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Author Topic: IAT (tans) effect on rlsol -> plsol  (Read 4179 times)
nyet
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« on: October 29, 2014, 04:26:32 PM »

roughly:

boost = load/fupsrl + pirg

fuprsl = KFURL * ftbr

ftbr = 273/(tans+273) * fwft

fwft = (tans+673)/731  ----- linear fit to ME7.1 FWFTBRTA

Roughly, this means that as tans goes up, fupsrl goes down.

which means as tans goes up, for a given requested load (post KFTARX correction), req boost goes up.
« Last Edit: October 29, 2014, 04:30:37 PM by nyet » Logged

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nyet
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« Reply #1 on: October 29, 2014, 04:32:05 PM »

So possibly one way to keep req boost constant with tans (post KFTARX) is to tune FWFTBRTA to follow (tans+273)/273
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phila_dot
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« Reply #2 on: October 29, 2014, 05:40:36 PM »

I see :

ftbr = FWFTBRTA * (B6h / (evtmod - 48 + 273))


actually B6h / (evtmod - 40h + 16Ch)
« Last Edit: October 29, 2014, 05:45:34 PM by phila_dot » Logged
nyet
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« Reply #3 on: October 29, 2014, 09:27:11 PM »

CWTEMPK bit 1 = false?

What about bit 0... is it true or false (input to FWFTBTRA is evtmod or tans?)
Bit 2?

in gbox, i see CWTEMPK=0.. have not looked for SWTEMPK in mbox yet.

that means that FWFTBRTA is addressed with tans, and dtbrt is always 0... and the FR is conflicting on the topic of tbrraum: The true/false box says bit1=false=>tans, bu the text in the FR says bit1=false means tbrraum=evtmod

I also can't find FWLKFTBR....


evtmod is a bit trickier than tans, it attempts to correct tans according to tmot

« Last Edit: October 29, 2014, 09:34:05 PM by nyet » Logged

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nyet
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« Reply #4 on: October 29, 2014, 09:59:11 PM »

I see :

ftbr = FWFTBRTA * (B6h / (evtmod - 48 + 273))


actually B6h / (evtmod - 40h + 16Ch)

the conversion to Degrees C is .75, correct?

00B6h = 136C
0040h = 48C
016Ch = 273C

so 136/(evtmod - 48 + 273)

Odd since the FR shows 273/(evtmod + 273), and that can't be right.... because the resulting ftbr would be ~0.5!

Also note that B6h<<1 = 16Ch... maybe you are missing a shift somewhere?
 
« Last Edit: October 29, 2014, 10:10:57 PM by nyet » Logged

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nyet
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« Reply #5 on: October 29, 2014, 10:04:45 PM »

also, just to make sure, is this right?

evtmod =  tans + (tmot-tans)*KFFWTBR
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nyet
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« Reply #6 on: October 29, 2014, 10:26:47 PM »

And.. what is SY_BDE, etc.... ugh.
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phila_dot
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« Reply #7 on: October 30, 2014, 12:02:36 AM »

CWTEMPK is 0.

Tans is used for FWFTBRTA.

The conversion to degrees celcius is 0.75 -48.

It's weird because 40h is subtracted from tans and tmot at the beginning of the function. Then added and subtracted throughout the function and finally added before being written to RAM in evtmod.

The B6h threw me off as well. The only shift operation is after the output is multiplied before writing to RAM in ftbr_w and only if the output is unsigned <= 7FFFh. Otherwise ftbr_w = FFFFh. This just rescales it to a 16 bit factor.

shl (FWFTBRTA * (136/(evtmod - 48 + 273)), 1)
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nyet
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« Reply #8 on: October 30, 2014, 02:31:52 PM »

I can't explain what you are seeing, but in my simulations, the FR is correct: 273/(evtmod+273) IN C (not native cpu var scaling)

The  conversions are weird because of the +48... it means that 48 has to be added/subtracted all over the place, depending on the operation. I haven't got my head around it yet...

BUT

What would help would be M box ram locations for all those variables...

evtmod
fupsrl
ftbr
fwft

the following would help me decypher the rest of FUEDK:
pirg
rlfgs
rfges
rfagr
psagr (this one is a real mystery, even the FR doesn't say where it comes from)

and finally
fpbrkds
vplsspls

I know this is a tall order... I also am starting to question how important it is to me to figure this all out ... for marginal gain lol.

Oh and my assumption is that SY_BDE=false and SY_AGR=true... is that right?
« Last Edit: October 30, 2014, 02:34:25 PM by nyet » Logged

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phila_dot
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« Reply #9 on: October 30, 2014, 03:02:11 PM »

273/(evtmod+273) gives you a percentage factor from 0* celcius. This makes sense.

The code is what it is though, even when it doesn't make sense right away.

The offset doesn't need to added or subtracted when using like variables. It is -48, unsigned so 0h is -48 c.

I take a look at the rest when I get a chance.
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nyet
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« Reply #10 on: October 30, 2014, 03:52:48 PM »

273/(evtmod+273) gives you a percentage factor from 0* celcius. This makes sense.

AGREED 100%!!

Quote
I take a look at the rest when I get a chance.

Thanks in advance, as usual, you rock.
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biela
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« Reply #11 on: April 03, 2015, 05:20:02 AM »

Hello
Sorry for the old topic post, but the S4wiki states that:
ftbr = 273/(evtmod-273) * fwft

May be the minus signus is a plus signus?

Thanks
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nyet
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« Reply #12 on: April 03, 2015, 09:56:05 AM »

Hello
Sorry for the old topic post, but the S4wiki states that:
ftbr = 273/(evtmod-273) * fwft

May be the minus signus is a plus signus?

Thanks


Yea, looks like a typo.
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