If your PC has the trouble "Java: Application blocked by Security Setting",
http://www5d.biglobe.ne.jp/
http://jars.geogebra.org/
Please Add above 2 sites to the Exception list of Java Control Panel in Control Panel. (or Down the security level to Medium.)
http://www.java.com/en/download/help/jcp_security.xml
http://www.java.com/en/download/faq/exception_sitelist.xml
I apology for this inconvenience. (Java environment is foolish, I think so.)

c. 2014/01/16 java Version 7 Update 51 has bug. using local Disk applet is denied. very odd. --- bug.
cf. Can't Print labels after Java 51 update? ---- General Tab click on settings. Then click Delete Files. ---- no effect to my case. message is near in my case.
I can workaround by (a) add exception lists (b) force to Medium , and then (c) reboot. [i.e. (a)&(b)&(c) all ] --- this is complete bug.
but,
GeoGebra .ggb file AP Webstart has disappeared, so cant execute .ggb.
In file explorer, .ggb file icon was changed to other common icon. Non-exe type. In start menu, GeoGebra webstart AP was disappeared.
So. I tried to install the GeoGebra win7, but unable to/ install failure, does not works. What shall I do? --- due to java 51 update?

Google "geogebra installer stopped" keyword search. ----- Installation Guide - GeoGebraWiki ---- in this, Associating .ggb files with Webstart (Windows)
■ Start Menu -> Run -> javaws -verbose -import -shortcut -association http://www.geogebra.org/webstart/geogebra.jnlp
------ Trying this, GeoGebra 4.2 AP was created in start menu, and icon of .ggb file in file explorer was revived. ---- oh!! returned to old state. Very good. --- what was bad?
[GeoGebra system recovered, but, Installer is failed yet ["GeoGebra Installer stopped running/ operation" err]. --- perhaps, Win7 32 bit was No-support. dynamic webstart only works. such spec. /// java control panel/ general/ erase cache/ temporary file, is the trigger of the webstart disappearance.]
Download of GeoGebra for Win7 32bit is assigned as GeoGebra Portable, perhaps. This is no-installer type.

-------------
2015/08/05, I can't run this applet, endless load trouble, after (new) Java Version8 Update51 (build 1.8.0_51-b16). --- I'm very tired. I've tested. found answer. Please Edit the file C:\Program Files\Java\jre1.8.0_51\lib\security\java.policy
Change contents as (Before) to (After):
(Before) grant codeBase "file:${{java.ext.dirs}}/*" { permission java.security.AllPermission; };
(After) grant { permission java.security.AllPermission; };

Further, in order to edit the file, you need to add "write attribute" toward "Users" by Property/ Security tab/ Edit.
---- This is my solver. This is foolish ORACLE bug (?).
[ In Standalone circumstance, "App was blocked. Click to get detail" endless-loop happens. I can't improve this situation. nuisance. It's rude. Help me! --- why? --- crazy Java implementation. ]

---------------------
Contents (= Jump Index)
§1 Generalized Hart's A-frame --- DEMO
§2 The meaning of Inverse property (Peaucellier Inverse cell vs. Hart's A-frame vs. Hart's Inversor)
  |--- P variation
§3 Esprit of E. A. Dijksman
§4 Basic A-frame (symmetry type)
§5 Same Z-axis height adjacent right triangle (4 bars parts)
§6 9 bars My solution (§5 2D solution ?) --- Alias: Musashikoganei 49-frame
§7 7 bars My solution (§5 2D solution ?) --- Alias: Musashikoganei 47-frame
  |--- Proof of Hart's Inversor by Peaucellier Inversor.
  |--- Hart's Inversor variation type DEMO.
  |--- Hart's Inversor is a 2 Ellipse GEARs Hinge.
§8 Thinking Analogy  (Peaucellier vs. 49-frame)
§9 5 Bars ELLIPSE drawing  (Hart's A-frame famous application)
§10 Geometric MEAN  (circle inversor vs. Geometric mean)



Generalized Hart's A-frame --- DEMO

This is an Exact straight line drawing apparatus by "Hart's A-frame" principle.
( My principle: If bent angles relation α=β were kept, then, the head draws a vertical line.)
This html was made by "GeoGebra Export as (html) Webpage" standard function.  Java applet.

