Originally Posted by toly
This is my old post from famous deleted MTM thread.
Letís assume a player rotates straight arm around shoulder joint with locked bend back wrist in horizontal plane and without arm pronation, see please picture below.
[refer to post #324 for picture]
The angle (ϕ) between long axis of the racquet and axis of the arm is 45į.
The arm rotation creates around impact linear racquet speed Vshoulder. Accordingly to the linear algebra, every vector of the racquet velocity Vshoulder can be decomposed as a sum of two orthogonal (perpendicular to each other) components (vectors). The normal component VshouldNorm is the perpendicular to the strings and the tangential component VshoulTang is the parallel to them.
The normal component (VshouldNorm) mostly defines ballís direction and ballís translational speed. Tangential component (VshoulTang) creates mostly clockwise sidespin.
If ϕ=0 there will be no sidespin at all.
Thus, the larger the angle ϕ, so there will be more sidespin. Apparently that sidespin has nothing to do with pulling across or whatever the name is, it is defined by position of the wrist relative to the arm/forearm.
On winner attempts, that are usually flatter, we have to decrease ϕ angle to minimize useless sidespin and thus increase translational ball speed.
The angle ϕ being non-zero is not the reason for sidespin. Just think, if what you are saying is true, there will be sidespin whenever contact is made with a laid back wrist - and this is definitely not so. On the other hand, even with ϕ=0, it is possible to have sidespin if the path of the racquet head at contact has a component that is inward or outward.
Also, note that the picture is a little misleading because it doesn't show the low to high component.