Converting Five Bar Trend indicator to Pro Order

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Viewing 15 posts - 1 through 15 (of 17 total)
  • #105482

    Good morning!

    I tried to rewrite the latest great indicator of Vonasi, Five Bar Trend Dashboard, to pro order, but failed. Any one care to elaborate what I am doing wrong, thanks!

     

    #105487

    What is yours not doing? Maybe you are blowing your Account with Lot = 100?? 🙂

    Attached with Lot = 1

    1 user thanked author for this post.
    #105491

    Just like GraHal it works for me. I have removed all references to customclose and referenced m5 instead so that it will work with the various custom prices by // out the prices you don’t need.

     

    2 users thanked author for this post.
    #105497

    Thank you Vonasi you solved it. I missed out to remove all references to customclose and to refere m5 instead, thanks!!

    #107886

    Dear Vonasi,

    Now I am trying to figure out how to calculate the upper line, which I thought was dcount (down, green) or ucount (up, red). I made a simple exit condition, BearCount that you will find in the end of the code. What am I doing wrong?

     

    #107902

    Sorry soulintact but I don’t fully understand what you are trying to do. The top line is just the count of bars in a row where all tests have been either all up or all tests have been down.

    The only thing I can say is that if a test results in not all tests being up or not all tests being down then the following code does not reset dcount or ucount to zero.

     

    1 user thanked author for this post.
    #107911

    Thank you Vonasi for your prompt reply. Sorry for not being clear enough.

    As of the included file, I want to be able to calculate the numbers 3(green) 1 (green) 1 (red) 8 (red) 1 (red) 1 (green) 4 (green) 1 (green).

    By the code I wanted to exit the position when the number of red/down was greater than 7. I enclose the code again with your suggested modification.

     

     

    #107922

    That wasn’t a suggested modification – that was the bit of your original code that I cut and pasted in where values are not set to zero at any time!

    In the original code it was like this:

    If we remove the graphics and replace them with variables then we can get something like:

    So now we have a variable ‘direction’ that can be either -1, zero or 1 and the variable ‘a’ which is the quantity of the same direction in a row or zero if all directions are not the same.

    1 user thanked author for this post.
    #107952

    Thank you Vonasi for your patience!

    What expression would I use to be able to “catch” the red 8, or rather 8 lines of green dots and one red dot of the following line?

    I tried a few, and the following is not correct, but with the best performance: BearCount=(direction=1) and (a < 1)

    What am I missing? Thanks!

     

     

    #107957

    There is a critical bit of my original indicator code that does not seem to have found its way into your code. The calculation of dcount and ucount!

     

    1 user thanked author for this post.
    #107978

    Sorry about that! Although it does not work with the edited code as enclosed. Strange! Any more ideas? Thanks!

     

     

    #107996

    It works fine for me.

    I took your code and just added ‘buy at -close limit’ to turn it into a dummy strategy. I rem’d out all references to L1, L2 etc to make it work and then graphed BEARCOUNT  – having changed it to ‘BearCount=(direction=1) and (a > 3)’ and every time there is a run of more than 3 red candles it returns a positive value for BEARCOUNT.

    1 user thanked author for this post.
    #107998

    Dear Vonasi,

    Sorry for being so slow in mind and persistent, but I still do not follow fully. Do the statement below really correspond to the definition in words?

    BearCount=(direction=1) and (a > 7) := (direction is now negative with a least one red dot) and (previously more than 7 of the same direction in a row of green dots)

     

    Cheers!

    #108000

    No

    BearCount=(direction=1) and (a > 7) would be that we have just had a run of 8 (sets of 5) red dots and this set is not a set of 5 red dots nor a set of five green dots.

    1 user thanked author for this post.
    #108001

    Aha, thanks!

    And how would we formulate the expression to “catch” the red 8 as enclosed.

Viewing 15 posts - 1 through 15 (of 17 total)

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