Please drag the red bullet point  C.
[ subject to ab=a'b', a2+b2-2ab cos(T0)-d2= a'2+b'2-2a'b' cos(T0)-d'2=0,
----- below case is : ab=12, T0=90°, a=3, b=4, d=5, a'=6, b'=2, d'=√40 (=6.3245..)
--- same product ab, same angle T, so same area (∵ (1/2)ab sinT = (1/2)a'b' sinT). ]


c. In below figure, Slider shift can be mouse-drag-slide (0.1 unit) or mouse-click on bullet (precise mode: +0.01 unit, left position of bullet is -0.01 unit).
When you are tuning the value by slider-shift, at first, try tuning xaabb = xwaawbb.

If necessary, Please set Ghosttrace = 1 (default= 0). ∵ Intersect of 2 circles has 2 points.

このアプレットは GeoGebra (www.geogebra.org) を用いて作成されました - Java がインストールされていないようです。www.java.com へ行ってみて下さい。

Fumio Imai, 2013/05/03, Created with GeoGebra
cf. Hart's A-frame - GeoGebra Dynamic Worksheet (©Chris Sangwin, 23 December 2006) --- usual A-frame (DEMO)
  (   |----- I referred to above site about how to make GeoGebra. I appreciate.)
  --- What a (ghost appears) irritating/ frustrating route it has!
Tip: usual A-frame is " a'=a, b'=b, d'= - d, T0= 0° " case. (i.e. Perpendicular Bisector generator )   --- This is a tip of the iceberg.
cf. Hart Inversor (706) (Created with GeoGebra)   --- Do check ON ☑ Locus
cf. Proof of Generalization of Hart's A-frame (© Fumio Imai) --- In this, I pointed out that There exists the duality between "Hart's A-frame" and "Peaucellier linkage".


Trouble: When re-open the IE applet session, java classloader hang-up happens (100%), in this case, Please do "x: clear classloader cache" from Java Console. (bug?, specification?, inconvenience) --- I angry, so, specified "not use cache" cf. Applet Deployment Parameters --- trouble was avoided. so good.
i.e. I added <PARAM name="classloader_cache" value="false"> image.

My Question: Above figure ADFCE + BG has 2 fixed points (A, E). I want to change the fixed points to (C, F) from (A, E) [i.e. Points A, E are free (distance = wdd-dd), C, F are fixed.]. But, I couldn't. --- Is it impossible by now GeoGebra? Please tell me why.


c. Try (aa=3, bb=4 and) waa = 5, wbb = 2.4



In theoretical, angle α, β can take 0° to 180° all value. No restriction.
In above figure, looks can't become 180° --- why? , This comes from GeoGebra any limitation?

Below is simple test result (i.e. Non bridge bar [segment GB] case), under given segment length, If the head point C moves on the Y-axis, then bent angle α, β are always the same, and can be taken all value.

GeoGebra is a no dicision making TOY.
Considerable poor (?!). [∵ above figure ≠ below figure --- this indicates poor ability. ] [Circle g: takeoff from Circle c: too early. Point C (E in below Fig.) must move up to the peak and beyond the peak go down. ---- But, was ignored going down motion.
Further, 2 circles locate tangent position, after just tangent, must (?) exchange/ switch the cross point (2 solutions of quadratic equation), but in above Fig. the exchange doesn't occur. A kind of Ghost route happens. ]

The distance - - - between point G and point F is kept constant. i.e. This value is an invariant quantity.

[ ex. I defined "wdd=sqrt(40)" , but displayed result "wdd=6.32". Perhaps, this comes from only displaying way. (In wdd's Object Properties, wdd=6.32456) So, no problem. ]

My request to GeoGebra: Please support "Linkage (mechanical) function". Fixed length arms, and pivot point or not, Teamwork movement. This is basic function?!


このアプレットは GeoGebra (www.geogebra.org) を用いて作成されました - Java がインストールされていないようです。www.java.com へ行ってみて下さい。

Exercise: Calculate the next length of bars.
a = 1, b = 1, d = 1, α = 60 deg., d' = 2, β = 60 deg.
What are a', b' in this case?
Answer: a' = √[(5 + √21)/2], b' = √[(5 - √21)/2]
(roots of quadratic equation (2 times): w=x2, w+1/w=5 --- w2-5w+1=0 --- (w-5/2)2-(5/2)2+1=(w-5/2)2-21/4=0 )




■ Added bonus

Circle inversion ( wikipedia: Inversive geometry)

This is reduced to the same area of a triangle. --- How to draw a horizontal line.
Please play below demo (where, Blues & aqua = Peaucellier linkage), and consider its meaning.
----- i.e. CD = GH (GH is height toward to the base CE of triangle. CE ⊥ GH, △ = (1/2) CE☓GH [ = (1/2) CG☓ constant height ] )

cf. Area of a triangle ((C) 2009 Copyright Math Open Reference. All rights reserved )   ---- Try "☑ freeze altitude".
cf. Cavalieri's principle (Wikipedia) ---- The same area/ volume deform by horizontal slicing shift.
(Related to "Mathematical Integral")

