[Atomic-S (OP)]

What is the factor that most limits the distance over which transmission of electrical power is practical:

Maximum practical voltage?
Maximum practical conductor size?
Acceptable frequency?
Other?

[another_someone]

What is the factor that most limits the distance over which transmission of electrical power is practical:

Maximum practical voltage?
Maximum practical conductor size?
Acceptable frequency?
Other?

Cannot see that increasing the frequency would be of much benefit.  An increase in frequency may make the transformers more efficient, but that will be offset by increases in loses through inductance and other transmission losses.

Conductor diameter and voltage are interrelated, although the greater significance regarding voltage is the level of insulation (whether this be simply an air gap or actual insulation around the cable).

Thicker conductors would allow higher currents (but aside from cost, one also has to consider weight implications), while higher voltage allows lower current to carry the same power.

[lightarrow]

What is the factor that most limits the distance over which transmission of electrical power is practical:
Maximum practical voltage?
Maximum practical conductor size?
Acceptable frequency?
Other?

Tell me the truth, you want to send electric energy from a plant on the moon to earth with cables, isnt'it? []
However I think it's not possible to answer your question unless you specify the distance and the way of power transmission: you mean with conductors on cables? In the ground or suspended in air? Is it allowed to use superconductors?

[Soul Surfer]

It is probably the maximum practical voltage because for every watt of power you want to send the higher the voltage you use the lower the current and the less the effect of the cable resistance on the loss.  400Kv is the highest generally used in the UK but I think up to a million volts has been used elsewhere for extreme length runs

[lightarrow]

It is probably the maximum practical voltage because for every watt of power you want to send the higher the voltage you use the lower the current and the less the effect of the cable resistance on the loss.  400Kv is the highest generally used in the UK but I think up to a million volts has been used elsewhere for extreme length runs

Yes, but this is true for AC currents only, not for DC currents. However, if you have very little currents and high voltages, the power dissipated away as EM field is reduced, but the power dissipated as Joule effect is increased: W = V²/R.
W = power dissipated
V = voltage
R = resistance.

[Atomic-S (OP)]

However I think it's not possible to answer your question unless you specify the distance and the way of power transmission:

I was thinking chiefly of terrestrial use. For example, in the United States, where I am, the question is relevant to whether or not power from the east coast could be brought to the west coast, a distance of some 3000 miles. I don't think that is currently possible. Or if California, which has environmentally sensitive politics, would want to import power from someplace far away such as Brazil. (Importing it from Arizona has been discussed, but is controversial).

[BillJx]

It is probably the maximum practical voltage because for every watt of power you want to send the higher the voltage you use the lower the current and the less the effect of the cable resistance on the loss.  400Kv is the highest generally used in the UK but I think up to a million volts has been used elsewhere for extreme length runs

Yes, but this is true for AC currents only, not for DC currents. However, if you have very little currents and high voltages, the power dissipated away as EM field is reduced, but the power dissipated as Joule effect is increased: W = V²/R.
W = power dissipated
V = voltage
R = resistance.

I'm not sure what you mean. Resistance losses are equal for AC and DC.   AC is used so that transformers can convert between high transmission voltages and lower point of use voltages.  For a given resistance, resistance loss (heating of the wire) increases with the square of the current.

[lightarrow]

It is probably the maximum practical voltage because for every watt of power you want to send the higher the voltage you use the lower the current and the less the effect of the cable resistance on the loss.  400Kv is the highest generally used in the UK but I think up to a million volts has been used elsewhere for extreme length runs

Yes, but this is true for AC currents only, not for DC currents. However, if you have very little currents and high voltages, the power dissipated away as EM field is reduced, but the power dissipated as Joule effect is increased: W = V²/R.
W = power dissipated
V = voltage
R = resistance.

I'm not sure what you mean. Resistance losses are equal for AC and DC.   AC is used so that transformers can convert between high transmission voltages and lower point of use voltages.  For a given resistance, resistance loss (heating of the wire) increases with the square of the current.

I mean what I have coloured in red.

[BillJx]

I mean what I have coloured in red.

You seem to have gotten it backward somehow.  Resistive losses increase with the square of the current.  Electromagnetic field strength increases with voltage, that's why some people think that living near high voltage power lines might be unhealthy.