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Movement above 1,508.33 invalidated both hourly wave counts. I had expected downwards movement for Wednesday’s session which did not happen.

Both wave counts remain valid while no new high has been made.

Click on the charts below to enlarge.

Main Wave Count.

S&P 500 daily 2013

Wave iv pink should exhibit choppy overlapping downwards movement. It may be over already (an alternate wave count on the hourly chart below) or it may continue further.

Because wave ii pink was a shallow 36% correction of wave i pink then wave iv pink may end about 1,440 at 0.618 the length of wave iii pink, given the guideline of alternation.

Wave ii pink was a zigzag so we may expect wave iv pink to be a flat, double or triangle. A flat is most likely.

The most common type of flat is an expanded flat which has a B wave that makes a new price extreme beyond the end of the A wave. A new high above 1,530.94 is possible if wave iv pink unfolds as a flat.

If wave iii pink is over it lasted a Fibonacci 55 days. Within it wave (i) green lasted 14 days, one more than a Fibonacci 13, wave (ii) green lasted a Fibonacci 8 days, wave (iii) green lasted 20 days, one less than a Fibonacci 21, wave (iv) green lasted a Fibonacci 3 days, and wave (v) green may have had no Fibonacci duration. So far wave iv pink has lasted six days.

Wave (v) pink may move price to a new high (although it does not have to) and may end again on the upper edge of the wider parallel channel containing the zigzag of wave (Z) black.

At 1,548 wave C blue would reach equality in length with wave A blue. When wave iv pink is complete I will recalculate this target at pink degree.

Wave iv pink may not move into wave i pink price territory. This wave count is invalidated with movement below 1,409.16.

If this main wave count is invalidated then the alternate wave count will be strongly confirmed.

S&P 500 hourly 2013

Because wave ii pink was a shallow 36% zigzag correction I will expect alternation in either or both depth and structure for wave iv pink.

Within wave (a) green the structure may have been complete at 1,485.01. There is no Fibonacci ratio between waves a and c orange.

Ratios within wave c orange are: wave 3 purple is 2.07 points short of 1.618 the length of wave 1 purple, and wave 5 purple has no Fibonacci ratio to either of waves 1 or 3 purple.

If wave iv pink is unfolding as a flat then wave (b) green must reach up to at least 90% of wave (a) green at 1,526. Wave (b) green may make a new high. Wave (b) green should breach the channel containing wave (a) green. It must subdivide into a three and so far it looks like it may be a zigzag.

When wave (b) green is complete then wave (c) green should take price below the end of wave (a) green to a new low and is most likely to be either 46 or 74 points in length.

There is no upper invalidation point for this wave count.

If price fails to reach up to 1,526 then wave iv pink cannot be a flat correction. It may be a double combination or double zigzag, however, and I will look at this possibility if price fails to reach 1,526.

S&P 500 hourly 2013

Alternation is a guideline and not a rule. Wave iv pink may have been over in a Fibonacci 5 days as a single shallow 32% zigzag. If this is the case there would be no alternation in either depth or structure between waves ii and iv pink. This is unlikely, but it is possible. They exhibit some alternation within each zigzag; the A and C waves are different lengths in relation to each other.

At 1,551 wave (v) green would reach equality in length with wave (i) green. At 1,548 wave C blue would reach equality in length with wave A blue.

Within wave v pink wave (iv) green may not move into wave (i) green price territory. This wave count is invalidated with movement below 1,498.99.

Alternate Wave Count.

S&P 500 daily alternate 2013

While price remains below 1,530.94 it remains possible that we have seen a big trend change at primary wave degree. The bearish engulfing candlestick reversal pattern indicates a trend change and price may have bounced off the upper edge of the wider parallel channel.

If primary wave B triple zigzag is complete then it is a 160% correction of primary wave A. This is longer than the common length, so it is a bit unusual, but it is not unrealistic. I have seen B waves within flats that are this long.

We should always assume the trend remains the same, until proven otherwise. We should assume that the main wave count is correct, the trend remains upwards, until price moves below 1,409.16 to invalidate the main wave count. At that stage this alternate would be my only wave count and I will calculate downwards targets for you.

Further downwards movement below the parallel channel about the zigzag for wave (Z) black would provide confidence in this wave count.

Because primary wave B is a triple zigzag, and the maximum number of structures in a multiple is three, once the third zigzag is confirmed as complete then the entire correction must be complete. At cycle degree I cannot see another explanation for primary wave B which would allow for further upwards movement.

Within the new downwards trend a leading diagonal for a first wave may be unfolding.

S&P 500 hourly alternate 2013

If we have had a big trend change then there are two possible structures which may be unfolding for wave i pink. An impulse has been invalidated so the only other possible structure is a leading diagonal.

Within a leading diagonal subwaves (i), (iii) and (v) are commonly zigzags, but they may also be impulses. Subwaves (ii) and (iv) must subdivide as zigzags.

Wave (iii) green is not be the shortest wave; it is longer than wave (i) green so the diagonal is likely to be expanding, with wave (iv) green being longer than wave (ii) green.

Within a diagonal wave (iv) green should overlap wave (i) green price territory, but may not move beyond the end of wave (ii) green. This alternate is invalidated with movement above 1,525.84.

Within a leading diagonal wave (v) green may not be truncated. It must make a new low below the end of wave (iii) green at 1,485.01.

At 1,479 wave (v) green would reach equality in length with wave (iii) green. It is likely to be longer than this because the diagonal is expanding, but sometimes the third wave is still the longest.