Single Ended Output Transformer

[mensur]

Hi Guys,
I want to build EL34/6L6 SE guitar power amp, but I also want to wound OT.
According to Valveheart, full calculation goes like this(for my example):
Power P= 10W(3.6K EL34-6L6)
Low freq response fd=82Hz
Primary Impedance Ra=3600 Ohm
Secondary Impedance Rs=4/8 Ohm
Flux density B - 5000Gauss

According to Valveheart core dimension is determinated by the eq:
Q=sqrt( (Pout*G*10^6) / (fmin*B*Is)) - where:
- Pout is the output power
- G is coefficient of weight between copper and iron /G=1.5 - 1.0/
- f min is the lowest desirable frequency
- Is is the density of current in the wire - Ampere/sq. millimeter (1.5 - 2.5 A/sq. mm).
- B is Magnetic Field Intensity in Gs
I don't get this equation cause every time I calculate it I think it's wrong.
Here's mine example:
Q=sqrt( (10*1.5*10^6) / (82*5000*1.5))=4.9cm^2
I'm confused because of the lower equation, cause for the same input I get:
Q=20*sqrt(Pout/fmin)
Q=20*sqrt(10/82)=6.98cm^2
I wonder how they got 20?Can someone explain this to me

I will take 6.25cm^2 core for the rest of an example:

Primary turns:
n1=10^8/(4.44*B*fmin*Q)=10^8/(4.44*5000Gauss*82Hz*6.25cm^2)=8.78
U1=sqrt(Ra*Pout)=sqrt(3600*10)=190VAC
Np= n1*U1=1667 turns - 1665 (5layers - 111turns, primary is divided in 3 parts, so (5*111)*3=1665

Secondary turns:
Ns4=1667/(sqrt(3600/4))=56 turns
Ns8=1667/(sqrt(3600/8))=79 turns

Primary current:Ip=Ia+Ia'
Ia=SQRT(Pout/Ra))*1000=SQRT(10/3600))*1000 = 52.70mA
Ia'=73mA(this is bias for tube)
Ip= Ia+Ia'=125.70mA

Secondary currents:
Is4=SQRT(Pout/4)=1.58A
Is8=SQRT(Pout/8)=1.12A

Wire diameter:
wp=SQRT(Ip*0,001/3)*1,13=SQRT(125.70*0,001/3)*1,13=0.23mm - 0.3m wire is chosen
ws=SQRT(Is4*0,001/3)*1,13=SQRT(1.58*0,001/3)*1,13 = 0.82mm - 0.7m wire is chosen

Can someone who is experienced in winding OT's can confirm this, before i start wounding the transformer.
Thanks