Musashikoganei Square Wheel

(subtitle: Peaucellier Linkage Wheel)

Bipedalism --- bipedal walking linkage mechanism

, Leg simulation Linkage/ Robot

Walking Bicycle

Leg by Chebyshev Linkage

Leg by A-frame crank

Ultimate Hart's A-frame Leg

Exact straight stroke Hart's A-frame Leg

Imai's Hart's A-frame Leg

Subtitle: Hart's A-frame having 360° crank is itself a walking Leg apparatus.
( --- I declare that This is one of the great discovery in 21th century.)

Imai's Hart's A-frame 7.5 Leg for Non-stop record (Vertical Leg)

Imai's Hart's A-frame 5.5 Leg for Non-stop record (Vertical Leg)

How to make the vertical leg from Imai's Hart's A-frame Leg

Imai's Inversor Leg

Imai's Inversor +2 Leg

Imai's Inversor out-crank Leg/ Decussation Bipedal

Imai's hiccup sector Leg

Hart's A-frame hiccup leg sample

2 bars crank amplifier

☞ Someone Please, prove that the below orange curve and Chebyshev Linkage curve are the same group, mathematically. (somewhat Chebyshev has sharp edge-end-angle.)

Centipede_circle_mapping.gif

Slider Curve-mapping Length law

45 degree slider Amplifier

Chebyshev-crescent-45 degree-slide-mapping

5 bars Chebyshev Leg (small span)

5 bars Chebyshev Leg (2 tier mapping model)
  --- Hypothesis

5 bars Chebyshev Leg (quasi-linear)

Find Symmetric bottom crescent !!!

Imai's linkage (alias, Crescent mapping leg.)

c. I tried to search the small INTEGER ratio environment, I compromised this result. B(1.41,0) is near (√2 [=1.41421356 ※], 0) (?). if so, this sample result is all 2 relatives.
Much much better another environment are there whose value isn't integer.

B(√2, 0) estimation is very, very good value.
ex. when D(1,0) --- F is on just middle of roof arc (√2, 3.98) , D(-1,0) --- F is on just middle of bottom arc (√2, 3.19) [where, 3.19 = sqrt(13-2sqrt(2)) ] . F most left is (-1.32,2.74) , F most right is (4.16,2.74) same Y-value, width =5.48. D(0,1) --- F(3.5,2.94) [where, 3.5 = sqrt(2) + sqrt(13/3), ], D(0,-1) --- F(-0.67,2.94) same Y-value [where, -0.67 = sqrt(2) - sqrt(13/3), 2.94 = sqrt((29+2sqrt(78))/3)-1 ---(= sqrt(26/3) --- what is this? why sqrt((29+2sqrt(78))/3)-1 = sqrt(26/3) ??? Nested radical )].
ex. x=(1/2)(sqrt(2)+1/sqrt(2)) = 1.060660172---- = (3/4)sqrt(2) = 1.060660172,,, magic?, difficult!!
ex2. sq((29+2sq(78))/3)-1
=sq((29+2sq(3*2*13))/3)-1
=sq(29/3+2sq(2*13/3))-1
=sq(29/3+2sq(26/3))-1 [here, 29/3=1+26/3]
=sq((1+sq(26/3))^)-1
=(1+sq(26/3))-1=sq(26/3)

Tip: Crank circle A and Crescent moon figure are in symmetric mapping relation each other.
i.e. Circle A +y, -y symmetry is corresponding to Crescent moon -x (=left hand), +x (=right hand) [here, x = sqrt(2) is center line symmetric line] symmetry precisely.
i.e. Circle A and Crescent are converted by y=-x line (- 45° line) .
c. Crescent curve: f(x,y) = 0 equation function, if so, f(sqrt(2)-x, y) = f(sqrt(2)+x, -y) is always true. ---- Please, prove.
c. Suppose Circle A : x^+y^=1 , Crescent: (X,Y), if So, next equation is true (= satisfies next). [ in above fig. D(x,y), F(X,Y), E(0.5(X+x),0.5(Y+y)) ]
• (√2,0) 2 circle is a middle point of Circle A and Crescent, so,
[0.5(X+x)-√2]^+[0.5(Y+y)]^=2^  i.e. [(X+x)-2√2]^+[Y+y]^=4^ --- ①
• length 4 bar, So, (X-x)^+(Y-y)^=4^ --- ②
From ①, ② we can get X,Y.
① - ② = 0; Y=(√2-x)(X-√2)/y ----- ③ ---- this indicates, Y=(√2-x)(-X+√2)/(-y) --- So, Crescent is X=√2 line symmetry. (Remark) † (∵ (x,y) to (√2-X,Y) mapping, (x,-y) to (-(√2-X),Y) mapping has the same Y.)
(if so, in above fig. Return_stroke = 90.14° description is wrong. real angle span is more smaller than this. and ± symmetric angle figure.)
X is next quadratic equation solutions.
[y^+(x-√2)^]X^-2√2[y^+(x-√2)^]X+[y^^+{(x-√2)^ -14)}y^+2(x-√2)^]=0 --- ④
(④ indicates, depends on y^, so +y, -y is indifferent.)

X^-2√2X+[y^^+{(x-√2)^ -14)}y^+2(x-√2)^]/[y^+(x-√2)^] = (X-√2)^ -2 +[y^^+{(x-√2)^ -14)}y^+2(x-√2)^]/[y^+(x-√2)^] =0
So, X=√2 + sqrt(2- [y^^+{(x-√2)^ -14)}y^+2(x-√2)^]/[y^+(x-√2)^] ) = √2 ± sqrt(y^[-1 + 16/[y^+(x-√2)^]] )
So, x^+y^=1 apply, then --- X= √2 ± sqrt((1-x^)[-1 + 16/(3-2√2x)] ) --- ⑤

