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to the water which adheres to the hall, and is subsequently 
thrown off. 
Now whenever a drop leaves the ball there will be a 
tangential reaction in a direction opposite to that in which 
it leaves, and if many drops are leaving the ball at the same 
time, there will be a force equal to the sum of all their 
reactions opposing the rotation of the ball, and a force equal 
to the resultant of all their reactions acting on the centre of 
the ball. If the water be thrown off equally all round, this 
latter force will be zero ; but if more drops leave in one 
direction than in another, the resultant force will be opposite 
to this direction. As this force is essential to the equilibrium 
of the ball, the question arises, is there any reason why 
most of the water should leave the ball in one direction ? 
and if so, in which direction will that be ? Now the water 
comes on the ball at p, and as it passes over the top of the 
ball, the action of gravity or the weight of the water will 
be to keep it on the ball, but after it has passed the top, the 
conditions for its leaving become more and more favourable — 
so that it appears as though the water would begin to leave 
as soon as it had passed over the top of the ball and go on 
until it was all thrown off In this way most of the water 
would leave between the top of the ball and that side which 
is opposite the jet, which on examination I find to be 
the case. 
It was the discovery of this fact which has enabled me 
to explain the phenomenon, for this water causes a resultant 
reaction which is the additional force necessary to maintain 
the equilibrium of the ball. 
Let this resultant reaction be called Q, it will act upwards 
and towards the jet, and its effect will be, first, to force the ball 
into the jet and so will help to counteract the obliquity of P ; 
secondly, it will assist in supporting the ball ; and thirdly, 
since it opposes the rotation, it will balance the tangential 
force R caused by the friction at p, and provided it have 
