Tuesday, September 29, 2015

Two Possible Targets

Because of yesterday's price action, the current position of the Elliott Wave Oscillator, and the lengths of the various waves, we are now able to provide two price projections, either of which will constitute satisfactory Elliott Wave relationships. Referring to the chart, below, we note that if wave 5 = 1, it would not make a new low. That is because wave 5 simply wouldn't be long enough given the length of wave 4.

First SP500 Target

So that must mean that the next most common relationship provides the first target. That target is at 1855, basis cash, where a wave 5 would equal 0.618 times the net distance traveled by waves one - through - three. This  is often abbreviated 5 = 0.618 x net (1 - 3). So, the first anchor of a Fibonacci ruler is placed on the 0 point, a second anchor is placed on 3, and the pivot anchor is taken to wave 4. This provides a 0.618 extension at 1855.46. Clearly, such a point will provide a new lower low in the S&P500 - really the only remaining requirement for a true impulse wave lower.

But, if this target begins to get exceeded, then the target would shift to wave 5 = 3, following the proposition that "two Elliott waves are often similar in length, sometimes remarkably so." We will allow the reader to perform this measurement for themselves, since this is relatively straight forward.

Should the 5 = 3 target come into play, this may mean that the 'degree' of wave labels we have chosen for the sub-divisions of wave 5 may have to be lowered one degree. In other words, instead of the minute degree labels chosen, they may, more accurately, be only minuet degree labels and minute wave i may not have formed in it's entirety yet. This is a situation we will deal with only if the first target is exceeded.

Until then, cheers and enjoy the chart!

2 comments:

  1. Thank you for your time and postings E-T! They are much appreciated!

    ReplyDelete