Contact us: office@geogebra.org

© 2018 GeoGebra

# Chebyshev N=2 Polygon Wheel

- Author:
- asifsound

Chebyshev Wheel
If finish shape were satisfied, middle form is any.
So, Add a bar H₁I₁ +--+, where EH₁=FI₁ [= r₂ ] (No need symmetry.)
[ H₁I₁ is a good lead very simple bar. ]
~~N=4 too~~. ---- N=4 case has no merit. why? )
cf.

**I recommend Purple color solution.**------ This is a remarkable method, I think so. I think N=3 is very affordable. cf. Chebyshev-like N=3 Polygon Wheel (Official ver.) [**2016 version**] [ But most basic case is N=2, human leg is biped, this is N=2 case. so, no need N>2 (??). Basic question: biped is low energy than N=2 wheel? Which is efficient? ] cf. N=4 case: Chebyshev Linkage Wheel --- typical.**---- Someone, Please convert this apparatus (i.e. N=2 or N=3 Chebyshev wheel) into a commercial reality.**This is no joking.**2018 version**of this N=3 toy is Polygon-head Chebyshev N3 Wheel** .**( This is better than above 2016 version. I recommend this N=3 implementation.**Chebyshev couple N=2 (couple version) 2018/03****■ Analogy :**This is virtual**r=∞**wheel. [here, r = radius] ( like as virtual storage/ virtual memory space [we say,**vs**] in computer: i.e. in virtual world: memory size =∞, in real world: assigning limited memory size/ disk dynamically. )**■ Line symmetry case was r₂ = 1 case in purple color shape.**H₁I₁ = sqrt(4(1/5)^2+(1+3(1/5))^2)=sqrt((4/5)^2+(8/5)^2)=sqrt(80/25) =sqrt(16*5/25)=(4/5)sqrt(5)=0.8√5 (=1.788854382) ----- or = 4/√5

**Tip:**By my test, this easy line symmetry simulation by constant length bar exists in not only Chebyshev ratio case, but also in other ratio case . ex. upper base=2, X-bar=√7, lower base=2, height =√3, crank range span is from -60° to +60° (range 120°) case, r₂ = 1.51 , H₁I₁ = 2 constant length bar supports

**exact line symmetry**. ---- WHY?, very interesting. What natural law exists in this special phenomena? cf. Chebyshev-like N=3 Polygon Wheel (GeoGebra) (upper base = lower base linkage (i.e.

**Antiparallelogram**) case is exact symmetry. ----- Hart's Inversor.)

**My simple Question:**Is this N=2 case line symmetry exactly or approximately? which?

**Answer:**It's approximately. not exactly. D(0,4), D(2,4), D(4,4) on 3 points, it's equal to exact symmetry shape. --- but, result is very good approximation.

**--- 99% = exact symmetry.**( but, You should be careful of "H₁I₁ ≠ 2 (≒1.79)" property. )

**■ About Other special condition**① r

_{2}= 1 causes line symmetry. It rotates 2 times angle. move at the constant speed [i.e. nertial law ] add same movement. So, 2 times movement is achieved. ② r

_{2}= 2 causes shrunken [i.e. angle= 0°, initial form mode.] . t

_{2}= 2. This special ratio length is useful, perhaps. Basically, If foot doesn't touch the ground, its linkage frame form is free. To save energy, I recommend r

_{2}= 2 setting. ----- always heel angle = 90° (⊥). ex. N legs wheel is easy?.

**★ r**Please set r

_{2}= 1.95 case is very interesting._{2}= 1.95 and check its behavior carefully. Axis D on (0, 4) case: foot is in left below. D on (0.1, 4) case: foot is in left upper. has jumped in a moment. and its shape is right triangle. The heel shape already exists. It looks like a biped lifting the heel highly. Totally, this case is near the biped mode. That is " biped motion and wheel motion is equal ultimately."

**In that meaning, this N=2 Chebyshev wheel is superior than other biped method apparatus.**

**★ r**r

_{2}= 2.46 case is very interesting/ honest._{2}= 2.46 is near the biped mode, too. [more than 1.95 case (?) ,,, honest foot trace. what happens in next cycle? ] ----- in next cycle, 2 feet touch the ground, and heels take off from the ground. ----- this is not allowable movement. So, r

_{2}> 2 value is not allowable. we should inhibit.

**this is bad pattern.**Please check this, too. we can easily step beyond over/ on the high barrier ___|

^{--- }by this (?). . heel angle is obtuse angle. this is better.

**■ Real implementat ---- Fact: in 3D real world, It doesn't need this arrangement. straight bar is OK/ no conflict.**Coordinator bar H₁I₁ conflicts with the red axis D ◯. We can avoid this. In r

_{2}≦ 1 case, point D is inner in H₁I₁ diameter circle (as purple colored circle in above Fig.). So, If you use the bar of H₁I₁ of hollowed-out bar (or circle etc.), you can crank the pedal. If you make Leg AE, BF as 凸 (partially convex leg form, upper 1 length part), then, you can get the No-confliction with the axis D.

**■ Simple question to YOU from me.**----- this is educational question. From my experience, Chebyshev linkage and Hart's A-frame are near these forms/ shape. α=∠AEF, β＝∠BFE then,

**sin(α/2) + sin(β/2) ≒ constant ----- eq. 1**is true. why? (if trace is exact straight line, = is true. Chebyshev trace is not exact straight line, so, ≒ .) Please Prove eq. 1 cf. Hint ----- N Polygon Wheel (N=2 shape) Hart's A-frame wheel sample QR ∽ RV (∵ △QSR ∽ △TRV)

**■ Noisy roller bag (/ wheeled bag/ luggage trolley etc.)**Generally, roller bag is so noisy. It's very annoying. It's all defective product. Please replace it with above caster on the world. --- if necessary, please put on the shoes to feet. (ex. in Venice) cf. Who is the best "Quiet wheel"? ... cf. DIY: Fixing That Squeaky Luggage Wheel cf. Venice isn't banning wheeled suitcases to stop noise pollution and protect its historic streets cf. Wheelie Suitcases Ban (Fodor's Travel) |---- in this. They say that "Rolling bags are one of the great inventions of the 20th century !! " ---- I think " it's somewhat so, but noise of caster is defective." ---- Above walking wheel is one of inventions of the 21th century !!, I think so. At least, it solves the Venice problem. Please enhance/ improve more. cf. wheeled luggage ban in venice on or off? (on 2015/03) cf. Venice Proposes Noisy Luggage Ban for Tourists (on 2014/11) ----- I propose, Italy should make a New caster named "

**Venice Chebyshev wheel**" as one solution for Venice problem. ---- this is no joking. cf. Venice To Ban Noisy Wheeled Suitcases (on 2014/11) ((C) dailymotion) vtr