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N=3 2R-Virtual Wheel (Reuleaux triangle-Omusubi)

Simplest Reuleaux triangle wheel implementation in the world, perhaps. ----- This method concept matches the original Reuleaux triangle concept essence, perhaps. ( My request: Please improve this method more, if possible. ) [ Now, I recommend straight bar interface than hinge (= bent bar) interface. --- from 2016/09/28 ] Detail is N=3 2R-Virtual Wheel (Reuleaux triangle)  (GeoGebra) This logic can also apply N= even number case. [It has not the constant width property.] See --- N=4 Reuleaux concept Square Wheel ■ variant: dolly caster (straight bar, bent bar) --- same concept, but different implementation. N=3 2R-Virtual Wheel (Reuleaux triangle-Frame) ------ There is a bicycle sample. cf. N=3 2R-Virtual Wheel (Reuleaux triangle-Frame3) ---- auto vertical arm ⇒ This is MY last answer.
Interval Z0H: radius = 4+1= 5 (Center is B), Interval HI: radius = 1 (Center is D), Interval IW: radius = 5 (Center is C). Cyan bullet MM ● is center of triangle [ (x,y) = ( (x(B)+x(C)+x(D))/3, (y(B)+y(C)+y(D))/3 ) ] . --- It draws wave trace. Tip: radius = 4 CD arc (Center is B) is slipping on y= 1 line. ■ circumference Z0H = 5 (π/3), HI =1 (π/3) , so, 180° is 6 π. ----- this is the same as r=3, 2 π r = 6 π exact circle. height Z1 = 6, same as 2r = 6 circle. ■ Which is better, merit & demerit About r=3 circle and above Reuleaux triangle wheel, Which can save energy? Which is mild to the road? ■ Why this works? Under earth gravity, Pushing point B is equivalent to pushing point A. Point A is kept on the lowest position (as possible as). Each vertex of Reuleaux triangle is circle sector top, so highest top vertex is real center of current circle, its just beneath point is always touching the ground. System keeps the balancing, if you pull/ push the point A, so , total balance will be broken, but the system automatically adjusts or recovers the unbalanced state into balanced state. This method uses this homeostasis property, indirectly. The point A does not lie beneath the Point B. If such situation occurs, this is an accident/ trouble/ disorder time. It's a time, that the wheel can not rotate, by any reason. It's an exception event, so, we don't need to consider such case. At least, its reason is not due to this wheel algorism. The trouble might be delayed than existing other method. it's a good property, perhaps (?). ---- i.e. Having a shock absorber/ one cushion. ■ When Chassis/ body takes off from the ground. The wheel will be hanging/ suspending from the chassis. Please improve this demerit. ex. Attach the stopper to the height of point B, C, D.  ---- but, in the arm is slant case, this is out. point MM height restriction is bad, because bar confliction occurs. ---- This problem is little matter. It's no problem. Don't care. ---- Is it true? One Answer: Making inner circumference fence/ wall [ = small Reuleaux triangle shape] which is between the point A and outer circumference [ex. 0.5 length inner from outer circumference] , and set the stopper to the arm upper position [ex. between wall and Reuleaux triangle through B/C/D.] . ■ 3D non-conflict layout Please consider/ design how to layout with no bar conflict in 3D (3 dimensional) space. --- This is like a kind of puzzle. bar vertical layers To avoid bar confliction, the bent bars must be laid constant distance away from vertical wheel face. bars are slant from vertical line. ■ \ || pink --- point B (x=1 to 5) ■ \ || blue --- point C (x=1 to 5) ■ \ || green --- point D (x=1 to 5) ↑ vertical wheel, || is chassis arm ★ old [= not best] point A is sphere/ cubic ball/ binded mass/ node [ become stopper] (x=5), and sphere can spin which top has 3 points mark ∴ (= seat of a swing). --- image: like a pencil cup |\|/| 3 points mark ∴ of Point A is corresponding to the end of pink/ blue/ green (x=5). --- So, point A can be located on point B/C/D.  ---- As a result, it's like a radish shape. When point A is just on C, ∴ mark spins/ rotates corresponding C become to be the most inner position, automatically. Like a Ball shaped pendulum. | /◉\ | A part (triangle roof) |/◉ ◉\| (3 ◉ are independent/ separated.) Real implementation is not so easy (??). New problem happens. Please challenge. ------ ★ old [= not best] ■■ Another 3D non-conflict method ■■ No twist rope/ pole implementation logic exists. i.e. Apply the next. ① To change the fixed point B/C/D into slidable B/C/D. (more corner side movable) ② To change 2 or 3 bent BA/CA/DA into N bent BA/CA/DA.  ---- like an accordion curtain. ex. image: A ●-/\/\/\/\/\/\/\(-◎B/C/D ) :corner, ( ) is slider interval. /\                 \------/  -\ / \- shape is OK. \ / \/ -\ /- shape is OK. (The piano wire/ spiral spring is OK.) ------- \ / ◉-| \--------◎A , (spiral image: here B◉ is slidable ⇔ in margin space.) \--- / This logic is valid under earth gravity environment. ex. Under new method, Green bent bars rotate clockwise between A and D space, after point A over D in above Fig. Not counterclockwise. (① can be realized by bar linkage.) (twist rope/ pole implementation logic is simpler than this. less friction than this.) ------- ★ latest solution (3 sets of 2 bars Hinge, and slidable (⇕) B/C/D points, and A is slidable 3 axes.) B/ C *** Below figure shows when node A crosses on D. *** □   / ■ \   □ □   / ■ \ □ □  / ■   \ □ □ /   ■   \ □ □ / /\ ■ /\ \ □ □┛ / \ ■ / \ ┗□ } A slidable 3 axes from arm. □━┛ | ■ | ┗━□ } A slidable 3 axes from arm.  ⇔   D ⇔ (Node A has 3 slidable axes. If node A crosses B/C/D , the corresponding axis will be slided to inner. slide degree ∽ bent bar angle. [i.e. 180° is no-slide]) How to bent bar control? left to right vertical view: K K K'//.\ | | |. K' (= 3 layers) / \ | | \ / \ ↓ spring slider | | \ D   ◉////.|A D ◉ A'---. A_ (D can cross the A' without conflict.) A' A_ set the wire K'A_ (or K'A'A_) : length = K'K + KA' (point . is K', . is A_ in above Fig.) ∠K'KA' ≠ 180°; K'A_ > K'A' ∠K'KA' = 180°; K'A_ = K'A' (i.e. A'A_ ∽ [K'K + KA'] - K'A' = 0) Tip: feeling: length of A'A_ ∝ height of B/C/D --- i.e. if upper then =0, if lower then ≠ 0. ------- ■ Hinge ( 0.5 bar + 0.5 bar, or, (1/3) bar + (1/3) bar + (1/3) bar, or , rope ) When A is on C, E and G conflicts (i.e. G is on E). So, "Σ(1/3) bar" is possible. I don't prefer "rope", because it's too flexible. ex. My recommended Hinge format (pink hinge parts sample format/ shape) (◉ [◉) ]── ◯ B G A , where ◉ is connector/ elbow which can move 3-dimensional angle. i.e flat bar ( ) ,and, flat bar + pole (or rope) [ ]── ■ SAME FUNCTION = Hedge inner arounding wheel/ gear /\ // \\ // \\ / -☉--- \ ☉ is a inner gear/ wheel moving along the outer triangle edge. This ☉ is the same function with the point A● in above Fig. But, having much friction. ■ Critical length : BA +CA +DA = 4×4 + δ (here, δ≒ 0) i.e. 3 sets of Hinge real working length is enough about 2 sets. About 1 set length is margin/ allowance/ non-used.
■ Horizontal wheel arm. (Especially in wheelchair, for Paralympic (?) ) Above figure shows vertical wheel arm. But this is bad in handling a wheel. Arm is horizontal and lies in lower position of the wheel, is good. /\ | / \ | chassis | / A \ | ----/-----◉-----\---- ← real arm (horizontal [= max slant]) /________________\ ________ ground line This type wheelchair is better than circle type wheelchair (?). height is r=R type, but wheel arc is r=2R type. operability is better than circle (?). wheel speed is better than circle (?). low body floor. Please check it out. ■ No slidable axis (⊿ cover/ roof) --- most easy/ simple trick. ★ old [= not best] Point A structure ------ If ◉ pulling up, it moves upper-right position automatically along the roof. /◉/| A for B (B position is upper) B is pulling up ◉. ---- this BA case is straight & critical. /◉// | A for C (C position is middle) /◉/// | A for D (D position is lower) D is not pulling up ◉. Here, /// is spring. ■ 3 parallel vertical spring sliders are more simple/ honest. ★ latest [= best] __ B ≡ spring \ ≡ spring /\\ ↓ same vertical line (i.e. arm line itself.) ◉ ↕ C/ /\\\. A for B --- BA is critical. / \\. A for D --- CA, DA is not critical. D (in Z0H interval, BA ,CA and DA are not conflict. bent CA does not cross the bent DA. ) (in HI interval, both CA & BA are critical. DA is not critical. BA line crosses over the bent DA. but BA does not conflict the DA. ∵ BA and DA is different/ saperated layer. --- So, total, no problem. very good.) ■ Non conflict best solution is straight bar (hinge is bad). --- on 2016/09/28 found In above Fig. Pink/ Blue/ Green hinge is not wise, straight bar through node A is the best and simplest. Remark: In this case, B/C/D must be slidable to the corner. Please enhance to reduce the friction of this method, more. (Although friction occurs on A hole, but this friction is very light. Straight bar isn't pushing on the Node A hole surface [i.e. no pressure]. ) (Straight bar does never touch the ground. --- why? --- God makes it so. ) N=4 Square wheel case is true, too. cf. N=4 Reuleaux concept Square Wheel (i.e. Square wheel can be made easily by straight bar method.) ★ Straight bar structure   ====================== ◉ B/C/D ◉ A⇔ | end  ---- i.e. bar with inner wide hole. bars are in 3 layers. ====================== If necessary, A can be a roller. roller is independently about B/C/D, respectively. Roller is touching on the upper roof of hole. roller axis is common with B/C/D (i.e. the same axis). ■ ∠CAD = 150°---- always true ( ∵ Inscribed angles ) ■ width of 凹凸 (in Cyan trace curve) Always 凹 width < 凸 width Proof: BC = d, DD'= e, 凹 = (π/3)(d+e) - d/√3 凸 = (π/3) e + d/√3 So, 凸-凹 = 2d/√3 - (π/3)d =(2/√3 - 1/3)d =(2√3 - π)(d/3) Then, 2√3=2*1.732..=3.464.. π=3.14.. So, (2√3 - π) > 0 So, 凸 > 凹