Here is the 5 Wave Model wave count from the March 2009 lows. As you can see, each wave breaks down into a smaller five wave sequence. This numbering is not arbitrary, or tentative, but is based on the precise, and unique, mathematical relationships between the waves themselves. These relationships have held true at every level through over 80 years of data.
Many of the waves are straight forward sequences, but others, such as Wave 4, are more complex. A series of inverted waves occurred during Wave 2, leaving the termination of that wave below the termination point of Wave 1. The same thing occurred during Wave 2 of 5. Such inversions are normal, usually appearing at least once in a given sequence. These inversions adhere to the same wave relationships as normal waves, and thus can be identified as they occur.
We are now approaching Wave 5 of 5. Waves 3, 4, and 5 have yet to complete, but due to the wave structure these should be short, quick waves. As mentioned earlier, these should terminate somewhere slightly above the 1378.79 level. Once this has been completed, a five wave corrective sequence will begin.
Since this corrective sequence will be a wave 2, an inverted wave is possible. This can take on several shapes, but could very well carry the market to new highs after a short corrective phase. Only time will tell.
You say: "This numbering is not arbitrary, or tentative, but is based on the precise, and unique, mathematical relationships between the waves themselves." Are you literally computing the numerical labels with formulas (e.g. a computer program), or are you assigning them manually or visually based on certain desired relationships (e.g. Fibonacci percentages)? Does your method allow for alternate counts?
ReplyDeleteHi Paul.
ReplyDeleteThe wave counts are determined by computation. Our model requires six inputs. Let's say a downtrend terminates at some point X. That becomes our first input. As the market moves up with alternating highs and lows, those also become inputs. When our equation is satisfied, the five points besides X that solve the equation become waves 1, 2, 3, 4, and 5. Wave 5 is then labeled Wave 1 of the next higher degree. Then we can use Point X, and the termination point of Wave 1 as the first two inputs for the next higher sequence.
Our model does not allow for alternate counts. Each wave sequence has one unique answer. When the criteria for our model is met, the wave has terminated.
Steve