apply my personnal indicator to graph

• 01/22/2019 at 9:14 PM #89371
  Vonasi

  multiple = 1 //(or any value you choose to tune the indicator to the market) x = 0 -(????) //( ???? = output of your indicator usually between -15 and + 15) z = medianprice * (1 + (x / 100) * multiple) return z

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  multiple = 1 //(or any value you choose to tune the indicator to the market) x = 0 -(????) //( ???? = output of your indicator usually between -15 and + 15) z = medianprice * (1 + (x / 100) * multiple) return z

  If your indicator output is normally between -15 and +15 then the above will do what I was suggesting. Tune the multiple to match individual markets.

  

  Vonasi

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  01/23/2019 at 9:35 AM #89399
  Nicolas

  Porion wrote:

  i can still use it only as support/resistance when it crosses zero but i wanted to improve it a little bit more.

  oh, easier to understand with a picture! Indeed, what you want is possible. Please share the indicator code in order to incorporate all the necessary code to it. I’m sure we’ll be able to add the graphical component you miss.

  

  Nicolas

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  01/23/2019 at 1:39 PM #89432
  Porion

  a = RSI[30](close) - RSI[100](close)

  1	a = RSI[30](close) - RSI[100](close)

  it’s a really simple code, but it can give you pretty decent signal if you use it correctly.

  

  Porion

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  01/23/2019 at 3:32 PM #89458
  Vonasi

  Porion wrote:

  a = RSI[30](close) – RSI[100](close)

  That is a bounded indicator as RSI is bounded between 0 and 100 so your combination of two RSI’s is now bounded between 100 and -100 as those are the maximum and minimum results possible.

  If you apply your indicator to my code you get a line that flips above and below price.

  multiple = 1 x = 0 - (RSI[30](close) - RSI[100](close)) z = medianprice * (1 + (x / 100) * multiple) return z

  1 2 3 4

  multiple = 1 x = 0 - (RSI[30](close) - RSI[100](close)) z = medianprice * (1 + (x / 100) * multiple) return z

  

  2 users thanked author for this post.

  Nicolas, robertogozzi

  

  Vonasi

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  01/23/2019 at 3:43 PM #89461
  robertogozzi

  Nice job Vonasi, but line 3 should be replaced by

  z = medianprice * (1 - (x / 100) * multiple)

  1	z = medianprice * (1 - (x / 100) * multiple)

  to replicate the true indicator.

   

  

  robertogozzi

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  01/23/2019 at 3:53 PM #89463
  Vonasi

  Porion wrote:

  when my indicator turn positive, it should be below the price and vice versa.

  I think mine is correct if I understood the above comment from the OP correctly.

  

  Vonasi

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  01/23/2019 at 4:06 PM #89465
  Vonasi

  Here is a version that draws a buy or sell arrow when the line crosses medianprice as Porion requested this originally I seem to recall.

  multiple = 1 x = 0 - (RSI[30](close) - RSI[100](close)) z = medianprice * (1 + (x / 100) * multiple) if z crosses over medianprice then drawarrowdown(Barindex, medianprice + range) coloured(128,0,0) endif if z crosses under medianprice then drawarrowup(Barindex, medianprice - range) coloured(0,128,0) endif return z

  1 2 3 4 5 6 7 8 9 10 11 12 13

  multiple = 1 x = 0 - (RSI[30](close) - RSI[100](close)) z = medianprice * (1 + (x / 100) * multiple)   if z crosses over medianprice then drawarrowdown(Barindex, medianprice + range) coloured(128,0,0) endif   if z crosses under medianprice then drawarrowup(Barindex, medianprice - range) coloured(0,128,0) endif   return z

   

  

  Vonasi

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  01/23/2019 at 4:55 PM #89472
  Porion

  Thanks a lot guys for your help, it seems quite encouraging. i used Vonasi’s first code. i think it would be better if i was able to smooth it a little bit but it return me an error message.

  

  Porion

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  01/23/2019 at 5:09 PM #89475
  Porion

  Quite interessing as i’m now able to see where i should short (late december, as you can see on the chart) but with the old indicator i wasn’t.

  

  Porion

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  01/23/2019 at 6:38 PM #89478
  Vonasi

  Porion wrote:

  i think it would be better if i was able to smooth it a little bit

  Try changing line for something like this:

  x = exponentialaverage[3](0 - (RSI[30](close) - RSI[100](close)))

  1	x = exponentialaverage[3](0 - (RSI[30](close) - RSI[100](close)))

   

  

  Vonasi

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  01/23/2019 at 6:44 PM #89479
  Vonasi

  I think I messed up the code in post #89465 as I was playing with Roberto’s suggestion. I have edited the post and corrected it now so you might want to try it again.

  

  Vonasi

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  01/23/2019 at 7:47 PM #89489
  Porion

  Thanks Vonasi,

  i added

  a1 = ((highest[3](z))+(lowest[3](z)))/2

  1	a1 = ((highest[3](z))+(lowest[3](z)))/2

  to your code to smooth the indicator a little bit(in yellow)

  

  Porion

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  01/23/2019 at 8:05 PM #89494
  Vonasi

  Increasing the period from 3 upwards will smooth it more but at the expense of lag.

  

  Vonasi

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