that the portion of the hose in the air will not be in a straight line? It will have a different angle at the drone end than at the ground end.

Can you see what the angle at the ground end must be?

There is a well known name for the shape that a uniform rope section under tension in uniform gravity will relax to.

 

[ethanesh]

Do we know the total length of the hose, counting both the portion on the ground and the portion in the air? I assume that this length is fixed. So we are not unreeling hose from a spool. I assume that we are looking for an equilibrium situation, dragging the host over the ground at a fixed speed.

You do realize that the portion of the hose in the air will not be in a straight line? It will have a different angle at the drone end than at the ground end.

Can you see what the angle at the ground end must be?

There is a well known name for the shape that a uniform rope under tension in uniform gravity will relax to.

The total hose length is 150 feet. Yes, I was assuming equilibrium motion, sum of forces = 0.

I do realize that the hose would sag slightly rather than form a straight line. I assume the angle at the ground end would approach horizontal. I was approximating it as a straight line because I wasn't sure how to calculate the shape it sags to, and since its being pulled taut I figured it would be fairly close to a straight line. If there is a way to calculate the exact parabolic shape it takes, I have not thought of it, but I would love to know. Thank you!

 

[jbriggs444]

If there is a way to calculate the exact parabolic shape it takes, I have not thought of it, but I would love to know. Thank you!

https://en.wikipedia.org/wiki/Catenary

 

[ethanesh]

https://en.wikipedia.org/wiki/Catenary

Well, today I learned that that exists. Thank you!

 

[haruspex]

The moment will be equal to the force multiplied by the horizontal distance between the CG of the drone ad the point where the force acts: in this case, that will be 30*tan(θ)

It is generally important to be clear what axis you are taking moments about and to what system it applies.

If you take the drone plus the raised section of rope as the system, and the drone as the axis, yes that is the distance to use, but the corresponding force is only the friction on the grounded rope. This is countered by torque arising from the weight of the raised portion, but the distance to use for that has to deduced from the catenary equation.

If the system is only the drone, the forces are lift, gravity and the pull from the rope, but none of these exert a torque on it.

Since the length of the raised section will matter, I recommend you work with the Whewell equation form.

 

[ethanesh]

It is generally important to be clear what axis you are taking moments about and to what system it applies.

If you take the drone plus the raised section of rope as the system, and the drone as the axis, yes that is the distance to use, but the corresponding force is only the friction on the grounded rope. This is countered by torque arising from the weight of the raised portion, but the distance to use for that has to deduced from the catenary equation.

If the system is only the drone, the forces are lift, gravity and the pull from the rope, but none of these exert a torque on it.

Since the length of the raised section will matter, I recommend you work with the Whewell equation form.

OK! To make sure I understand what you're saying: if I want to get a realistic model of the moments acting on the drone, I should consider the system as the drone + the raised section of the hose, with the drone as the axis of torque. My assessment of the moment caused by the friction is correct, but the moment from the hose weight is not, since I didn't consider the curvature of the hose (which can be found from the catenary equation.) Once I know the true length and shape of the raised hose, I assume its still correct to take the weight force as acting on the midpoint of the raised hose section. Did I get that right? Thank you!

 

[haruspex]

OK! To make sure I understand what you're saying: if I want to get a realistic model of the moments acting on the drone, I should consider the system as the drone + the raised section of the hose, with the drone as the axis of torque. My assessment of the moment caused by the friction is correct, but the moment from the hose weight is not, since I didn't consider the curvature of the hose (which can be found from the catenary equation.) Once I know the true length and shape of the raised hose, I assume its still correct to take the weight force as acting on the midpoint of the raised hose section. Did I get that right? Thank you!

I forgot to say that I don’t think moments help here. Just consider the forces.

 

[Tom.G]

Wikipedia has an extensive page about the catenary.
https://en.wikipedia.org/wiki/Catenary

See especially the sections:

Anchoring of marine objects​

and

Determining parameters​

It looks like your drone better be a rather strong one!

Cheers,
Tom

p.s. please let us know how the project works out... or doesn't!

 

[ethanesh]

I forgot to say that I don’t think moments help here. Just consider the forces.

I am not sure I understand, I want to know the moments so that I can avoid potentially flipping the drone.

 

[haruspex]

I am not sure I understand, I want to know the moments so that I can avoid potentially flipping the drone.

