The correction may have ended 4.46 below the expected low. The Elliott wave count remains the same.

Summary: There are two possible scenarios for Monday: either a short sharp upwards thrust to 1,235 (which would be confirmed with a new high above 1,213.45), or one more spike down to 1,168 – 1,172 before the third wave up begins (confirmed with a new low below 1,186.29). While price remains below 1,213.45 and above 1,186.29 both scenarios will be possible. Overall, the minor degree trend remains up and I still have confidence in the mid term target at 1,262 which remains the same for both scenarios.

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Primary wave 4 is complete and primary wave 5 is unfolding. Primary wave 5 may only subdivide as an impulse or an ending diagonal. So far it looks most likely to be an impulse.

Within primary wave 5 intermediate wave (1) fits perfectly as an impulse. There is perfect alternation within intermediate wave (1): minor wave 2 is a deep zigzag lasting a Fibonacci five days and minor wave 4 is a shallow triangle lasting a Fibonacci eight days, 1.618 the duration of minor wave 2. Minor wave 3 is 9.65 longer than 1.618 the length of minor wave 1, and minor wave 5 is just 0.51 short of 0.618 the length of minor wave 1. I am confident this movement is one complete impulse.

Intermediate wave (2) is an incomplete expanded flat correction. Within it minor wave A is a double zigzag. The downwards wave labelled minor wave B has a corrective count of seven and subdivides perfectly as a zigzag. Minor wave B is a 172% correction of minor wave A. This is longer than the maximum common length for a B wave within a flat correction at 138%, but within the allowable range of less than twice the length of minor wave A. Minor wave C may not exhibit a Fibonacci ratio to minor wave A, and I think the target for it to end would best be calculated at minute degree. At this stage I would expect intermediate wave (2) to end close to the 0.618 Fibonacci ratio of intermediate wave (1) just below 1,281.

Intermediate wave (1) lasted a Fibonacci 13 weeks. If intermediate wave (2) exhibits a Fibonacci duration it may be 13 weeks to be even with intermediate wave (1). Intermediate wave (2) has just completed its eighth week, and so it may continue now for another five weeks.

So far within minor wave C the highest volume is on an up day. This supports the idea that at this stage the trend is up.

The upper maroon trend line is the upper edge of the Elliott channel drawn about this impulse on the weekly chart, and copied over here to the daily chart. See the most recent Gold Historic Analysis to see how this trend line is drawn on the weekly chart. If minor wave C lasts long enough it should find extremely strong resistance at that trend line, I would not expect that trend line to be broken while primary wave 5 is incomplete.

The aqua blue trend line is a simple TA trend line which is showing where price is currently finding resistance. The power of the middle of the upcoming third wave should break above this trend line.

The target for primary wave 5 at this stage remains the same. At 956.97 it would reach equality in length with primary wave 1. However, if this target is wrong it may be too low. When intermediate waves (1) through to (4) within it are complete I will calculate the target at intermediate degree and if it changes it may move upwards. This is because waves following triangles tend to be more brief and weak than otherwise expected. A perfect example is on this chart: minor wave 5 to end intermediate wave (1) was particularly short and brief after the triangle of minor wave 4.

Intermediate wave (2) may not move beyond the start of intermediate wave (1) above 1,345.22. I have confidence this price point will not be passed because the structure of primary wave 5 is incomplete because downwards movement from the end of the triangle of primary wave 4 does not fit as either a complete impulse nor an ending diagonal.

To see a prior example of an expanded flat correction for Gold on the daily chart, and an explanation of this structure, go here.

*Note: I am aware (thank you to members) that other Elliott wave analysts are calling now for the end of primary wave 5 at the low at 1,131. I am struggling to see how this downwards movement fits as a five wave impulse: I would label the second wave within it (labelled minor wave 2) intermediate wave (1), and the fourth wave intermediate wave (4) (labelled as a double zigzag for minor wave A). Thus a complete impulse down would have a second wave as a single zigzag and a fourth wave as a double zigzag, which would have inadequate alternation. Finally, the final fifth wave down would be where I have minor wave B within intermediate wave (2). This downwards wave has a cursory count of seven, and I do not think it subdivides as well as an impulse as it does as a zigzag. If any members come across a wave count showing possible subdivisions of a complete primary wave 5 I would be very curious to see it.

