Physics Forums

Physics Forums (http://www.physicsforums.com/index.php)
-   Classical Physics (http://www.physicsforums.com/forumdisplay.php?f=61)
-   -   About Hot Air Balloon shape (http://www.physicsforums.com/showthread.php?t=658802)

Poseidonho Dec14-12 12:23 AM

About Hot Air Balloon shape
 
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 cant 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?

K^2 Dec14-12 02:49 AM

Re: About Hot Air Balloon shape
 
1 Attachment(s)
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, lets 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.

http://physicsforums.com/attachment....6&d=1355474952

Yachtsman Jul9-13 07:00 PM

Quote:

Quote by K^2 (Post 4196090)
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, lets 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.

http://physicsforums.com/attachment....6&d=1355474952

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.

Poseidonho Jul10-13 09:41 PM

Quote:

Quote by Yachtsman (Post 4440638)
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.


All times are GMT -5. The time now is 04:55 PM.

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums