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This question is related to a previous question (link to question) asked. The image below describes two situations (Cases):

enter image description here

Notations:

$y_b(y_g)$ = y-axis of the green plane and the blue plane.

$e_b$ = intersection of major-minor axis of the ellipse on the blue plane from the cone with vertex at $P$

$e$ = intersection of major-minor axis of the ellipse on the blue plane from the cone with vertex at $O''$

$(x^r_O,y^r_O,0)$ = Superscript is the color of the plane, subscript is point.

$O_b$ = Intersection point of the line $PO_b$ on the blue plane.

First Case with the axis of a cone lying along the $OO''$ line (z-axis of the coordinate system), forming a right circular cone with the reference plane $(x^r_O,y^r_O,0)$ (ideal unrotated red plane) and the vertex of the cone at point $O''$. This situation forms an ellipse in the rotated blue-plane (rotated by $\alpha$ about the $x^r_O$ axis of the ideal plane and by $\beta$ about the $y^g_O$ axis of the first rotated plane) with the intersection point of the major and minor axis being $e$. Now to relate the ellipse formed on the rotated plane, I use the ellipse equation $$\frac{(x-x_o)^2}{a^2}+\frac{(y-y_o)^2}{b^2}=1$$

and plug in the rotation of the plane to get:

$$\frac{(xcos\beta -x_o)^2}{a^2}+\frac{(xsin\alpha \cdot sin\beta+ycos\alpha -y_o)^2}{b^2}=1 \tag1$$

(here I guess, I have applied a translation by $x_o$ and $y_o$ first then a rotation)

I then equate another equation of an ellipse, given as

$$\frac{(xcos\theta - ysin\theta-x_o)^2}{a^2}+\frac{(xsin\theta+ycos\theta -y_o)^2}{b^2}=1 \tag2$$

(which includes the ellipse parameters i.e. major and minor axis), in the 2D blue plane (that is translated by $x_o$ and $y_o$ then rotated by $\theta$) with equation $(1)$. On equating (1) and (2) and solving by substituting $x=0$ and taking the $y$ sample point from the ellipse on the blue plane I get $\alpha$ and then further $\beta$. These equations work for the first case when the cone axis is orthogonal to the circle $CC'$ and coincident with the line $OO'$ or specifically when the cone axis contains the point $O$ (rotation center).

Second Case is when the cone vertex is at the point $P$, which is contained in the plane $O''$, here the cone becomes an oblique cone and the green line $PO_b$ is no more the axis of the cone. This is because the set constraint is that the circumference of the circle $CC'$ must contain (intersect) the surface of the cone and circle is fixed (stationary). This condition leads to the bigger ellipse visible in the image. In this situation the above equations do not work since (I guess), the center of rotation is no more lying on the cone axis.

I have applied the transformation and rotations but this does not work with the second case, what am I doing wrong here?. I think I have done something wrong in order of rotation and translation or is it something else?. What could be a possible solution?.

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I suspect there are a number of errors in the equations, but it's unclear because the entire idea behind all these transformations of coordinates is unclear.

Yes, the ellipse projected onto an oblique plane can be rotated and translated in 3-D space back onto a plane parallel to the circle. But what's the purpose of that? Or you could rotate and translate some ellipse in the parallel plane onto the projected ellipse, but how do you construct the correct ellipse in the parallel plane to begin with?

It seems to me a much simpler approach is to write out the equations of the cone and the plane in three dimensions ($x,$ $y,$ and $z$ coordinates) and solve the equations. If the circle is parallel to the $x,y$ plane of the first set of coordinates, it may be easiest to write the equations first in that system and then transform the coordinates (all three coordinates, not just $x$ and $y$) before solving the equations.