test: x=0 case; X= √2 ± sqrt((1-0^)[-1 + 16/(3-2√2*0)] ) =√2 ±sqrt(-1+16/3)=√2 ± sqrt(13/3) -- oh! valid value!
test: x=+-1 case; X = √2 ± sqrt(0) = √2 --- true!, then Y = sqrt(13 +- 2√2 ) ---true. (not available ③)
----
----
Please find theoretical arm3 (now =7), arm4 (now =6.41) value. [ arm4 = 6.0 is better than 6.41 ? --- YES. ]
[ arm3 = 7.0, arm4 = 6.0 (all INTEGER) is the best.
ex. in tri-pedal case; when D(-1,0) case, 3 legs are colinear and each y value is the same -2.81, i.e. all on the ground with no height gap.   but bottom line total has a up-down wave; y = -2.8107(=left) to -2.7752 to -2.8107(=center) to -2.7752 to -2.8107(=right) [gap 0.0355] in stride x span 11. [error rate: 0.035/11 = 0.32%], lowest 3 points are corresponding to -180°, -180±120° 3 points. ---- (-2.81 is near -2√2 ?), ----- these 3 points are local minimum.
Chebyshev linkage case; y = 4.0000(=left) to 4.0098 to 4.0000(=center) to 4.0098 to 4.0000(=right) [gap 0.0098] in stride x span 4 [error rate: 0.0098/4 = 0.25%] --- has the same error order (?) and same up-down wave trend. lowest 3 points [= local minimum] are corresponding to -180°, -180±90° 3 points. and -180±120° point are corresponding the most left/ right point y= 4.062 [gap 0.0620] , So, angle trend is different. Chebyshev is fit to 4-pedal. And, Imai's is fit to tri-pedal.
Chebyshev case 120° is already in return stroke mode. but Imai's Crescent case 120° is in forward stroke mode yet.
---   as of 2014/06/10 Fumio Imai. ]

(Remark) † : symmetry property is generalized as below
• center B(a, 0) radius b circle [(x-a)^+y^=b^] is a middle point of Circle A and Crescent coupler/ bridge bar, and radius b length is the same as half coupler bar length.
i.e. (1/2(X+x)-a)^+(1/2(Y+y))^=b^ ---- eq.ⓐ
then ((X+x)-2a)^+((Y+y))^=(2b)^ --- eq.ⓐ'
(X-x)^+(Y-y)^=(2b)^ --- eq.ⓑ
(2X-2a)(2x-2a) +(2Y)(2y)=0, So, (X-a)(x-a)+Yy=0 ---- Y = (X-a)(x-a)/ y ----- eq.ⓒ
So, arbitrary center (a, 0) radius b circle ensures the symmetry property.
i.e. when x^+y^=1 is the crank and center (a, 0) radius b circle is middle point of arm from crank to objective trace node.
then, Crescent moon (or other shape) mapping image is (always) X = a line symmetric. (not dependent to b length.)

(√2,0) radius 2 circle is only a one typical crescent sample. it's just the tip of the iceberg.
Also, (2,0) radius 2.5 circle of Chebyshev Linkage for approximation of line drawing is a tip of iceberg. Many other ratio value set exists.
c. eq.ⓒ also indicates "Vector (x-a, y) inner product Vector (X-a,Y) = zero." i.e. 2 vector are in ⊥ relation. tanθ = -1/tanΘ; ,,,, ∠ABF = ∠ABD + 90°, So, point F most right point is corresponding to point D which BD is tangent to th circle A case [--- = returning point, angle summit point.] .
So, return stroke narrow crank angle is corresponding to this tangent +- angle.
i.e. center B(a, 0) value is quick return velocity barometer. i.e distance |a-1| is smaller is tangent angle is smaller, ----- So, if you want to achieve the more quick return, set value |a-1| smaller. ---- this is important.
My Crescent shape case B(√2, 0), Chebyshev Linkage case B(2,1) , so, √2 < 2 then My crescent Linkage achieve more quick return than Chebyshev.

[ witness; Chebyshev B(2,0) circle case: AB=2, tangent AD=1, ∠ADB = 90°, then angle is ∠BAD = 60° (∵ cosθ=1/2), so +- sum is 120° -- so, forward span: return span = 240:120=2:1 (i.e. 2 times fast)
B(√2, 0) circle case; cosθ=1/√2 --- ∠BAD = 45° so +-sum 90° -- so, forward span: return span = 270:90=3:1 (i.e. 3 times fast) ]


-----------------------------------

Here is the Anchor name = "Cheby_sym_mapping".

(quasi) Vertical straight line slider doesn't need long long radius bar.

In above Fig. Vertical slider is realized long long ∞ length bar (i.e. radius = ∞ 's circle's arc.), it's approximated by long radius long bar. --- I called this result line as "quasi linear".
In general this is true. but, above Imai's Linkage case, short radius is OK, I found by experience/ test.
Why? Short is ok. ----- I can explain its reason now.
This trick comes from the analogy of above section, +- y symmetry (input circle)--- black-box mapping (by constrained linking bar)--- +- X symmetry (output).
The same principle occured, black-box is a circle, middle of linking bar is on circle, and radius length = 0.5 linking bar.
input is MN bar circle. FI ≒ FJ, and crescent bottom curve is ≒ FJ radius circle (this environment is the same as before section story.), output is Leg trace point J.
So, Even short MN ensures output x symmetry. So, MN is short is OK.
----- See the DEMO, and change the length MN by yourself. and get feeling. MN = 2 is OK, MN = 1 NG ( too short).
(my explain might be mistaken. but, short MN is OK. this is FACT.)


This is related Watt's Linkage.
cf. Robotic Mechanisms Linkages ? Watts Straight Line 51016
Watt's Linkage is equivalent to Evans Linkage.
cf. Robotic Mechanisms ? Linkages ? Evans Straight Line 51013
cf. Chapter 3 Linkage Synthesis ( famous linkage) ( (C) Mechanism Simulation in a Multimedia Environment )

I've inspired/ ignited the new leg method/ idea. 3 bars/ leg ----- perhaps,
The smallest number of bars Leg linkage system (in the world) (?).
(2 straight bars (crank bar, r length bar) + 1 bent bar ( 2r length bar + foot part); half Lemniscate curve trace shape. )

---- Please check.

cf. Optimization of Watt's Six-bar Linkage to Generate Straight and Parallel Leg Motion
(.pdf, (C) Hamid Mehdigholi and Saeed Akbarnejad @2012) --- Figure. 10. d) is good. but, orthodox (? -- somewhat, yes.).
    |--- BYU - Idaho ME: Walking Mechanism with Straight and Parallel Leg (Youtube)

My comment: Figure. 10. d) is 5 bars/ vertical leg. this is considerable good. 2 legs are 10 bars. --- excellent.
This is superior to my Imai's Linkage (tri-pedal). Because less number of bar, 10 < 15. and perhaps, less slipping. simplicity and robustness comparison needs the real apparatus test.

