Excellent answers by Elizabeth!
Electrical signals (as opposed to individual electrons, whose movement is irrelevant to the question) propagate through wires at roughly 60% to 90% of the speed of light in a vacuum, which in turn is about 186,000 miles per second. If we assume 75% as being typical (the actual number depends mainly on the dielectric constant of the cable's insulation), the corresponding propagation time is about 1.36 nanoseconds per foot of cable. 1.36 nanoseconds is 0.00000000136 seconds.
The speed of sound in room temperature dry air is approximately 1126 feet per second.
Based on the 1.36 ns/foot figure, a 20 foot difference in cable length between the two channels, for example, would result in an arrival time difference at the speakers of 27.2 ns.
Based on the 1126 ft/sec figure for the speed of sound in air, 27.2 ns corresponds to the arrival time difference at the listener's ears that would result from his or her head being 0.00037 inches closer to one speaker than to the other. In other words, not likely to be audible!
Regards,
-- Al
How different would wire lengths have to be for the acute listener to notice a difference in the arrival time in a pair of a stereo signals at the speakers?Several miles or more, by which point timing differences would of course have long since been swamped by other effects (frequency-dependent signal losses caused by resistance, inductance, capacitance, among many other effects).
Electrical signals (as opposed to individual electrons, whose movement is irrelevant to the question) propagate through wires at roughly 60% to 90% of the speed of light in a vacuum, which in turn is about 186,000 miles per second. If we assume 75% as being typical (the actual number depends mainly on the dielectric constant of the cable's insulation), the corresponding propagation time is about 1.36 nanoseconds per foot of cable. 1.36 nanoseconds is 0.00000000136 seconds.
The speed of sound in room temperature dry air is approximately 1126 feet per second.
Based on the 1.36 ns/foot figure, a 20 foot difference in cable length between the two channels, for example, would result in an arrival time difference at the speakers of 27.2 ns.
Based on the 1126 ft/sec figure for the speed of sound in air, 27.2 ns corresponds to the arrival time difference at the listener's ears that would result from his or her head being 0.00037 inches closer to one speaker than to the other. In other words, not likely to be audible!
Regards,
-- Al