Here's an attempt via the second approach. I make a few assumptions based on an interpretation of the diagrams of the cone and the intersecting plane. Assume the coordinates of the point $O$ in all three coordinate systems are $(x^r_O,y^r_O,z^r_O) = (x^g_O,y^g_O,z^g_O) = (x^b_O,y^b_O,z^b_O) = (0,0,0).$ Assume the coordinates of $O'$ in the "red" system are $(x^r_{O'},y^r_{O'},z^r_{O'}) = (0,0,z^r_{O'}).$ Assume the circle $CC'$ has radius $R$ and is parallel to the red plane, so the "red" coordinates of points on that circle satisfy the simultaneous equations \begin{align} (x^r)^2 + (y^r)^2 &= R^2,\\ z^r &= z^r_{O'}. \end{align}

Now to find an equation of the cone whose vertex is at $P$ and whose sides pass through the circle $CC',$ consider an arbitrary cross-section of the cone parallel to the red plane. The cross-section is a circle with center on the line $PO'$ and radius proportional to the distance from the parallel plane through $P.$ In particular, the center of the cross-section has "red" coordinates $(x^r,y^r,z^r) = \left(h(z^r - z^r_{O'}), k(z^r - z^r_{O'}), z^r \right)$ where $$h = \frac{x^r_P}{z^r_P - z^r_{O'}} \quad\text{and}\quad k = \frac{y^r_P}{z^r_P - z^r_{O'}},$$ and the radius is $\left\lvert\dfrac{z^r_P - z^r}{z^r_P - z^r_{O'}}\right\rvert R.$ The equation of the cone in "red" coordinates is therefore $$ \left(x^r - h(z^r - z^r_{O'})\right)^2 + \left(y^r - k(z^r - z^r_{O'})\right)^2 = \left(\frac{z^r_P - z^r}{z^r_P - z^r_{O'}} R\right)^2. \tag1 $$

Now to find the equation in "blue" coordinates, we need to work out the conversion of coordinates. The point with "blue" coordinates $(x^b,y^b,z^b)_b$ has "green" coordinates $$(x^g,y^g,z^g)_g = (x^b\cos\beta + z^b\sin\beta, y^b, -x^b\sin\beta + z^b\cos\beta)_g.$$ (This assumes that a small positive rotation angle $\beta$ would bring the positive $z$ axis of the blue plane closer to the positive $x$ axis of the green plane; if the positive direction of rotation is in the other direction, just reverse the sign of $\sin\beta$ in the formula.) The point with "green" coordinates $(x^g,y^g,z^g)_g$ has "red" coordinates $$(x^r,y^r,z^r)_r = (x^g, y^g\cos\alpha - z^g\sin\alpha, y^g\sin\alpha + z^g\cos\alpha)_r$$ (assuming the positive direction of rotation takes the positive $y$ axis toward the positive $z$ axis; if it goes the other way, reverse the sign of $\sin\alpha$).

Now suppose a point on the cone has "blue" coordinates $(x^b,y^b,z^b)_b.$ The "red" coordinates of that point, $(x^r,y^r,z^r)_r,$ have the formulas \begin{align} x^r &= x^g = x^b\cos\beta + z^b\sin\beta,\\[6pt] y^r &= y^g\cos\alpha - z^g\sin\alpha \\ &= y^b\cos\alpha - (-x^b\sin\beta + z^b\cos\beta)\sin\alpha \\ &= x^b\sin\beta\sin\alpha + y^b\cos\alpha - z^b\cos\beta\sin\alpha,\\[6pt] z^r &= y^g\sin\alpha + z^g\cos\alpha \\ &= y^b\sin\alpha + (-x^b\sin\beta + z^b\cos\beta)\cos\alpha \\ &= - x^b\sin\beta\cos\alpha + y^b\sin\alpha + z^b\cos\beta\cos\alpha. \end{align}

That is, the "red" coordinates of the point with "blue" coordinates $(x^b,y^b,z^b)_b$ are \begin{align} x^r &= a_{11}x^b + a_{13}z^b, \tag2\\ y^r &= a_{21}x^b + a_{22}y^b + a_{23}z^b, \tag3\\ z^r &= a_{31}x^b + a_{32}y^b + a_{33}z^b \tag4 \end{align} where \begin{align} a_{11} &= \cos\beta, & & & a_{13} &= \sin\beta, \\ a_{21} &= \sin\beta\sin\alpha, & a_{22} &= \cos\alpha, & a_{23} &= -\cos\beta\sin\alpha,\\ a_{31} &= -\sin\beta\cos\alpha, & a_{32} &= \sin\alpha, & a_{33} &= \cos\beta\cos\alpha. \end{align}

