How to draw a great circle on a sphere?

The idea is quite simple: Since any great circle can be parametrized as $\cos(\theta)u + \sin(\theta)v$ where $u$ and $v$ are two orthonormal vectors. One can start with $u=\{1,0,0\}, v=\{0,1,0\}$ and use RotationTransform to get out of the xy plane, then use RotationTransform again to spin around the z-axis to get all great circles with desired inclination.

Manipulate[ (* Rotation deg° out of the xy plane *) rx = RotationTransform[deg Degree, {0, 1, 0}]; (* Spin around z axis *) rz = RotationTransform[ϕ Degree, {0, 0, 1}]; {u, v} = rz @ rx @ {{1, 0, 0}, {0, 1, 0}}; Show[{ Graphics3D[{ {Opacity[0.4], Sphere[]}, {Opacity[0.5], Polygon[{ {-1, -1, 0}, {1, -1, 0}, {1, 1, 0}, {-1, 1, 0}}]}, {Arrow[{{0, 0, 0}, u}], Arrow[{{0, 0, 0}, v}]} }], ParametricPlot3D[{ Cos[θ] u + Sin[θ] v, (* The great circle in question *) {Cos[θ], Sin[θ], 0}, (* Normal unit circle *) RotationTransform[θ, {0, 0, 1}] /@ {u, -u} (* The red dashed stuff *) }, {θ, -Pi, Pi}, PlotStyle -> {Directive[Blue, Thick], Black, Directive[Red, Dashed]}] }, Axes -> True, AxesLabel -> {"x", "y", "z"}], {{deg, 15, "Inclination"}, -180, 180}, {ϕ, 0, 360}] 

Here is a static solution to the problem. It shows a mesh on the sphere that represents the normal lat-long coordinate system.

A function representing the equator.

equator[θ_] := {Cos[θ], Sin[θ], 0} 

A function and a plot representing the inclined circle. Note that the inclination is accomplished by a rotation of the equator about the x-axis.

inclinedCircle[θ_, i_] := RotationTransform[i, {1, 0, 0}][equator[θ]]; circlePlot[i_] := ParametricPlot3D[ Evaluate[inclinedCircle[θ, i °]], {θ, 0, 2 π}, PlotStyle -> {Thickness[0.005]}] 

A function and a plot representing the sphere.

sphere[u_, v_] := {Cos[u] Cos[v], Sin[u] Cos[v], Sin[v]}; spherePlot = ParametricPlot3D[ Evaluate[sphere[u, v]], {u, 0, 2 π}, {v, -π, π}]; 

A great circle inclined at 60°.

Show[{spherePlot, circlePlot[60]}, Axes -> False, Boxed -> False, PlotRange -> All, SphericalRegion -> True, ViewPoint -> {5, 0, 0}, ViewAngle -> 12 °] 

Update 1

With a little modification to inclinedCircle and circlePlot to handle nodal point shift, it is easy to construct an interactive version. A rotation about the z-axis has been added to accomplish the nodal shift.

inclinedCircle[θ_, i_, offset_] := RotationTransform[offset, {0, 0, 1}] [RotationTransform[i, {1, 0, 0}][equator[θ]]]; circlePlot[i_, offset_: 0] := ParametricPlot3D[ Evaluate[inclinedCircle[θ, i °, offset °]], {θ, 0, 2 π}, PlotStyle -> {Thickness[0.005]}] 

The parameter ι is the inclination and the parameter α is the equatorial ascending node.

Manipulate[ Show[{spherePlot, circlePlot[ι, α]}, Axes -> False, Boxed -> False, PlotRange -> All, SphericalRegion -> True, ViewPoint -> {5, 0, 0}, ViewAngle -> 12 °], {{ι, 60.}, 0., 90., 5., Appearance -> "Labeled"}, {{α, 0.}, 0., 355., 5., Appearance -> "Labeled"}, SaveDefinitions -> True] 

Update 2

Rather than describing a great circle as the composition of rotations applied to the equator, those rotations can be used to derive a more conventional description, a parametic function built from elementary trigonometric functions. To do this, first apply inclinedCircle to some value-free symbols.

Clear[θ, ι, φ]; inclinedCircle[θ, ι, φ] 

{Cos[θ] Cos[φ] - Cos[ι] Sin[θ] Sin[φ], Cos[ι] Cos[φ] Sin[θ] + Cos[θ] Sin[φ], Sin[θ] Sin[ι]}

From the output expression, write a function generator, inclinedCircle, that when given the inclination ι and the ascending node φ, returns a pure function producing the points on the desired great circle.

inclinedCircleF[ι_, φ_] = {Cos[#] Cos[φ] - Cos[ι] Sin[#] Sin[φ], Cos[ι] Cos[φ] Sin[#] + Cos[#] Sin[φ], Sin[#] Sin[ι]}&; 

Next write a plotting function that will use the function generator rather than the rotations.

circlePlot2[ι_, φ_: 0.] := ParametricPlot3D[ inclinedCircleF[ι °, φ °][θ], {θ, 0., 2. π}, PlotStyle -> {Thickness[0.005]}] 

Finally, substitute the new plotting function for the original one in the Manipulate.

Manipulate[ Show[{spherePlot, circlePlot2[ι, φ]}, Axes -> False, Boxed -> False, PlotRange -> All, SphericalRegion -> True, ViewPoint -> {5, 0, 0}, ViewAngle -> 12 °], {{ι, 60.}, 0., 90., 5., Appearance -> "Labeled"}, {{φ, 0.}, 0., 355., 5., Appearance -> "Labeled"}, SaveDefinitions -> True] 

The output from evaluating this Manipulate expression is exactly the same as with the one using rotations, so I won't repeat it here.

