Many of you would have seen various illistrations that show side by side a spheric polytope {3,4}, a tiling like {3,6} and {3,7}. The usual way of dealing with these solids, is to pretend {3,5} and [3,6] are euclidean shapes and that {3,7} must be dalt with with hyperbolic trignometry. but we dont deal with {3,5} as spheric trignomety.
What if i said we ca deal with {3,7} like a euclidean polytopes. The intersection of any two isocurve (circle, horocycle is an isocuve of lesser radius, this allows us to deal with surtopes like x3o5x, which requires a isocurve that can contain a decagon. The dintersection of a horocycle and an isocurve is a circle (or sphere), one notes that euclidean geometry is preseved in this case, as well as rescaling. the pentagons in x5o3x are just the ordinary pentagons, the trun length is called arc length,we are useng chordal length. if you have the gimbal , the half circle maked to show not angles, but the chord lengths, you will find that any regular pentagon has chords in the ratio of 1:f or1:74.20(1.618033),
The idea is to work with euclidean geometry and not worry that the numbers dont fit expectations. It more a nature of how cosh and cos work.After bumbling with ordinay length, it was found that if the squares of the chords were used, things are a lot easir to handle.
The impotant figure is to find the diameter of a triangle of side 1:1:s sincs s is the vertix figure (s is the shortchord of the polygon inn question), with a little fiddling in algbera we find that d²=(4/(4-s²), generally the chord squar is represent by just a power, like s.
if the shortchrds square ofA and Bare a, b we can find out the d2 of {A.B}by noting that the verf is a B-sized ppolygon as s=4b/4-a, in squared units , d=4/(4-s) gives d=(4-a)/(4-a-b).
more tomorrow.