Mathematical model for drag on tether

[Tobi9242]

Homework Statement

Make a model that describes the aerodynamic drag on a tether holding a kite:

The kite is moving crosswind in a regular flight pattern
The tether is fixed at the ground
The model should only describe the drag the tether experiences while moving across the wind, and therefore wind speed isn't relevant, only the speed of the kite

Relevant Equations

F_drag = 1/2 * (\rho * V^2 * C_d * A )

I have a model for airs density as a function of height
I would imagine the speed can be describes as the angular velocity times length
The coefficient of drag can be found online, seems to be around 1.17 for a cylinder
It seems to me that im going to need an integral somewhere, but can't quite figure our how and where.

 

[pbuk]

Can you write down an equation for the drag on a small length of the line $\delta l$ at a distance $l$ from the anchor?

 

[Tobi9242]

Thanks for the reply!

F_drag=1/2*C_d*rho*(delta_l*diameter of line)*(angular velocity * l)

This is my best guess

 

[pbuk]

Oh, I forgot:

F_drag=1/2*C_d*rho*(delta_l*diameter of line)*(angular velocity * l)

That looks OK: are you familiar with writing mathematical expressions in $\LaTeX$ - it makes them much easier to read? You could try Replying to my message and it will include what I have posted for you to copy and edit like this: [Edit: this includes an error left for the OP to correct]
$$ \delta F_{drag} = \frac{1}{2} C_d \rho D \omega \ l \ \delta l $$
I have used $D$ as the diameter and $\omega$ the angular velocity: it would probably be better to replace $\omega$ with an expression involving the velocity of the kite - use $V$ for this, and $L$ for the length of the line.

 

[pbuk]

it would probably be better to replace $\omega$ with an expression involving the velocity of the kite - use $V$ for this, and $L$ for the length of the line.

... but you can do that after the integration (it is constant of course).

To get the total drag, you simply need to integrate the expression you wrote over the length of the line! [Edit: this includes an error left for the OP to correct]

$$ F_{drag, total} = \int_0^L dF_{drag} = \int_0^L \frac{1}{2} C_d \rho D \omega \ l \ dl $$

 

[Tobi9242]

Oh, I forgot:

Thank you!

Yes, I am familiar with LaTex, so i'll do that moving forward

And thank you very much, you are a lifesaver!

 

[pbuk]

And thank you very much, you are a lifesaver!

Oops, I've just noticed that although the equation in your first post was OK:

Relevant Equations: F_drag = 1/2 * (\rho * V^2 * C_d * A )

... when you substituted $\omega l$ for $v$ you lost the squaring. This has carried through to the integral in my last post, can you fix this?

 

[Tobi9242]

Oops, I've just noticed that although the equation in your first post was OK:

... when you substituted $\omega l$ for $v$ you lost the squaring. This has carried through to the integral in my last post, can you fix this?

Oh, my bad. I hadn't noticed the missing square on here, but I have the velocity squared in my own document, so no worries!
Thanks again!

 

[haruspex]

The model should only describe the drag the tether experiences while moving across the wind, and therefore wind speed isn't relevant, only the speed of the kite

That would be true if the drag were linear, but it is quadratic.
If the wind speed is w and the kite's speed v, the relative speed is $\sqrt{v^2+w^2}$ and the magnitude of the drag is $\frac 12\rho (v^2 +w^2)C_d A$. The component of that in the kite's direction of movement is $\frac 12\rho v\sqrt{v^2 +w^2}C_d A$.
As the kite tracks back and forth across the wind, there will at least be some parts of the trajectory where $w>v$, making the drag component closer to $\frac 12\rho vwC_d A$ than to $\frac 12\rho v^2C_d A$.

In reality, $w$ is probably greater than $v$ all the time.

 

[pbuk]

That would be true if the drag were linear, but it is quadratic.

No. Two points:

1. Drag force exists parallel to motion. When the kite is directly downwind the wind direction is orthogonal to the kite's direction of travel and so does not increase drag so your equation
   

   $\frac 12\rho v\sqrt{v^2 +w^2}C_d A$

   cannot be correct.
2. The question tells us to ignore wind speed (probably because if we do not the model becomes inappropriately complicated for the level of learning the OP is at).

 

[haruspex]

Drag force exists parallel to motion.

Relative motion.
See e.g. https://www.roadbikereview.com/threads/the-physics-of-crosswinds.372183/

The question tells us to ignore wind speed (probably because if we do not the model becomes inappropriately complicated for the level of learning the OP is at).

Then the text should simply say ignore it. The given reason for ignoring it misinforms the student.
Besides, the correct integral is no harder!

