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 F_{1}K_{1} should be removed, and set it as VK_{1}.)
---- Cyan colored leg/ foot is VK_{1} 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. )
VF_{1} = 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)√(4^{2}-2.5^{2})=(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 VF_{1}, Bold Cyan bar VP_{1} are needed. Bold Pink bar is no-needed.
3 Points V, F_{1}, P_{1} 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.)