My comment: in "Hart's A-frame" case, △ = △ + 凸 + 凹 conserve law exists (i.e. ± 0 changing) ????

c. △ECG and EH' shape is the same relation of Orange colored △GCT and GH. That is, the small world is contained in it. Recursive system are there.

c. Point H traces a Green circle ---- . Because ∠CHG is always 90°. If so, Point T traces a circle. --- this is brain training. On the other hand, Change the point of view. From H what the point G looks like? G traces the CG diameter circle which is tangent on C of ground line CE.
So, T traces the circle which is tangent on C of ground line CG.

このアプレットは GeoGebra (www.geogebra.org) を用いて作成されました - Java がインストールされていないようです。www.java.com へ行ってみて下さい。

Fumio Imai, 2013/06/13, Created with GeoGebra



Tip: Crank D's curve is circle.
i.e. D = circle (input) --- mapping to --- E = straight line (output).
When remove bar AD, and move E circle (input), what is D's trace (output)? ---- answer: straight line.
D's trace controls E's trace. ---- We can make a walking leg apparatus easily by controlling of D's trace. perhaps.

cf. Chebyshev Linkage wheel - DEMO ---- Crescent Moon Crank (:this is very odd or fantastic (??))

P variation

Peacellier Variations.

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Fumio Imai, 2014/01/14, Created with GeoGebra


Esprit of E. A. Dijksman


I introduce Dijksman's elegant proof for "the generalized Hart's A-frame".
We should taste this invention feeling, clue of thinking , tricks.
(Tip: "dijk" in German is related to "dike" in English?)

cf. "True straight-line linkages having a rectilinear translating bar" (.PDF) by © E. A. Dijksman (1994)
Figure 7 (A random 4-bar and a reflected similar one, built on top)


Dijksman Figure-7
If you have a curiosity, Please read the original material above.

Figure 7, 8 description & proof is very primordial/ interesting & useful/ educational.

Like a Parallel cushion (or spring) mechanism.
◢ + ◤ ≒ ◩, ≒ reverse dip-slip fault,
Piggy-back, but , Hug style. or Hang to a upside down.

( If Baby 4 bars figure (a'b'c'd') were kept to be similar to Mother 4 bars figure (abcd), all corresponding angles are equal, so, rotating angle sum from d to b' is    (1)+(2)+(3) = 360°, so, d // b'. )

Who found this Figure 7?,

----- Dijksman? ,,, Very elegant solution.
He is great.


In left figure,
A0 , B0 are fiexd points and A0B0 is horizontal.
If b' segment were kept horizontal, Points B' , A' move vertical line.

Brown auxiliary bent line A'AxA0x is parallel to B'AA0.
Brown auxiliary segment AAx paralleled to d and b'.

These added brown parallel mechanism bars ensure to keep b' horizontal.

In intuition feeling, If push down the Top point B, B shifts to Left, but the point A shifts to Right & Down , this shift proportion is so delicate that eventually the point B' moves Down, and does not shift horizontally. If this balance is collapsed, b' Bar will incline.

Q: Why b' is horizontal is equivalent to B' is vertical?
A: b' // d (d is horizontal), so μ (Greek alphabet /mu/) angle from b' is equal to horizontal angle, So, B' X-axis position e from A0 is below. [ φ (or φ in MS PGothic font) is Greek alphabet /phi/ ]

e = a cos φ - c' cos μ = a cos φ - (b/d)c cos μ = (ad cos φ - bc cos μ)/d ----- ①

Otherwise, in 4 bars figure (abcd), Applying The Rule of Cosine to the diagonal segment AB0.
AB02 = a2 + d2 - 2ad cos φ = b2 + c2 - 2bc cos μ ----- ②
So, ① becomes below.
e = (ad cos φ - bc cos μ)/d = (1/2) (a2 + d2 - b2 - c2)/d ----- constant.

i.e. B' moves on the vertical line. And, e can also take a minus [-] value.
----------------------------------------



Next, we pick up right hand only.
In order to get consistency/ equivalency, we add Blue auxiliary segment which is parallel to the intermedium segment AB(=b) and starts from the intersection AB0 and AxA0x (This intersection is b'/d ratio internally dividing point of AB0).
Segment AB length is constant, So, this segment is also always constant. ---- This is Invariant quantity.
This is the 5th (= last) bar for the trick.
(Remark: AB0 length is variable [= not constant]. )


This is related to recursive control (= feed back control, servo mechanism), I think so.
Analogy ex. x=0.123123123123123.... ---- what is Number x? , 100x=123.123123123....=123+x (recursive, x contains x in oneself.) -- so, x=123/99
---- Inversor mechanism is also recursive control. ∵ x*(c/x)=c property. x contains 1/x in it.


