How to make N Polygon Wheel.[br]It can be made by straight line drawing mechanism, easily.[br]I want to name this apparatus as [b]"Moonsault Wheel"[/b]. Below figure is [b]N = 2[/b] sample.[br][br](in Chebyshev case: heel is not only heel but also toe.[br] in Hart's A-frame case: heel is heel, toe is toe. )

1. Peaucellier Linkage --- 7 bars (exact straight-line) vertical[br]2. Hart's Inversor --- 5 bars (exact straight-line) vertical[br]3. Hart's A-frame --- 5 bars (exact straight-line) vertical[br]4. Chebyshev Linkage --- 3 bars (approximate straight-line) horizontal[br][br]I forgot Pentagon (360°/5=72°) case. 　it's easy, of course.[br][br][b]■ Good property[/b][br]This apparatus is better than Chebyshev wheel logic. Each unt independence is high. [br][br]There exist multiple feet on the ground at same time, this is not so bad. Like a centipede.[br]Foot takes off from the ground in shape 3/4/5 length edge right triangle. [br][br]QR= 0 at RT angle 0°[br]QR= √7 -1 (= 1.645751), at RT angle 90°[br]QR= 2√3 (=3.464102), at RT angle 180° --- I use this value as ≒ [b]3.5[/b] later.[br]QR= 4 (3/4/5 triangle)[br][br]To shorten from QR = 4 to 0 by spring coil is feasible?[br]At least, when next foot taking on the ground, PQRT must be one line shape.[br]Please consider.[br]How to always y(Q)=3 keep? (--- Q=R indirectly.)[br]Probably, by weak pull tension & self gravity, automatically resolved.[br][br]Restriction rule needed More.[br]ex. When foot is on ground, its distance QR is increasing. but, other unit's QR = 0.[br]So, adding, "one QR + most-opposite QR = (or ≦) constant" restriction.[br]　　----　ex. connect constant length (= 4) string/ thread. [Q---R~~~Q, where, -- + ~~~ = 4][br][br]■ Ready-to-assemble Leg-type Wheel[br]This is Collapsible Leg-type Wheel.[br]portable/ compact.[br]for travel.[br]useful in army work. (?)[br]needs no tire/ rubber.[br]on the carpet/ sand beach/ desert/ snow[br][br]N = 2 case --- interesting. sufficient? [br]( At R(3.5,3) [ ∠QRS = 30°, so ∠QRS ≦ 30°another ∠QRS should = 90°, angle spring coil ] [br] RT is vertical, so, another foot starts to touch the ground.[br]0.5/4 = 12.5%, 88% of 4, at this point, opposite QR must be 0.[br]So, constant length = 4*88%, [br]but, if you want to exist Both feet on ground, string must be flexible which extends to 4*112%. )[br][br][b]■ Green figure [/b]---- sum of bent bar length = constant.[br]N=2, QR =x(R), RA[sub]1[/sub]=3.5-x(R) [ i.e. constant length 3.5 ] case: this fig. is true. [br][ Here, --- RA[sub]1[/sub]=2√3-x(R) is OK, too. of course. ] [br][b]Tip: [/b][br]Restriction QR + RA[sub]1[/sub]= 2√3 --- ①[br]is the same/ equivalent to[br]Restriction VR + RE[sub]1[/sub]= √3 --- ②[br](∵ △QRS ∽ △VRT, △RA[sub]1[/sub]B1 ∽ △RE[sub]1[/sub]T')[br]∠VTE[sub]1[/sub] is not acute angle. ---- this is better. 　So, [b]In real implementation, [/b][b]② is better than ① [/b].　[br](See [b]Brown color[/b] bent segment [by wire/ chain/ flexible bar implementation? ] in above Fig.)[br][br]drawing figure was easy.[br]∵ RA[sub]1 [/sub]⊥ A[sub]1[/sub]Z , so, RZ = sqrt((3.5-x(R))^2+3^2), TZ = 4[br]like [b]a backward somersault[/b] [gymnastic exercises][br][br]cf. [url=https://www.youtube.com/watch?v=ni3kgHn212k]Cody Rhodes' amazing moonsault from the top of a cage at MSG![/url] (YouTube)[br][br]The bigger N, the easier for implementation. ex. N = 5, Pentagon wheel, that's easy thing to make.[br]Suppose saying an extreme example, this method doesn't need to divide equally.[br][br]ex. 30°+50°+100°+140°+40° = sum 360°,　such Pentagon hub wheel is OK. It is easy thing.[br][br][b]■★★ Purple figure ★★ ---- This is HOME RUN !!!!! , I think so.[/b][br]Purple color implementation is extended ② condition.[br]Restriction VH[sub]1[/sub] = √3 (No-bent, No-through R, a bar) 　---- ③[br][b]This is Feasible.[/b][br][b]I recommend this implementation.[br][/b][strike]This is simpler implementation than N=2 Chebyshev Wheel.[/strike] ----- Exists the same solution in Chebyshev, too.[br][b]---- We can use this in real use world.[br][br][/b]drawing figure was easy.[br]H[sub]1[/sub] is given by H[sub]1[/sub]T'=1, H[sub]1[/sub]V=√3[br]I[sub]1[/sub] is given by 4 times segment T'H[sub]1[/sub][br]J[sub]1[/sub] is given by I[sub]1[/sub]J[sub]1[/sub] = 3, RJ[sub]1[/sub] = sqrt(RI[sub]1[/sub]^2 -9)[br][br][br][b]Tip:[/b] VH[sub]1[/sub] lead hand bar is key. in N = 2 case, TV = T' H[sub]1[/sub] = 1.[br]So, in N ≠ 2 case, TV = T' H[sub]1[/sub] = 1 & VH[sub]1[/sub] is more smaller.[br][br][br]■ constant control (landing restriction)[br]Left and right adjoining foot constant length restriction has options. [br]If both feet have touched the ground, restriction can be removed. [br](but, its control is not easy.)[br][br]My conclusion about constant value control.[br]I checked many patterns, I've gotten next conclusion.[br]--- Simple is the best.[br]I recommend, the constant distance value ZERO.[br]i.e. After taking off from the ground, you should make QR = 0 as soon as possible. ---- this is most easy. [br]i.e. Shrinked form. Its shape is only one bar, like a closed umbrella.[br][b][b][br]■ N=2 Other property[/b][br]∠QRT' ≒ ∠A[sub]1[/sub]RT[br]i.e. bisector of an angle ∠QRA[sub]1[/sub]　⊥ TT' ---- we can use this.[br][br]■ Compare[/b], [url=https://www.geogebra.org/m/ARCcx49T]N=2 Chebyshev Wheel[/url] (GeoGebra).[br]Which is simple?[br]Please do implement. ex. caster/ dolly. [br]-------------------------[br]| |[br]| _ _ | [br]| / / |[br]--/------------------/--[br] /_ /_ [br][br]cf. [url=https://www.geogebra.org/m/VNGQ5Z5U][b] Chebyshev N = 3 Polygon Wheel[/b][/url] (GeoGebra).