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原帖由 5575338 于 2008-7-31 17:48 发表 ![]() & k' U' i: g' A& B8 b' {
9方图计算公式
w* \: s# ?5 f; R6 M# c1 n: Y在下面几个角度线上的：! E$ [: @: c h7 X* f
0 degrees: (2n + 5/4)squared
7 E0 H4 _+ Y7 @/ A4 _; _45 degrees: (2n + 6/4)squared; D# a8 _# x* F. h+ J
90 degrees: (2n + 7/4)squared7 _9 ^3 q# L' E T
135 degrees: (2n) squared: h: B2 @7 o" ^ M4 n$ H
180 degrees: (2n + 1/4)squared
- @& A, j2 f2 V. S, J225 d ...
& q, D4 W! Q, a0 l6 t/ I" c+ x p
' C) a; i6 j6 q, s& _是这个吗?: Q" k$ \1 Z- f- J& {+ _
7 q. u/ l. d9 @- A$ \: f
" S& J2 n1 k$ W- |. GSquare of Nine Essentials
' U3 O: h, q' R0 d; g. S8 [9 ^Daniel Ferrera, 2002
- X {. D V' gIn my experience with working with this method, price &  must balance on a hard aspect.
) `$ i4 R/ q. J9 w7 VThe hard aspects are 45, 90, 135, 180, 225, 270, 315 and 360 or 0 degrees. The most. p H7 n* A2 L0 C+ M) F; k
important being the squares or 90-deg harmonics (0, 90, 180, 270).! X9 W7 {) ^' X' s
In terms of selecting a past date and price to start from, I have found that the lowest low over2 \* q; u9 \' x& ~
the past 365-days and the highest high over the past 365-days have the greatest influence on$ f Z N0 s/ S
these balance points. This technique can be used to generate the horizontal support &8 l% B6 ~6 h& h: }2 ?% S
resistance levels for intraday trading. This is extremely useful when you anticipate that a
5 n3 d+ g. j7 L) m: r4 }; {particular day will be a trend change as the result of cycles or counts, etc.; U1 ?, \% v8 j5 P. m
Carl Futia's formula for this reads9 j+ g2 i. {* N8 h
=MOD 360 ((price distance or Time change)^0.5*180-225): Q; z" U. V$ b+ \
This formula assumes that the Squares of Even numbers fall on the 135-deg angle and that the
" A4 x9 r, r e& qSquares of Odd numbers fall on the 315-deg angle, which is not true on Gann's actual Square$ u# ~/ ?) Y$ r6 C) p
of Nine chart.& K+ B' b; S/ E' T& f
If you start with a "1" in the center, the Squares of Odd numbers will fall on the 315-deg angle,' n; o! ]+ N$ K9 T: f7 N7 L* y d
but the Even Squares (16, 36, 64, 100, 144....) will gradually float towards 135-degrees. For
e+ c5 Q: m! mexample, on the actual Square of Nine
( N/ Z9 d5 V9 |16 is on the 112.50-deg angle,
Q7 U# z; D* a6 I0 f36 is on the 120-deg angle,+ z$ q& U" D/ i1 @6 v* `
64 is on the 123.75-deg angle,5 j# _; D% g1 i+ x- h1 f
100 is on the 126-deg angle and
~, h, j/ T: S }5 X! M! @) V144 is on the 127.50-deg angle9 V% D7 Z7 G3 {/ n( G5 p
and so on.
