linear vibration analysis using screw theory - helix - Georgia Institute ...

34where MV x1 , MV x2 , MV x3 are the three roots of (3.24). Next, apply the same procedure to (3.13)which gives the coe¢cient of MV y as,b 1 = MV y1 MV y2 + MV y1 MV y3 + MV y2 MV y3 (3.26)The sum of those two coe¢cients represent the sum of all three dot products between the vectorsfrom M to all three V ,a 1 + b 1 = X i;j³ ¡¡! MVi ¢ ¡¡! MV j´i; j = (1; 2); (1; 3); (2; 3) (3.27)Resolving a 1 and b 1 from (3.11) and (3.13) yields a negative number,Xi;j³ ¡¡! MVi ¢ ¡¡! MV j´= ¡3m °m < 0 (3.28)Finally, consider the conditions (3.23) that put M at the orthocenter of the triangle formed byV 1 , V 2 and V 3 . The conditions can be written as,¡¡!MV 1 ¢ ¡¡! MV 2 = ¡¡! MV 1 ¢ ¡¡! MV 3¡¡!MV 2 ¢ ¡¡! MV 3 = ¡¡! MV 2 ¢ ¡¡! MV 1 (3.29)which means that all three dot products between two ¡¡! MV i are either all positive or all negative.Obviously, if the dot products are positive then their sum is also positive which would violate thecondition of (3.28), hence the dot products are all negative. If the dot product between two vectorsis negative then the angle between the two vectors has to be greater than 90 ± . It is then impossibleto place two vibration centers in the same shaded quadrant of Figure 3.2 and hence one can statethe following theorem.Theorem 13 When there are three vibration centers, there can only be one per shaded quadrant ofFigure 3.2.