Main Hourly Wave Count

This wave count follows on from last hourly wave count. It is slightly adjusted.

Minuette wave (ii) may have completed not as a double combination but as a double zigzag. Within the second structure of the double, subminuette wave y, its B wave is less than 90% the length of its A wave so this structure cannot be a flat correction. Unfortunately, micro wave A is ambiguous and can be seen as either a three or a five.

Double zigzags are different in purpose to double combinations. While double combinations exist to take up time and move price sideways, double zigzags exist to deepen a correction when the first zigzag does not move price deep enough. Subminuette wave w did not reach down to the 0.382 Fibonacci ratio and so that certainly fits the description of not moving the correction deep enough.

If minuette wave (ii) is a double zigzag then it is over now. The second structure in the double is a complete 5-3-5, and has deepened the correction.

Within minuette wave (iii) no second wave correction may move beyond the start of its first wave below 1,186.29.

At 1,235 minuette wave (iii) would reach 0.618 the length of minuette wave (i). This would be a slightly unusual relationship for a third wave, but it is entirely possible (I have seen third waves which are shorter than the first waves before). The core Elliott wave rule which must not be broken is that the third wave may not be the shortest of the actionary waves (actionary waves are waves 1, 3 and 5 within an impulse). This is not the same as the third wave must be the longest, which is a common misunderstanding of Elliott wave rules.

However, this would be a little unusual and so for that reason I do not have full confidence in the target at 1,235. If it is wrong it will be too low.

I do have confidence in the target for minute wave iii. At 1,262 it would reach 1.618 the length of minute wave i. This is a very common ratio for a third wave.

Minuette wave (iii) must move far enough above the end of minuette wave (i) at 1,221.34 to allow room for movement for a subsequent fourth wave correction for minuette wave (iv) which may not move back into minuette wave (i) price territory. All Elliott wave rules for an impulse must be met for minute wave iii.

Alternate Hourly Wave Count

This wave count sees the subdivisions within minuette wave (ii) differently. What if it is a single zigzag and not a double?

Subminuette wave a does fit very nicely as a five wave impulse. There is perfect alternation between its second wave, a flat, and its fourth wave as a zigzag.

Subminuette wave b is seen as a complete zigzag in the same way as the main wave count sees this movement.

If minuette wave (ii) is a single zigzag then within it subminuette wave c must subdivide as a five wave structure. I have considered both an impulse and ending diagonal. It is difficult to see how the third wave would subdivide for an impulse, and so an ending diagonal is much more likely and has a better fit.

Within an ending diagonal all the sub waves may only subdivide as zigzags, and the fourth wave must overlap back into first wave price territory.

Within the possible ending diagonal of subminuette wave c micro wave 1 fits well as a zigzag. Micro wave 2 is a 0.88 correction of micro wave 1, only just longer than the maximum common length which is between 0.66 to 0.81.

Micro wave 3 fits well as a zigzag, and within it sub micro wave (A) is a leading contracting diagonal.

Micro wave 4 so far has not overlapped back into micro wave 1 price territory. It must move above 1,201.39. The rule for the end of a fourth wave of a diagonal is it must not move beyond the end of the second wave above 1,213.45.

This ending diagonal is expanding because micro wave 3 is longer in length with micro wave 1. This means micro wave 4 should be longer than micro wave 2, so it should reach above 1,198.35. It is most likely to end within the common range of 0.66 to 0.81 the length of micro wave 3, between 1,204.22 to 1,208.29.

Thereafter, a final zigzag downwards to the target would complete an ending expanding diagonal for subminuette wave c.

At 1,168 subminuette wave c would reach 1.618 the length of subminuette wave a. At 1,172 minuette wave (ii) would reach the 0.618 Fibonacci ratio of minuette wave (i). This gives a $4 target zone with a reasonable probability. If this target is met in one more day with one more candlestick then minuette wave (ii) may complete in a Fibonacci five days total.

Minuette wave (ii) may not move beyond the start of minuette wave (i) below 1,142.88.

If the main wave count is correct then it should be confirmed quickly on Monday. If the alternate is correct then it too should be confirmed on Monday as well.

This analysis is published about 08:58 p.m. EST.