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So a tick below 4896.35 would invalidate above labels and then we may have to consider the following Alternative wherein (v) of Ya is still in play:
.png)
So 4896.35 should be watched carefully for choosing our labels:
However even if we go below 4896.35, we would not breach 4953.9-102=4851.
9.10 AM
Pre-open session has ended at 4875 that means our Ist Count has been invalidated and we have now to follow the alternative shown above.
As per this alternative also, (v) of Ya is playing out as a Diagonal wherein sw1>sw3>sw5
as sw1=133
sw3=102
sw5 from 4954 downwards can not do more than 102 points so my advice is to accumulate longs below 4871 with stop loss at 4851.
10.15 AM
So our labels have survived the Ist onslaught.
What if our SL of 4851 is hit?
I would sit on the sidelines and wait till EOD to come-up with fresh labels.
However,a break-out above today's high of 4883 would confirm my labels---
10.48 AM
There was a small arithmatic mistake:
Sw1=133
sw3=4976-4871.5=104.5
sw5 boundary=4953.9-104.5=4849.9
So our SL should be 4849.5
EOD
Labels have been failing quite consistently.
I am putting up a chart showing the downmove 5342-4837* and still wondering as to what labels should be ascribed now:
.png)
Ya having played out in following manner:
(i)=5342-5154=188
(ii)=5154-5279
(iii)=5279-4988=291
(iv)=4988-5125
(v)=5125 downwards as extended wave making a 3x3x3x3x3 diagonal Target min 4820
Guruji, I am a novice... going through the EW theory and enjoying yours posts regularly following how you are using different timeframes with EW and initiate trades based on completion of set of waves in a direction.
ReplyDeleteRules of wave 4 indicate that they should not overlap with wave 1, however you comment is suggesting to wait till low of wave 2. pls clarify :
"this c of sw4, travelling downwards from 4953.9 cannot go below the low of sw2 i.e. 4896.35"
Q
ReplyDeleteRules of wave 4 indicate that they should not overlap with wave 1, however you comment is suggesting to wait till low of wave 2. pls clarify
UnQ
In case of diagonal, sw4 overlaps with sw1.
thanks for your prompt reply sir.
ReplyDeleteguruji,
ReplyDeletesw3 is 105 i guess so sl 4849?? 4954-105
Thanks
Delete:)
my pleasure guruji :-)
ReplyDeleteSir
ReplyDeleteA(a) – 5630 – 5171
A(b)– 5171 – 5499
A(c) – 5499 – 5136
B – 5136 – 5378 [Retraced 0.49 of A]
C(a) – 5378 – 5191
C(b) – 5191 – 5343
C(c) – 5343 – 4906[5 waves]
X - 4906-4976
1 4976-4868 (3)
2 4868-4954 (3) 0.80 of 1
3.1 4954-4837
3.2 4837- running upto 4930
3.3 4930 upto 4785 and so on
Sameerji,
DeleteWe have two major issues here( not listing the minor ones):
Whenever we are talking of A as a 3 wave form( as you hve shown here) we are ruling out the probability of a zig-zag ABC i.e. If A is 3 wave then ABC has to be either a Flat or the Corrective will take the form of ABCDE triangle(wherein all the arms will be 3 wave forms).
As you are not considering the Corrective as a triangle and zig-zag is already ruled out so the only option left for ABC as a Corrective is FLAT.
And in a FLAT, B has to retrace more than 90% of A.
As the B mentioned by you is retracing only 49% so this Count is not a valid one.
This was the one major issue.
Secondly, the C mentioned by you from 5378 downwards is playing out as a 3 wave form which is not the basic tenet of C which has to be a 5 wave form(Impulse or Diagonal)
Sir,
ReplyDeleteJust finished EW book by Robert Prechter.
Cloud you please share/suggest an excel sheet for wave & time calculations. puravih@gmail.com