Java applet Online-DEMO (GeoGebra) ; See (13-j)

Imai's Watt's linkage Leg (3 bars/ Leg)
--- Why exists Arches of the foot? ---

I found the reason the existence of Arches of the foot.
Bi-pedal system needs such shape foot. For return stroke support, 180° span angle of crank needs for forward/ return stroke, but 120° span is easily achieved by simple foot shape, more 60° end phase of forward stroke (this 60° phase is lift mode ?) are needed a special shape foot, i.e. toe part are needed. and the area between the heel and the toe is flat or 凹 or hollow shape is allowable. (ex. in below Fig., between tF and F' area is hollow is ok. size zero is ok, of course.)

Below Leg model of linkage is the most primitive/ basic style 3 bars/ leg. this model teaches that such fact.
( Please try DEMO, and feel this fact. please try your design of foot.)

c. Condition, segment length BA = 2r - long, is satisfied. ∵ at crank angle = 0°, G₁ mapping to A (G'₁).

c. Why the lift mode foot bottom is such wide? almost stride size. too big. it's bad.
foot length ≒ leg stride . ----⇒ this is not a leg, isn't it? ,,, similar to a roller, wheel??

Someone, please tell me why.

Q: This odd leg is useful, ??? --- A: as, low height large radius wheel.

------- Anyway, if dislike this, apply the 120°-120°-120° tri-crank, tri-pedal/ set. i.e. 9 bars/ set. ---- sufficient it is!


cf. related site: 佐藤ロボット研究所不定期日誌 (Language is in Japanese, @2008) --- see 2 animations (Please Click the figures in page, animation will be spawn.).


c. This apparatus foot curve was approximated by a complete circle arc. it was easy work to find the best fit curve. Because, most easy curve is circle, and it looks very very good curve.
Someone, please prove that " the best fit curve of this apparatus is circle (?).".
→ This hypothesis is true. I can prove.
Please check of the trace of center of this fit large circle in DEMO. its trace is drawing the horizontal line. So, this hypothesis is true.
this large circle radius length depens on the "r, long, AB =2r - long" relations.
Chebyshev linkage case "r= 2.5 long, long, AB = 2r-long= ignore = 2 long", large fit circle is not needed.
Further, this large circle center trace rotates the same rotaton direction of crank. and diffrent angle tuning has symmetric trace curve, like a chebyshev curve, but rotion is same sa crank.
So, this trace curve mechnism is superior to Chebyshev mechanism. I think so. This is related to the story of 45° angle shift mechanism previous described.





- Centipede_Watt.gif -


bold bars only.

AA' part of trace is almost straight curve. but BC radius ▲―● is short.

Orthogonal dashed line ( ― - ― - ― - ) is symmetric line of input and output figure.

( Mapping from G_i to G'_i [ i = 1 to 4 ] )

This is related to " Y-axis slide mapping to X-axis slide by mediator circle x² + y² = r² : --- Scott Russell linkage. ".


• leg stroke under the crank. --- good.
• but, forward interval span angle is about 120° --- bad/ small.
(at least 3 legs needs to walk.)

I tried 120° up to 180° by adding improved foot. Then , I found that foot shape has the arch of the foot.
i.e. This foot is not Flat feet.
---- This is very intersting.


- Centipede_Watt.gif -

-------------------------------------

Imai's Watt's linkage Leg walking Toy on the slope

I designed a below bi-pedal apparatus. flat feet (distance between tF and F' is zero).
Please make below toy (if possible), and challenge the GUINNESS Book non-stop records.
Pink ball ● is a heavy ball and is rolling on the foot plate. generates unbalance state. Switch/ transfer to another foot at the time of both feet exchanging.
A kind of The Tumbling Toy.

What's the meaning as foot size > stride size ?
slip occurs? --- No. and, moved distance is just stride [= x(G'₃)-x(A) ].
A'G'₃ curve is achieved by tF-I' long curve bar rolling result.
feeling: stride = length of tF-I' curve.
quick speed: ( tF-I') / [( tF-I')-(AA')] times velocity.

Tip: Low energy tuning;
From a view point one leg to other leg, forward stroke leg lifts a other leg by tops arm CC'₂ ( length = 2*long ), moves angle from 0° to 180° + ∠G₃AG'₃, and other leg is in hanging state.
So, this energy of lifting is not little, in order to decrease this energy, set short the length arm bar. i.e. long is short is low energy consuming.
Who lifts arm ? ---- line of (AB - angle₂) is vertical is implicit condition.
( Gravity makes effect/ effort to become vertical. ? )
Changing the point of view is needed.

Conclusion (?):
in order to ensure AB line vertical, I recommend to use 4 long toe legs system, i.e. 90°-90°-90°-90° crank system. 2 legs always on the ground, so AB is vertical. total 3*4 = 12 bars/ system. but structure is simple. one leg is light ( not heavy). perhaps, better than Chebyshev 2 vertical legs 5*2=10 bars/ system.
(Groud slip problem will occur? --- Can't imagine. Needs check by DEMO. )

Java applet Online-DEMO (GeoGebra) ; Enjoy (13-k)

- Centipede_Watt_toy.gif -


- Centipede_Watt_toy.gif -



-------------------------------------

Imai's Watt's linkage Vertical Leg (5 bars/ leg)

Related site is below.
cf. Optimization of Watt's Six-bar Linkage to Generate Straight and Parallel Leg Motion
[ Figure 10. c) or d) are the same figure to the below "Tchebycheff's Engineers' Table" (up/ down is reverse).  i.e. The above paper has no originality at all. but, useful. ---- I think so. ]
cf.
【Tchebycheff's Engineers' Table】 Another bit of excess - but a valid, real-life application.
http://www.mfdabbs.pwp.blueyonder.co.uk/Maths_Pages/SketchPad_Files/Mechanical_Linkages/Mechanical_Linkages.html
Below figure's Copyright is in above site all. © Mark's Maths and Excel Homepage

Here is the Anchor name = "Chebyshev_parallel_table".


