Equation for great circles of a sphere [closed]

Background: I have a sphere of some radius R. What I'm trying to do is essentially create a wireframe consisting of great circles that run along the sphere.

I want an equation to represent the great circles so that I can find points that run along those trajectories.

I don't have enough credit to post links, but there a post in the Mathematica stack exchange titled: How to draw a great circle on a sphere?

Which is what I want to achieve.

However, I want to implement it in python and require a further breakdown of the equations that are happening behind the scenes.

I will be plotting these equations in python.

closed as off-topic by Xander Henderson, Namaste, Shaun, Theoretical Economist, TaroccoesbroccoAug 21 '18 at 3:50

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• Seems more like a programming question to me, as the answer depends on your graphics library, if it is 2D or 3D, about how fast or smooth the rendering should be etc. – mvw Aug 20 '18 at 20:47
• What attempts have you made at this problem yourself? Do you know the equation that points on a sphere must satisfy? What about the equation for points lying on a plane? A great circle is the intersection of your sphere with a plane which passes through the origin. Can you see what two equations the points satisfying these conditions must satisfy? – Matt Aug 20 '18 at 20:50
• Did you mean this Mathematica link: How to draw a great circle on a sphere? – Weather Vane Aug 20 '18 at 20:51

The equation of equally placed great circles can be written as following in the spherical coordinates$$R=1\\\phi=\dfrac{i}{n}2\pi$$which by translating to Cartesian coordinates gives us $$x^2+y^2+z^2=1\\y=x\cdot\tan \dfrac{i}{n}2\pi$$you can use the following parametric equation for this$$\left(\sin t\cdot\cos \dfrac{i}{n}2\pi\quad,\quad\sin t\cdot\sin \dfrac{i}{n}2\pi\quad,\quad\pm\cos t\right)$$here is an image of what this can generate using MATLAB • \left( ... \right) would help... I would also suggest using parentheses to group the arguments of the trig functions, as this would enhance readability. – Xander Henderson Aug 20 '18 at 21:06
• Why not write $$\left( \sin(t)\tan\left(2\pi i/n\right), \sin(t), \pm\sqrt{1- \sin^2(t)\, /\cos^2\left(2\pi i/n\right) }\right)$$ – mvw Aug 20 '18 at 21:18
• Sorry i don't get you. What is the difference in your post with my answer? – Mostafa Ayaz Aug 20 '18 at 21:24
• For clarification are you defining phi to be the angle between the z-axis and the XY plane and theta to be the angle of an arc in the XY plane? Where I'm defining the z-axis to be pointing up. – NoviceCoder Aug 20 '18 at 21:35
• Well i defined them exactly reversed :) $\phi$ is the angle of arc – Mostafa Ayaz Aug 20 '18 at 21:37

The top answer over at the Mathematica.SE question is effectively doing the following:

1. Construct two unit vectors $\hat{u}, \hat{v} \in \mathbb{R}^3$ such that $\hat{u} \cdot \hat{v} = 0$. These two vectors lie in the plane of the desired great circle. If one wants a "random" great circle, one could simply construct two random vectors and then apply the Gram-Schmidt process.

2. Plot points of the form $$\vec{r}(\theta) = (R \cos \theta) \hat{u} + (R \sin \theta) \hat{v}.$$ As $\theta$ runs from $0$ to $2 \pi$, this traces out a parametric circle of radius $R$ lying in the plane spanned by $\hat{u}$ and $\hat{v}$.

• This partially worked for me, except I could not rotate the points about the y-axis to create great circles around the sphere. Instead, I could rotate about the x-axis and get small circles instead. I may use this for a different application. Thank you regardless I upvoted this, but I do not have enough rep to change your score. – NoviceCoder Aug 20 '18 at 23:59

Presumably you are using the usual spherical coordinates, Thus define the origin as the center of the sphere, $P_0 =(\theta_0, \phi_0)$ as the angular coordinates of any fixed point P on the sphere, and $N = (\theta =0,\phi =0)$ as the coordinates of the "north pole" where the positive $z$ axis meets the sphere. We choose $P_0$ as the "pole" with respect to which the great circle would be the "equator", situated $90°$ from the pole $P_0$.
Draw $\triangle PP_0N$ on the sphere, where $P$ is any point on the great circle that is $90°$ from $P_0$. By hypothesis $|PN|=\theta, |P_0N|=\theta_0, |PP_0=90°,|\angle N|=|\phi-\phi_0|.$ Then the Law of Cosines for arc $PP_0$ immediately gives
$\cos\theta\cos\theta_0+\sin\theta\sin\theta_0\cos(\phi-\phi_0)=0$.