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原帖由 5575338 于 2008-7-31 17:48 发表 ![]() ; @' L! \/ L P4 j* j# v' R
9方图计算公式
' e5 N9 n( O, D# a在下面几个角度线上的：
. y0 o0 J# u0 d8 p3 |5 }0 degrees: (2n + 5/4)squared
& v4 r$ l- S9 G45 degrees: (2n + 6/4)squared9 K% \ _' U; b0 i
90 degrees: (2n + 7/4)squared v* a, t( ]8 B0 r' @9 u
135 degrees: (2n) squared- ]' @$ S8 N# _9 e0 }
180 degrees: (2n + 1/4)squared$ e; c% t8 n4 t2 z, z1 m
225 d ...
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是这个吗?, q% W! W" p% @4 p" o. A$ D
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Square of Nine Essentials
: M, j( T+ w2 a$ b/ z3 o. wDaniel Ferrera, 2002
. D- F4 o3 G" @In my experience with working with this method, price &  must balance on a hard aspect.
0 }/ v: X3 d7 R( n; w' `$ N- nThe hard aspects are 45, 90, 135, 180, 225, 270, 315 and 360 or 0 degrees. The most
* q! F0 O$ ~- M2 Ximportant being the squares or 90-deg harmonics (0, 90, 180, 270).9 S: N, J/ C; b; H% P
In terms of selecting a past date and price to start from, I have found that the lowest low over- ?$ ~& `. x/ w+ i' Z/ B0 D
the past 365-days and the highest high over the past 365-days have the greatest influence on
5 E: j' d8 u( l, M" {/ ^7 ]these balance points. This technique can be used to generate the horizontal support &+ _, O: [& q1 G$ z
resistance levels for intraday trading. This is extremely useful when you anticipate that a- ?( Q+ v. S! t
particular day will be a trend change as the result of cycles or counts, etc.% }! P3 R- e' F" _5 |3 |
Carl Futia's formula for this reads
# x, _! c8 R4 d! B3 x0 S* |=MOD 360 ((price distance or Time change)^0.5*180-225)
3 w+ S! ?7 S) x+ e! AThis formula assumes that the Squares of Even numbers fall on the 135-deg angle and that the
4 c; r, W' N+ U8 i/ VSquares of Odd numbers fall on the 315-deg angle, which is not true on Gann's actual Square
3 I7 {7 ?" v p7 C! v% o9 S t" N8 E2 Cof Nine chart.
2 k5 I# e, P' Y% s1 GIf you start with a "1" in the center, the Squares of Odd numbers will fall on the 315-deg angle,
k0 l1 l2 ]' J _8 T! qbut the Even Squares (16, 36, 64, 100, 144....) will gradually float towards 135-degrees. For
* W9 h8 v4 w6 L1 }$ kexample, on the actual Square of Nine
! g9 a2 g0 p6 v2 m16 is on the 112.50-deg angle,
2 U8 u6 G. ]( K/ S; h36 is on the 120-deg angle,/ L5 R! b' X5 [: j. k
64 is on the 123.75-deg angle,
# `" h h3 ^ G; v100 is on the 126-deg angle and X. s" K' \1 G0 R {
144 is on the 127.50-deg angle( z) M% n# a4 n, r
and so on.
& I3 x: c7 ?+ C7 O, FStarting with "0" in the center, the Squares of Even numbers will line up on the 135-deg angle
$ A8 ~4 y# f. f- c- q4 a, R) e: Sand the Squares of Odd numbers will Float.5 I8 S) M" ^5 g1 ~8 Q
Could this amount of inaccuracy or "Lost Motion" be important? After all, it is impossible to draw* \% M4 R% O z0 B) q& ~5 e/ K
or actually build a Square of Nine Chart based on the MOD 360 formula above. If you want to
) l0 Y1 A) o/ L/ X: {! xwork with calculations that are based on W.D. Gann's printed Square of Nine chart, the# y' {! Q) f+ ~9 Q
following formulas will be of great use to your research:
/ m! H* i: T( ]: pRing# = Round(((SQRT(Price)-0.22 / 2),0)
: O3 V: p x* O$ T. o{This rounds to the nearest whole number, i.e. it eliminates the decimals}
/ Y0 f& X: \( j$ u% TExample: The number 390 is in Ring #10 if you crunch the above formula.
