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. ]
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.
N=4 too. ---- N=4 case has no merit. why? )
cf. 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 r2 = 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 r2 = 2 causes shrunken [i.e. angle= 0°, initial form mode.] . t2 = 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 r2 = 2 setting. ----- always heel angle = 90° (⊥). ex. N legs wheel is easy?.
★ r2 = 1.95 case is very interesting.
Please set r2 = 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.
★ r2 = 2.46 case is very interesting/ honest.
r2 = 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, r2 > 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 r2 ≦ 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