I've tried multiple searches but have been unable to find answer or guidance on this question (that I can follow). I also apologize in advance if my terminology is not quite correct.

Consider a right had coordinate system where y is up (i.e. x forward, y up, z out / towards you).

Now for example, consider a plane that is rotated about the x axis by 30 degrees and the z axis by -15 degrees and defined by the point P0(x0,y0,z0) as (5000, 3500, -500). Given these rotations, a normal N(a,b,c) to this plane can be defined as (-48.3, 83.65, -25.88). So in point / normal notation we can define this plane as:

$ a * (x-x0) + b * (y-y0) + c * (z-z0) = 0 $

$ -48.3(x-5000) + 83.65(y-3500) - 25.88(z+500) = 0 $

Now consider a 2d circle (r/theta notation) that lies on the xz plane at a known height y(3000) with a know radius r(5000). You can find the x / z points that lie on the circle as follows:

$ x = r * cos(\theta) = 5000 * cos(\theta)$

$ z = -r * sin(\theta) = -5000 * sin(\theta)$

Substituting the height/y of the xz plane circle and the formulas for x and z into the formula for the 3d plane you now have:

$ -48.3((5000 * cos(\theta))-5000) + 83.65(3000-3500) - 25.88((-5000 * sin(\theta))+500) = 0 $

Now, if the circle intersects the plane, there should be two values of $\theta$ that cause the above equation to equal $0$. Using an iterative solver to two decimal places, the values for this example are $\theta(18.85, 284.78)$.

In other situations there could be no solution if the circle does not intersect, one if it just touches and I suppose infinite if it lies exactly on the 3d plane.

My hope is to arrive at a series of equations that allow me to solve for $\theta$ rather than using the iterative approach.


2 Answers 2


If you expand your equation for $\theta$, you'll get $$A \cos \theta + B \sin \theta = C$$ with some values of $A$, $B$, $C$ that are known from your input data.

Such type of equation can be solved analytically.

I believe $A^2+B^2 > 0$ otherwise equation is $0=C$ which has eiher no solutions or infinitely mane solutions depending on $C$

First, divide both parts by $\sqrt{A^2+B^2}$: $$\frac{A}{\sqrt{A^2+B^2}} \cos \theta + \frac{B}{\sqrt{A^2+B^2}} \sin \theta = \frac{C}{\sqrt{A^2+B^2}}$$

Then we can find $\beta \in [-\pi/2;3\pi/2)$ such that $\sin\beta=\frac{A}{\sqrt{A^2+B^2}}$, $\cos\beta=\frac{B}{\sqrt{A^2+B^2}}$: $$\beta=\frac{\pi}{2}+\left(\arcsin \frac{A}{\sqrt{A^2+B^2}}-\frac{\pi} {2}\right)\cdot k,$$ where $k=-1$ if $B<0$ otherwise $k=1$

Then equation transforms to $$\sin \beta \cos \theta + \cos \beta \sin \theta = \frac{C}{\sqrt{A^2+B^2}}$$ $$\sin (\beta+\theta) = \frac{C}{\sqrt{A^2+B^2}}$$ $$\beta+\theta=\frac{\pi}{2}\pm\left(\arcsin \frac{C}{\sqrt{A^2+B^2}}-\frac{\pi}{2}\right)$$ $$\theta=\frac{\pi}{2}-\beta\pm\left(\arcsin \frac{C}{\sqrt{A^2+B^2}}-\frac{\pi}{2}\right)$$


Substituting circle into the plane we can get equation for $\theta$:
$ a (r\cos \theta-x_0) + b (h_y-y_0) + c (r\sin \theta -z_0) = 0 $ or $a\cos \theta + c \sin \theta =\frac{ax_0+cz_0-b (h_y-y_0)}{r}$
This can be rewritten as $$\sin(\theta+\alpha)=\frac{ax_0+cz_0-b (h_y-y_0)}{r\cdot\sqrt{a^2+c^2}}$$ where $\alpha=\tan^{-1}\frac{a}{c}$. The last equation has one or two solutions when right hand side is between $-1$ and $1$.

  • $\begingroup$ Aren't we still left with a situation where finding the results requires an iterative solver? $\endgroup$
    – Justin
    Jan 27, 2022 at 17:21
  • $\begingroup$ @Justin: not really, we can find solution using inverse sine if the solution exist $\endgroup$
    – Vasili
    Jan 27, 2022 at 17:25
  • $\begingroup$ Thank-you for the response. I feel I'm still confused. I've worked through your explanation and basically follow what you are saying. I've tried to work through this using my example above (adjusted for clockwise vs. counter-clockwise theta as in your summary. So in this case, our theta answers should be approx. 341.149 and 75.2177, but I'm not able to achieve that result. Honestly, I feel it's a misunderstanding on my part, I'm not comfortable with this math... are you able to provide more guidance? $\endgroup$
    – Justin
    Jan 27, 2022 at 18:18

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