Keltner Channels on Force Index

[dragon]

Hello All,

When I drag the Bollinger bands on Force Index it is working nice without any problem. But, on the same way when I dropped the Keltner Channels on Force Index it is not displaying properly, rather than it is looks like a Moving average on the force Index. May be I think it is different calculation methods in Bollinger Bands and Keltner Channel. After that I have tried to replicate the code after researching the codes from forum. But not succeed. I have tried the following, I assume the code can be calculated on Force Index as centerline in the Keltner channel. Later on I will change the force index parameters to 13 from parameters menu.

_SECTION_BEGIN("Force Index");
periods = Param("Periods", 2, 1, 100, 1);
FI = EMA(((Close - Ref(Close, -1)) * V), periods);
FI_kol = IIf(fi < 0, colorRed, colorBrightGreen);
Plot(Close, "Close", colorRed, 2);
Plot(FI, "Force Index", FI_kol, styleLine|styleThick);
Plot(0,"", colorViolet, styleLine | styleThick | styleNoLabel);


KTop2  = FI + 2*ATR(21);
KBot2 = FI - 2*ATR(21);

Plot( KTop2, "KBTop2", colorRed, styleLine ); 
Plot( KBot2, "KBBot2", colorRed, StyleLine );

I request you to please help me in this regard.

Thank you very much.

[fxshrat]

Please use forum search which would lead to here.
Is it so difficult to type Keltner Channel into search field?

ATR uses H,L,C!
Your FI multiplies volume with price.
So you are adding apples to oranges.

If it makes you feel better:

periods = Param("Periods", 2, 1, 100, 1);
FI = EMA(((Close - Ref(Close, -1)) * V), periods);
FI_kol = IIf(fi < 0, colorRed, colorBrightGreen);
Plot(Close, "Close", colorRed, 2);
Plot(FI, "Force Index", FI_kol, styleLine|styleThick);
Plot(0,"", colorViolet, styleLine | styleThick | styleNoLabel);

CenterLine = MA( FI, Periods );
H = L = C = FI;
KTop2  = CenterLine + 2*ATR(21);
KBot2 = CenterLine - 2*ATR(21);
RestorePriceArrays();

Plot( KTop2, "KBTop2", colorRed, styleLine ); 
Plot( KBot2, "KBBot2", colorRed, StyleLine );

[dragon]

Thank you very much.