[ARCHIVE]Any rookie question, so as not to clutter up the forum. Professionals, don't pass it by. Can't go anywhere without you - 5. - page 200

 
Inimitable charm, that's very much our way. She stayed.
 
Nature is still patient! How long will her patience last?
 
I'm going to bed and you don't need to come here yet.
 

Yeah, I'm going to bed now too, I'm not turning off the computer, positions are open waiting for movement. Good night!

 

Good afternoon.

Can I find a manually opened order out of all the open orders?

Does it have a zero magic number (or is there just no magic number)?

 
Stells:

Good afternoon.

Can I find a manually opened order out of all the open orders?

Maybe its magician is zero?


Yes, he will have zero magic
 
Vinin:

Yes, he will have zero magik.
Thank you, Victor.
 
r772ra, thank you.
 
Thank you all... :))))
 
borilunad:

You will notice that when you switch on MT, all the charts have a vertical spread, depending on the extremes of the bars that fit in the chart. You adjust this swing with the mouse, only squeezing vertically. From here you jump in your calculations. The number of bars is constant if you do not change the chart width. And the number of pips, count from the upper extremum subtracting the lower one but this difference will change, when one of the extrema is hidden behind the left border of the chart.

I can't go into details, but why should you waste your precious time on it. At least I am practicing logic. Why would you make such a mess?


Here it's simple - I want to bind the stop size (and lot) to a square of the grid - say 1-2 squares (I like the algorithm of its construction as a volatility indicator). But MT has some very tricky algorithm that probably cannot be used in the indicator, because the number of squares of these changes from the size of the chart, not from bars in it, but from the size of the chart.

PS. I think I know what you mean. I need to calculate the number of bars in a chart, the size I need, the amplitude to calculate for that number of bars and divide it by the number of autocubes in that chart. How simple it turned out to be. Thanks for the tip.

Reason: