hard · Elliott Wave Theory fibonacci
An impulse advances from a wave-1 origin where wave 1 spans 100 points. Wave 4 ends at the 0.382 retracement of the entire wave-1-through-3 distance, and waves 1 and 3 together cover exactly 261.8 points. An analyst must project the most common Fibonacci target for wave 5 measured from the wave-4 low.
Using the standard convention that wave 5 frequently equals the net traveled distance of waves 1 through 3 multiplied by 0.618 when wave 3 is extended, what is the projected length of wave 5?
- Approximately 161.7 points, taking 0.618 of the 261.8-point waves-1-through-3 net distance
- Approximately 100 points, on the basis that wave 5 equals wave 1 when wave 3 is the extended wave
- Approximately 38.5 points, applying 0.382 to the 100-point wave 1 as the wave-4 retracement implies
- Approximately 161.8 points, taking the 0.618 expansion of wave 1 plus the wave-4 retracement depth
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