 

[pbuk]

Relative motion.
See e.g. https://www.roadbikereview.com/threads/the-physics-of-crosswinds.372183/

I understand your mistake: the equivalent calculation for a cyclist would be for a "wind" coming from above, not from the side.

 

[Tobi9242]

No. Two points:

1. Drag force exists parallel to motion. When the kite is directly downwind the wind direction is orthogonal to the kite's direction of travel and so does not increase drag so your equation
   
   cannot be correct.
2. The question tells us to ignore wind speed (probably because if we do not the model becomes inappropriately complicated for the level of learning the OP is at).

Yes, this would be how I understand it as well

 

[haruspex]

I understand your mistake: the equivalent calculation for a cyclist would be for a "wind" coming from above, not from the side.

I would like to think that's intended as a joke. It's quite simple: if an object is moving at velocity $\vec v$ in the ground frame and the air is moving at $\vec w$ in that frame then the relative velocity is $\vec v-\vec w$. The drag is $k|\vec v-\vec w|(\vec w-\vec v)$. The component of that in the direction of $\vec v$ is $k|\vec v-\vec w|((\vec w-\vec v)\cdot\vec v)\vec v/v^2$. If the two velocities are orthogonal that reduces to $k|\vec v-\vec w|(-\vec v)$.

 

[haruspex]

Yes, this would be how I understand it as well

Wrt @pbuk's point 1, try googling it.
Wrt point 2, see post #11.

 

[pbuk]

I would like to think that's intended as a joke.

No, it was intended to illustrate the difference between the two models. I'll try a diagram instead:

if an object is moving at velocity $\vec v$ in the ground frame and the air is moving at $\vec w$ in that frame then the relative velocity is $\vec v-\vec w$.

That's all well and good, but if you want to calculate drag you need the surface area orthogonal to that relative velocity, and furthermore you will need to consider that (unless the kite is at right angles to the wind) some of that drag will increase the tension in the line rather than slow down the kite. If you do the math you will find that when the kite is approximately downwind then the contribution of the ground wind to horizontal tangential drag is small (and of course intuitively when the kite is exactly downwind it must be zero).

 

[Lnewqban]

Yes, this would be how I understand it as well

The problem is not realistic, as it disregards the lift and drag forces on the kite.
The author seems to be interested only on the drag induced by the lateral swinging of the kite.
That flow-induced vibration makes the kite deviate from its most stable position (center), acquiring more and more energy from the incoming wind (disregarded here) until reaching a resonant state (just like a fluttering flag).

In that case, the drag would change with the angular velocity, being its value zero at the left and right extreme points of the oscilation, and maximum as the kite reaches its highest point (swinging in both directions).

Perhaps I am wrong, but the situation seems similar to an inverted pendulum, which plane of oscillation is tilted backwards certain angle from the vertical line crossing the anchor to ground.

Please, see:
https://kitelife.com/2012/09/26/working-the-wind-window/

 

[pbuk]

The problem is not realistic, as it disregards the lift and drag forces on the kite.

Nevertheless, that is the problem that was asked - to consider only the drag on the line (tether).

Please don't confuse people by attempting to produce answers to questions that have not been asked, that is not helpful in itself and also demonstrates an attitude to answering questions that is likely to be detrimental to progress in most learning environments.

 

[haruspex]

No, it was intended to illustrate the difference between the two models. I'll try a diagram instead:

View attachment 335329

I think we can take the wind as being horizontal. Its velocity will have a substantial
component normal to the line.

The point of my intercession in the thread is to counter the argument in the problem statement that because the wind is orthogonal to the string's motion it is irrelevant to the drag. That does not depend on its being a kite rather than a bicycle.

some of that drag will increase the tension in the line rather than slow down the kite.

Which is why I took the component of drag in the direction of the line's motion.

If you do the math you will find that when the kite is approximately downwind then the contribution of the ground wind to horizontal tangential drag is small (and of course intuitively when the kite is exactly downwind it must be zero).

Now you've lost me. Even when the kite is directly downwind it is not lying on the ground!

 

[Lnewqban]

...
Please don't confuse people by attempting to produce answers to questions that have not been asked, that is not helpful in itself and also demonstrates an attitude to answering questions that is likely to be detrimental to progress in most learning environments.

What in the world are you talking about?

 

[berkeman]

Thread closed temporarily [edited]

 

[berkeman]

Thread is reopened provisionally; let's try to keep the discussion civil. But to be honest, this question is very confusing to me, especially the "simplifying" assumption (which seems wrong to me).

@Tobi9242 -- When is this assignment due? Do you have a Teaching Assistant that you can visit for clarification?

