# Generalized Hart's A-frame

Author:
asifsound
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.) 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=a'b', same angle T, so same area △ (∵ (1/2)ab sinT = (1/2)a'b' sinT). ]
■ Bold Pink comment is very important: Pink --- image is a kind of shadow of its origin image. Always exists. It's complementary property (or duality). Please feel next. (here, Point F is pale-green color top point.) △EHF ∽ △BDF , & ratio HF : DF = CE : DF △AIF ∽ △GCF , & ratio IF : CF = DA : CF ・・・ same ratio ( ∵ CE×CF = DA×DF --- so, CE/DF = DA/CF ⇒ q) we define GB = q × AE , on purpose　( then, △AFE ∽ △GFB is true. [△GFB is rabatment type.]) (Now "α = β" is not proved yet. but , "△CFE ∽ △DFB [by 3 edges same ratio]" and "△DFA ∽ △CGF [by 3 edges same ratio]" are proved. --- what will happen? ∠FDA = α ∈ △DFA, ∠FDB = β ∈ △DFB, ⇒ double character　∠DFA = ∠FDB = α = β. conclusion "「α = β」 is always true." ) ( γ = ∠EFG,　δ = ∠AFB --　always, γ = δ ) [ AE in pink figure will be rotated by rabatment. i.e. orange color, AE → AE', E' is on same circle. FE=FE', AE=AE' , So figure FGBDC is miniature of rotated FAE'H'I. ] IF you have this knowledge, it's easy to answer to next question. Q1: What is DB length? Q2: What is GB length? Tip: Hart's A-frame top point F ● traces the exact straight vertical line to (horizontal) base segment AE or GB. There exists 2 apparatus in it. ---- i.e. recursive structure. ■ Comparison number of bars: 1. Peaucellier Linkage --- 7 bars (exact straight-line)  vertical 2. Hart's Inversor --- 5 bars (exact straight-line)  vertical 3. Hart's A-frame --- 5 bars (exact straight-line)  vertical 4. Chebyshev Linkage --- 3 bars (approximate straight-line)  horizontal