If $(x^b,y^b,z^b)_b$ are the "blue" coordinates of a point on the cone, then the "red" coordinates of the same point must satisfy Equation $(1),$ above. That is, we can use Equations $(2),$ $(3),$ and $(4)$ to make substitutions for $x^r,$ $y^r,$ and $z^r$ in Equation $(1).$ The resulting equation is \begin{multline} \left(a_{11}x^b + a_{13}z^b - h(a_{31}x^b + a_{32}y^b + a_{33}z^b - z^r_{O'})\right)^2 \\ + \left(a_{21}x^b + a_{22}y^b + a_{23}z^b - k(a_{31}x^b + a_{32}y^b + a_{33}z^b - z^r_{O'})\right)^2 \\ = \left(\frac{z^r_P - (a_{31}x^b + a_{32}y^b + a_{33}z^b)} {z^r_P - z^r_{O'}} R\right)^2. \tag5 \end{multline}

But we are only interested in the intersection of the cone with the blue plane, where $z^b = 0.$ So we can substitute $z^b = 0$ in Equation $(5),$ with the result \begin{multline} \left(a_{11}x^b - h(a_{31}x^b + a_{32}y^b - z^r_{O'})\right)^2 + \left(a_{21}x^b + a_{22}y^b - k(a_{31}x^b + a_{32}y^b - z^r_{O'})\right)^2 \\ = \left(\frac{z^r_P - (a_{31}x^b + a_{32}y^b)} {z^r_P - z^r_{O'}} R\right)^2. \end{multline}

Now, this may still look daunting, but everything in this equation except $x^b$ and $y^b$ is a known constant. You can multiply out the products and squares of the expressions in parentheses until everything is just individual terms, each of which is some kind of constant times $x^b,$ $y^b,$ $(x^b)^2,$ $(y^b)^2,$ or $x^b y^b.$ Collect all the terms together on one side of the equation so that it looks like $$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 ,$$ and then you can find the center, major axis, minor axis, and angle of the ellipse by following one of the procedures in the answers to
these questions:

Note that the center of the ellipse will not usually be at the same point as the projection of $O'$ onto the blue plane.

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  • $\begingroup$ Lets say the blue plane is what I have in hand and no information of any other plane, this plane is rotated by an angle $\alpha$ (unknown) along $x^r_O$ (x-axis of a reference plane normal to the circle) and by an angle $\beta$ (unkown) along $y^g_O$ (y-axis of the $\alpha$ rotated plane). I need to find $\alpha$ and $\beta$ using the projection of cone on the blue plane. I have 2 conditions, first with cone vertex at $O''$ and second with cone vertex at $P$. I use 3D software to simulate the ellipse on the plane and find the parameters. First case works with the equation above, 2nd does not. $\endgroup$ – radk May 16 '17 at 13:10
  • $\begingroup$ since I just have the info of ellipse parameters, I first write the equations of the ellipse in the plane $(x_O,y_O,0)$ parallel to the circle, apply 3D rotation (using rotation matrices with $\alpha$'s and $\beta$'s) and apply translation using the intersection point of ellipse major-minor axis (using the information from the blue plane). Then I write equations for an ellipse in the blue plane with in-plane rotation and translation using the angle $\theta$ between the major axis and the $y_b$ axis of the blue plane. I equate them thinking they would overlap and solve for the angles. $\endgroup$ – radk May 16 '17 at 13:37
  • $\begingroup$ I see no reason to think that your method would work in general. You assume that the projected ellipse, if rotated back into the red plane by the rotations you describe, will have axes parallel to the $x$ and $y$ axes. Why should it? $\endgroup$ – David K May 16 '17 at 13:44
  • $\begingroup$ On the other hand, if you write the equation of the cone in the "red" coordinates, then write the transformed equation in "blue" coordinates using the parameters $\alpha$ and $\beta,$ you should have a correct equation for the ellipse in the blue plane. That seems relatively straightforward. $\endgroup$ – David K May 16 '17 at 13:56
  • $\begingroup$ I guess it should work for the first case. First, say the cone forms a circle (ellipse with $a$=$b$) on red plane, I rotate it to the blue plane and I get the ellipse with point $e$ away from $O$. If I rotate back in same order I guess I would reach the starting point?. $\endgroup$ – radk May 16 '17 at 13:59

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