Physics demonstrations: Magdeburg hemispheres

Sometimes one can demonstrate very profound and remarkable physics with very simple, even mundane, tools.  Last week I received the tools to perform one such demonstration by mail:

This pair of iron hemispheres, with handles attached and a valve on one side, are a small scale model of one of the earliest and most dramatic displays of the power of atmospheric pressure.  They are now known as the Magdeburg hemispheres, and they still work as a great demo to this day.

Otto von Guericke

Curiously, the hemispheres are not named for a person, but a place: Magdeburg, Germany.  The spheres were designed by the German scientist (and mayor of Magdeburg) Otto von Guericke (1602-1686) in 1654 to show off his newly-invented vacuum pump and the concept of atmospheric pressure.

The premise is simple: with the hemispheres pressed together, air is pumped out of the interior, creating at least a partial vacuum.  This seals together the hemispheres with a remarkable force.

Guericke first demonstrated this force in 1654 for the Emperor Ferdinand III.  Thirty horses, in two teams of 15, were unable to pull apart the evacuated hemispheres!  He performed a smaller scale performance in 1656 in his hometown of Magdeburg, using two teams of eight horses.  This latter performance was immortalized in a famous sketch by the scientist Gaspar Schott in a 1657 book, Mechanicahydraulica-pneumatica.

So how do the spheres stay together?  Simply put: atmospheric pressure!  All objects within the atmosphere are under constant bombardment from air molecules traveling every which way; this atmospheric pressure is not noticeable to us because our bodies have an internal pressure that matches and balances it.  (Though you won’t explode if exposed to a vacuum, movie tropes notwithstanding.)

When the hemispheres are first placed together, the air pressure within them balances the air pressure outside, and they are easily pulled apart.  When air is removed from the interior of the hemispheres, however, there is no longer any force pushing outward: the atmospheric pressure outside dominates, pushing the hemispheres together and keeping them from being separated.

An attentive reader may have already noticed a similarity between the Magdeburg hemispheres and a very common, everyday object: a suction cup!

A suction cup gets its sticking force by the same action as the Magdeburg hemispheres: when one pushes the cup against a smooth wall, one forces the air out of the cup, allowing atmospheric pressure to hold it to the wall.  We could say that the Magdeburg hemispheres are an early primitive form of suction cup (though suction cups apparently have a history that goes back much further)!

A good set of Magdeburg hemispheres, however, can provide suction forces significantly stronger, as Guericke’s original horsepower demonstration shows.  In mathematical terms, the force holding together a pair of hemispheres of radius a can be shown to be

F = \pi a^2 \Delta P,

where \Delta P is the pressure difference between the interior and exterior of the hemispheres.  My pair of hemispheres have a radius of 5 cm; assuming conservatively that my hand vacuum pump can reduce the interior pressure to half of atmospheric pressure (atmospheric pressure being 1\times 10^5 \mbox{ N}/\mbox{m}^2), I find that the hemispheres will stick together with a force of 392 Newtons, or 88 lbs!  Guericke’s hemispheres had a radius of about 25 cm; if he had almost completely evacuated them of air, they would provide a force of about 4400 lbs!

For those not willing to spend some $60 for a metal set of hemispheres, it is possible to perform a simple version of Guericke’s demonstration at home with readily available supplies.  You will need:

  • two identical glasses
  • a candle
  • a damp piece of paper that covers the opening of a glass

Put the candle in one of the glasses, and light it.  Place the other glass on the top of the first, with the wet piece of paper between them; this paper will help form an airtight seal between the glasses.  The candle will slowly consume all of the available oxygen in the glasses, and be extinguished.  If you have done things correctly, you will find that the glasses are strongly stuck together due to the drop of atmospheric pressure inside!  If you have set up a really good seal, it might be impossible to pull the glasses apart by hand — they can be twisted apart in this case.

Obviously, the seal between the two glasses is key: I’ve only been able to get a weak seal on my preliminary home experiments; a heavier piece of paper should work better than a lighter one.

How does the home experiment work?  It looks like several factors can play a role in reducing the pressure within the glasses.  For one, a candle burns by combining the oxygen in the air with hydrogen present in the candle to produce water vapor and carbon dioxide.  Less CO2 gas is produced than the amount of O2 consumed (thanks to the creation of water, as well); when the water vapor condenses, this results in a pressure drop inside the glasses, serving as a crude form of vacuum pump. Another aspect: the candle heats up the air within the glass, causing it to expand.  When the candle goes out, the temperature drops, causing the pressure to drop within the glass as well.  The more I look at this simple setup, the less simple it seems to be!

The Magdeburg experiment is a delightful glimpse of the hidden and powerful atmospheric forces that are with us at all times.  It is just as fascinating today as it was some 350 years ago!

Update: revised my calculations for the force of attraction of the hemispheres!  Also, related to comments below, I revised my explanation of the “candle effect”; it looks like it is another experiment in which a number of factors can play a role!

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19 Responses to Physics demonstrations: Magdeburg hemispheres

  1. The only correction I would make to your description is about how the candle creates the vacuum. I doesn’t do it by consuming the oxygen, at least not directly, since as it is consuming the oxygen, it is producing carbon dioxide. Instead, as it is burning (consuming oxygen) it is heating the air in the glasses, causing it to expand and be forced out between the paper and glasses. When it extinguishes, the remaining air cools and contracts, pulling the glasses tightly together, causing the drop in pressure. A simple experiment will show that immediately after the candle goes out, their isn’t yet a vacuum. It will form over the next few seconds, minutes.

