By N F Barber; G Ghey

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8 . Transverse velocity can cause a change o f ' flow' . In this iCaS() the element has zero flow a t first. But a flow soon develops ; the reason is that the element changes its attitude as the particles move. At one end the velocity may be v and at the other end v + ov. Then in unit time one end will move out more than the other by a distance O V x 1 second. The direction of the element will have changed by a small angle that I can write as (ov x 1 second! OZ) in radian measure. To find the component of velocity along the element in this new position, multiply this angle by the average velocity v + iov.

3 z I tried to sketch this part at a reduced scale. I would need to make the slope smaller in proportion. If at z = 1 , I chose the reduced scale so that the ordinate was as high as it was originally at z = 0, then the slope would need to be just the same as it was at z = O. On this reduced scale the curve between z = 1 and z = 2 would exactly duplicate the previous part. So when I use the same scale throughout I see that the ordinate increases by the same factor in the two steps. At z = 0 the ordinate is 1 , at z = 1 the ordinate is V, so at z = 2 the ordinate must be greater again by a factor V, that is to say V 2, and by going to z = 3 , I would produce another factor of V and the ordinate becomes V3.

I will try to show this. I must first see what the waves of our formula look like. The profile of the waves First we can deduce the velocities. These are : horizontal : u = - d4>/dx = - c + kC exp (kz) sin kx, vertical : w = - d4>/dz = - kC exp (kz) cos kx. These show the velocities at any position x and z. Now I want to find where the particles go to. 45 You can see that if there were no waves at all, which would be the situation if I took the number C to be zero, then the horizontal velocity would be the same everywhere and equal to - c while the vertical velocity would be zero.