Madeburg’s Hemispheres from 1657 Revisited

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In 1657, von Guericke demonstrated atmospheric pressure's strength using hemispheres that couldn't be pulled apart due to a vacuum. The discussion explores a modern variation using a horizontally mounted cylinder with two pistons, separated by a 1-inch cavity under a 95% vacuum. To determine the force required to pull the pistons apart, the area of the pistons is crucial, as the force exerted is calculated by multiplying pressure by area. The confusion arises regarding the failure point of the pneumatic seals, which depend on the pressure difference across the seals rather than the absolute pressure. Ultimately, pulling the pistons apart would result in air rushing in, rather than a failure of the seals at lower pressures.
Roger900
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Hello Specialists in Physics!
Way back in 1657, von Guericke in Madeburg Germany demonstrated the tremendous force of atmospheric pressure by placing two large hemispheres together, evacuating the air from inside, and hitching a team of horses to each hemisphere. The horses were unable to pull the hemispheres apart. In fact, it took four teams straining on each side to break the vacuum seal.
What if von Guericke's experiment was performed today, but instead the set-up was a horizontally mounted cylinder with 2 pistons in the cylinder. Assume the pistons have pneumatic seals that provide zero vacuum leakage, but the seals do have a maximum PSI point before failure. Assume the pistons are separated 1” apart, and in this 1” cavity a 95% vacuum was drawn, and then the cavity is sealed. How much force would be required to “pull” the pistons further apart? Can they even be pulled apart? At what force would the pneumatic seals fail?
Thanks,
Roger
 
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I think the area of the pistons must be given. Only then, you can calculate the force being exerted due to the air. (ie, F=P*A)
 
Let's assume (1) the surface area of each piston is one square inch, (2) the piston seals have a 100 PSI rating, and (3) a 95% vacuum exists in the cavity which is equal to (0.95 * 14.69 PSI) = 13.95 PSI.

Here is where I get confused:

If each piston is receiving a vacuum of -13.95 PSI, but the seals will not fail until 100 PSI, would it not require a total force of 100 pounds to "pull" the pistons apart by making the seals fail?
 
No. The seals fail when there is a 100 psi difference from one side of the piston to the other. Run them at anything less than 3100 psi in a 3000 psi environment, and they're fine.

All you get when you pull a 1 sq. in. piston from a sealed cylinder when in a normal atmosphere environment is a "pop" as the air moves into the evacuated volume of the cylinder.
 
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