    • J Thomas says:

      You’re right. If hot air is forced out of the container and then an airtight seal keeps cooler air from getting in, then you’ll get reduced air pressure.

      But isn’t it also true that a mole of CO2 takes less space than a mole of oxygen at the same temperature?

      Because each molecule is heavier. Just like a mole of helium gives less lift than a mole of hydrogen.

      • J Thomas says:

        oops, you’re right.

        The density at sea-level and 0 °C for air and each of the gases is:

        Air (ρair) = 1.292 kg/m3.
        Hydrogen (ρH2) = 0.090 kg/m3
        Helium (ρHe) = 0.178 kg/m3

        Helium is twice as heavy so it has twice the weight per volume.

        So for methane, half the O2 would turn into water, while for very long chain paraffins 1/3 the O2 would go to water. If you burned, say, magnesium, you could remove all the O2 from the air and not replace it with another gas.

        I wonder if there’s a way to remove nitrogen? The obvious slow approach would be to add a solution containing nitrogen-fixing bacteria and everything they need to grow. They could convert the nitrogen to ammonia and then nitrites, but in the process they would produce their own waste products which might evolve gases. A separate container of potassium hydroxide could absorb CO2 and also water. But it’s all complicated and slow, for the potential payoff of removing part of the 80% of nitrogen in air.

        Probably better to start out with pure oxygen as the gas, and burn it with something slower than magnesium so you don’t make too much heat at once.

      • For an ideal gas, it turns out that the pressure isn’t related to the mass of the particles at all, only their number! PV = NkT However, as described in the link added above, the combustion of a candle produces both CO2 and H2O, and half as many CO2 molecules are produced as O2 molecules originally existed.

        2 O2 + C H 4 = C O 2 + 2 H 2 O

        Less gas molecules = less pressure. However, it looks like there are a number of factors that can come into play in this sort of experiment. The chemistry can play a role, as I originally attributed everything to, but the physics probably plays a comparable role. Without having a lot of time to fiddle with the experiment, I’ll just leave my explanation a little open! In the past few months, I’ve come to realize that many “simple” experiments have explanations that are not necessarily so simple.

    • Hmm… are you sure about that? I’d love to see some references to the explanation, but most of them I find are particularly vague. This website discusses a similar experiment and argues that it really is due to the imbalance of less CO2 being produced than O2 consumed. I think I’ll revise my post and take a more “balanced” explanation, as this related site gives.

  2. Kees says:

    Maybe the original statement in the blog about the oxygen being removed isn’t wrong. Candles are usually made from paraffin, (CH2)n

    First this happens:
    CH2(s) + 11/2 O2(g) –> CO2(g) + H2O(g)

    When the candle is still burning, the water won’t condense, so the reaction produces more moles of gas than it consumes. When the candle is extinguished, water may condense, and a vacuum is created.

    • I don’t think the explanation was wrong, as much as a bit incomplete! You’re right that, in the end, there should be fewer molecules of gas floating around than there were in the beginning. In fact, the extra site I linked to argues that the two effects — reduction of gas molecules and air heating — cancel each other out while the candle burns to make it look like there’s no effect. It’s hard to say for certain what is going on without more detailed studies, though!

  3. ColonelFazackerley says:

    I can recommend a pressure demonstration.

    Apply sun screen at the bottom of a mountain. Ascend 1000m. Shake bottle upside-down so that there is cream over the bottle’s opening. Hold the bottle pointing up, tilted at 45 degrees, and open. I achieved a cream-glob range of around 2m

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  5. katarina says:

    thanks for the tips and info needed for my homework badly!!

  6. Pingback: Vacuum hemispheres | Cartographix

  7. Joss Thompson says:

    Hi Skullsinthestars,

    I have a quick question; presumably the larger the spheres (assuming a perfect vacuum created inside) the greater the sticking force?

    Is it possible to simply double the original calculation, based on the increased radius of the spheres (again, I am assuming a near-perfect vacuum could be achieved inside the spheres) – i.e. 50cm radius provides 8,800lbs (40,000 N?) of resistance? I’m also assuming this test would carried out as close as possible to sea-level, thus ensuring the greatest achievable atmospheric pressure.

    Would be fascinated to know!


    • J Thomas says:

      Joss, when you try to pull the spheres apart, you are pulling against air pressure pushing them together. What you are pulling against is the surface area where the air is pushing against you, when the air pushes sideways it doesn’t matter.

      So that fits the formula. The force should be proportional to the radius squared. You ought to get the same effect with two shallower halves, what matters is the cross section, not the depth.

      Any tiny irregularity in the rim might let air in and destroy the vacuum. The steel is kind of elastic and it will stretch some to reduce that, but the more force you pull with the more the steel is already stretched and so it might let air in sooner than you’d expect.

      If your steel hemispheres need to withstand 4 times as much pressure they may need to be more than twice as thick. Likely steel fractures in complicated ways and it probably isn’t obvious how thick it would need to be. It doesn’t have to be spherical to get a lot of force, but a sphere is the obvious shape to make it to balance the forces when you aren’t trying to pull it apart.

  8. Christelle Merrie Estrada says:

    Where can I buy the metal set of hemispheres?

  9. Pingback: Vacío – clickonphysics

  10. Akshay says:

    What will happen if we will take a cylinder instead of sphere of the height same as the diameter of sphere and radius same as radius of sphere?

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