My comment:
This is a table moving linkage apparatus with keeping almost exactly parallel to the floor.
(non-pantagraph parallel moving technique.)

----- proof -----
Naming of node:
 ( D m F)  G
     M
A        B
(M, m is middle point)
[ where, AB = 4, height = 4 (= distance from AB to mG), AF = 5, BM = 2.5, MD = 2.5, MG = 2.5, Dm = 1, mG = 2 ]

I will prove AB // mG and AF // MG.
length condidition: mG = 0.5 AB, MG = 0.5 BD, AF = BD
and M, m is middle point.

Vector sum; AB+BD+DF+FA = 0 --- ①,
(Here, under bar mark is Vector mark.)
if so, 0.5AB+0.5BD+0.5DF+0.5FA = 0 --- ②,

0.5 AB = mG, 0.5BD = MD, 0.5DF = Dm, 0.5FA=GM

mG, GM is size is OK, but direction is not sure yet.
Next, check △MDm. ----- △MDm ≡ 0.5 △BDF
and all direction is the same.
So, Mm = 0.5 BF
So, △MGm ≡ 0.5 △FBA
i.e. mG // AB, MG // AF are true/ proved.
So from ② is equevalent to
mG+MD+Dm+GM = mG+GM+MD+Dm = 0 --- ③
(proof end/ Q.E.D)

(The heght is almost constant is a different property. ----→ Chebyshev Linkage.)

Tip: Also, This figure indicates that Chebyshev Linkage is equivalent to Hoekens Linkage. i.e. These draw the same curve.
Tip₂: If this table is set on the slope, what will happen?
Upper part will Automatically move/ slide to downward. ---- this property can be applied to the passive walking biped.



c. Watt's linkage and Chebyshev linkage are in the same family.
i.e. (Crank bar) + (2r + r , 90° generate coupler bars) sum 3 bars system
c. Recursive propety exists.
c. Below figure trick (Vector sum = zero, trick) is near the generalized Hart's A-frame trick a bit.
cf. "True straight-line linkages having a rectilinear translating bar" (.PDF) by © E. A. Dijksman (1994)
cf. Kempe's (Focal) Linkaget Generalized, particularly in connection with Hart's second straight-line mechanism (.pdf) (© E. A. Dijksman , 1974)

c. Below vertical leg is 120° crank angle cycle, so, tri-pedal is needed.
tri-pedal is bad/ not clever.
Compromising method is adding long toe (long arc toe described in previous section). By this, 180° crank angle cycle can realize. ---- i.e. vertical leg + long arc toe (OR mode/ both function mode) ---- this is enduring to GUINNESS Book challenge. (?)

Java applet Online-DEMO (GeoGebra) ; (13-l)

- Centipede_Watt_vertical.gif -


Please use as tri-pedal element/ component.

----- Applying to Chebyshev Linkage (bi-pedal) vertical leg version is the same 5 bars/ leg, of course. (∵ in same family). So, I recommend Chebyshev.

nodes A'₁ and C'₁ are under control/ fixed, So, node E is dependent/ automatically uniquely fixed.

(∵ In creating a triangle, if base edge end 2 nodes position and between of those nodes 2 edge length are given, such triangle figure is uniquely fixed/ decided. )

Another easy proof:
CC'₁ and A'₁A are same length and X crossed at middle point, so ACA'₁C'₁ is rectangle, all corner angle is 90°, then AC // C'₁A'₁, further △ABC and △C'₁EA'₁ have same length 3 edges, so △ABC ≡ △C'₁EA'₁, then AB // C'₁E.

(i.e. 2r, r bars are an implicit parallel mechanism. so, number of bars can be decrease.)


- Centipede_Watt_vertical.gif -


----------------------------------
Here is the Anchor name = "Centipede_Watt_Leg_180deg".

Imai's Watt's linkage Vertical Leg (180° cycle)
Subtitle: Competitor method against Chebyshev Linkage

This is a method for Watt's linkage Crank angle 120° cycle extending 180° or more.
(and, Leg stroke rotates as the same crank rotation direction. --- I think this is important.)
i.e.
(a): Imai's Watt's Linkage Leg is mapping crank to X-axis bottom stroke.
but, bottom part crank angle < about 120°.
(under condition; crank center B, r/ 2r bar center A, AB = 2r -crank radius [ perhaps this restriction is big/ influenced. ])

(b): (a) environment has half stroke ( corresponding to crank 180° span) can be simulated by large radius rotating circle.
[ This large radius circle is corresponding to rr in previous Fig. - Centipede_Watt_toy.gif -. ]

(somewhat incredible matter. because (a) half stroke curve has flat bottom and acute standing curve, this can be mapping by all flat line. but, this is in real world. )
How to find such large radius circle? Answer is below.
By this we can make a great environment from ordinary environment of (a).

above-the-crank type 3 bars/ leg flat bottom crescent .
This can be shifted under-the-crank type by Chebyshev trick by only +2 bars adding. and vertical leg can get, too.
[ ADCINM crimzon color frame is 5 bars Leg apparatus. Also, ADCIKJA (including yellow-green color) is variant frame. ]

(Please review by DEMO. try to changing parameter value.)

c. This method is related to Hart's A-frame Leg and Chebyshev linkage Leg. both.
It seems as if hybrid case.

Theoretical bar length:
Theoretically below estimation is true.
Below Fig. describing at crank 180° position situation, Node I (Please tuning --- Slidable) position is decided by tuning.
but, estimation value is given as below.

