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Stonehenge

Stone Shifting 2

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Hi everyone and welcome to stone shifting 2. It's true, I do have two brains! One is under my desk and runs at 2.4GHz, the other is in my skull and works at a tiny fraction of that speed, but is reasonably adept at telling the first one what to do.

Anyway, just to prove that I am fallable, I think I got the explanation wrong for the side forces on the Tower. Here is the revised version: The frictional force (which acts along the surface of the block) and the reaction force of the tower (which acts perpendicularly to the block) resolve to a vertical component that exactly balances with the weight of the block, so there is no side load. However, when the block begins to slip, the frictional force is no longer sufficient to balance the reaction force and the weight. This results in a net side load being imposed on the tower.

To verify this theory, I placed a long wooden lath on top of a tall, narrow support block so that it would just overbalance. As the lath rotated the support block remained upright until the point of slipping was reached. It then imediately fell over.

I'm happy with this new explanation for two reasons. Firstly, the side forces should be a lot less than I had feared and secondly, I couldn't get the maths to work properly before, but now it all falls into place quite nicely.

Good, so slippage is our only enemy (A.) because it elongates the arc (rather unpredictably) and (B.) because it strains the tower.
Perfect pivoting produces perfect positioning.

And the thing is, contrary to my previous worrying about the roughness of the stone, it's only one small spot on the stone that should affect the motion, i.e. the point in contact with the edge of the tower.

I think the Beeb man talked about making a notch, but rejected it on the grounds the ancients didn't.

So our only way to minimize slippage is to maximize turning speed? (subject to depositing the stone vertical of course)?