Then you need to consider exactly how each force acts on the drone.
The lift force, presumably, acts up and to the left, through the centre of the rotor.
Gravity through the centre of mass of the drone. Drag, roughly likewise?
Tension, down and to the right, through the attachment point, which is surely below the mass centre.

Not apparent to me how it could flip except by wind gusts, but maybe I'm missing something.

 

[Tom.G]

But the center-of-support is at the geometric mean of the (usually four) rotors, and the drag vector (hose attachment point) is well below that point and likely offset by some amount.

One possible workaround (that I'm not sure of) could be to attached the hose near the Front edge of the drone. This would orient the drone for forward travel. --- Oops! On 2nd thought that seems like it would give positive feedback to flipping the craft.

 

[ethanesh]

Then you need to consider exactly how each force acts on the drone.
The lift force, presumably, acts up and to the left, through the centre of the rotor.
Gravity through the centre of mass of the drone. Drag, roughly likewise?
Tension, down and to the right, through the attachment point, which is surely below the mass centre.

Not apparent to me how it could flip except by wind gusts, but maybe I'm missing something.

Wouldn't the weight of the hose cause a moment that could rotate/flip the drone, since it isn't acting through the drone's CG? If I attached some heavy cargo like this

that would surely cause to drone to rotate. Is the hose situation different? Am I misunderstanding what you're trying to say?

EDIT: realized my mistake was not considering the difference between a hose, that only pulls in tension, and another rigid body like the picture above that is fixed to the drone. So if I want to find the moment caused by the hose, I need to find the tension in the hose that is caused by the weight and the friction forces. Seems obvious now but I have a tendency to overlook things like that :p

 

[haruspex]

View attachment 359813

But why is it arranged that way? Why don’t the lift and tension forces act near the mass centre?
If they are offset for some reason, those are the details needed that I mentioned in post #11.

 

[haruspex]

But the center-of-support is at the geometric mean of the (usually four) rotors,

I assume you mean geometric centre or arithmetic mean

 

[ethanesh]

But why is it arranged that way? Why don’t the lift and tension forces act near the mass centre?
If they are offset for some reason, those are the details needed that I mentioned in post #11.

Yes, the hose has to be offset slightly towards the back of the drone. So if I want to find the moment, I need to find the tension in the hose and use that for the force the drone experiences (outside of aerodynamic drag/thrust)... yes? That leaves me with another question. Say I have a setup like this

where the hose starts off horizontal but quickly gains an angle. What direction would the tension force act? Surely it wouldn't be horizontal, although that's how it is attached, it makes no physical sense. Would it be at something like the average angle of the hose? EDIT: or could it be the resultant of the two forces that are acting on the hose (weight and friction)?

 

[Lnewqban]

EDIT: realized my mistake was not considering the difference between a hose, that only pulls in tension, and another rigid body like the picture above that is fixed to the drone. So if I want to find the moment caused by the hose, I need to find the tension in the hose that is caused by the weight and the friction forces. Seems obvious now but I have a tendency to overlook things like that :p

The drone-hose connection must be free to rotate in all directions.

No moment should be transferred to the flying machine, which should remain more or less horizontal, while the hose changes angles as needed.

The closer the anchoring point can be to the CG, the less power you will waste to compensate for unnecessary pitch or pitch down induced by an off-set pulling force from the load.

Please, see this practical application:
https://www.tdworld.com/overhead-tr...8/save-time-by-pulling-lead-lines-with-drones

 

[ethanesh]

The drone-hose connection must be free to rotate in all directions.

No moment should be transferred to the flying machine, which should remain more or less horizontal, while the hose changes angles as needed.

The closer the anchoring point can be to the CG, the less power you will waste to compensate for unnecessary pitch or pitch down induced by an off-set pulling force from the load.

Please, see this practical application:
https://www.tdworld.com/overhead-tr...8/save-time-by-pulling-lead-lines-with-drones

I do not think I understand what you're saying here. You say no moment is transferred from the load, but then talk about the pitch caused by the load, which is a moment. Is there some terminology difference I am not comprehending?

 

[Lnewqban]

I do not think I understand what you're saying here. You say no moment is transferred from the load, but then talk about the pitch caused by the load, which is a moment. Is there some terminology difference I am not comprehending?

I wrote that no moment should be transferred from the load.
The less, the best.

There are two things that you could do about that:

1) Avoid a rigid connection, so your drone only has to pull.

