Interpolating a three point curve at any angle using cubic splines

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MC
MC 2017-10-17
评论： MC 2017-10-17
I'm trying to interpolate a curve using cubic splines and three points in the x-y plane. I have some troubles finding the equation for the middle point such that the normal vectors in point P0 is always perpendicular to the x-axis, given any angle between the yellow line and the x-axis. See figure
P0 is always known, P1 is given by:
P_1 = [r1*cos(beta); r1*sin(beta)], where r1 is the the length of the yellow line and beta is the angle between the yellow line and the x-axis. In the image above I have experimentally found the equations for P2, which is:
P2 = [0.75*P1_x; 0.25*P1_y]. But this only works if beta is 45 degrees. I've tried to reverse engineer it but I failed miserably. How would I go about finding the equations for P2?
Best regards MC
  2 个评论
Jan
Jan 2017-10-17
Start with defining the problem exactly: What are the given inputs? What are the conditions for the output? How does the wanted output look like?
MC
MC 2017-10-17
编辑：MC 2017-10-17
Sorry I'll try and make it more clear.
Given inputs: Beta and P0. Conditions for the output: Define P2 such that the normal vectors in P0 are perpendicular to the x-axis, given any input angle Beta. Wanted output: Well, I want it to fulfill the conditions above.

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回答（1 个）

John D'Errico
John D'Errico 2017-10-17
编辑：John D'Errico 2017-10-17
Simple. Work in polar coordinates, centered around the location (0,0.4).
Now you will fit a curve for radius (thus distance from the point (0,0,4)), as a function of polar angle theta.
I don't have your points, so I cannot show you how to solve the problem better than that. As I said, simple, even trivial.
  1 个评论
MC
MC 2017-10-17
I am not sure I follow you. So if we give P2 in polar coordinates, centered at (0,0.4) then generally it is given by:
P2 = [0;P1(2)] + [r2*cos(theta);r2*sin(theta)], where from the figure P1(2)=0.4.
I don't understand entirely what you mean by fit a curve for radius, as a function of theta.
The points are: P0 = [0;0] and P_1 = [r1*cos(beta); r1*sin(beta)], where r1 = sqrt((2*w)^2+(2*w)^2), where w = 0.2. And Beta is any angle, in the picture above it is 45 degrees.

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