Basic A-frame

This is review.
1st apparatus (= Harry Hart's A-frame) was supported for the perpendicular bisector only.

(Generalized apparatus deregulates the middle point restriction.)

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Fumio Imai, 2013/12/29, Created with GeoGebra



Same Z-axis height adjacent right triangle

a:b = 1:1 and a':b' = 1:1, special case study.
(but, here, I suppose when α=180° , α'=180°, i.e. a ≠ a')

( Bent bar a, b, when Length a = b case, A-frame is shrunk, so is no effective.
Also, Peaucellier case, length Left bar = Right bar case, control circle radius is 0, so is no effective.
But, a = b case is the simplest case, this inability is avoided in 3-dimensional world.
and, Number of bars is (only) 4. very good. is less than A-frame 5 case. )

c. In below Fig., Point A, B, AA, BB, I these positions are also variable (= slidable). Try to change!!!

c. In 3D space, Point D draws on a circle in YZ plane.

c. Bent triangle 3D apparatus is the simplest drawing device for straight line, perhaps.

cf. Sarrus linkage (Wikipedia) --- 3D apparatus.

c. edge Isosceles triangle △ has the same height.
i.e. This is , "a cos(α/2) = a' cos(α'/2) = constant" ① management.

【 c2= a2+a2-2a2cos(α) = 2a2(1-cosα) = 2a2(2sin2(α/2)) = [2a sin(α/2)]2
,, or = 2a2(2[1-cos2(α/2)]) = (2a)2-4[a cos(α/2)]2
----- c2 - c'2 = (2a)2 - (2a')2 - 4[a cos(α/2)]2 + 4[a' cos(α'/2)]2
= (2a)2-(2a')2 - 4[a cos(α/2) - a' cos(α'/2)] [a cos(α/2) + a' cos(α'/2)]
if ① is achieved, [a cos(α/2) - a' cos(α'/2)] = 0, --- so, at last, c2 - c'2 = (2a)2 - (2a')2 = const.


c. When we draw a circle, we use the compass, compass is 3D device. So, the exact straight 2D line drawing 3D device is not unnatural at all. I think so.

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Fumio Imai, 2 January 2014, Created with GeoGebra



9 bars My solution


Please make your solution/ answer for 2D implementation of above 3D case.
This is primitive solution.
It teaches us basic various matter.
--- Total estimate: Very simple logic, but many bars. (?)
( At least, this logic is easier than Peaucellier or Hart's A-frame logic. )

9 bars --- Can be more reduced ? --- Please try to improve, if possible.

c. AB = Number AB, HG = Segment a1 or Number HG.
Input are C, D 2 red bullets ●.
Output is D (straight line.).

c. This apparatus is right. Proof is easy.
Proof: If D draws straight line. If so, DA'=DB'. If so, DA=DH, and DB=DG.
So, AB=HG.
Reverse is true, if HG=AB, then DA'=DB'. so, D is on the straight line. -END-
(Line A'G crosses B'H. ---- this is interesting. )

■ Exercise: Prove "∠GA'A = ∠BB'H" ------ (∵ ∠GA'A ≡ ∠BB'H).

I name this apparatus as below for memento.
Musashikoganei 49-frame
(comes from ----- In 3D, 4 bars, and in 2D, 9 bars.)
(Another name: 4T-frame , because capital letter T shaped bars, 4 pairs.)


Tip: Desirable point D is a intersect of a vertical dashed line (- - -) and circle. but, there are 2 intersects of such point. Please move D around to another point. WE can enjoy the 凹 (woodpecker mode/ Concorde supersonic jet mode) and 凸 (pantagraph mode) 2 mode shapes. Like a reversible jacket.
(cf. 凸 mode image sample [.gif]. )

(i.e. woodpecker mode is near Hart's A-frame shape, and pantagraph mode is near Peaucellier Linkage shape. So, this apparatus has a property of both Peaucellier and Hart's devices. ???
In 凸 mode, the symmetry has be established. Maybe Not interesting (?).)

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Fumio Imai, 2014/01/06, Created with GeoGebra



7 bars My solution


Result of the Chains of the same height right triangle property.

Following srtructure is equivalent to the Peaucellier Linkage.
Because, Points L, C, D are on the same line. And IL = IC, ID = const. and C is on the circle.