Crank angle 0° corresponding position on trace is W (= crank angle 0°)
△WCI ∽ △ACD [isosceles triangle] , ( i.e. start-end crank angle 180° span triangle mapping, WI // AD. this approximation is rough but considerable good. )

WC: CI = AC : CD --- (IC + 2long) : IC = (2r - 2long) : r ---- so, IC = 2long*r / (r - 2long) [very simple]

( r = 2long case is infinite, but in real world, no problem. please try, and tuning.
--- As a matter of fact, long < 0.5r and Try N * IC ( N is Integer [N = 1, 2, 3, ...]), please.
Someone, please give me more precise formula.)

Java applet Online-DEMO (GeoGebra) ; (13-m)

- Centipede_Watt_vertical_180deg.gif -
)

Please extend the rectangle diagonal line bar.
( toward upper left direction )

△CBA and △IKJ relation is in complement relation.
In general △ABC and base-edge AB and left edge b (= opposite to B), right edge a (= opposite to A), ------- bABa is complement to aABb triangle. [exchange a, b --- such trick]

This trick is also used in Hart's A-frame.


- Centipede_Watt_vertical_180deg.gif -



----------------------------------

Imai's Watt's linkage Leg (180° cycle) Judge

In general, Imai's Watt's linkage Leg always has its 180° type flat crescent, especially r > 2* crank radius case.

[ r = 2 crank radius, seems a border line criteria, 凹凸 switching border. Typical flat bottom crescent is r=2, u=0.9/0.8/0.7 etc. case. Now, the rotation direction of the crank and the leg stroke are the same direction. This is ordinary/ natural event. we can say that, Chebyshev linkage reverse rotation is unnatural event, now. ]

I introduce the best-fit bar length estimation tool.

Please set the input r, u [u is coming from u of unit radius]. And, find best fit XC output by yourself tuning.
(Automatic XC from r, u. formula doesn't exist yet. ----- Someone, please make it. ※)

below tool, is checking 4 points on input crank circle curve, mapping to output crescent flat bottom line.

Tip: Epsilon parameter ε is a adjustment parameter by my experiment. [ Now 0.01 is specified. but 0.005 is OK. 0.001 is BAD.]
round off the horny corners.
( --- this is corresponding to the C' trace turning angle at point A is very acute/ sharp (= zero, tangent), to reduce this sharpness. --- to shift the node A position to right bit. )
cf. Sans-serif. (wikipedia)

※ I found by chance,
(1) Before section formula is : XC₁ = 2ur / (r - 2u)
(2) Long ago, I proposed formula: XC₂ = (1/u)(sqrt(u^+r^)-u)(sqrt(u^+r^)+r)
(cf. Slider Curve-mapping Length law in this site. [spawns new tab])

By my experiment, both estimation result is 5030% true and 5070% false, case by case respectively.

But by my observation, mean of both, i.e. XC₃ = 0.5 (XC₁ + XC₂) is almost 100% true.
Why? --- I don't know. --- but, very useful. ( I am a Pragmatist. (?))

[ Remark: XC₂ estimation value is very precise value in FH interval. completely fit. but, this value is broken for HW interval coordination. For Compromise, this value is decreased. ]

Tip: r=4.8, u=1.6 ---- this sample is great. 4 points are colinear completely. 4.8 = 1.6 ☓ 3 --- is related?
((r=3, u=1) (6,2), (9,3), (2.1, 0.7) --- all cases, almost exact straight line, too. )
I can declare that

Imai's Watt's linkage Leg,
In ratio r:u = 3:1 case [and XC = (10/3) r or π r (π = pi=3.14..) ],
bottom is completely flat.

[ in this case, even ε = 0 is OK. but ε > 0 case is better (but, large ε is worse. ε ≒ 0 is needed.). I recommend ε > 0 applying. 3:1 case result is prominent than other case. ]

i.e. AD, DC hinge length r = 3, BC crank radius u = 1, AB = 5 (= 2r-u), CX = 10 (i.e. DX=13) [, and stride is about CX 10. ],

crank angle at K case, △ABM is just 5-4-3 triangle [ ,where M is tangent point]. and corresponding to this, △W'BK' is about 10-8-6 triangle.
(Like as Chebyshev Linkage, this Watt's Linkage, right-triangle 345 is related.)

(Please play DEMO. Perhaps, new discovery happens.)

Java applet Online-DEMO (GeoGebra) ; (13-n)

- Centipede_Watt_180Leg_Judge.gif -

By left figure case, I show how precise this exact line is.

length of F'W' = 8.95,
and define the line through F' and W', and define the intersection of line F'W' and CX line as R.

the length segment XR (= gap) is ,
from F' to H', 0 to 0.007 ,
from H' to W' , 0 to 0.047.

i.e. error rate = 0.047/8.95 = 0.005251 = 0.5251%
---- this indicates very precise / almost no error.

Chebyshev Linkage bottom line error rate was 0.25% order.

0.5% is sufficient to use in real life work.

c. Adjustment ε = 0.02 case; from F' to H': 0 to 0.013, from H' to W': 0 to 0.015, So, 0.015/8.95 = 0.001676 = 0.167% --- better than Chebyshev Linkage.


- Centipede_Watt_180Leg_Judge.gif -


----------------------------

Imai's Watt's Linkage curve

Watt's linkage (wikipedia) has 3 bars linkage, length a, b, a , and the trace of middle point M of bar b, is.
Imai's Watt's Linkage curve is variant of Watt's Linkage, 3 bars a, b, a , this is the same, middle bar b, extend the both end node by crank radius length.

ex. In below Fig. (pink 3 bars H'H'''A'A is Watt's Linkage), bar b is H'''A' → (extend both to outside) X₂H'''A'X (here, X₂ and A'X length are crank radius length)
node X₂, X draws the closed loop, bottom is almost flat.
large radius AA' circle, small radius A'X circle, and the center of small circle moves on the large circle, and is self-spinning.
[ i.e. This is orbit of planet system. large radius is Orbital revolution, small radius is Earth's rotation. and it's related the Cycloid on 凹 curve road. i.e. especially, Hypocycloid. small-radius/ large-radius = r/R = 1/2 case is straight line (= diameter),,, --- this is in same concept.
large R length, small r length, R rotate velocity, r rotate velocity, these 4 parameters combination problem.]

large radius 凹 surface, +, 凸 curve by small radius circle (= like a Hypocycloid 凸 curve) ----- makes straight line curve. considerable good straight line. We are happy.

Java applet Online-DEMO (GeoGebra) ; (13-o)

- Centipede_ImaiWatt_curve.gif -

condition;
AC = u =1, AB = 4.99, CD = DB = r = 3, CX = about 10
(key: AB = 2r - u = 5. and r/u = 3.)

AA'MH'''H' is similar to AD'M'DB.
extend ratio is, r = 3 mapping to CX, so, ratio = (CX/3).


- Centipede_ImaiWatt_curve.gif -


-------------------------
Here is the Anchor name = "Centipede_Watt_slope_toy".

Imai's Watt's linkage vertical Leg walking Toy on the slope

Please make this concept toy. Long stride non-stop walking linkage.

By the heavy rolling-balls, shifts its weight back and forth, predictively, for the inertia control.
[ At least, steep slope and short rr (= crank radius) case, the rolling ball trick doesn't need. ]
(i.e. the center of the gravity of the apparatus body is in right hand area of suppoting pivot point. both phase 1 and phase 2 cases. So, rectangle automatically expands and shrinks. --- = automatically walks.)
[ and, In below Fig., suppose the ball ● and ● were very heavy, if so, the lifting leg 100% goes ahead, so this logic is true. ]
( This phenomenon is not queer at all, as if the ball rotates on the slope automatically, or, the water wheel rotates by water stream. )

Iteration motion of expanding/ shrinking the X shaped rectangle Pantagraph (like a pair of scissors).
( Like a Geometer moth walk [see this picture.   Not Marilyn Monroe - Walking (YouTube)] )

Keeping vertical standing position. And, 6 bars/ system, very compact system, without departing from the straight course.


Please challenge by this, GUINNESS Book records, the non-stop walking for one year. (Not exaggeration at all.)


cf. MEAM 248: Passive Dynamic Walker (YouTube) --- usual concept.
cf. 2 Legged Robot PLANETARY GEAR System 11102 (Posted on Jun 14, 2013, © ROBOTS ROBOTICS.) --- somewhat akin to.


c. Another condition is pending now. Under consideration. --- Found.
How to manage the heavy ballancing balls, is not difficult. there is a law/ rule.
Law: The horizontal distance between the ball and another leg is constant (or more).
( This is similar to the action of swing one's arms in walking. (?) )

[ connect the ball and another leg by stick (no rolling ?), or , pushing/ pulling the ball from the side by stick, or, magnetic repulsion force (weight slide method), or, switch the weight position by any method (possible?), or, get its Envelope curve of the ball rolling, etc.
---- but, within the same (relative) system pushing each other is in vain. needs the power/ trigger/ influence from out of (absolute) system. So, the method which searches for (absolute) more lower place is the last true method. (perhaps)
---- but, stick pushing method is easy to implementation. So, we should use this (??). ]

--- Please see the DEMO (15-b) ----- but, the simplest implementation is (15) original method.


My Request:
I don't get the best solution by this apparatus, yet. The management of center of gravity is difficult (?). but, there must be existed a solution. Please tell me its result, if you could solve it.

Java applet Online-DEMO (GeoGebra) ; (15), (15-a), (15-b)

- Centipede_Watt_slope_toy.gif -

phase 1: left leg on the ground, and right leg is lifting and leaping over the left leg.

phase 2: same as right leg.

foot trace shape is upper half part of the lemniscate shape.

Key point is, whether Crank C can rotate or not.
---- that's all problem/ matter.
---- I'm afraid that stopping the crank from the first and remains in its status for ever.


Key is the design of phase 2 action.
phase 1 is no problem. Because center of gravity is already in the right hand area of the pivot point A (0,0) (= triangle corner).

In phase 2, the pivot point is F (in phase 1 C'), we must turnover/ tumble FF' bar clockwise by the kinetic moment power.
Pink color segment is horizontal angle line upper and lower limit. i.e. crank angle 0° (= phase start point) and 180° (= phase end point)...., the meaning in detail can be understand by DEMO. Please see/ play DEMO.

In Phase 2 case, crank I rotates upward direction, so, to manage causing this action is difficult a bit. ---- balancing ball rolling (= changing the center of gravity position) is needed to manage this.


- Centipede_Watt_slope_toy.gif -

-------------------------
Simple variant

Imai's Watt's linkage vertical Leg walking Toy on the slope
(with horizontalg balancing bar)

Alias; Passive walking biped toy (long stride)

Comparison Before fig. and below fig.:
Before fig. thinking is very honest. but, tuning angles are dependent to the slope angle, it's not generalized.
On the other hand, below has independent angles, but it's not intuition.

For example: in below, suppose, I is just on N', if Add E₁ more weight, what will happen?
Does triangle rotate to the right? ------ I can't predict the future by my intuition.
but, the center of gravity is in right hand area of tumble criteria line, so it will tumble.

Which is true? Before or Below? ------ Both (?)

Java applet Online-DEMO (GeoGebra) ; (15-c)

- Centipede_Watt_slope_toy_balance.gif -

Adding brown horizontal balancing bar to the middle of vertical bars.
(--- The more higher position, the more effective.)


By this, in phase 2, Total center of gravity is in the right-hand area of the rectangle tumble criteria line, so, FF' line rotates automatically to the right (i.e. the tumble can be realized).
------ this brings about happy end.

This is a symmetric solution.

Tip: This is a dynamic balancing.
So, It can be applied to right-down/ left-down mixed slope.

i.e. backward walking (switch back) is OK.


- Centipede_Watt_slope_toy_balance.gif -


----------------------
Variant: Above is equivalent to the next.

Imai's Watt's linkage vertical Leg walking Toy on the slope ( with Pendulum )

Alias; Passive walking biped toy with Carrot and stick (long stride)

I wonder if this works really? --- Please check it out !!!
( In theory, my logic is true. ---- Evident.