 

[pbuk]

I will draw a diagram to get us all on the same page. I have based this on an image retrieved from https://www.researchgate.net/figure...be-kite-flying-crosswind-while_fig1_258351980

The tether is anchored at the origin. The tether is at a (constant) angle $A$ above the horizontal and the [kite at the] end of the line is moving horizontally at $| \vec v | = L \dot \theta = L \omega$. The wind is horizontal and wolog assumed parallel to $y = 0$.

 

[haruspex]

I will draw a diagram to get us all on the same page. I have based this on an image retrieved from https://www.researchgate.net/figure...be-kite-flying-crosswind-while_fig1_258351980
View attachment 335385
View attachment 335387

The tether is anchored at the origin. The tether is at a (constant) angle $A$ above the horizontal and the [kite at the] end of the line is moving horizontally at $| \vec v | = L \dot \theta = L \omega$. The wind is horizontal and wolog assumed parallel to $y = 0$.

Thanks for taking the trouble to do this, but as I keep explaining, my issue with the problem statement is that it implies, as a general principle, that crosswinds don't contribute to drag. That misleads students, and it seems the OP took it as true.

Whether the wind would contribute to the drag meaningfully in this specific case is somewhat secondary, so we should not really have taken up so much of the thread on it. So here I'll just say that to match your diagram (except A is already in use in the drag equation, so I'll call the tether slope $\alpha$) I would change my equation in post #9 to

The component [of drag] in the kite's direction of movement [at $\theta=0$] is $\frac 12\rho v\sqrt{v^2 +w^2\sin^2(\alpha)}C_d A$.

If you are happy with that, we can move on. Otherwise we can take it a PM thread or some new public thread.

 

[pbuk]

If you are happy with that

Not quite. I hope we can both be happy with $\frac 12\rho v \ \sqrt{v^2 +w^2 \sin^2 (\alpha)} \ C_d(\alpha) A(\alpha)$ which recognises that the drag coefficient of a long thin oblique cylinder is different from that of one perpendicular to the wind direction, and that the surface area A must be corrected for the presentation to the apparent wind (I think $A(\alpha) \approx A \cos (\alpha)$).

We can then note that for sufficiently small $\alpha$ (i.e. the kite near to downwind) $w^2 \sin^2 (\alpha)$ reduces to zero and so the ground wind has no effect.

 

[haruspex]

We can then note that for sufficiently small $\alpha$ (i.e. the kite near to downwind) $w^2 \sin^2 (\alpha)$ reduces to zero and so the ground wind has no effect.

Now I'm confused again. Will respond by PM.

 

[Tobi9242]

Thread is reopened provisionally; let's try to keep the discussion civil. But to be honest, this question is very confusing to me, especially the "simplifying" assumption (which seems wrong to me).

@Tobi9242 -- Wen is this assignment due? Do you have a Teaching Assistant that you can visit for clarification?

The problem is part of a larger project, so the due date is in a couple of weeks, and no, unfortunately there aren't any teaching assistants available.

I have however been informed that if we are able to, we are allowed take take more thing into consideration when making the model. So if a more accurate model is achievable, without complicating things to a level I would have difficulty explaining, that would be great.

 

[haruspex]

The problem is part of a larger project, so the due date is in a couple of weeks, and no, unfortunately there aren't any teaching assistants available.

I have however been informed that if we are able to, we are allowed take take more thing into consideration when making the model. So if a more accurate model is achievable, without complicating things to a level I would have difficulty explaining, that would be great.

You mentioned in post #1 you thought you needed an integral, but the integral of interest depends on why you want the answer.
As @Lnewqban noted, the system resembles a pendulum with axis at the tether. When the kite moves to the right the drag is to the left, and this exerts a force to the left on the tether. Correspondingly, the reaction force at the tether has a component to the right.
If you care about how the drag affects the motion of the kite, what matters is not the total drag force but its total torque about the tether; if you care about the sideways pull the drag creates on the tether, it's reversed: you want the drag's torque about the kite.
In neither case does the simple integral of the drag force seem to be interesting.

Another extension, if you care to challenge the problem setter, is to incorporate the contribution the wind makes to the drag, as described in post #9 (except, as @pbuk noted, I had not considered the angle the string element makes to the horizontal; if that is $\alpha$ then the drag component opposing v is $\frac 12\rho v\sqrt{v^2 +w^2\sin^2(\alpha)}C_d .dA$, where dA is the element's effective area).
While incorporating w in this way might not make much difference at peak v, it will be significant towards the extremes of the trajectory as the kite swings left and right.

Another complexity is that the string will not be straight but form a catenary. If we assume it at least always lies in a vertical plane, the velocity of the element is proportional to the horizontal distance from the tether, not distance along the line.

Whatever combination(s) you choose to model, please post your integral and solution.