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