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原帖由 5575338 于 2008-7-31 17:48 发表 2 h" A$ _2 g& T# `; n9 i1 D
9方图计算公式
0 l( R4 Q; u8 e. @1 \* I" `在下面几个角度线上的:! ^8 X! j8 I' k
0 degrees: (2n + 5/4)squared6 k$ E; I! j. f( C, m
45 degrees: (2n + 6/4)squared% }0 \" }8 |. ?* q" I! q1 T
90 degrees: (2n + 7/4)squared0 G _5 ^5 U& ?7 S
135 degrees: (2n) squared
/ @' O$ G/ T; U: U/ w; t- ?180 degrees: (2n + 1/4)squared
& X8 v1 h: i. V3 q, {225 d ... 2 r. J) v# p3 M' U5 h5 w7 ~. d
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是这个吗?2 @' d( c2 T6 `! S( Q
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Square of Nine Essentials
3 ?7 v/ y( E0 M: Z# w9 q- GDaniel Ferrera, 2002) r1 x5 z9 x( \- L |
In my experience with working with this method, price & must balance on a hard aspect.; L7 H' e" p1 {* \( |
The hard aspects are 45, 90, 135, 180, 225, 270, 315 and 360 or 0 degrees. The most1 e- v) _. p" w3 T- S6 |# l5 w
important being the squares or 90-deg harmonics (0, 90, 180, 270).% L4 Q' ^# Y, A$ b
In terms of selecting a past date and price to start from, I have found that the lowest low over
4 V4 z( h& J7 U# o# J7 d$ r4 Uthe past 365-days and the highest high over the past 365-days have the greatest influence on9 i; V9 E: @8 q! Y
these balance points. This technique can be used to generate the horizontal support &
3 X [6 h* y6 _( U: z3 h5 uresistance levels for intraday trading. This is extremely useful when you anticipate that a
4 X: |. Y5 i2 n7 d2 M* ^particular day will be a trend change as the result of cycles or counts, etc.
/ \% E1 r8 q/ s" {; b6 b9 LCarl Futia's formula for this reads) d9 Z7 t% o- z! Q7 V0 S5 P% L( H
=MOD 360 ((price distance or Time change)^0.5*180-225)1 n1 ?5 r1 o4 S* ]4 V. u
This formula assumes that the Squares of Even numbers fall on the 135-deg angle and that the
- r) `* ~( y) W7 H& I$ Z6 {Squares of Odd numbers fall on the 315-deg angle, which is not true on Gann's actual Square+ d! P2 G, U: u) q' T
of Nine chart.
" S/ W5 P1 h2 m3 ^: B/ q6 IIf you start with a "1" in the center, the Squares of Odd numbers will fall on the 315-deg angle,* y$ q7 A/ d R- i
but the Even Squares (16, 36, 64, 100, 144....) will gradually float towards 135-degrees. For" o+ I) ?1 p0 E
example, on the actual Square of Nine
$ H/ v# Y+ [4 R/ t8 g16 is on the 112.50-deg angle,
' w$ P$ k2 L0 i* `7 p36 is on the 120-deg angle,
9 i1 k. [* E) q: z0 N8 N" V. J64 is on the 123.75-deg angle,
0 ^; h8 M% M! H% ?. Q100 is on the 126-deg angle and- g& u+ d" u# H4 B
144 is on the 127.50-deg angle* J7 P: s- }/ B8 ]0 V4 z: \
and so on.
! U$ a/ ^" H, f. C* JStarting with "0" in the center, the Squares of Even numbers will line up on the 135-deg angle
# F9 }' q0 `. z8 pand the Squares of Odd numbers will Float.
; d* |& q! ^* v9 _+ rCould this amount of inaccuracy or "Lost Motion" be important? After all, it is impossible to draw3 ]- v8 i9 w9 R) Q
or actually build a Square of Nine Chart based on the MOD 360 formula above. If you want to% }0 F' b2 y' v5 [! M( P, y
work with calculations that are based on W.D. Gann's printed Square of Nine chart, the9 e* Z* ~) d) l7 D- ], @. H2 H
following formulas will be of great use to your research:7 N% {4 z! U( \1 ?5 F9 P; d( j% p/ V0 ?
