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These Prague street performers have pulled off one of the finest executions of the levitating man illusion you will ever see. For those of you who hate fun and wonder, modern science has revealed the secret and the physics behind this ancient trick.
Lost Art
The art of levitation has long fascinated travelers, as street performers claim to have conquered gravity.
The special wrap-around configuration, along with a solid base, provides the stability and strength required to pull it off. The three components include a solid base (3), the staff (2), and seat (3).
Afloat
After all, if they could really float, couldn’t they find a more lucrative profession than street performing?
The two-man version is slightly different, but it borrow from the same solution. Writing in the James Randi Educational Foundation, Kyle Hill explains:
To fit underneath the carpet, the plate must be rather small. And from the looks of the picture [below], the sitting man isn’t that elevated therefore the plate is not too thick. A thin plate can be heavy, so you may be thinking that the weight alone is balancing the whole apparatus, but a ~180 pound man on a steel rod can produce quite a bit of force. There is more physics to it than this, though the additional weight certainly helps. It could also be the case that the sitting man is helping to balance the levitating one, but considering that this trick can also be done with one person, we’ll assume that the sitter isn’t contributing to the stabilization. [image credit: Michael Shermer]
I’ll do some engineering calculations in a process called statics to sort this out. I’ll assume a 180 lb man whose center of gravity is directly above the bar connecting to the plate.
Now, the most efficient way to do this trick would be to sit directly above where the steel rod connects to the plate. This way, there is no torque in the plate-rod connection created by the levitator. This looks to be how the two-person levitation trick is set up.
The calculations are easy. To be in equilibrium, all the forces and moments (rotational forces) must cancel out. And what we are really concerned about, if we were the tricksters, is whether or not the apparatus will tip over. We can find this out by calculating how much rotational force is acting at the edge of the plate assuming nothing is moving. In the set-up above, the moment at point A is calculated by multiplying the downward force by the perpendicular distance (I guessed again here) to the point. This results in 180 ft-lbs that push the plate edge down into the ground (a clockwise force).
