# Generalized Hart's A-frame

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**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', a^{2}+b^{2}-2ab cos(T0)-d^{2}= 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

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