how to calculate this equation?
4 ビュー (過去 30 日間)
古いコメントを表示
sebastiano della gatta
2022 年 1 月 4 日
編集済み: Walter Roberson
2022 年 1 月 5 日
i have:
P3=[l1 - (l3y*(2*c3 - 2*c1*c2))/c_delta + (l3z*(2*c2 + 2*c1*c3))/c_delta + (l3x*(c1^2 - c2^2 - c3^3 + 1))/c_delta;
(l3x*(2*c2 + 2*c1*c2))/c_delta (l3y*(c1^2 - c2^2 + c3^2 - 1))/c_delta - (l3z*(2*c1 - 2*c2*c3))/c_delta;
(l3y*(2*c2 + 2*c2*c3))/c_delta - (l3x*(2*c2 - 2*c1*c3))/c_delta - (l3z*(c1^2 + c2^2 - c3^2 - 1))/c_delta]
P4=[p4x+0.9972s;p4y-0.0712s;p4z-0.0216s];
l4=P4-P3;
i have to do this product:
(P4-P3)⋅(P4-P3) - l4^2=0.
when I calculate this matlab equation it gives me back the conjugate complexes that I don't want, why?
6 件のコメント
Star Strider
2022 年 1 月 5 日
Put each line in parentheses (the trailing semicolons are outside the parentheses). That generally solves the problem that spaces create, and preserves the readability of the code.
採用された回答
laurent jalabert
2022 年 1 月 5 日
Hello, maybe I did not understood your problem cause l4=P4-P3; and you want to solve:
(P4-P3)⋅(P4-P3) - l4^2=0, which means (P4-P3)⋅(P4-P3) - (P4-P3)^2 = 0
Do you mean (P4-P3).*(P4-P3) - (P4-P3).^2 = 0 ? which is 0=0
その他の回答 (0 件)
参考
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!