Area comparison



  • The challenge is to remove the equation (local points) of the circumference between the centre and the circumference plane. I cleaned the centre's location and took it (0.0.0) for simplicity (this would just add to the coordinates of my center).

    What's the problem? Area comparison x2+y2+z2=R2 Equation of Ax+By+Cz=0 where N(A,B,C) is the normal vector.

    My equation is the system of the last two equations.

    Then I have a shuttle... I need to find a series of circumferential points clockwise through step α=360°/N.

    I understand that this equation must not just be resolved, but also lead to spherical coordinates. But how? Mathematician's fine, but he's definitely drowning.

    UPD: There's still a solution, knowing (X.Y) the right circumference points on the plane, getting (X,Y,Z) by the "return" of the N(A,B,C) interdependency, but how to deal with this. ♪



  • You've gone quite complicated. It's easier.

    First, we need to build an orthodox coordinate system. You need to find any vector perpendicular to the vector. It's the most unspeakable part of the decision. And in this place, you don't have to be afraid to put the contingent operators - according to the "gea comb" from the topology, you don't have to do it without them.

    The simplest way to turn the vector into a straight corner is to multiply it into something. But if the multiplied vectors are close to the direction, the accuracy will be affected (and if they are collinears, zero).

    Therefore, I suggest that three base vectors be considered.ij and kand multiply on the one who's the most unlike. For that. We'll find the coordinate with the smallest mole.♪ May, for certainty, .A. À åå .B. and .C.. So we multiply the vector. i and receive the first of the vectors of the local coordinate system:

    i1 = n x i = (Ai + Bj + Ck) x i = -Bk + Cj

    Now we'll re-model him to the vector and get the second vector of the base:

    j1 = n x i1 = (Ai + Bj + Ck) x (-Bk + Cj) = (please open the brackets yourself)

    Now, the base vectors have to be normalized so that they become single, and the parametric equation of the circumference can be used:

    r = r0 + R cos φ i1 /i1 + R sin φ j1 /j1




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