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Apologies for the super basic question, but we all have to start somewhere right?

Can somebody please explain exactly how you would calculate the number of helium balloons it would take to lift an object of mass $m$ here on earth, the variables I would need to take into account and any other physics that come into play.

I think I can roughly calculate it using the method below but would love somebody to explain how this is right/wrong or anything I have negated to include. This model is so simple, I am thinking it can't possibly be correct.

  • 1 litre of helium lifts roughly 0.001kg (I think?)

  • Assumption: an inflated balloon is uniform and has a radius $r$ of 0.1m

  • $\frac{4}{3}\pi r^3 = 4.189$ cubic metres $\approx$ 4 litres capacity per balloon

  • Lets say $m = 1$kg, therefore $\frac{m\div0.001}{4} = 250$ balloons to lift that object?

As you can tell, I haven't touched Physics since high school and would really appreciate any help. It seems like an easy question, but actually is probably more complex than I thought.

Thanks a lot.

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  • $\begingroup$ $\frac{4}{3}\pi*0.1^3=0.004189 \text{m}^3$, Although this is 4L ($1 \text{m}^3 =1000 \text{L}$) $\endgroup$
    – nivag
    Commented Oct 23, 2014 at 10:13
  • $\begingroup$ I must have changed units accidentally, thanks for pointing that out! $\endgroup$
    – Ed Holloway-George
    Commented Oct 23, 2014 at 10:56
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    $\begingroup$ You're not planning on moving a house to Paradise Falls, are you? $\endgroup$
    – LarsTech
    Commented Oct 23, 2014 at 18:24
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    $\begingroup$ @LarsTech: Not until I get the mathematics right no $\endgroup$
    – Ed Holloway-George
    Commented Oct 23, 2014 at 22:48

3 Answers 3

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You need to take density of the air into the question. And weight of the balloon itself.

The Archimedes' law says:

Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.

Therefore a balloon can support the same weight as equal volume of air has (reducing gravitational acceleration from the equation as it appears on both sides). Which includes weight of the baloon itself. And note that weight of the rubber will probably be considerably higher than weight of the helium inside.

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  • $\begingroup$ Thanks for your answer, the quote of Archimedes law is especially helpful. $\endgroup$
    – Ed Holloway-George
    Commented Oct 23, 2014 at 9:23
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The net upward force is, according to Wiki buoancy:

$$F_\mathrm{net}=\rho_\mathrm{air}V_\mathrm{disp}g-m_\mathrm{balloon} \cdot g$$ For helium, the $m_\mathrm{balloon}=\rho_\mathrm{helium}V_\mathrm{disp} + m_{shell} $, thus $$F_\mathrm{net}=\rho_\mathrm{air}V_\mathrm{disp}g-\left(\rho_\mathrm{helium}V_\mathrm{disp} + m_{shell} \right)\cdot g=\left(\rho_{air}-\rho_\mathrm{helium}\right)V_\mathrm{disp} \cdot g - m_{shell} \cdot g$$ With $V_\mathrm{disp}=N_\mathrm{balloon} V_\mathrm{balloon}$ and $F_\mathrm{net}=m_\mathrm{load} \cdot g$, you're able to calculate the number of balloons necessary.

EDIT: Some more steps how to actually solve the problem.

To isolate the value of $N_\mathrm{balloon}$, we plug in the volume expression to obtain:

$$F_\mathrm{net}=\left(\rho_{air}-\rho_\mathrm{helium}\right)N_\mathrm{balloon} V_\mathrm{balloon} \cdot g - m_\mathrm{shell} \cdot g$$

We can then isolate the value of $N_\mathrm{balloon} $ by adding $m_\mathrm{shell} \cdot g$ on both sides: $$F_\mathrm{net} + m_\mathrm{shell} \cdot g=\left(\rho_{air}-\rho_\mathrm{helium}\right)N_\mathrm{balloon} V_\mathrm{balloon} \cdot g $$

And then divide both sides by $\left(\rho_{air}-\rho_\mathrm{helium}\right)V_\mathrm{balloon} \cdot g$ to get:

$$\frac{F_\mathrm{net} + m_\mathrm{shell}\cdot g}{\left(\rho_{air}-\rho_\mathrm{helium}\right)V_\mathrm{balloon} \cdot g}=N_\mathrm{balloon} $$

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  • $\begingroup$ Thank you, this makes a lot more sense. Just to clarify, is the $N_{balloon}$ in the last equation the Number of balloons? $\endgroup$
    – Ed Holloway-George
    Commented Oct 23, 2014 at 9:20
  • $\begingroup$ Yes, that's exactly what it is. The total net force is simply the forces of all the volumes (balloons) added. $\endgroup$
    – ROIMaison
    Commented Oct 23, 2014 at 10:11
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    $\begingroup$ Shouldn't $F_{net}=m*g$ were $m$ is the mass of the object we want to lift? $\endgroup$
    – Michal
    Commented Oct 23, 2014 at 15:48
  • $\begingroup$ You're neglecting the mass of the balloon itself (your $m_\mathrm{balloon}$ appears to be the mass of the helium within it). That seems like it might be significant. $\endgroup$
    – David Richerby
    Commented Oct 23, 2014 at 16:51
  • $\begingroup$ I presume you mean the rubber/material of the balloon? @DavidRicherby - is it as easy as just adding it to the $m_{balloon}$ definition above? Or are there additional changes that would need to be made? $\endgroup$
    – Ed Holloway-George
    Commented Oct 23, 2014 at 22:54
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Your approach is pretty much correct.

First you calculate the buoyancy generated by 1L of Helium. This can be calculated by the difference in the mass of 1L of Helium and 1L of air, which you can look up on the internet. The answer is ~0.001kg, as you got.

Next you estimate the volume of a balloon. A sensible assumption, followed by some questionable maths, but you get to the right answer (1$\text{m}^3$ = 1000L).

I'll break the next step into two for more clarity. First you calculate the volume of helium required to lift the object. This is given by $\frac{\text{mass of car}}{\text{Buoyancy of He}}$. Then you calculate how many balloons this volume of helium requires.

The only other thing you could consider the mass of the balloons. However, I expect this will be negligible unless you need a really accurate calculation.

Another point to consider is that the volume is proportional to $r^3$. It is, therefore, quite sensitive to changes in the radius. For example changing the radius to 15cm gives a volume of 14L compared to 4L for 10cm. Therefore it is important to consider the accuracy of your assumption as it has a significant effect on the result. You could try filling a balloon with water to get a better estimate of the volume.

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