Calculating voltage, current, and impedance values on an old transformer

I'm working on an old valve amplifier, with the intention of reusing the transformers in a new project. I've used what I know of transformers to calculate the ratio of each winding the corresponding voltages however what I calculate does not agree with what is specified in the schematic so I must be making an error somewhere. The output transformer has one primary and three secondary windings; the measured resistances are as follows:

P = 7.9
S#1 = 28.4
S#2 = 3.60
S#3 = 0.1

Given that the ratio of a transformer can be calculated as:

Vp/Vs = Ip/Is = a

The impedance of either winding can be deducted as:

Zp = (a*Vs)/(Is/A) = (a^2)Zs
Zs = (a
Vp)/(Ip/A) = (a^2)*Zp

So far, I know the impedance of each winding, but not the turns ratio or voltages however by rearranging the above formulas I can use the impedances to deduct the ratio:

a = Square Root (Zp/Zs)

Therefore

a(S#1) = 0.53
a(S#2) = 1.48
a(S#3) = 3.36

Using these values of a, we can predict the voltages on the secondary:

V(S#1) = 250/0.53 = 474.01
V(S#2) = 250/1.48 = 168.76
V(S#3) = 250/3.36 = 74.42

However these values are not what is specified by the schematic, which is available here. The schematic specifies 470V, -37V, and 6.3V for the first, second, and third windings respectively and even allowing for a considerable voltage drop during rectification the calculated figures do not match the specified figures for the second and third windings.

Where am I going wrong in this process?

Thanks.

• @JoshTaylor : No. If you are going to reuse these transformers just take them out. Apply an ac voltage to the primary and measure the secondary voltage $\frac{V_1}{V_2} = \frac{N_1}{N_2}$ Apr 11, 2018 at 19:31