N Polygon Wheel
How to make N Polygon Wheel. It can be made by straight line drawing mechanism, easily. I want to name this apparatus as "Moonsault Wheel". Below figure is N = 2 sample. (in Chebyshev case: heel is not only heel but also toe. in Hart's A-frame case: heel is heel, toe is toe. )
1. Peaucellier Linkage --- 7 bars (exact straight-line) vertical 2. Hart's Inversor --- 5 bars (exact straight-line) vertical 3. Hart's A-frame --- 5 bars (exact straight-line) vertical 4. Chebyshev Linkage --- 3 bars (approximate straight-line) horizontal I forgot Pentagon (360°/5=72°) case. it's easy, of course. ■ Good property This apparatus is better than Chebyshev wheel logic. Each unt independence is high. There exist multiple feet on the ground at same time, this is not so bad. Like a centipede. Foot takes off from the ground in shape 3/4/5 length edge right triangle. QR= 0 at RT angle 0° QR= √7 -1 (= 1.645751), at RT angle 90° QR= 2√3 (=3.464102), at RT angle 180° --- I use this value as ≒ 3.5 later. QR= 4 (3/4/5 triangle) To shorten from QR = 4 to 0 by spring coil is feasible? At least, when next foot taking on the ground, PQRT must be one line shape. Please consider. How to always y(Q)=3 keep? (--- Q=R indirectly.) Probably, by weak pull tension & self gravity, automatically resolved. Restriction rule needed More. ex. When foot is on ground, its distance QR is increasing. but, other unit's QR = 0. So, adding, "one QR + most-opposite QR = (or ≦) constant" restriction. ---- ex. connect constant length (= 4) string/ thread. [Q---R~~~Q, where, -- + ~~~ = 4] ■ Ready-to-assemble Leg-type Wheel This is Collapsible Leg-type Wheel. portable/ compact. for travel. useful in army work. (?) needs no tire/ rubber. on the carpet/ sand beach/ desert/ snow N = 2 case --- interesting. sufficient? ( At R(3.5,3) [ ∠QRS = 30°, so ∠QRS ≦ 30°another ∠QRS should = 90°, angle spring coil ] RT is vertical, so, another foot starts to touch the ground. 0.5/4 = 12.5%, 88% of 4, at this point, opposite QR must be 0. So, constant length = 4*88%, but, if you want to exist Both feet on ground, string must be flexible which extends to 4*112%. ) ■ Green figure ---- sum of bent bar length = constant. N=2, QR =x(R), RA1=3.5-x(R) [ i.e. constant length 3.5 ] case: this fig. is true. [ Here, --- RA1=2√3-x(R) is OK, too. of course. ] Tip: Restriction QR + RA1= 2√3 --- ① is the same/ equivalent to Restriction VR + RE1= √3 --- ② (∵ △QRS ∽ △VRT, △RA1B1 ∽ △RE1T') ∠VTE1 is not acute angle. ---- this is better. So, In real implementation, ② is better than ① . (See Brown color bent segment [by wire/ chain/ flexible bar implementation? ] in above Fig.) drawing figure was easy. ∵ RA1 ⊥ A1Z , so, RZ = sqrt((3.5-x(R))^2+3^2), TZ = 4 like a backward somersault [gymnastic exercises] cf. Cody Rhodes' amazing moonsault from the top of a cage at MSG! (YouTube) The bigger N, the easier for implementation. ex. N = 5, Pentagon wheel, that's easy thing to make. Suppose saying an extreme example, this method doesn't need to divide equally. ex. 30°+50°+100°+140°+40° = sum 360°, such Pentagon hub wheel is OK. It is easy thing. ■★★ Purple figure ★★ ---- This is HOME RUN !!!!! , I think so. Purple color implementation is extended ② condition. Restriction VH1 = √3 (No-bent, No-through R, a bar) ---- ③ This is Feasible. I recommend this implementation. [strike]This is simpler implementation than N=2 Chebyshev Wheel.[/strike] ----- Exists the same solution in Chebyshev, too. ---- We can use this in real use world. drawing figure was easy. H1 is given by H1T'=1, H1V=√3 I1 is given by 4 times segment T'H1 J1 is given by I1J1 = 3, RJ1 = sqrt(RI1^2 -9) Tip: VH1 lead hand bar is key. in N = 2 case, TV = T' H1 = 1. So, in N ≠ 2 case, TV = T' H1 = 1 & VH1 is more smaller. ■ constant control (landing restriction) Left and right adjoining foot constant length restriction has options. If both feet have touched the ground, restriction can be removed. (but, its control is not easy.) My conclusion about constant value control. I checked many patterns, I've gotten next conclusion. --- Simple is the best. I recommend, the constant distance value ZERO. i.e. After taking off from the ground, you should make QR = 0 as soon as possible. ---- this is most easy. i.e. Shrinked form. Its shape is only one bar, like a closed umbrella. ■ N=2 Other property ∠QRT' ≒ ∠A1RT i.e. bisector of an angle ∠QRA1 ⊥ TT' ---- we can use this. ■ Compare, N=2 Chebyshev Wheel (GeoGebra). Which is simple? Please do implement. ex. caster/ dolly. ------------------------- | | | _ _ | | / / | --/------------------/-- /_ /_ cf. Chebyshev N = 3 Polygon Wheel (GeoGebra).