Great answers have already appeared. In the spirit of demonstrating multiple solutions to a problem with Mathematica, I would like to offer one using a different approach.

First, some geometric analysis. This great circle bounds a hemisphere lying in a half-space determined by a normal direction to the circle's plane. Letting $\theta$ be the latitude at which the circle crosses the equator, we find that $(\cos(\theta), \sin(\theta), 0)$ is a direction on the circle. The perpendicular direction $(-\sin(\theta), \cos(\theta), 0)$ has to be elevated by an angle $\alpha$, whence it equals $(-\cos(\alpha) \sin(\theta), \cos(\alpha)\cos(\theta), \sin(\alpha))$. The cross-product of these vectors is the desired normal direction.

One expedient--and straightforward--solution uses Contour3D to depict the surface of the sphere as the contour $x^2+y^2+z^2=1$ and exploits its RegionFunction option to limit that contour to the desired hemisphere.

For reference I use a world globe taken directly from the help page for Texture. It requires an image of the world's surface, such as

im = Import["http://www.ngdc.noaa.gov/mgg/topo/pictures/GLOBALeb6colshadesmall.jpg"]; 

This is plastered onto a slightly shrunken parametric plot of a sphere (or ellipsoid, if you want greater realism), stripped of all detail except this texture:

globe = ParametricPlot3D[0.99 {Cos[u] Sin[v], Sin[u] Sin[v], Cos[v]}, {u, 0, 2 π}, {v, 0, π}, Mesh -> None, TextureCoordinateFunction -> ({#4, 1 - #5} &), PlotStyle -> Directive[Specularity[White, 20], Texture[im]], Lighting -> "Neutral", Axes -> False, RotationAction -> "Clip"]; 

Over that we may drape the hemisphere semi-transparently.

The code allows dynamic manipulation of the elevation and orientation ("azimuth") of the great circle.

Manipulate[ Block[{normal = Cross[{Cos[θ], Sin[θ], 0}, {Cos[α] (-Sin[θ]), Cos[α] Cos[θ], Sin[α]}]}, Show[globe, ContourPlot3D[x^2 + y^2 + z^2, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, Contours -> {1}, ContourStyle -> Opacity[0.5], Mesh -> None, RegionFunction -> Function[{x, y, z}, normal . {x, y, z} ≥ 0]], Boxed -> False]], {{α, 0, "Elevation"}, -π/2, π/2}, {{θ, 0, "Azimuth"}, -π, π} ] 

Include additional 3D graphics as you will to indicate the normal direction, the Equatorial Plane, a latitude-longitude graticule, etc.

As noted in this thread, it's often more convenient to manipulate B-splines instead of lines.

Thus, using the B-spline representation derived by Piegl and Tiller in this article, we have the following routine:

greatCircle[φ_, θ_, r_: 1] := BSplineCurve[ Composition[RotationTransform[θ, {0, 0, 1}], RotationTransform[-φ, {0, 1, 0}]] /@ (r {{1, 0, 0}, {1, 1, 0}, {-1, 1, 0}, {-1, 0, 0}, {-1, -1, 0}, {1, -1, 0}, {1, 0, 0}}), SplineDegree -> 2, SplineKnots -> {0, 0, 0, 1/4, 1/2, 1/2, 3/4, 1, 1, 1}, SplineWeights -> {1, 1/2, 1/2, 1, 1/2, 1/2, 1}] 

Try it out:

With[{ε = 1*^-3 (* shrinks sphere slightly *), φ = 30° (* inclination *)}, Graphics3D[{{Opacity[2/5, Blue], Sphere[{0, 0, 0}, 1 - ε]}, {Directive[AbsoluteThickness[3], Red], greatCircle[φ, 0]}}, Lighting -> "Neutral"]] 

Spin things around a bit:

With[{ε = 10^-3, φ = 30°}, Animate[Graphics3D[{{Opacity[2/5, Blue], Sphere[{0, 0, 0}, 1 - ε]}, {Directive[AbsoluteThickness[3], Red], greatCircle[φ, θ]}}, Lighting -> "Neutral", SphericalRegion -> True], {θ, 0, 360°, 15°}]] 

Adding the fancy stuff (e.g. arrows and planes) is left to you as an exercise.

Just use MeshFunction.

Manipulate[ParametricPlot3D[{Sin[\[Theta]] Cos[\[Phi]], Sin[\[Theta]] Sin[\[Phi]], Cos[\[Theta]]}, {\[Theta], 0, \[Pi]}, {\[Phi], 0, 2 \[Pi]}, PlotStyle -> Opacity[0.5], Mesh -> {{0.}}, MeshStyle -> {Red, Thick}, MeshFunctions -> {Sin[a] Cos[b] #1 + Sin[a] Sin[b] #2 + Cos[a] #3 &}], {a, 0, \[Pi]}, {b, 0, 2 \[Pi]}] 

Let me explain my code.

Since the great circle is the intersection of the sphere and a plane passing through the origin. So all you have to do is to

1. Find or define the normal vector of the plane. In my case, $(\sin{a}\cos{b},\sin{a}\sin{b},\cos{a})$
2. Translate the mathematical plane equation into MeshFunction, i.e. Sin[a]Cos[b]#1+Sin[a]Sin[b]#2+Cos[a]#3 &

Mathematica will do the rest. Enjoy:)