Now I know why I am confused...
Sorry about that. Pictures might help but I'm not at a location where I can do that right now. I'll try with text.
So what about a single phase 240V system delivered to a residence. It is single phase but we get 240V between the legs. Is that the same situation as you stated or do the phase angle change.
It works the same way for any set of signals.
It works the same way for 3-wire 120/170 quadrature (the traditional "two-phase") as well except the phase angles change to 90 degrees insead of the 120 degrees we have with the 3-wire 120/208 network.
It works the same way with 3-wire 120/240 single-phase except the phase angles change to 180 degrees instead of the 120 degrees we get with the 3-wire 120/208 network.
For all three systems, I have used leg voltages with a common reference (the neutral) to define the signals, which is a common definition for measuring a common rise/fall between signal points (I'll put in a numerical example below if you want to use a different signal definition reference).
These would have the voltages L1 & L2 with a N common point. The voltage signals are L1 minus N for the first and L2 minus N for the second. Keep in mind that you can define the signals as L1 minus N for the first and N minus L2 for the second (you will change the phase differences between the defined signals).
120/208 volt numerical examples:
For the leg voltages:
L1-N = 120@0d = L1 (N is signal reference)
L3-N = 120@120d = L3 (N is signal reference)
At any point in time, L3 leads L1 by 120 degrees (or L1 leads L3 by 240 degrees). At any instant, a reference location (like the maximum or minimum or rising zero crossing, falling zero crossing, etc.) on signal L3-N will reach a value that the L1-N signal location will not reach until some time later (5.56 ms later for a 60 Hz signal). This separation at any instant can be measured as a phase difference of 120 degrees (it takes 360 degrees of phase change for the signal to complete a cycle).
However, since we normally only use either L1 or L3 separately for a device, we normally think of this as single-phase as we only serve single phase loads with the 120 volt signals. But we could serve a load that uses the 120 degree phase difference the same as we can serve a two-phase load that uses the 90 degree difference from a two-phase source. We would not call the 120 degree case "two-phase" as that name is reserved for the 90 degree case.
For the leg-leg 208 voltage we take the difference across the two windings (note that the sum would give us 120 volts at 60 degrees)
L1-L3 = (L1-N)-(L3-N) = 120@0d - 120@120d = 208@-30d
Note that the L1 signal leads this signal by 30 degrees, not that we normally compare them.
Only one signal of this 208 volt magnitude means only one phase position is present at any instant.
There is another thing to keep in mind and that is that we can use something different as the reference other than "N" for both.
120/208 volt numerical examples for the different signal definition:
For the leg voltages:
L1-N = 120@0d = L1N (N is signal reference)
N-L3 = 120@300d = NL3 (L3 is signal reference)
At any point in time, L1N leads NL3 by 60 degrees (or NL3 leads L1N by 300 degrees). At any instant, a reference location (like the maximum or minimum or rising zero crossing, falling zero crossing, etc.) on signal L1N will reach a value that the NL3 signal location will not reach until some time later (2.78 ms later for a 60 Hz signal). This separation at any instant can be measured as a phase difference of 60 degrees (it takes 360 degrees of phase change for the signal to complete a cycle).
However, since we normally only use either L1N or NL3 separately for a device, we normally think of this as single-phase as we only serve single phase loads with the 120 volt signals. But we could serve a load that uses the 60 degree phase difference the same as we can serve a two-phase load that uses the 90 degree difference from a two-phase source. We would not call the 60 degree case "two-phase" as that name is reserved for the 90 degree case.
For the leg-leg 208 voltage, we can now take the sum across the two windings instead of the difference (note that the difference would give us 120 volts at 60 degrees):
L1N+NL3 = (L1-N)+(N-L3) = 120@0d + 120@300d = 208@-30d
Note that the L1 signal leads this signal by 30 degrees, not that we normally compare them.
Again, only one signal of this 208 volt magnitude means only one phase position is present at any instant.
Finally:
You might wonder if we can physically add and subtract these winding voltages the same way we do with math and the answer is yes, and we do exactly that in real world applications.
I can go through the same numerical examples for 120/240 or 120/170 but it works the same way for any set of signals.