Diagonal Degrees and the Avoidance of Conflict

If you did not read the Friday or Saturday posts and comments, you should. There is a lot of information there. Today, though, we want to talk about a topic that you do need to know about and one that will help clarify a lot of things about wave counting for you. That topic is "how do diagonals avoid degree conflicts?".

Don't worry. We're going to make it short simple and take it step-by-step so it is worth the read. So let's begin.

In order to make this as simple as possible, we will start with the case of the contracting diagonal first, with reference to the diagram, below.

Contracting Diagonal - No Inherent Degree Conflicts

If what we mean by the term 'lower degree' is that the next sub-wave after a higher degree wave has completed must be shorter in price and time than the previous wave of the higher degree, then we can see that a contracting diagonal inherently avoids a degree conflict.  Why?

So, since wave (iii) is shorter overall than wave (i), then we know by logic that wave a, within wave (iii) must also be shorter than (i). This is provided that wave (ii) is shorter than (i) – which it must be if it is a second wave.

Let's be clear. Wave a does not have to be shorter than the first 'a' within (i), it just has to be shorter than wave (i), overall. Wave (i) is the higher degree wave. Wave a is what we are calling the first lower degree sub-wave after wave (i). And so, the same occurs within wave (v). And, since wave (v) is shorter than (iii), then all of the same considerations apply, and inherently degree conflicts are avoided.

It should now also be clear why contracting diagonals usually, almost always, also contract in time, too. That is because the sub-waves must be shorter in each section. In fact, I contend that “counting bars” can be an effective way to determine whether one is in a contracting diagonal or not. If the next sub-wave gets “too long in time”, then one most likely will not be in a contracting diagonal.

The time considerations of degree violations also help determine why wave (v) is always shorter than wave (iii) in a contracting diagonal. Most people tend to think of it as, “wave (iii) can never be the shortest wave”. And that is true, too. But, in fact, if wave (v) is adhering to time degree considerations, then it should also not be longer in time than wave (iii), either.

Now, let's move on to the somewhat more diabolical Expanding Diagonal. In the case of the Expanding Diagonal, we expect wave (iii) to be longer in price and time than wave (i). But serious Elliott Wave students want to know, “just how does this happen without degree violation?!”.

Avoiding Degree Conflicts in an Expanding Diagonal

So let's assume we have made first wave down, (i). This wave (i) establishes the “Field of Play” as it were. Then, the first sub-wave a of the next wave must be shorter only than wave (i). It does not need to be shorter than wave 'a', of wave (i) – just shorter than (i).

So now follow this logic. Since, wave a can be almost as long as (i), then if you put two of them together in an a-b-c wave, then the whole wave can be longer than wave (i), particularly if there is a shorter b wave in the middle. And now the wave (iii) can be longer in price & time than wave (i) without degree violation! And, it is OK for wave (iii) to be longer than wave (i), as they are the same degree wave, and that is the usual course of events – to have the third wave longer than the first wave in a sequence.

Again, the only restriction is that the sub-waves of (iii) are shorter in price and time than all of wave (i). Well, the same thing occurs for wave (v), except now it is wave (iii) that has become the 'longest wave on the board' and it sets the new “Field of Play” as it were. Then the sub-waves of wave (v) must only now be shorter in price and time than all of (iii).

The next consideration is wave (iv). Keep in mind there is no Elliott Wave rule that says a wave four can not be longer in price than a second wave. And as long as the sub-waves in wave (iv) are shorter in price and time than all of wave (ii), then wave (iv) is going to take full advantage of the lack of a rule regarding fourth wave price and travel as far as it can - provided it does not travel beyond the end of wave (ii). That, of course, is something no fourth wave is allowed to do, by rule!

When you consider time and price degree like this, you can also answer for yourself, “what is the maximum extent of travel” of wave (v) in an Expanding Diagonal?

And, if you have ever seen an Expanding Diagonal in real time, you know that the last upward move can be especially aggressive in an Up market, and the last downward swoosh can be quite devastating in a Down market. How far will the move go? A person caught trapped in such a wave might have a vital interest in knowing.

Well, it should be clear by now that wave (v) can not proceed farther in time than it's c wave would become longer in time than all of wave (iii)! That may not be much consolation if money is lost – but at least it is a LIMIT that would otherwise be undefined. Otherwise, how do you know? And you know, the c wave – wherever it starts is limited to the maximum price travel of wave (iii), too!

Now the above example was for a 3-3-3-3-3 diagonal. By extension, the serious Elliott Wave student should undertake specifying the limits of travel of the various sub-waves of a 5-3-5-3-5 diagonal, as well. We think if you will do that it will also crystallize for you the concept of degree violations in impulses, too. 

Does this make sense? We hope so. After reviewing this article, you may be able to answer more questions about price and time than many professional Elliott analysts can. More importantly, we hope it makes your wave counting that much more accurate.

Have a good start to your Sunday.
TraderJoe