14. starptautiskā konference 2012 - Latvijas Jūras akadēmija

Proceedings of 14th International conference „Maritime Transport and Infrastructure - 2012”where n – the amount of measurements; X i = t i and Y i = lny i are calculated using the resultsof measurements.Parameters a and k in the formula (1) are found, so that the sum of squared deviations is thesmallest. Demonstrated that this will be in the case if the partial derivatives are equal to zero:ns 2kXi ln Xi b YiXi 0;ki1n(6)s 2 kXiln Xib Yi1 0. bi1The system is transformed:nnnn 2k Xi b Xi XiYiXi1 i1 i1 i1 nn nk Xi b1 YilnXi.i1 i1i1iln X ;i(7)All sumes can be calculated using the results of experiment (Table 1). The two-equationsystem (8), that can be easily solves usin Gaussian elimination, is obtained.Table 1Results of the experiment and sumes for the method of least squarest[sek} = X y[V] Y=ln(y) X*X X*Y lnX XlnX0,01 3,6 1,2809 0,0001 0,0128 -4,6052 -0,04610,2 31 3,4340 0,04 0,6868 -1,6094 -0,32190,4 34 3,5264 0,16 1,4105 -0,9163 -0,36650,6 29 3,3673 0,36 2,0204 -0,5108 -0,30650,8 26 3,2581 0,64 2,6065 -0,2231 -0,17851,0 23 3,13550 1,00 3,1355 0,0000 0,00001,2 20 2,9957 1,44 3,5949 0,1823 0,21881,4 18 2,8904 1,96 4,0465 0,3365 0,47111,6 15 2,7080 2,56 4,3329 0,4700 0,75201,8 14 2,6391 3,24 4,7503 0,5878 1,05802,0 13 2,5649 4,00 5,1299 0,6931 1,38632,2 11 2,3979 4,84 5,2754 0,7885 1,73462,4 10 2,3026 5,76 5,5262 0,8755 2,10112,6 9,5 2,2513 6,76 5,8534 0,9555 2,48432,8 7 1,9459 7,84 5,4485 1,0296 2,88293,0 7 1,9459 9,00 5,8377 1,0986 3,29583,2 5,5 1,7047 10,24 5,4552 1,1632 3,72213,4 5,1 1,6292 11,56 5,53942 1,2238 4,16083,6 5,1 1,6292 12,96 5,8653 1,2809 4,61143,8 4 1,3863 14,44 5,2679 1,3350 5,07304,0 4 1,3863 16,00 5,5452 1,3863 5,5452Sumes42,01 294,8 50,3797 114,8001 87,3412 5,5417 38,278076