. h/ N% D& B: k4 O8 x ]# TStarting with "0" in the center, the Squares of Even numbers will line up on the 135-deg angle
, a* c: g( F2 Dand the Squares of Odd numbers will Float.( |. ?3 n3 `8 m, T
Could this amount of inaccuracy or "Lost Motion" be important? After all, it is impossible to draw
, f& i- z+ J+ G, ?, J/ m! nor actually build a Square of Nine Chart based on the MOD 360 formula above. If you want to6 a: Q3 ?9 ^- w' D) Z
work with calculations that are based on W.D. Gann's printed Square of Nine chart, the
( n% t8 @* t p2 |& D8 a. zfollowing formulas will be of great use to your research:
; Y( ?7 r! l1 Y* P* q$ a9 S, ^Ring# = Round(((SQRT(Price)-0.22 / 2),0)# X+ e+ b* [3 X
{This rounds to the nearest whole number, i.e. it eliminates the decimals}, x: q7 A. @7 A" Q9 O0 J: R) {
Example: The number 390 is in Ring #10 if you crunch the above formula.7 ~6 V, P5 Y) j# a* ^9 j# T
315-deg Angle: This is the most accurate angle of the entire chart and is used to calculate all
3 F6 B' \ g+ j K9 lother values. The Squares of Odd numbers are all on this angle.5 s0 W4 q* L$ U6 g
315-deg Angle = (Ring# * 2 +1)^2
4 z9 L) P( I# V4 R3 k4 {Example: 390 was in ring# 10 so the 315-deg number is (10 * 2+1) ^2 or simply (21)^2 = 441" h. q k. Z( M' \1 H' M
The Zero Angle on this Ring = ((Ring# * 2 + 1)^2) - (7* ring#). So you would get 441 - 70 =' j5 m: @/ Y# H8 y9 Q& N4 ]
371 This number is needed to calculate the angle that the 1st value of 390 is on.
. }4 Z& W2 k3 C" M. N6 _Angle = Sum ((Price- Zero Angle) / (Ring/45)). So we have ((390 - 371) / (10/45) = 85.50-deg
* D; |5 c; u* y% z4 eYou may have to occasionally adjust the Angle calculation because sometimes you will get a
% N+ M! p4 A( n, wnegative value when you have a number that is approaching the 0-deg angle of the next ring.0 I" {" f$ ]7 o& Q2 O l
For example: We know that 371 is a zero-deg number. If you try to find the angle of the number
. e& y( }; z2 i! E) [% R370.5, which is a number in the previous ring approaching the next ring, you get Sum ((370.5 -0 n/ m3 l/ e2 F8 q% o
371) / (10/45)) = -2.25-deg. If you get a negative number, just add 360 to correct it. So this
+ m# r5 S2 K5 E0 Hwould actually be 357.75-deg.
. B, T7 s" `7 A3 a) c& QA simple formula to correct this is If Angle<0 then +360 else Angle = Angle.9 I! K6 i' q6 z$ w
To generate other values on the Square, use this formula: (Ring# * 2+1)^2) - (7* Ring#) +1 y8 A- V( ^ U. I' P8 v/ V; D
((Ring# / 45) * Angle)6 j. i6 C9 ^: v+ W
Angle is this formual is your input value. For example, we know that 390 is on the 85.50-deg0 t! ]8 G1 k- B4 e! H8 ?% r
angle. If we want to know the value of the number that is 45-deg to this number, we would be% A3 k6 }+ ^; }0 t# e
interested in the angle of 130.50-deg (85.5 + 45). Inputing this in the above formula gives us:; x0 J5 T) m R
(10 * 2+1)^2 - (7 * 10) + ((10 / 45) * 130.5). Simplified a little, we have 371 + (28.99971) =9 M) F2 D* q! ~( N# C: P
399.99 is 45-deg to 390.% b" k: S( m2 r
Keep in mind that if you add or subtract an amount that will change the original angle (85.5-deg)
2 A9 u) C! G6 c- Mto an amount greater than 360 or less than 0, that you JUMP rings. For example, if you subtract1 R/ x( {( V% s! n. n4 {
90-deg from 85.5 to potentially find a square aspect, you get -4.5-deg. Add 360 gives 355.50-
8 Z$ `) x( n) J% A& bdeg in the previous ring. We were using Ring# 10 in the formula, but for this calculation, we K3 E! n/ E9 H F& K6 o
would have to use Ring# 9. Similarly, if you added 315-deg to 85.5-deg, you get 400.50, which" H1 w9 z: @* K: N) Z0 N
is 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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