# j/ w+ S, i$ r$ v315-deg Angle: This is the most accurate angle of the entire chart and is used to calculate all
) @. F: s/ S0 T1 b% s6 G* Cother values. The Squares of Odd numbers are all on this angle.
/ T& n& T# j' |, f/ T" a315-deg Angle = (Ring# * 2 +1)^2
$ ]$ V( g T! pExample: 390 was in ring# 10 so the 315-deg number is (10 * 2+1) ^2 or simply (21)^2 = 441/ u, Z# y, K' q4 u' v
The Zero Angle on this Ring = ((Ring# * 2 + 1)^2) - (7* ring#). So you would get 441 - 70 =) A+ Q, I0 y$ _9 L2 b/ J% p% c2 I
371 This number is needed to calculate the angle that the 1st value of 390 is on.
( M* W) d& S$ J: j) r) N. G; K: J4 rAngle = Sum ((Price- Zero Angle) / (Ring/45)). So we have ((390 - 371) / (10/45) = 85.50-deg8 K! b( j( S% K' U& Q/ W: u
You may have to occasionally adjust the Angle calculation because sometimes you will get a7 r1 G# m, j! q O7 i$ _) v
negative value when you have a number that is approaching the 0-deg angle of the next ring.
: ^/ r+ [2 l* p0 H4 { mFor example: We know that 371 is a zero-deg number. If you try to find the angle of the number
) g% u# x8 Y! [" D$ i370.5, which is a number in the previous ring approaching the next ring, you get Sum ((370.5 -$ h4 R) U# Q4 F Z& `" O
371) / (10/45)) = -2.25-deg. If you get a negative number, just add 360 to correct it. So this) i6 n. ~* q9 K& N
would actually be 357.75-deg.
) b) d+ O- p* g4 \A simple formula to correct this is If Angle<0 then +360 else Angle = Angle./ A v/ T/ T% h" l
To generate other values on the Square, use this formula: (Ring# * 2+1)^2) - (7* Ring#) +0 T4 ]* e, j" i3 o" T
((Ring# / 45) * Angle)
1 t0 y, ?% C# w5 t- ~6 z! P$ jAngle is this formual is your input value. For example, we know that 390 is on the 85.50-deg
+ `# v+ w& F! j2 z# h: gangle. If we want to know the value of the number that is 45-deg to this number, we would be
1 O& F" f( L/ winterested in the angle of 130.50-deg (85.5 + 45). Inputing this in the above formula gives us:7 J+ l7 f5 U% I2 b
(10 * 2+1)^2 - (7 * 10) + ((10 / 45) * 130.5). Simplified a little, we have 371 + (28.99971) =
8 n8 O, v5 Z0 @+ e K. a& L4 a399.99 is 45-deg to 390.7 U3 U* J; N3 w2 U& n# P4 G; @
Keep in mind that if you add or subtract an amount that will change the original angle (85.5-deg)
7 L9 Z+ W! |8 x! ?# ?to an amount greater than 360 or less than 0, that you JUMP rings. For example, if you subtract
b4 H- q! \* x. p4 T6 S) T k+ ]" V90-deg from 85.5 to potentially find a square aspect, you get -4.5-deg. Add 360 gives 355.50-
& @4 u' T" W; M/ j; X3 c8 sdeg in the previous ring. We were using Ring# 10 in the formula, but for this calculation, we
0 g& v1 z2 Y a, Owould have to use Ring# 9. Similarly, if you added 315-deg to 85.5-deg, you get 400.50, which
; |" P* S) |: @3 C3 ` w+ k9 Wis 40.5-deg in the next ring. So you would have to use ring# 11 for this calculation |
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发表于 2008-7-31 17:48:41
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