EMA 45 degres Uptrend

This topic has 5 replies, 3 voices, and was last updated 5 years ago by robertogozzi.

Viewing 6 posts - 1 through 6 (of 6 total)

05/12/2018 at 6:57 AM #70237

Inertia

Hi dream team !

How can we code this “EMA (x°) uptrend” please ?

Thank you 😉

Inertia

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05/12/2018 at 10:29 AM #70242

robertogozzi

No, we CAN’T!

As you can see from the attached pic made with the help of eXcel, all of the 3 examples are uptrending, but their grade is different.

You can only tell uptrends as Higher Highs & Higher Lows and downtrends as Lower Highs & Lower Lows!

You may detect a downtrending EMA comparing its value, say, now and 10 bars ago and 20 bars ago…. if it’s always decreasing you may assume there is a downtrend ongoing (in that TF), but you cannot grade it.

robertogozzi

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05/12/2018 at 12:40 PM #70254

Inertia

Grazié à Lei Roberto for your complete answer.

Appreciated.

Inertia

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05/13/2018 at 10:58 AM #70279

Vonasi

Similar I think to what Robertogozzi said you can apply a gradient score to the average by subtracting the average level x bars ago from the average level now and dividing by x. By comparing different values of x you can confirm whether the steepness of gradient is increasing or decreasing.

Could you not then also return this as degrees by the following formula?

Period = 200 //Average period
x = 1 //Number of bars to look back

Gradient = (ExponentialAverage[Period](close) - ExponentialAverage[Period](close[x]))/x

Angle = ATAN(Gradient)

Return 0, Angle

Period = 200 //Average period

x = 1 //Number of bars to look back

Gradient = (ExponentialAverage[Period](close) - ExponentialAverage[Period](close[x]))/x

Angle = ATAN(Gradient)

Return 0, Angle

Vonasi

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05/13/2018 at 11:25 AM #70280

Vonasi

Sorry – please ignore the code in the last post as I just realised that one side of the gradient calculation is an unknown distance in pips so the calculation will not return a true angle. Back to school for me I think!