【 In rigid body dynamics, the condition to tumble or not is dependent to the position of its center of gravity, not related to the inner structure. inner structure can be regarded as black box. 】 )
cf. Lever/ Law of the lever (wikipedia)
cf. overturning moment (in Japanese page,,, Letter maybe broken, but Fig. is international.)
--- See, last figure. left moment = 5t × 3m = 15 [tm], right A moment = 3t × 4m = 12 [tm], right B moment = 3t × 6m = 18 [tm] , ---- So, case B, the crane turnover will happen (∵ 18 > 15).

Tip: This implementation has many variations. If biasing the center of gravity , any method is OK.
----- Center of gravity of this 6 bars original apparatus is in about almost K position, So, adding stick and carrot causes to shift its center of gravity to the right direction. --- this is valid. It can be easily understood.
( i.e. this apparatus, phase 1 and 2 both are always in overturning mode. it has good successful structure. )

Tip2: Imai's Hart's A-frame 5.5 Leg for Non-stop record (Vertical Leg) is also passive walking biped itself. I think so.

(but, the Watt vertical is simpler than the A-frame vertical. why. because, adding the weight is corresponding to one specific leg, so, after the pivot leg switched , the weight effect doesn't work. this property is very useful. (?) ---- i.e. Even if adding very heavy weight were done, it doesn't effect/ influence another leg. it's Like a orthogonal axis property.
To realize this overturning moment trick, the A-frame needs adding to the vertical bar the longer stick and carrot bar. but the A-frame case, its top position (= crank center = waist position (?)) trace draws exactly the straight line parallel to the slope.)

Java applet Online-DEMO (GeoGebra) ; (15-d)

- Centipede_Watt_slope_toy_pendulum.gif -

right down slope case sample.

Carrot and stick (wikipedia)

cf. YouTube: Carrot Stick Riding


As a matter of fact, wire length ZERO is OK.

- Centipede_Watt_slope_toy_pendulum.gif -

----------------------------

Imai's Chebyshev linkage vertical Leg walking Toy on the slope ( with Pendulum )

Application: I provide this for studying.
Please brush up more.
Chebyshev case, perhaps more improve/ more tuning is needed. I don't satisfy now.

impression/ feeling: Chebyshev bottom and roof shape is near the flat, so, it works by lower energy.

Java applet Online-DEMO (GeoGebra) ; (15-e), (15-f)

- Centipede_Cheby_slope_toy.gif -
phase 1 is pivot vertical leg is on E₁ ▲.
crank angle interval is from th' to th (anti-clockwise direction). foot trace is arch.

phase 2 is on K ▲.
crank angle interval is from th' to th (large angle). foot trace is \^{―――――}/ shape.

It looks like no need adding weight.
I don't know.
I can't explain by intuition.

length: AB = 1 (crank radius),AH' = 2, H'C = 2.5, BC = 2.5, CB' = 2.5, DB' = 4, B'₁D = 5, M (mid of DB' ) trace bottom from A = 4.

Someone, please show the true answer.

------------
right down direction force by gravity is fit.
but in phase 2, Can you explain the first up "\" action ?
this action is caused by N'E (N' is fixed, E moves to the right) rotates anti-clock. and this is caused by NO (N is fixed, O moves to the right) ----- these are induced by heavy C₁ ●. ------ this is somewhat odd (?).
I can't understand by my intuition.
Something another trick/ concept is needed, perhaps.

c. ---- phase 2 , real pivot of lever is N' or N, and hanging lever. So. additional weight must be in left hand area. so, C₁ right hand picture is mistaken.

- Centipede_Cheby_slope_toy.gif -
------------------
the same. revese type.

Imai's Chebyshev Biped Toy

I am satisfied, now.

c. new variant: It looks Ok another simpler method (?). i.e. Not adding L letter weight to D or O', lengthen PN to PNP' and add P' position the weight, this causes PN rotates clockwise power (here, N is fixed point), so induces the crank radius counterclockwise easy rotation. --- please try.

c. the L letter bar weight logic is mistaken. feeling is OK, but, the shift of center of gravity doesn't happen. its gravity center is on vertical line of just under the hanging D or O'. This mobile sculpture is (only) the same as one weight. --- But, feeling is important.

c. If one motor were supported, applying to m (or N') point, the biped can be made easily.




- Centipede_Cheby_slope_biped.gif -

AB (= 4) is leg foot itself.
Simpler than before.

in phase 1, By Pivot B, bar FB overturns clockwize by the weight A₁ ◆.

in phase 2, swinging motion of the bar. but at the beginning, trigger force is needed to lift the foot.

( my feeling: this swing motion is corresponding to the cycloid motion ( under 3 distance), π r = 3.14 (180°) is to 4.0 (=AB), i.e. crank rotation is to circle rotation on the flat road. i.e. Chebyshev crank rotation is 1 to 1 mapping to circle rolling on the road.)

the L letter bar weight ● on the crank.
(almost horizontal.)

Keeping the heavy crank in the left hand area of the circle.
( induces counterclockwise crank rotation. )

- Centipede_Cheby_slope_biped.gif -


------------------

Roberval balance

- My simple question about Perpetual motion -

I produced next apparatus. Is this a perpetual rotation machine, isn't it?
( and Please apply this crank to the passive walking biped. )

feeling:
Crank A (A is an axis/ axle of this crank), diameter BB', both end B, B', set the heavy balls D, D' to the horizontal left hand of B, B' (where, distance DB is constant.).
[ BC, CC', C'B' additional bars are making horizontal bar mechanism. A' is mid of CC', and is the fixed point on Y-axis. ]

What will happen?
( Rigid body of this apparatus (= crank bar system). The supporting pivot point is A (0, 0). but, center of gravity is M (= different constant point from A) (-d, 0). i.e. A ≠ M. ----→ usually, it automatically rotates by unbalancing.)

c. I tried many other structure, but all were failed. but below is stable and successful (?).
other cases, pedal downward is Ok, but upward supporting force was insufficient.

