Generalized Hart's A-frame --- DEMO
This is an Exact straight line drawing apparatus by "Hart's A-frame" principle※.
Fumio Imai, 2013/05/03, Created with GeoGebra 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?! 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. 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 variationPeacellier Variations.
Fumio Imai, 2014/01/14, Created with GeoGebra |
Basic A-frameThis is review.1st apparatus (= Harry Hart's A-frame) was supported for the perpendicular bisector only. (Generalized apparatus deregulates the middle point restriction.)
Fumio Imai, 2013/12/29, Created with GeoGebra |
Same Z-axis height adjacent right trianglea: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.
Fumio Imai, 2 January 2014, Created with GeoGebra |
9 bars My solutionPlease 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 (?).)
Fumio Imai, 2014/01/06, Created with GeoGebra |
7 bars My solutionResult 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???
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 DEMOThis 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. )
Fumio Imai, 2014/01/29, Created with GeoGebra |
Hart's Inversor is a 2 Ellipse GEARs HingeHart's Inversor is the hinge which is composed by 2 Ellipse GEARs.
Fumio Imai, 13 February 2014, Created with GeoGebra |
Thinking AnalogyI 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.
Fumio Imai, 2014/01/12, Created with GeoGebra |
5 Bars ELLIPSE drawingcf. Guida rettilinea di Hart (2°) ---- This is famous apparatus. Please compare to other existing ELLIPSE linkage. (© http://apollonius.math.nthu.edu.tw/)
Fumio Imai, 2014/02/17, Created with GeoGebra |
Geometric MEANCompare"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). (?)
Fumio Imai, 2014/02/28, Created with GeoGebra |