Let's start with the power supply:

The resistors in the mainline are for reducing the voltage in the moment of heating up the tubes. It's better to do that slowly, especially when they draw much current. It's like a shock for them
and the filament could break. Just like you go out of a warm water swimming pool and you jump into a cold one. You could drown because of the cold shock.

The resistor is then gets bridged by a relays timed at 4-5 seconds. The circuit for that consists of a RC element, a zener diode and a mosfet. For calculating the resistor you need to know how
much voltage one the primary(prim) needs to be cut off, to have the wanted voltage at the heater elements.

When you switch on, only the filaments draw current. Therefor you need to know the total ampere they draw. Calculate the power needed for all the filaments using P=UI (heater current*heater voltage) for each filament and than add the values.You may add 20% to your result for the prim power because of the power loss in the transformer.

Done that, calculate the prim current with I=P/U (prim power/main voltage). You will need it later. After that you need to know the transformation ratio from the prim to the sec of the heaters. For that dived the prim voltage through the sec voltage. Choose your desired voltage with witch you want to heat up the tubes at the start, multiply it by the transformation ratio and you have the required prim voltage for that sec voltage. The voltage to be burned is your main voltage divided through the above calculated voltage.

Then it's child's play. Use Ohm's law to find the value of the resistor. You already have the prim current in the heating up moment. The power loss(P=UI, voltage to burn*prim current) will be
high for the moment, but don't worry. I'm using only one 17W resistor and it's enough. Three of them took to much space. It can stand that power for a short time.

The power supply is a very important part of an amplifier. If not build properly, it will affect the sound of the amp. When you have hum problems look at the power supply immediately. It might be not filtered enough.

The first filter capacitor after the rectifier tube(s) is calculated this way:

I found those formulas in an old book I bought on eBay.

The maximum value of the capacitor right after the rectifier tube is written in the datasheet. If you take more capacity, the tube get's damaged. When you use tubes like EY88 and EY500A, look on
the internet witch capacitance is used. These tubes were used for rectifying the fly-back impulse in old TV's. You won't find anything about a maximum value of a filter capacitor in the
datasheets. For a Greatz-bridge of EY500A 100µ are OK. The formulas of a dual wave rectifier apply here too.

The filter itself is calculated with this formulas:

Bleeder resistor parallel to the filter capacitor should be added. After the amp is switched off, they discharge the capacitors. It's a safety measure. For you experimenting and working with the
amp and for the components inside.

For the amplifier itself, only Ohm's law and the equation f=1/2piRC is needed. The voltage at the control grid(G1) needs to be negative respect to the cathode, else it could damage the tube. Only certain tubes and conditions allow positive voltages at G1. This voltage can be gained with the voltage drop a the cathode resistor.

The cathode resistor Rk for the preamp is calculated dividing the desired cathode and therefor G1 voltage through the amount of current that flows through the tube, when this voltage is applied
to G1. There are diagrams in the tube datasheets were you can pick out these values. At first you need to know in witch class your amp should work later. For my EL34 amp I've chosen Class A for
the preamp's and Class AB1 for the power stage.

The anode resistor gets determined by dividing the chosen current through the difference of the supply voltage and before determined anode voltage. Only U = RI ;).

If you have a penthode or a tetrode or a power beam tube, you also have a screen grid(G2). That needs a resistor, otherwise it could melt. Using Ohm's law here too, you need to calculate the value of the resistor at in the datasheets given values of voltage and current of G2.

The above calculations are working for every amplifier stage with a variable bias or when the bias at G1 is gained through the voltage drop at Rk.

To calculate the coupling capacitor or resistor you need the equation f=1/2piRC. You pick either the cap or the resisotr value. I reccomend the capacitor value, because its easier to find a
matching resistor with the calculated value or one near it. Then you rewrite the equation the have as a result R or C. For f you select the frequency you want to be the cut-off frequency of your
amp to be. The frequency needs to be the same across all stages in your amp. Capacitor values tend to increase and resistor values to decrese when you go towords the output stage of your amp.
Unfortunately I never found something where it explains with wich values you start. From experimenting and looking at a lot of schematics, I found the capacitance for the first cap to be at
around 22-47n in audio amps for speakers. The resistor values is 1M. I saw those values the most and they worked fine in my amps. I

When you have a amplifier stage with a fixed bias, like my power stage of the EL34 amp, you have to use two very easy formulas. They'll give you a range between you should operate. Look in the
datasheets for the maximum plate dissipation.

Min(mA) = (Pa*500)/Ua

Max(mA) = (Pa*700)/Ua

Pa = max anode dissipation in W.

Ua = anode voltage supplied by the power supply.

The 500 and 700 are the percentage of the anode dissipation, at witch the tube will be operating. Of course you can choose other values. But be aware, if you choose higher ones the tube life will
be shorter, although I've never seen tubes die because they were driven at an excessive current. I think it's unlikely that I experience that.