Why Is the Tear Drop Shape Optimal for Hot Air Balloons?

In summary, the hot air balloon's ideal shape is a tear-drop shape due to the physics of pressure and tension. By considering a 2D slice of the balloon and assuming the material weighs nothing, the tension throughout the balloon is constant and half of the gondola's weight. The pressure difference inside and outside the balloon, with the highest pressure at the top and lowest at the bottom, causes the balloon to have a curved shape with the greatest curvature at the top and decreasing as it goes lower. This shape is supported by the tension in the material, which is necessary to support the weight.
  • #1
Poseidonho
12
0
I had read an article about a hot air balloon, when coming into balloon shape design, I had question on it.
For best shape of the air balloon was the tear drop shape, but I do not know what is the physics behind it. From the article I read is about the pressure vector, and he is using an example of the water pressure and the depth. But I can't found any mathematical proving on it or the detail.

So any suggestion where I can find more detail information about the shape of hot air balloon?
 
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  • #2
It's fairly simple. First, you need to understand that you can give it almost any shape at an expense of extra tension. I'm sure you've seen all these crazy-looking novelty hot air balloons designed to look like castles, parade floats, or whatever. So the goal should be to have only necessary tension. That means, all of the tension is there to support weight. The most important conclusion you can get from this is that you can consider just a 2D slice across the center of the balloon. So we are looking for that tear-drop shape in a 2D model.

Second, for simplicity, let's assume that the material itself weighs nothing. It isn't true, but you get same qualitative result either way. If you just want to understand why the shape is tear-drop, it's not important. So all of the weight is due to the gondola, and the tension throughout material is constant. Since we are looking at the 2D case, the tension throughout is just half of the gondola's weight. Half, because in 2D the gondola is supported from two sides.

Finally, we look at what's happening to air pressure in the balloon. The hot air balloon is open at the bottom. So at the bottom, the pressure is the same inside and out. However, because density inside is lower, the pressure gradient is lower. That means pressure at the top of balloon is higher. We are going to ignore the fact that as pressure changes, so does density. That's a small correction. So we have now a pressure difference inside and out that's highest at the top and drops to zero at the bottom. That pressure difference will push outward on the balloon at every point.

Now, suppose you have a string under tension. How much force can it apply sideways? That will depend on the curvature. A straight stretch of string will not apply any sideways force. So to support a load, the string must bend. The load perpendicular to the string that stands for the skin of the balloon in the 2D case is due to the pressure. So the surface of the balloon will be most curved where pressure differential is highest. As a result, the top of the balloon will look like the top of a sphere. As you go lower, the curvature will decrease. At the bottom, where pressure differential is almost zero, this wil make a straight line to the gondola.

Running all of this through a numerical diff eq solver, I get the following shape. Positive x is "up". Sorry, I didn't rotate it or scale it better.

attachment.php?attachmentid=53906&d=1355474952.png
 

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  • #3
K^2 said:
It's fairly simple. First, you need to understand that you can give it almost any shape at an expense of extra tension. I'm sure you've seen all these crazy-looking novelty hot air balloons designed to look like castles, parade floats, or whatever. So the goal should be to have only necessary tension. That means, all of the tension is there to support weight. The most important conclusion you can get from this is that you can consider just a 2D slice across the center of the balloon. So we are looking for that tear-drop shape in a 2D model.

Second, for simplicity, let's assume that the material itself weighs nothing. It isn't true, but you get same qualitative result either way. If you just want to understand why the shape is tear-drop, it's not important. So all of the weight is due to the gondola, and the tension throughout material is constant. Since we are looking at the 2D case, the tension throughout is just half of the gondola's weight. Half, because in 2D the gondola is supported from two sides.

Finally, we look at what's happening to air pressure in the balloon. The hot air balloon is open at the bottom. So at the bottom, the pressure is the same inside and out. However, because density inside is lower, the pressure gradient is lower. That means pressure at the top of balloon is higher. We are going to ignore the fact that as pressure changes, so does density. That's a small correction. So we have now a pressure difference inside and out that's highest at the top and drops to zero at the bottom. That pressure difference will push outward on the balloon at every point.

