N=3 Polygon Wheel

Hart's A-frame Application sample Trigonal wheel Detail is N Polygon Wheel  (GeoGebra) Pending: in N = odd number !!. (not easy to control.)
Above fig. contains inconsistency. When N is odd number case, length restriction by rigid bar is bad. (N-1 bars, another 1 is empty. --- this means , one restriction is missing. --- bad.) all bars should be replaced into thread or hinge. Slack thread is needed. and, N threads should be supported. A・---B・  \ /   C・  3 points A,B,C must be connected by thread (not rigid one bar) or hinge. [ each distance is variable. i.e. constant max. ] i.e. change method: from Old: = restriction, to New: restriction [ from the bar -- change to the hinge /\ . ] (No bent thread. so, there's no slide friction through Axis point. ) Tip: in N = 3 Chebyshev wheel, above method was failed by my 1st test. i.e. not feasible. But After more learning, I found there exists above feasible solution in Chebyshev case. ----- above is logic miss. exact rule is " grounding leg's 2 restrict bar is rigid bar, other not grounding leg's bar is thread/ hinge" . i.e. dynamic changing is needed. it is so difficult for the automatic control. ---- from this, N = 2 case is easy, but N=3 case is not easy. dynamic hinge lock mechanism is needed. (In above fig. bold pink bar F1K1 should be removed, and set it as VK1.) ---- Cyan colored leg/ foot is VK1 replaced restriction sample. this leg behavior is good. ---- inspiration has occurred. 3D lock mechanism is feasible perhaps. xyz axis space, offset z=0 or not is controllable. See. Chebyshev N = 2 Polygon Wheel (GeoGebra). Chebyshev N = 3 Polygon Wheel (GeoGebra). Above has yet logic miss. 2 restrictions is enough number of restrict. 3 restriction is odd. grounded foot/ leg is independent, and other 2 legs is dependent by one restriction, so, number of 2 restriction is enough number. Please find the best tuning result. Answer: 3 constant length restriction approach has contradiction. 1 constant length restriction + 2 constant angle movement approach has no contradiction. So, implementation is not so difficult. ■ N=3 Coordinator length (at just 120° span point, another foot has touched on grouned. ) VF1 = 9/8 = 1.125 ∵ 0.5=1/2 (3-0.5)*(1/4)=2.5/4=5/8 1/2+5/8=4/8+5/8=9/8 (=1.125) ------ Stride is (3/8)√39=2.341874249 ∵ (3/4)√(42-2.52)=(3/4)√(16-[5/2]2)=(3/4)√([64-25]/4)=(3/4)(1/2)√(39) -------------------- Black, Purple, Cyan are 3 components. Pink is no-needed ( or redundancy). (Bold Purple bar VF1, Bold Cyan bar VP1 are needed. Bold Pink bar is no-needed. 3 Points V, F1, P1 are completely controlled by 2 bold bars. F1-P1 is not connected directly, but is connected indirectly.) 180° and then -60°, +60° method is needed the thick sandal under the foot in this case. --- why? (result of Chebyshev-like N=3 is different from this result.)