Line Symmetry Linkage (Alias: Imai's Butterfly)

Author:
asifsound
Topic:
Symmetry
Sub title: How to make Non-slide symmetric umbrella ? Line Symmetry Linkage apparatus. Line symmetry (or, line of symmetry)/ rabatment/ line reflection is an elementary function than drawing exact straight line. You don't have to know Peaucellier Inversor. cf. Peaucellier–Lipkin linkage (wikipedia) This tool is basic & very useful in real world. cf. N=4 2R-Virtual Wheel (Vers.-B) cf2. N=3 2R-Virtual Wheel
Remark: Cross bar must be long than outer edge. Line symmetry line is assigned to outer edge. If outer edge length were long than a cross bar length, please reassign a short length. From my experience, long: short edge ratio is 2:1 is good for easy calculating. cf. Chebyshev-like N=3 Polygon Wheel ---- in this fig. I presented the exact straight line tool sample. ( This tool logic/ concept is simpler than the logic of Peaucellier–Lipkin linkage [on 1864] / Hart's Inversor [on 1874–5]/ Hart's A-frame [on unknown year]. ) ( I challenged many times to make exact straight line by line symmetry concept from many years ago, but all failed.　I've been abandoned, but I found this by chance. God is so volatile/ capricious. ) Above 1,√2,√2,1 ratio Antiparallelogram is one of typical case (i.e. □ form is square), other ratio Antiparallelogram  (i.e. □ form is rectangle) is OK, of course. ■ Is this an invention? ----- NO, it was famous theory "Kempe's reversor". I searched this apparatus on internet web, but I couldn't find it. So, this apparatus is new one, perhaps. So, I named after my name as a memorial to finding. [ "Imai" is pronounced /imai/ (not /aimai/ etc.) . Japanese Surname. ] I found the article. about Kempe works. cf. Kempe's Linkages and the Universality Theorem (.pdf) ～ Kempe's work of 1876, now known as the Universality Theorem, has a distinctive standing in kinematics. ～ ----- p223 2.2 The Reversor ～～ For this reason, Kempe's reversor is also called the 'angle-doubler'. --- i.e. adding the same angle. ---- multiple repeat is OK. +α, +α, ... cf. How to draw a straight line ; a lecture on linkages by Kempe, A. B. (Alfred Bray), 1849-1922 Published 1877 Topics Links and link-motion ------- See. Fig. 30. ( page moving operation= page dragging from right to left. or, Use/ click "fullscreen view" 「↖↗ ↙↘」 icon [= nenu shown].) ■ In a nutshell It's 2 Butterflies which a one wing is overlapped. ■ Orange trace shape is interesting. ----- like a leg/ foot trace. This trace is similar to Chebyshev Linkage foot trace (by 3 bars), but above is a exact straight line, and, crank rotates same direction of forward movement (by 7 bars), and, has more flexible length ratio. crank range 180° to 360° (in return stroke): After passing the shrunk point, Antiparallelogram switched to Parallelogram form (same as Chebyshev case). Tip: Between 180° and 360°, we can realize Antiparallelogram, too. In real world, we can realize Antiparallelogram and Parallelogram, both. In GeoGebra's default specification leads the mixture. Suppose a, b are circles. if so, Intersection a and b have 2 points. Intersect[a, b, 1], Intersect[a, b, 2] ---- default. This does not reflect the real world. Intersect[a, b, if[condition, 1, 2] ] is real world solution. ------ Please try. -------------------------------- Memo: cf. Related: Origami (wikipedia) rabatment or rabattement /rəˈbætmənt/ --- Definitions Collins English Dictionary noun: geometry the act of rotating a plane to align it with another ■ For educational material See Linkage (related to "inversion geometry") in Whistler Alley Mathematics Please remember Next. Antiparallelogram C◯　　 D◯ /| ＼ X. ／ |\ 　 / | ／ ＼ | \ A◯ H･ 　　　 J･ B◯ Here, Suppose Antiparallelogram, AC=a, AD=b, CD=cc, AB=dd, CH=h (= vertical) then AH=0.5 (dd-cc), AJ=cc+AH=0.5(dd+cc) So h2=a2-(1/4)(dd-cc)2, h2=b2-(1/4)(dd+cc)2 ∴ a2-(1/4)(dd-cc)2 = b2-(1/4)(dd+cc)2, So, b2-a2 =(1/4)[(dd+cc)2 -(dd-cc)2) =(1/4)(4 cc×dd)=cc×dd i.e. Antiparallelogram upper base CD, lower base AB, AB // CD, and AB × CD = b2- a2 [= constant] Tip: ∠DAB=θ, ∠BCD=∠θ, CXA=2θ, (from Parallelogram, AB is rabatment line, from ∠A to ∠A - 2θ, ∠C is the same, ∠X is born +2θ) Parallelogram C◯---------- D◯ / | \ ／ / | 　 / | ／ \ / | A◯ H･ -------B◯ J･ Suppose, AC=a, CD=b, AD=dd, CD=cc, CH=h (= vertical), AH=α So h2= a22, h2= dd2-(b+α)2, h2=cc2-(b-α)2, So, from last 2 lines, sum, 2 h2 = dd2 +cc2 - 2 b2 - 2 α2, then 1st image is including in it. So, dd2 + cc2 = (2 h2 + 2 α2 )+2 b2 = 2 (a2 + b2 ) [= constant] ---- Diagonal line property: Sum of square is constant. Not Pythagoras Theorem.