I tried to calculate the strut lengths using Nsolve.[br]Unfortunately, Geogebra does not come up with a solution.[br]I therefore use Chris Kitrick's calculations.[br][br][url=https://groups.google.com/g/geodesichelp/c/RFBZ6uFmsjU/m/QfiMqboLBAAJ][b][color=#0000ff]New Nexorade/Rotegrity project[/color][/b][/url] : 19 jan 2019 18:58:43[br][br][b]Chris Kitrick,[/b][br][br]I will explain the solution to the next simplest nexorage: nex21 (4,1) [Taff's 2v{1,1}].[br][br]A diagram that illustrates the spherical triangle composition is attached.[br]Now there are six (6) unique spherical triangles to consider.[br]Due to symmetry triangle 0 is a right triangle at the pentagonal vertex[br](same as the first case).[br]Triangle 1 is an isosceles spherical triangle with side a equal to side b.[br]There are two large blue dots indicating the two unique places[br]where the joining corners of spherical triangles must add up to 180 (π).[br]Keeping the 1/3 arc division you see there are only two arcs[br]one composed of (3x) and the other (3y).[br]Even though we now have two unknowns (x,y) there is really only one[br]unknown (x), since (y) has a direct relationship to (x).[br][br]Vertex (0) will sum up to 180 (π) degrees by default based on construction.[br]The highlighted blue and red colors at vertex (1) must sum to 180 (π).[br]Essentially the three red corners must sum to the complement of the blue corner. [br][br]Here are the angular relationships:[br][br]Since the right spherical triangle 0 is at the pentagon,[br]the B face angle is always 36 degrees.[br]Solve a right spherical triangle with one edge and opposing face angle.[br][br] b0 = x / 2[br] B0 = 36[br] solve triangle 0[br] [br]Triangle 1 is isosceles and the 'C' angle is the complement[br]of the pentagon corner that sums to 180 (pi).[br]Solve a spherical triangle with two edges and included face angle.[br][br] a1 = b1 = 2x[br] C1 = 180 - A0 * 2[br] solve triangle 1[br] y = c1 / 2 [br] [br]Since triangle 1 is isosceles the resultant face angles A and B will be the same.[br][br]Construct spherical triangles 2, 3, and 4 which are identical.[br][br] a2 = a3 = a4 = y ( = green_x ) [br] b2 = b3 = b4 = x ( = red )[br] C2 = C3 = C4 = 180 - A1[br] solve triangle 2, 3, 4[br][br]cos(c) = cos(a)*cos(b) + sin(a)*sin(b)*cos(C)[br]cos(C) = (cos(c) - cos(a)*cos(b))/( sin(a)*sin(b) )[br] [br]Finish spherical triangle 5 which is equilateral:[br][br] a5 = c2[br] b5 = c3[br] c5 = c4[br] solve triangle 5 (three edges known)[br] [br]Finally the summation of face angles at vertex 1 must be 180 (pi).[br][br] B1 + A4 + B5 + B2 == 180[br] [br] or[br] [br] B1 - ( A4 + B5 + B2 ) == 0[br][br]NSolve(sum_x(x) - 180° = 0, x = 14°)[br] [br]Now the only question is what is the value of the edge angle x such that the summation at vertex 1 is 180.[br]By simply iterating the edge angle x and solving the spherical triangles to solve the 180 degree requirement at the blue vertex 1 you arrive at the value. Remember the value of edge y is not independent.[br][br]Here are the final spherical angle values for the two triangles:[br][br]x 14.495596141331275 0.252995879705616[br]y 16.289240170644682 0.284300873625873[br][br]TRI a b c[br]00 10.081227058633685 7.247798070665638 12.394269914560871[br]01 28.991192282662553 28.991192282662553 32.578480341289364[br]02 16.289240170644682 14.495596141331276 26.839102797641246[br]03 16.289240170644682 14.495596141331276 26.839102797641246[br]04 16.289240170644682 14.495596141331276 26.839102797641246[br]05 26.839102797641246 26.839102797641246 26.839102797641246[br] [br]TRI A B C[br]00 54.640124510181501 36.000000000000000 90.000000000000000[br]01 58.172482483059767 58.172482483059767 70.719750979636999[br]02 31.859581862437821 28.102009218011407 121.827517516940247[br]03 31.859581862437821 28.102009218011407 121.827517516940247[br]04 31.859581862437821 28.102009218011407 121.827517516940247[br]05 61.865926436490909 61.865926436490909 61.865926436490909[br][br]You'll notice the geometrically derived numbers differ slightly from Taff's. [br][br]Cheers, Chris