Paucellier vs. Hart's A-frame
- Author:
- asifsound
■ Very interesting
in above Fig,
AB×BD cos(∠ABD) has 2 results.
① AB×BH = variable × variable ---- = long × short
② AB cos(180ﾟ- ∠ABO)×BD = BO×BD = constant × constant
[ ① is the same ②. ]
Further,
AB2 = AD2 + BD2 - 2 AD×BD cos(∠ADB)
so, AD2 - AB2 = -BD2 + 2 AD×BD cos(∠ADB)
Here, AD×BD cos(∠ADB) = DA×DH' = variable × variable ---- ③ [where, DH' = BD cos(∠ADB) ],
or
= AD cos(∠ADB)×BD = DO×BD = constant × constant ---- ④ [ ③ is the same ④. ]
Then, Gap of ① and ③: ④ - ② = DO×BD - BO×BD = (DO-BO)×BD = BD2 = constant
----------
| ◆ More easy recognition:
| △DBH ∽ △ABO ---so, BH/DB = BO/AB, so, BH×AB = BO×DB --- ①②
| △ADO ∽ △BDH' ---so, DO/AD = DH'/BD, so, DO×BD = DH'×AD --- ③④
■ Trick
In above under Figure, distance ( from middle [= MM] to vertical [= HH]) is orange MMHH .
This is the same 0.5 AACC - HHCC = 0.5 (AACC - 2 HHCC)
So, 0.5 (AACC) × 0.5 (AACC -2 HHCC) = 0.25 AACC × AAC'C' [where, C'C' is mirror point CC against HH]
|--------| AAC'C'
|------------| AACC
--- this is orthodox Paucellier inversor result.
[AAMM×MMHH = MMCC×MMHH , is more honest.]
■ Hart's Inversor
butterfly SMMTHH is Hart's inversor.
ST // MMHH, and, ST × MMHH = constant,
So, horizontal line between S and MM, ---- ex. SS' = q SMM (; inner divided point, q = inner ratio)
S' M' H' T'
|---|----|---| any horizontal segment (S' is the cross node with SMM segment, node M'/H'/T' are ditto.)
ST × MMHH = constant --- So, S'M' × M'T' = constant. --- ∵ q(1-q) = constant, too.
and
S'M' = H'T' so, S'M' × S'H' = constant.
■ Power of a point
OA = aa, OB = aa, OC = bb,
H is middle of AB, and OH ⊥ AB,
A, C, H, B are colinear (= on the same line).
if so,
AC×CB = constant
Proof:
OB2 = HB2 + OH2
OC2 = HC2 + OH2
so, OB2 - OC2 = HB2 - HC2 = aa2 - bb2 = constant
and, HB2 - HC2 = (HB+HC)×(HB-HC) = BC×AC
Then AC×CB = constant