Hi Jacob,
in your other posting, you gave the example (using the notation of the
detwin documentation):
DANOTw(h1) = 52
DANOTw(h2) = 28
and if you use the formulas that you cite below, you get the true values
back:
DANOtrue(h1) = (0.6*52 - 0.4*28)/(1-0.8) = 100
DANOtrue(h2) = (0.6*28 - 0.4*52)/(1-0.8) = -20
So then, what's problematic?
Kay
Am 11.05.16 um 20:33 schrieb Keller, Jacob:
>> and that could be used to detwin the DANOs directly:
> DANOtrue(h1) = ((1-tf)*DANOTw(h1) -tf*DANOTw(h2)) / (1-2tf)
> DANOtrue(h2) = ((1-tf)*DANOTw(h2) -tf*DANOTw(h1)) / (1-2tf)
>
> This would be problematic, since many of the signs of DanoTrue would not be known per the Keller Theorem just posted. This would not be a problem, however, when detwinning I+/I-.
>
> A corollary: the twin fraction at which one reflection from a twin-related pair will flip sign is a function of the ratio (Dano/Danotwinrelated); higher ratios flip at lower twin fractions. All Dano twin-pairs with ratios < -2, for example, will have flipped sign with twin fractions > ~0.33. Working right now on pinning down the right function.
>
> JPK
>
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