Finally those trig lessons make sense For a triangle of sides a,b,c and angles opposite those sides A,B,C : a/(sin(A))= b/(sin() = c/(sin©) So if our circular orbit is 2500km then we add kerbin's radius 600km this gives us one of the two short sides at 3100km now if we choose to have 3 satellites then the angles between each will be 120deg as we know triangle angles add up to 180deg so the remaining two angles must total 60deg. Seeing as all the satellites are at the same altitude we can infer that this is an isosceles triangle and the two angles are equal therefore the corresponding angle to our side must be 30deg (half of 60deg). So our satellite spacing, the side of the triangle opposite 120deg, can be found using: Spacing = (3100/sin(30))*(sin(120)) This can be simplified for all 3 satellite constellations, with the a circular orbit and equal inclination, to: Spacing/sin(120) = (Altitude + 600km)/sin(30) (rearrange as required) " " " " " " 4 " " " " " " " " " " " Spacing/1 = (Altitude + 600km)/sin(40) (sin(90) = 1) " " " " " " 5 " " " " " " " " " " " Spacing/sin(72) = (Altitude + 600km)/sin(54) " " " " " " 6 " " " " " " " " " " " Spacing = Altitude + 600km (all the triangles here are equilateral with sides of the same length so the sin maths is not needed)