Ring# = Round(((SQRT(Price)-0.22 / 2),0)
* r4 |: ^ d6 `6 I{This rounds to the nearest whole number, i.e. it eliminates the decimals}5 |: d( V' U. V6 A
Example: The number 390 is in Ring #10 if you crunch the above formula.
3 b$ d( Y9 D1 g$ \8 ^- G315-deg Angle: This is the most accurate angle of the entire chart and is used to calculate all
3 B2 L6 I* n/ G- @6 O/ tother values. The Squares of Odd numbers are all on this angle.
, c8 U! d1 \3 ~3 G0 c4 x1 o0 A315-deg Angle = (Ring# * 2 +1)^2
) f. A2 x0 \1 o1 |+ L! R5 B8 X ? ^Example: 390 was in ring# 10 so the 315-deg number is (10 * 2+1) ^2 or simply (21)^2 = 441 c! X- ~. F, }6 j9 C/ K
The Zero Angle on this Ring = ((Ring# * 2 + 1)^2) - (7* ring#). So you would get 441 - 70 =2 I9 Y% c- i U# S
371 This number is needed to calculate the angle that the 1st value of 390 is on.9 T2 m5 E2 R# _- f
Angle = Sum ((Price- Zero Angle) / (Ring/45)). So we have ((390 - 371) / (10/45) = 85.50-deg
: k9 G5 @( v1 q: ~+ U5 QYou may have to occasionally adjust the Angle calculation because sometimes you will get a
7 k/ K6 t6 v- {negative value when you have a number that is approaching the 0-deg angle of the next ring.% n/ I/ S) e7 a- f( `
For example: We know that 371 is a zero-deg number. If you try to find the angle of the number
0 c; X: S% m5 k9 M" F370.5, which is a number in the previous ring approaching the next ring, you get Sum ((370.5 -
7 n; N: s* r" M3 ~ u( \, c371) / (10/45)) = -2.25-deg. If you get a negative number, just add 360 to correct it. So this* ` l8 T7 k+ D5 U. u- u
would actually be 357.75-deg.
5 ^, E; P2 L0 y( JA simple formula to correct this is If Angle<0 then +360 else Angle = Angle.
* I! x" I! q- V8 mTo generate other values on the Square, use this formula: (Ring# * 2+1)^2) - (7* Ring#) +6 y" Q- X. y9 Z
((Ring# / 45) * Angle)" ^! |' }) n6 G9 r; j/ I$ L
Angle is this formual is your input value. For example, we know that 390 is on the 85.50-deg
8 r- o" f- H E& g$ }+ H. nangle. If we want to know the value of the number that is 45-deg to this number, we would be' G( ^# \3 g+ J _$ ]' `
interested in the angle of 130.50-deg (85.5 + 45). Inputing this in the above formula gives us:
! r% H$ G* R6 U( d8 F(10 * 2+1)^2 - (7 * 10) + ((10 / 45) * 130.5). Simplified a little, we have 371 + (28.99971) =" y/ R9 R( m& b1 L
399.99 is 45-deg to 390.
: _- [9 \" v; m3 B0 vKeep in mind that if you add or subtract an amount that will change the original angle (85.5-deg)
" W$ I" `. I, Y% u2 j" _; Bto an amount greater than 360 or less than 0, that you JUMP rings. For example, if you subtract
0 F( F; x# p: g/ S90-deg from 85.5 to potentially find a square aspect, you get -4.5-deg. Add 360 gives 355.50-- @9 O: b# {" p& z
deg in the previous ring. We were using Ring# 10 in the formula, but for this calculation, we; x8 L, w1 [7 G& k
would have to use Ring# 9. Similarly, if you added 315-deg to 85.5-deg, you get 400.50, which% T0 \+ |, M2 P
is 40.5-deg in the next ring. So you would have to use ring# 11 for this calculation |
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