M is exactly fixed point. if d = 0 --- then M = A, this case is typical case, if no-friction, crank rotates forever by initial force with hand (but, automatical rotation doesn't happen). the Same analogy exists in d > 0 case; if no-friction, the rotation doesn't stop. and M is in the left, so automatical counterclockwise rotation will generate.
--- or Otherwise, if friction is big, make d big, perhaps can overcome the friction. (?)
(M position doesn't change. i.e. pedal downward force = other pedal upward force feedback is done at any time, this simplicity is good.)

Tip: About Roberval balance:
Gilles de Roberval (French mathematician, was born at Roberval, Oise, near Beauvais, France. His name was originally Gilles Personne or Gilles Personier, that of Roberval, by which he is known, being taken from the place of his birth.) Roberval is French province name.
In below figure, gravity is up to down = vertical direction.
Suppose D and D' weight hanging to B and B', by wire, if AB = AB' and D weight = D' weight, then this is balancing. long wire/ short wire is indifferent.
We think, this fact is of course/ self-evident matter.
If so, suppose, gravity, right to left direction (= horizontal gravity). D and D' is hanging from B,B'. BD // B'D', and length DB is indifferent.
I think so. this paradigm shift is occured/ simulated by Roberval balance parallel bars trick.

Java applet Online-DEMO (GeoGebra) ; (15-g)

- Centipede_perpetual_crank.giff -

Suppose, except D,D' all other weight can be neglected.

Node D, D', distance form Y-axis,
when x(B) ≤ 0;
2d ≥ |x(D)|-|x(D')| ≥ 0.
when x(B) > 0;
2d ≥ |x(D')|-|x(D)| ≥ 0.
So,
Always, counterclockwise rotation power is given.
(and, center of gravity of this rigid body is M ◇ (-d, 0). this is fixed point.)

Is there logic miss?
→ Yes, it's Logic miss
--- cf. Roberval balance (wikipedia),,, or cf. (2), (3) YouTube,


if So, D, D' orthogonal projection foot on radius BB' points as the new D, D' position works well?
i.e. D, D' weights slides on BB' bar. Even in this case, Roberval balance law is true?

cf. "Overbalanced Wheel (in wikipedia)" -- In this wikipedia sample, gravity average center is just under the axis (?).
but, in my left fig. case is not under, in the left. torque is not balanced.

and, the number of my metal balls is just 2.

To put it plainly, in the left hand of circle , radius is long a bit, in right hand, radius short a bit.

- Centipede_perpetual_crank.gif -

-------------------

variant of above.
Challenge again.

Imai's Perpetual Motion ( with proof )

Very simple structure.

cf. An Analysis of a Perpetual Motion Machine (WIRED.com © 2014 Conde Nast. )

Java applet Online-DEMO (GeoGebra) ; (15-h)

- Centipede_perpetual_crank2.gif -

A' is horizontal left side from Crank axle A.

C, C' are heavy slide weight.
if no-friction, this apparatus spins forever.

Just M is on A point is a critical point. vertical diameter becomes balancing (= stopping peak point). but, usually is passed through by inertia force.

Other x(M) < 0 point is stable counterclockwise power is caused.

Exercise: Please prove next.
(1) C trace is circle. ----- easy.
(2) M trace is circle. ----- not easy (?) [∵ △A'MA is right triangle.]
( (2) is a good problem. )

------------------------

lower picture is an equivalent frame.

pantagraph is middle point generator linkage.

I'm afraid " this is miss logic, too. the same illusion" (?)

c. This structure which is including pantagraph is similar to the Peaucellier Linkage a bit. --- why?
c. Near the below. ---- but, I'm in frustration. Please show me the trace of center of gravity of the apparatus body.
cf. Spring Engine. Пружинный двигатель (YouTube)
cf. overbalanced cross

c. My simple impression: upper fig. is unbalancing torque (pivot = A). but, lower fig. looks like a balancing torque (pivot = M₁). --- which is true?

(i.e. traces of upper/ lower cases are all the same. lower is pivot is M₁ and weights C₁, C'₁ are balancing. but the position restriction exists, (1) M is on the circle mid. (2) C₁ A'₁ distance is constant.
lower is balancing, so, automatic counterclockwise rotation torque is not existing. ----- this is against the upper fig. thinking property. --- paradox? ,,, which is true?
otherwise, both is true? ---- from M₁ view, balancing. but, from view A₁, unbalancing. ---- so totally, from the 3rd person view, unbalancing.
--- Someone, Please judge this.)

c. in below fig., point A₁ is not explicit. Suppose A'₁A₁ segment bar, and Please Put/ support A₁ on your finger top.
the Mobile sculpture will overturn to the left, perhaps.
• Put on mid on your finger top, the mobile ----- what happens?
• Put on M₁ on your finger top, the mobile is balancing.

Worst case: upper case (3 bars) is balancing. ---- I can't imagine. --- false, Absolutely.


- Centipede_perpetual_crank2.gif -






※ (About Japanese culture:) Japanese elementary school kid all know this value. Q. WHY? --- A. It's learn by heart. There is a famous pun.
another samples; √3 =1.7320508, √5 =2.23620679 ,not difficult things at all, for us.

(nuance of meaning; root2= one night by one night, grows good looking [about cherry blossoms (?)], root3= like other people, please buy me anything. root5= at the foot of Mt. Fuji, parrot sing.)

( Related my service; Let's(3 . 1) have(4) a(1) white(5) Christmas(9) in(2) HAWAII(6), Japan(5) Air(3) Lines(5) . --- 3.1415926535 -- This π (pi) learn by heart method is my invention. Christmas = 9 characters. [ My sample of 10 digits after decimal point. ]
(No <ruby> form:  Let's have a white Christmas in HAWAII, Japan Air Lines.  3.1415926535 )
---- cf. Pi Day [March 14 (or 3/14)], Albert Einstein's birthday )

Imai's linkage (tri-pedal)

arm4 Theoretical Estimation (arm3, arm4 rule);

Equivalent frame

Imai's Linkage (bi-pedal with foot)

What is Hart's Inversor?

Hart's Inversor is a 2 Ellipse GEARs Hinge

Inversor Mapping

5 Bars ELLIPSE drawing

The meaning of Inverse property

Above Fig. is interesting.