For the love of gawd, stop being so damned pedantic.
Just trying to insure that the information given is true.
If you could uncork your illusion that you are perfect for one moment, you will see that I am just telling this guy that a "50 amp service" does not mean he can only have 50 amps of demand at 120 volts.
But my friend this is just what a 50 amp service is.
He has two sides of the phase and each can carry 50 amps.
This is an untrue statement. It is one 50 amp circuit not two sides that carry 50 amps each.
So if he has two loads that each demand 45 amps at 120v he does not need a 100 amp service to accommodate them.
this is true, he will only need a 90 amp service
Do you enjoy being tiresome?
I am here to insure that any and all information given is true and safe.
A 50 amp feeder is just that, 50 amps. It is not 50 amps at 120 volts plus another 50 amps at 120 volts for a total of 100 amps at 120 volts.
If there was 50 amps at 120 volts on one side and another 50 amps at 120 volts on the other side the neutral would have 100 amps on it.
If you are going to say this is untrue then pray tell me how these amps are returning to the transformer.
Edited to add;
For the sake of this discussion let’s take away the trip curve of breakers and assume that a 50 amp breaker will trip at 51 amps.
I have a panel that is supplied by 240 volts at 50 amps.
I load the panel to 50 amps of 120 volt loads and all this load is on one side of the 50 amp breaker. I now load the other side of the breaker with a wall clock and the breaker trips. The total load on that 240 volt two pole breaker will equal 50 amps plus one wall clock.
What can’t be done is to install a two pole 50 amp breaker and load one side to 50 amps at 120 volts and then add another 50 amps at 120 volts to the other side of the breaker at 120 volts to where there would be a total of 100 amps of 120 volt load. It just doesn’t work that way.
Take this same feeder at 50 amps and install two 100 watt lights one on each side of the breaker. The total load on that feeder will be 200 watts with or without the neutral. As long as each side of that breaker has the exact same amount of current the neutral is not needed. A simple experiment you can do is use two lights at 100 watts each and connect them to a 240 volt circuit in series without a neutral and check the voltages at each bulb. You will find that each bulb has 120 volts of drop which equals a total load on that 240 volt circuit of 1.66 amps although each bulb only pulls .83 amps each. Now install the neutral and the load on the feeders will still be 1.66 amps.
Using this same circuit that has a amperage draw of 50 amps on one leg and a 40 amps on the other leg and the neutral will have 10 amps of current flowing through it. You will have a total of 40 amps of 240 volt series circuit and 10 amps of 120 volt parallel circuit but the very second that either side reaches 51 amps the circuit will open (this is if we take away any trip curve of the breaker).
Taking your analogy of a 50 amp breaker allowing 50 amps of 120 volts on one side and then adding another 50 amps to the other side would be a total of 100 amps of 240 volt series circuit.
I agree with you that a 50 amp feeder will be more than he will ever use but I don’t see any reason not to install 250 kcmil copper conductors and a 250 amp disconnect out there if that is what he wants to do.