Back map finite N-piped in spherical

Back map finite N-piped in sphericals PhL 2.22.12 This was just an exercise with spherical coordinates and Maple. As a result I added to parametric curve plot items into my Maple doc. First, some general comments about the N-pipeds. The finite x-space N-piped is a real world object it is true, having areas and volume and edges. But, the interior of this object does not map back to any useful object in x'-space. Certainly the interior does not map back into the interior of any orthogonal N-piped in x-space, and certainly not into the entire "office building" in x'-space. Only the entire x-space maps into that full interior! Having clarified the obvious, I now take a "moment" to see just what such a finite x-space N-piped does map into. The edges of this N-piped have equation xn(α) = x0 + α en(x0) where x0 is the tail meeting point of the en vectors in x-space. The locus of xn in x-space is a finite straight line segment starting at the point where parameter α = 0, which is to say, xn(α=0) = r0. As dimensionless α runs 0 to 1, each xn(α) tracks on of our N-piped edges. Surely this line segment maps into some kind of curve inside the office building in x'-space. We have x'n(α) = F(xn) = F(x0 + α en(x0)) We know that x' = F(x) r = (x2+ y2+z2)1/2 x = rsinθcosφ tanφ = y/x y = rsinθsinφ cosθ = z/r z = rcosθ and we know er(r) = (sinθcosφ, sinθsinφ,cosθ) |er| = 1 = h'r eθ(r) = r(cosθcosφ,cosθsinφ,-sinθ) |eθ| = r = h'θ eφ(r) = rsinθ(-sinφ,cosφ,0) |eφ| = rsinθ = h'φ So let's first look at xn(α) = x0 + α en(x0): [xn(α)]i = [x0]i + α [en(x0)]i [xr(α)]1 = r0sinθ0cosφ0 + α 1 sinθ0cosφ0 = (r0 + α 1) sinθ0cosφ0 [xr(α)]2 = r0sinθ0sinφ0 + α 1 sinθ0sinφ0 = (r0 + α 1) sinθ0sinφ0 [xr(α)]3 = r0cosθ0 + α 1 cosθ0 = (r0 + α 1) cosθ0 [xθ(α)]1 = r0sinθ0cosφ0 + α r0 cosθ0cosφ0 = r0(sinθ0 + α cosθ0) cosφ0 [xθ(α)]2 = r0sinθ0sinφ0 + α r0 cosθ0sinφ0 = r0(sinθ0 + α cosθ0) sinφ0 [xθ(α)]3 = r0cosθ0 – α r0 sinθ0 = r0(cosθ0 - α sinθ0) [xφ(α)]1 = r0sinθ0cosφ0 – α r0sinθ0sinφ0 = r0 sinθ0(cosφ0 – α sinφ0) [xφ(α)]2 = r0sinθ0sinφ0 + α r0sinθ0cosφ0 = r0 sinθ0(sinφ0 + α cosφ0) [xφ(α)]3 = r0cosθ0 + 0 = r0cosθ0 In the above, I write 1's because 1 means 1 cm if we are doing cm, and r0 means r0 cm. Now back to x'n(α) = F(xn) n = 1,2,3 means r, θ, φ which we translate to say rn = ( Σi[xn(α)]i 2 )1/2 tanφn = [xn(α)]2 / [xn(α)]1 cosθn = [xn(α)]3/rn The curve for n = 1 meaning r. So lets start with n = 1 meaning the r case rr = ( Σi[xr(α)]i 2 )1/2 = [(r0 + α 1)2 sin2θ0cos2φ0 + (r0 + α 1)2 sin2θ0sin2φ0 + (r0 + α 1)2 cos2θ0]1/2 = [(r0 + α 1)2 sin2θ0 + (r0 + α 1)2 cos2θ0]1/2 = [(r0 + α 1)2]1/2 = r0 + α 1 tanφr = tanφ => φ0 = φ cosθr = (r0 + α 1) cosθ0/ (r0+ α 1) = cosθ0 => θr = θ0 So this "curve in x'-space" has the following equation rr = r0 + α 1 θr = θ0 φr = φ So this curve is a vertical line segment inside the office building! This was the simple one. The curve for n = 2 meaning θ. So lets next do with n = 2 meaning the θ case rθ = ( Σi[xθ(α)]i 2 )1/2 = [r02(sinθ0 + α cosθ0)2 + r02(cosθ0 - α sinθ0)2 ]1/2 = r0[(sinθ0 + α cosθ0)2 + (cosθ0 - α sinθ0)2 ]1/2 = r0[ 1 + α2 ]1/2 Now keep going: tanφθ = [xθ(α)]2 / [xθ(α)]1 = [r0(sinθ0 + α cosθ0) sinφ0] / [r0(sinθ0 + α cosθ0) cosφ0] = tanφ0 => φθ = φ0 at least this one is simple! cosθθ = [xθ(α)]3/rθ = r0(cosθ0 - α sinθ0)/ { r0[ 1 + α2 ]1/2} = (cosθ0 - α sinθ0) / So this curve comes out being rθ(α) = r0 θθ(α) = cos-1[(cosθ0 - α sinθ0) / ] φθ(α) = φ0 This is a curve lying in the plane φ = φ0 . How would you plot it? Let x(α) = cos-1[(cosθ0 - α sinθ0) / ] y(α) = r0 With θ0 = π/6 and r0 = 2 ( the starting point for our curve), Maple says where x(α) = θθ(α) and y(α) = rθ(α) and we are in the φ0 plane of the office building. The curve for n = 3 meaning φ. Finally lets do the n=3 case which is φ. rφ = ( Σi[xφ(α)]i 2 )1/2 = [r02 sin2θ0(cosφ0 – α sinφ0)2 + r02 sin2θ0(sinφ0 + α cosφ0)2 + r02cos2θ0]1/2 = r0[sin2θ0(cosφ0 – α sinφ0)2 + sin2θ0(sinφ0 + α cosφ0)2 + cos2θ0]1/2 = r0[sin2θ0( 1 + α2) + cos2θ0]1/2 tanφφ = [xφ(α)]2 / [xφ(α)]1 = (sinφ0 + α cosφ0)/ (cosφ0 – α sinφ0) cosθφ = [xφ(α)]3/rφ = r0cosθ0 / { r0[sin2θ0( 1 + α2) + cos2θ0]1/2} = cosθ0 / { [sin2θ0( 1 + α2) + cos2θ0]1/2} So our parametric curve in this case is described by rφ(α) = r0[sin2θ0( 1 + α2) + cos2θ0]1/2 θφ(α) = cos-1{ cosθ0 / { [sin2θ0( 1 + α2) + cos2θ0]1/2}} φφ(α) = tan-1{(sinφ0 + α cosφ0)/ (cosφ0 – α sinφ0)} and finally we get a curve that does not lie in any of the main planes. Here is Maple on this one: where I took some odd values for r0, θ0, φ0. The curve is not planar. Question: Is there some way I could fill x-space with a sea of dots, and then see what this maps to in x'-space, just to get a feel for the general shape of the resulting 3D object? Suppose I write this x(α) = x0 + α1 e1(x0) + α2 e2(x0) + α3 e3(x0) I could make a loop to generate perhaps 103 values inside the N-piped in x'-space. Then for each of those points I have to compute x' = F(x) using this r = (x2+ y2+z2)1/2 x = rsinθcosφ φ = tan-2(y/x) y = rsinθsinφ θ = cos-1(z/r) z = rcosθ OK, that is a project for another day. It would take me many hours to debug in Maple.