2) Make the direction of the purely pulling force of the load go through the geometrical location of the center of lift.

 

[ethanesh]

I wrote that no moment should be transferred from the load.
The less, the best.

There are two things that you could do about that:

1) Avoid a rigid connection, so your drone only has to pull.

2) Make the direction of the purely pulling force of the load go through the geometrical location of the center of lift.

Thank you! A follow-up question if you don't mind: if I avoid a rigid connection, will the force on the drone caused by the tension in the hose be equal to the combination of the forces acting on the hose? For example, if the two forces on the hose are 5 lb caused by gravity in the -y direction and 3 lb caused by the friction of the hose on the ground in the x direction, will the drone also experience 5 lb in the -y direction and 3 in the x direction, acting at the point the hose is attached? Or does it not work like that, since the hose isn't a rigid body? (I have only worked with rigid bodies in my physics courses.)

 

[jbriggs444]

will the force on the drone caused by the tension in the hose be equal to the combination of the forces acting on the hose? For example, if the two forces on the hose are 5 lb caused by gravity in the -y direction and 3 lb caused by the friction of the hose on the ground in the x direction, will the drone also experience 5 lb in the -y direction and 3 in the x direction, acting at the point the hose is attached? Or does it not work like that, since the hose isn't a rigid body? (I have only worked with rigid bodies in my physics courses.)

Regardless of whether the hose is rigid, we are assuming that it is in an equilibrium state. Therefore, its shape is unchanging. No incremental bit of the hose is under any acceleration.

Accordingly, the sum of all forces acting on the hose aside from the force at the attachment point must be equal and opposite to the force at the attachment point.

The forces to sum would include:
* The force of gravity on the suspended rope section.
* The aerodynamic drag force on the suspended rope section.
* Any aerodynamic lift force on the suspended rope section.
* The drag force from the section of hose scraping over the ground.

We do not know, without testing or careful analysis, how the aerodynamic forces compare to the force of gravity. So we do not know what shape the hanging curve will take.

The equilibrium assumption may not hold good. We have not ruled out the possibility that the hose will flutter.

 

[Lnewqban]

... if I avoid a rigid connection, will the force on the drone caused by the tension in the hose be equal to the combination of the forces acting on the hose?

Yes.
A simple length of rope can comply with all the requirements of a non-rigid connection.

The flight attitude of the drone, as well as its responses to the control inputs (including yaw), will be independent of the angle adopted by the catenary formed by the hose at different heights and different horizontal velocities.

For example, if the two forces on the hose are 5 lb caused by gravity in the -y direction and 3 lb caused by the friction of the hose on the ground in the x direction, will the drone also experience 5 lb in the -y direction and 3 in the x direction, acting at the point the hose is attached?

Yes.
The drone (and its lift vector) will need to pitch (nose down) to be able to keep its spatial location; even more in order to move forward.

Or does it not work like that, since the hose isn't a rigid body? (I have only worked with rigid bodies in my physics courses.)

With our arrangement, the hose becomes a load, which the drone only pulls more or less, up, forward, or simultaneously in both directions.

 

[haruspex]

But the center-of-support is at the geometric mean of the (usually four) rotors,

I assume you mean geometric centre or arithmetic mean

 

[haruspex]

Yes, the hose has to be offset slightly towards the back of the drone. So if I want to find the moment, I need to find the tension in the hose and use that for the force the drone experiences (outside of aerodynamic drag/thrust)... yes? That leaves me with another question. Say I have a setup like this
View attachment 359837
where the hose starts off horizontal but quickly gains an angle. What direction would the tension force act? Surely it wouldn't be horizontal, although that's how it is attached, it makes no physical sense. Would it be at something like the average angle of the hose? EDIT: or could it be the resultant of the two forces that are acting on the hose (weight and friction)?

It is unclear what sections or joints in your diagram are rigid.
A rope can only transmit tension tangentially to its shape. If the blue line above is flexible then yes, the force it exerts on the black vertical rod is horizontal.

I note that more recent posts have brought up the issue of aerodynamic drag on the rope.
Ignoring wind, this will depend on the speed. In the steady state, the whole rope moves at the same speed. If we take the drag on a section of rope to be proportional to its length (in reality, it will be less for the more horizontal sections) then the drag just acts like a horizontal gravity. As a result, we still get a catenary, but tilted.

 

[Lnewqban]