02.15.2021 11:00 AM

How to Calculate the String Angle of a Kite vs. a Balloon

It's a beautiful day to go outside with a kite or a balloon, and compute how wind speed alters their flight.

blue balloons — Photograph: Jena Ardell/Getty Images

I'm reading Randall Munroe's book How To: Absurd Scientific Advice for Common Real-World Problems. I probably don't have to tell you this, but it's awesome (as is everything from Randall Munroe, the creator of xkcd comics). The whole idea of the book is to go use some crazy ideas to solve mostly common problems. One chapter focuses on how to cross a river. He gives you lots of options. You could change the course of the river or even evaporate all the water in the river (both ideas are silly and fun). Another option is to use a kite to get across the river. And here is the fun part—Munroe states that both a kite AND a balloon could extend over a river. As the wind speed increases, a kite gets higher in the sky. However a balloon gets lower as the wind increases.

So, at some value of wind speed the kite and the balloon would have a string at the same angle. Oh! I want to calculate this. That will be fun.

Let's start with a balloon. If you have a helium-filled balloon and there is no wind, it will float in the sky and the string will be completely vertical. There are just three forces acting on the balloon. There is the downward-pulling gravitational force that depends on both the mass of the object (m) and the gravitational field (g = 9.8 N/kg). Since the balloon displaces air, it has a buoyancy force that is equal to the weight of the air displaced (Archimedes' principle). If the balloon only had these two forces, the net force would most likely be upward and the balloon would accelerate away. Bye-bye balloon.

Of course you might want to keep that balloon. That's why you tie a string to it. This string exerts a downward tension force (T) with a magnitude to make the net force equal to zero. With a zero net force, the balloon is in equilibrium and stays at rest so that you can enjoy looking at your gravity-defying balloon. Here is a diagram representing these forces.

Illustration: Rhett Allain

Adding up just the vertical components (let's let the vertical be the y-direction) of these forces, I can write it as the following sum.

Illustration: Rhett Allain

We already have an expression for the gravitational force (m*g), and the tension will be whatever value it needs to be in order to make the total force zero (it's a force of constraint). So, if we have an expression for the force from the air (the buoyancy force), then we can get some stuff together. Since this buoyancy force is the weight of the air displaced, I need the volume of the balloon (V) and the density of air (ρ). Assuming the balloon is a sphere with a radius R, then the buoyancy force would be:

Illustration: Rhett Allain

OK, now let's add some wind. Suppose the wind is blowing horizontally with some velocity (v). This means there will be another force on the balloon, an air drag force. We can model this air drag as a force in the same direction as the wind with a magnitude that depends on the wind speed, the cross-sectional area of the balloon (A), the shape of the balloon (C), and the density of air (ρ). If you are the wind (yes, YOU are the wind), the cross section of the balloon looks like a circle with a radius R. That makes the area equal to πR² (the area of a circle).

Illustration: Rhett Allain

But now we have a problem. Since there is a horizontal force from the wind, there has to be some other horizontal force so that the net force in that direction is zero. Yes, this extra horizontal force comes from the string as it pulls at an angle. Here is a new diagram. It's a little bit more complicated.

Illustration: Rhett Allain

Notice that I added the wind—just for a fun visual effect. I labeled the string angle with the variable θ. If the balloon is still in equilibrium, the net force must be zero in the both the horizontal (x) AND vertical (y) directions. The tension in the string has a component of force in both the x and y directions such that the following two equations would be true.

Illustration: Rhett Allain

Since the tension is a constraint force, there's no direct way to calculate it. That's fine. I can just solve for T in the y-forces equation and substitute into the x-forces equation. Problem solved. Now I can get an expression for the lean angle of the balloon. Remember that the drag force depends on both the radius of the balloon and the velocity of the wind, but the buoyancy force also depends the radius (because of the volume). Putting all of this stuff in, I get this crazy-looking expression (but it's not as bad as it looks).

Illustration: Rhett Allain

Don't worry, I'm going to plot the leaning angle of a balloon for different wind speeds, but first let's look at kites. A kite isn't a balloon—just to be clear. However, it can still fly in the air AND it has a string. Just like the balloon, the kite also interacts with the moving air (also called "wind"). However, for the kite, the air pushes back (drag) and also up (lift). One way to model both the lift and drag force for a kite is to use the lift-to-drag ratio (it's a real thing).

It's not mysterious. The lift-to-drag ratio is literally just the lift force divided by the drag force. Every flying object that produces lift also produces drag. They are both due to the same interaction with the air. So if you fly faster (or have faster wind over a stationary kite), both the lift AND the drag will increase. Yes, this lift-to-drag ratio depends on the shape and size of the flying object as well as orientation with respect to the motion of the air (called the angle of attack). But for this kite, I'm just going to calculate the drag and then multiply by C_L (lift coefficient) to get the lift force.

I think we are ready for a diagram. Here is my kite with forces.

Illustration: Rhett Allain

What? This looks just like the forces for the balloon? OK, it does look similar—but there is a big difference. For the balloon, there is that upward-pushing buoyancy force, and it's just one value. It doesn't change when the wind speed increases. For the kite, the upward pushing force is the lift, and it DOES depend on the wind speed. So it's not the same. Just consider the case when there is zero wind. The drag force will be zero, which means the lift is zero. The kite won't fly—it just falls down and it's sad.

Again, I get two force equations that I can use to eliminate the unknown value of T. With that, I get the following expression for the angle of the kite (θ_k). Actually, I put a subscript k on a bunch of stuff so you could see it's different than the values for the balloon. Oh, air still has the same density for both objects.

Illustration: Rhett Allain

OK, I'm about to make a plot of the flying angle for both the balloon and a kite at different wind speeds. But before I do that, let's think about the minimum speed to fly this kite. In order to lift off the ground, the lift force must be at least equal to the weight of the kite. I can then solve this for the wind speed. Anything lower than this and you won't have a flying kite.

Illustration: Rhett Allain

Now I can pick some values for all the parameters for both the kite and the balloon. From that, I will calculate the minimum velocity and plot the string angle for both the balloon and kite. Then I just increase the velocity and look at the pretty graph. I'm just going to make some rough guesses for stuff like the mass of a kite and the lift-to-drag ratio. But don't worry. If you don't like my choices, you can change the values in the code below. Here is what you get.

Yes, that's actual Python code. If you click the pencil icon, you can edit it and run it again. But you should notice some important features for these two curves (the kite and the balloon).

As the wind speed increases, the kite's angle gets larger and the balloon's gets smaller. That's what we expect.
For some value of wind speed, the kite and the balloon are flying at the same angle (for my values, it's at about 2.19 m/s).
This kite will never be straight overhead (angle of 90 degrees). Instead, it gets to a maximum angle of about 61 degrees.

If you change all the values (mass and drag coefficients for the balloon and kite), you will get a different wind speed at which they have the same angle. Oh, and one last thing. It's true there was quite a bit of math in this post. But it could have been much worse. In all of these calculations, I assumed the strings had no mass. Just imagine how fun this problem would be with more realistic strings. I'll leave that for you as a homework assignment.