Finding lowest point on circle after rotating it in 3D Space.

Question

Say there is a circle in 3D space currently lying on the x-y plane. (Radius = 1). You rotate it about the y-axis (going into the screen) $\theta$ degrees. And you then rotate that about the x-axis by $\varphi$ degrees. How would I find the lowest point of the circle (in the $z$ axis)? Edit: As well as the (x,y,z) coordinate. CENTRE is at (0,0,0) Futhermore: Now that I have the coordinates, is there a way to convert these coordinates to where the point was on the original circle?

Answer 1

Let $P=(\cos\alpha,\sin\alpha,0)$ be a generic point on the original circle (I assume it is centered at the origin). The rotation matrix, for a rotation by $\theta$ about the y-axis, followed by a rotation by $\phi$ about the x-axis, is $R=R_x(\phi)R_y(\theta)$, where $R_x$ and $R_y$ are the basic rotations described here. $P'=RP$ is then given by: $$ P'=(\cos\theta\cos\alpha,\ \sin\phi\sin\theta\cos\alpha+\cos\phi\sin\alpha,\ -\cos\phi\sin\theta\cos\alpha+\sin\phi\sin\alpha). $$ To find the value of $\alpha$ which leads to minimum and maximum value of $z_P$ we can differentiate $z_P$ with respect to $\alpha$ and equate the result to zero: $$ {dz_P\over d\alpha}=\cos\phi\sin\theta\sin\alpha+\sin\phi\cos\alpha=0, $$ which has solutions: $$ \cos\alpha=\pm{\cos\phi\sin\theta\over\sqrt{1-\cos^2\phi\cos^2\theta}},\ \sin\alpha=\mp{\sin\phi\over\sqrt{1-\cos^2\phi\cos^2\theta}}. $$ Substituting these into $z_P$ we find $$ z_P=\mp\sqrt{1-\cos^2\phi\cos^2\theta} $$ hence the minimum value (negative $z_P$) corresponds to the upper sign ($+$ for $\cos\alpha$ and $-$ for $\sin\alpha$) and the maximum value to the lower sign.

Substituting $\cos\alpha$ and $\sin\alpha$ found above (with upper sign) into the expressions for $P'$ and $P$ will then give the coordinates of the lowest lying point and of its original position.

Answer 2

I assume that $0 \leq \theta \leq 90$ and that $0 \leq \varphi \leq 90.$ If we have to account for other angles we might have to work out more separate cases.

We also have to remove the ambiguity about which way each rotation turns. ("Into the screen", with no other clues, is no help, since if one part of a circle goes into the screen when you rotate one way around an axis, there is also a part that goes into the screen when you rotate the opposite way around the same axis--that is, both rotational directions go "into the screen" in some way!) I will assume that the first rotation takes the point $(1,0,0)$ to a negative $z$ coordinate, and the second rotation takes $(0,1,0)$ to a positive $z$ coordinate. This corresponds to "right-hand convention" rotations around the $y$ and $x$ axes.

Suppose we have a spherical model of the Earth using the Earth's radius as the unit of measurement (so the Earth is a sphere of radius $1\newcommand{deg}{^\circ}$), with the Earth's center at $(x,y,z) = (0,0,0),$ the equator in the $x,y$ plane, the north pole on the positive $z$ axis (at $(0,0,1)$), and the prime meridian passing across the positive $y$ axis (therefore passing through $(0,1,0)$).

Then the equator is your initial circle (radius $1,$ center at $(0,0,0)$). After the first rotation, this circle is taken to a new circle $O_1$ that crosses the prime meridian at the equator at an angle $90\deg - \theta\deg.$ The second rotation takes $O_1$ to another circle, $O_2,$ that still crosses the equator at an angle $90\deg - \theta\deg,$ but now the crossing point (call it $B$) is at latitude $\varphi\deg$ rather than at the equator.

Let $A$ be the north pole and $C$ the "lowest" point on $O_2$ (the point farthest south). Re-use the names $A,$ $B,$ and $C$ for the angles at vertices $A,$ $B,$ and $C,$ respectively, of the spherical triangle $\triangle ABC,$ and let the angular measurements of the "sides" of that spherical triangle be $a$ from $B$ to $C$, $b$ from $A$ to $C$, and $c$ from $A$ to $B$. In terms of our known angles, $B = 90\deg + \theta\deg,$ $b = 90\deg + {\lambda_\max}\deg$ where ${\lambda_\max}\deg$ is the greatest latitude reached by $O_2,$ $c = 90\deg - \varphi\deg,$ and $C = 90\deg.$ So $\sin b = \cos({\lambda_\max}\deg),$ $\sin B = \cos(\theta\deg),$ and $\sin c = \cos(\varphi\deg).$

Since $C = 90\deg,$ we can apply Napier's rules for right spherical triangles. So $\sin b = \sin B\sin c,$ that is, $$ \cos({\lambda_\max}\deg) = \cos(\theta\deg) \cos(\varphi\deg). $$ So $C$ is at ${\lambda_\max}\deg$ south latitude, which makes its $z$-coordinate $z_C = -\sin({\lambda_\max}\deg).$ Using the fact that $\sin^2 \alpha + \cos^2 \alpha = 1,$ $$ z_C = - \sqrt{1 - \cos^2({\lambda_\max}\deg)} = - \sqrt{1 - \cos^2(\theta\deg) \cos^2(\varphi\deg)}. $$

Also from Napier's rules, $\cos c = \cot A \cot B,$ which implies $\cot A = \tan B\cos c.$ Since $\tan B = -\cot(\theta\deg)$ and $\cos c = \sin(\varphi\deg),$ then $$ \cot A = -\cot(\theta\deg)\sin(\varphi\deg) < 0. $$ By inspection, the "lowest" point on $O_2$ has $x>0$ and $y<0,$ so that $90\deg < A < 180\deg,$ consistent with the fact that $\cot A < 0.$ So we can solve for $A$ using $$ A = 180\deg - \arctan\left(\tan(\theta\deg)\csc(\varphi\deg)\right). $$

Since $C$ is on the sphere, its coordinates $(x_C,y_C,z_C)$ satisfy the equation $$ x_C^2 + y_C^2 + z_C^2 = 1, $$ so $\sqrt{x_C^2 + y_C^2} = \sqrt{1 - z_C^2}$ and $$\begin{align} x_C &= \left(\sqrt{1 - z_C^2}\right) \sin A > 0, \\ y_C &= \left(\sqrt{1 - z_C^2}\right) \cos A < 0 \end{align}$$ (different from the usual convention for polar coordinates, since in this case $A$ is measured clockwise from the $y$ axis). And we have now found the complete coordinates of the "lowest" point on the final circle.

To find the point on the original circle that was rotated to $C,$ we can find the size of the arc $a$ using another of Napier's rules, $\tan a = \cos B\tan c.$ Now $\cos B = -\sin(\theta\deg)$ and $\tan c = \cot(\varphi\deg),$ so $$ \tan a = -\sin(\theta\deg) \cot(\varphi\deg) < 0. $$ By inspection, $90\deg < a < 180\deg,$ so we can solve for $a$ using $$ a = 180\deg - \arctan\left(\sin(\theta\deg) \cot(\varphi\deg)\right). $$ This again measures an angle clockwise from the positive $y$ axis, so the point on the original circle that was mapped to the "lowest" point $C$ has coordinates $$\begin{align} x_0 &= \sin a > 0, \\ y_0 &= \cos a < 0. \end{align}$$