Now, suppose you have a string under tension. How much force can it apply sideways? That will depend on the curvature. A straight stretch of string will not apply any sideways force. So to support a load, the string must bend. The load perpendicular to the string that stands for the skin of the balloon in the 2D case is due to the pressure. So the surface of the balloon will be most curved where pressure differential is highest. As a result, the top of the balloon will look like the top of a sphere. As you go lower, the curvature will decrease. At the bottom, where pressure differential is almost zero, this wil make a straight line to the gondola.

Running all of this through a numerical diff eq solver, I get the following shape. Positive x is "up". Sorry, I didn't rotate it or scale it better.

attachment.php?attachmentid=53906&d=1355474952.png

I am interested in the math involved to calculate the ideal natural teardrop shape for a hot air balloon. I want to learn the details of what is involved to calculate this accurately.

I read your reply which was a really nice start, but it unfortunately does not get into the details of how this is calculated and I really want to learn how this is done.

Regarding the pressure gradient within the balloon, I'm curious how to calculate the pressure change from the top to the bottom, and the math involved in calculating the shape.

I appreciate the help. Thank you.
 
  • #4
Yachtsman said:
I am interested in the math involved to calculate the ideal natural teardrop shape for a hot air balloon. I want to learn the details of what is involved to calculate this accurately.

I read your reply which was a really nice start, but it unfortunately does not get into the details of how this is calculated and I really want to learn how this is done.

Regarding the pressure gradient within the balloon, I'm curious how to calculate the pressure change from the top to the bottom, and the math involved in calculating the shape.

I appreciate the help. Thank you.

Ya, I also interested in the equation for the shape of balloon, pressure, height, temperature and gas density (AIR). If can involve in Vector is better. But so far what I found is just a simple equation, the ideal gas law related it.
 
  • #5


I would recommend starting by researching the principles of aerodynamics. The shape of a hot air balloon is designed to minimize drag and maximize lift, which are both important factors in flight. The tear drop shape is commonly used because it allows for a smooth flow of air around the balloon, reducing drag. This shape also creates a pressure differential between the top and bottom of the balloon, which helps lift the balloon into the air.

To understand the physics behind this shape, it is important to study the Bernoulli's principle and the Venturi effect. These concepts explain how the shape of an object affects the flow of air around it and the resulting pressure differentials.

You can also find more detailed information on hot air balloon design by researching scientific articles and publications from organizations such as the American Institute of Aeronautics and Astronautics (AIAA) or the Royal Aeronautical Society. Additionally, consulting with experts in the field of aerodynamics and hot air balloon design could provide valuable insights and resources.

It is also worth noting that the shape of a hot air balloon may vary depending on its intended use and size. Factors such as payload capacity, altitude, and wind conditions may also play a role in determining the optimal shape for a specific hot air balloon.

In conclusion, the tear drop shape of a hot air balloon is based on principles of aerodynamics and is designed to optimize flight performance. Further research and consultation with experts in the field can provide more detailed information on the specific physics behind this shape and its effectiveness in hot air balloon design.
 

1. What is the shape of a hot air balloon?

The shape of a hot air balloon is typically a teardrop or oval shape. This shape allows for maximum air resistance, which helps the balloon to stay afloat.

2. Why is the shape of a hot air balloon important?

The shape of a hot air balloon is important because it affects the aerodynamics and buoyancy of the balloon. The teardrop shape allows for better air flow, while the rounded bottom helps to distribute the weight evenly.

3. How is the shape of a hot air balloon determined?

The shape of a hot air balloon is determined by the design and construction of the balloon's envelope, which is made of a lightweight and durable fabric. The shape is also influenced by the amount of air inside the balloon and the weight of the basket and passengers.

4. Can the shape of a hot air balloon be changed?

Yes, the shape of a hot air balloon can be changed by adjusting the amount of air inside the balloon. By adding or releasing air, the pilot can control the shape and altitude of the balloon.

5. How does the shape of a hot air balloon affect its flight?

The shape of a hot air balloon plays a crucial role in its flight. The teardrop shape allows for better control and stability, while the rounded bottom helps to keep the balloon afloat. A well-designed shape also helps to reduce drag and improve overall performance.

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