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= a2-α2,
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.