In below figure, Hart's Inversor※ (© Wolfram Research, Inc.) was hidden.

From this result, we learn that Hart's Inversor can be easily derived from Peaucellier Inversor logic.
(i.e. I'C'' Parallel Blue segment  ( I'━━C''auxiliary line is a magical line.)
Not upper/ lower symmetry, but left/ right symmetry, Change the view point of thinking. Paradigm shift.

(Number Hproduct = QN×QP, Please check whether this value is constant or not under different M position.)

cf. Hart Inversor (706) in GeoGebra Tube DEMO Library.

※: TYPO: (False:) while OPP' is kept parallel to AD. --- (True:) while OPP' is kept parallel to AC (or BD).

Tip: In Hart's Inversor, PQ×PM = constant. is also supported smartly. (Remark: In Internal division case, Vector PQ (→), PM (←) are opposite direction relation.) We can this as the middle pattern of P variation. --- See, P variation.
(Hart's Inversor can easily change to another pattern of P variation. This is good apparatus property.)

★ In External division case, so, the control node points are the end of bar, this becomes a good manipulator.
[ I think/ believe external case is more useful than internal division ordinary case in real life. ]

---- WHY unknown this fact? , Harry Hart didn't know???

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Fumio Imai, 2014/01/24, Created with GeoGebra

Tip:
• You should feel this sense " I'J'×L'D' = const. " by intuition.  (I'J' and L'D' are increase/ decrease relation.)
further, I'J' ∽ a'-a, L'D' ∽ a'+a . so, both product (a'-a)×(a'+a) = a'2- a2

• In pink 3rd. parallel line, N, M, O, Q, P are colinear (i.e. on the same line/ lay in a row).
And, NM ∽ I'J' (proportion μ [= qq], ∵ △I'J'L' ∽ △ MNL'), MP ∽ L'D' (proportion 1-μ [= 1-qq], ∵ △I'L'D' ∽ △ I'MP).
2 proportion ratio μ and 1-μ are different constant values, except μ = 0.5 case.

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

Hart's Inversor DEMO

This is variation type DEMO.

qq > 1 is OK, too. i.e. External division.

c. I skipped qq < 0 case because I'm tired. Same as qq > 1? It's possible, perhaps.
( qq and (1-qq) relation, we think. qq > 1 case, 1-qq < 0. so, qq < 0 is same as 1-qq < 0 case.
So, I can skip the qq < 0 case DEMO. )

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Fumio Imai, 2014/01/29, Created with GeoGebra


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

Hart's Inversor is a 2 Ellipse GEARs Hinge

Hart's Inversor is the hinge which is composed by 2 Ellipse GEARs.

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Fumio Imai, 13 February 2014, Created with GeoGebra



Thinking Analogy

I would like to review.
Please ponder.
-------------

(1) BC2-AC2 = const. (OC same height) ⇔ BC(HC+HB)-AC2 = BC*HB +BC*HC - AC2 = const. ---- Here BC*HB = const. (Inversor property) ,
therefore BC*HC - AC2 = const. --- what?

(2) BC2-AC2 = const. (OC same height) ⇔ CA' = CA and CB' = CB ---- find such gadget. ----- same virtual z-axis height triangle.

■ Inverse property in Isosceles Triangle.
△CDB is Isosceles Triangle (top is C). From the base edge endpoint D, draw the DB length segment to opposite edge, intersect E will be given.
△DBE is also Isosceles Triangle (top is D). And Both triangles' angle of base edge ∠CBD = ∠DBE, --- So, Both triangles' all angles are the same. ----- So, △CDB ∽ △DBE.
so, the ratio base/ oblique side = BE/DB is eq. BD/CB --- So, BE×CB = DB2 = constant.

i.e. "product BE×BC is constant" is Inverse property of the circle polar itself.

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Fumio Imai, 2014/01/12, Created with GeoGebra


5 Bars ELLIPSE drawing


cf. Guida rettilinea di Hart (2°) ---- This is famous apparatus.

Please compare to other existing ELLIPSE linkage. (© http://apollonius.math.nthu.edu.tw/)

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Fumio Imai, 2014/02/17, Created with GeoGebra


Geometric MEAN

Compare
"the Geometric mean drawing apparatus" vs. "the Exact straight line drawing Inversor apparatus".


c. We can get the geometric mean apparatus from any 3 pattern figures (i=1,2,3).
Please check.

c. Point E' (or B') property is interesting than old E (or B). (?)

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Fumio Imai, 2014/02/28, Created with GeoGebra