(This is NOT homework) just my personal interpretation, because these are the formulas as you already know: r = √(x^2 + y^2 + z^2) φ = arctan(y/x) θ = arccos(z/r) using (x,y,z) = (∞,∞,∞) I come across a bit of a sinister problem: r = √(∞^+∞^+∞^) = √(3∞^2) which is right because if we just said that ∞ was x, then it would be: x^2 + x^2 + x^2 = 3x^2 right? (just the small proof) that means that (using surd rules) √(3∞^2) = √3 x √∞^2 = ∞x√3 = ∞√3 r = ∞√3 φ = arctan(∞/∞) = arctan(1) = 45° or ∏/4 radians. θ = arccos(∞/∞√3) <---- (z / r) ||| arccos(∞/∞√3) = arccos(1/√3) - because can I please say that we shouldn't forget that the ∞'s cancel out(???) so: θ = 54.73561032° <--- This is what I don't understand, why is it that if you use x,y,z as infinity, why is the angle of θ NOT 60°? (I assume it's because r is in equidistance from x,y,z as they are all infinity, and φ is 45°, it shows r is equidistant from x and y) - ^ That is my thesis, I don't know why θ doesn't equal φ if all the lengths are equal, (again because r is equidistant from the line x, line y and line z) they SHOULD be the same no? (again, I assume it's because of the 1/√3 - but there is 3 infinities (to find out r) so it should be √3∞^2 no?) Thank's for reading, I was just reading through the spherical co-ordinate theories and formulas, it's not homework, it's actually something I thought other people would want to see. Thanks for reading. Eggman100 Oh, -.-, I think I forgot that Phi is the angle with respect to r in the x and y axis, I think when I said about the equidistance, then I think I forgot that, "respect to r in the x and y axis" - I think I used it so that its the angle from the x axis and r which of course would have to implement a 'z' which ofcouse phi is only the x/y plane angle, sorry.