I once saw a video on youtube about how a man used a collection of balloons to lift himself along with his lawn chair from one side of Los Angeles to the other. My understanding of physics is a bit rusty, and I'm not well acquainted with aerodynamics, but I'm wondering how much helium is required to lift an object weighing a certain amount of grams. I know that lift is generated because helium is lighter than air, but I'm not sure how to determine how much lift each gram of helium generates. But, that's really it. Also, is the rate at which the object ascends into the sky constant? If I had to take an educated guess, I'd say that things like temperature, air pressure and humidity all might play a role. But, I'm not exactly sure how. Could someone point me to a set of formulas?

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    $\begingroup$ 8000 balloons to lift a 100 kg man, according to science.howstuffworks.com/science-vs-myth/everyday-myths/… $\endgroup$ – user81619 Jun 19 '15 at 18:23
  • $\begingroup$ i.ytimg.com/vi/gxF24ioN9I4/hqdefault.jpg :-) probobly not what you're looking for. $\endgroup$ – userLTK Jun 19 '15 at 19:42
  • $\begingroup$ "...how to determine how much lift each gram of helium generates..." Read the answers carefully. The helium does not generate the lift. On the contrary, the helium is dead weight, trying to drag the balloon down to Earth. The balloon would lift $n$ grams more weight if you could remove $n$ grams of helium from it without changing the shape and size of the balloon and, without letting anything else in to take the place of the helium that you removed. $\endgroup$ – Solomon Slow Jun 19 '15 at 20:22
  • $\begingroup$ 1 balloon is enough. Put size on it. $\endgroup$ – Helder Velez Jun 19 '15 at 20:28

The acceleration of gravity $|g| = 9.81 \frac{m}{s^2}$ since you are moving upward we set this to $g = -9.81 \frac{m}{s^2}$ for the force gravity generates on you is just $F = ma = mg$ where $m $ is your mass in kg what you need to calculate is the Buoyancy force $F_B$ that offsets the gravitational force $F_g$ here is a video that show this for a single balloon. Basically for a single balloon $F_B = \large\rho _{air} V_{displaced- air}\large g = ( m_{balloon}+ V_{balloon}\large\rho_{\small He} )g + Ten $ where $\large\rho _{air}$ is the density of air, $\large\rho _{He}$ is the density of helium, $V_{displaced- air}$ is the volume of the air displaced and is equal to the volume of the balloon $V_{balloon}$, $Ten$ and is the Tension in one thread the tension can also be thought of as the force generated by the balloon. So to isolate for $Ten$ we set $Ten = ( m_{balloon}+ \large(\rho_{\small He} + \rho _{air}) V_{balloon} )g$.

But remember the density of a fluid changes based on the following formula $\rho = \frac{pm}{k_B T}$ $p$ is the pressure in Pascals, $m$ is the mass , $T$ is the temperature in Kelvin,$k_B$ is the Boltzmann constant.

So to find the number of balloons you need $N$ we divide $NTen = m_{obj}g$ so $N = \frac{m_{obj}g}{Ten}$

And in the video she uses $T$ for $Ten$ do not confuse this with the $T$ I have used for temperature.

Be sure your threads are strong

P.S I am not responsible for any injuries incurred anyone who use this information does so of their own volition

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    $\begingroup$ the density of the atmosphere also changes (the baloon surface is in equilibrium with the atmosphere irt the pressure and temp. So..) $\endgroup$ – Helder Velez Jun 19 '15 at 20:27

The lift is equal to the weight of the displaced air. The net lift is the difference between the weight of the air, and the weight of the helium displacing it.

Air has a density of about 1.2 kg / m3, and helium has a density of about 1/7th of that. The lift of a balloon (ignoring the mass of the balloon) is therefore about 1 kg / m3.

Barometric pressure and air temperature play a small role, since


from which you can derive that the density will go down with higher temperature, and up with higher pressure. But the changes are typically just a few %.

As for the rate of ascent - this is typically limited by the drag of the balloon(s). Hard to estimate this, but drag for a sphere goes with the square of the velocity: this means you need 4x greater lift force to go twice as fast - but that means a 4x greater volume of balloon, which will have a greater area... Greater lift speed would best be achieved by using a pear-shaped balloon. In fact, for high altitude ballooning (think Felix Baumgartner's jump from the edge of space) you start with a balloon that is heavily _under_inflated as the helium will expand at higher altitudes (lower pressure). What is interesting is that the lift will be the same if you allow the balloon to expand - so the lift really is a multiple of the mass of the helium (in a ratio of about 6 to 1).

You would need approximately 17 kg of He to lift a 100 kg load (including the weight of the balloon etc).

  • $\begingroup$ The lift is equal to the weight of the displaced air minus the weight of the helium which displaces it. $\endgroup$ – WhatRoughBeast Jun 19 '15 at 18:32

In a nutshell, if we assume all gases are similar - which they're in the ballpark, Helium has an atomic weight of 4, Nitrogen-2, an atomic weight of 28 and Oxygen 32. So Helium is roughly 1/7th the density of air.

The cool thing is, 1 gram of helium in air, weights -6 grams, so in simplest terms, 1 gram of helium can lift 6 grams of stuff (including the mass of the balloon). That's ballpark. The balloon creates internal pressure which makes the helium more dense, so it might in reality be more like 1 gram can lift 5.

So to lift a person, say, 72 KG, you'd need a bit over 12 KG of helium, but that's a lot of helium. And, it may seem strange that 10 KG can lift 60 KG, but it's like holding a balloon under water. A balloon - which weighs very little in the air has significant lift in water because it's much lighter than the water surrounding it. If you could build a strong, very light container with a perfect vacuum inside, you'd in theory need very little mass to lift you off the ground. There's no theoretical limit to the lightness of an object that can lift you in the air, but it needs of a certain size.


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