Controlling helium balloon attitude

  • Thread starter Thread starter bbd001
  • Start date Start date
  • Tags Tags
    Balloon Helium
AI Thread Summary
Controlling the attitude of helium balloons can be achieved by using a tank, pump, and valve system, allowing for adjustments without cutting the balloons or dropping weight. Releasing helium from the tank effectively reduces buoyancy, similar to cutting balloons, while pumping air in or out can slightly adjust overall system weight. However, this method may not provide sufficient control for precise attitude adjustments. A bladder within the main balloon can also facilitate air movement, helping to manage pressure changes at higher altitudes. Overall, while some control is possible, limitations exist in achieving fine-tuned adjustments.
bbd001
Messages
6
Reaction score
0
Hi,

Is it possible to control the attitude of helium balloons without cutting the balloons or dropping the weight? (e.g. using a remote control to control "something" to adjust the attitude)

Thanks,
 
Physics news on Phys.org
thx for the reply russ.

Can you give me more information?

If I have a tank fill with helium, it is part of the system. Releasing the gas from the tank is like cutting some of the balloons. If I want to move up again, the system cannot get extra helium back from the sky.

If I pump/release air (instead of helium or other material that is lighter than air), it can adjust the over all system weight a little bit, but it should not enough to adjust the attitude. Is that right?

thx
 
bbd001 said:
Can you give me more information? If I have a tank fill with helium, it is part of the system. Releasing the gas from the tank is like cutting some of the balloons.
I'm pretty sure he meant to use the tank to move the higher density helium from the tank into the balloon, and a pump to move the lower density helium in the balloon back into the tank. The balloon would expand or contract depending on how much helium was in the balloon, and the tanks expansion or contraction would not be significant.

You'd have a side issue with the balloon expanding due to lower pressure of the atmoshpere at higher altitudes.
 
You have a bladder in the main balloon that you can pump air in or out off.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...

Similar threads

Replies
8
Views
2K
Replies
22
Views
3K
Replies
10
Views
2K
Replies
6
Views
3K
Replies
5
Views
2K
Replies
2
Views
3K
Replies
8
Views
3K
Back
Top