0
$\begingroup$

Suppose you had two balloons, one inside the other (a inside b).

Would filling balloon "a" with the 80% nitrogen/20% oxygen mix you find in the surrounding air at ground level, and then filling balloon "b" with 100% helium provide you with any benefit? In other words would you be able to use less helium to provide an equal lift compared to a third balloon that was only filled with helium?

I suppose another way to look at it: Would you be able to use less helium if the cargo was inside the balloon vs hanging from the bottom?

$\endgroup$
  • $\begingroup$ Balloon "a" would rest at the bottom (inside) of "b", adding weight and steeling volume. $\endgroup$ – Gyro Gearloose Feb 27 '16 at 21:47
2
$\begingroup$

No. The buoyant force on the balloon is a function of the difference in density inside and outside the balloon. If the density inside the balloon is higher, it sinks; if it's lower, it rises. Putting a plain-air balloon inside a helium balloon increases the overall density compared to a pure helium balloon of the same size: you can easily see this by taking the limit as the inner "air" balloon fills the outer helium balloon, bringing the whole thing to the same density as the surrounding air and thus zero buoyancy (ignoring the weight of the balloon itself of course).

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

It's a matter of volume and weight. Suppose the (outer) volume of the balloon is 1 cubic meter. The air inside that volume has a certain weight, roughly 1 kg.

The upward force of the balloon is just the difference between the weight of what's in it, and the weight of the same volume of air. The reason helium balloons float is because 1 cubic meter of helium weighs less than 1 cubic meter of air. (Assuming the same pressure and temperature.)

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.