FAQ
TL;DR: A 3×10^10 m wire shows a voltage step after about 200 s at 0.5c; "Wires have an approximate propagation delay of 1 ns for every 6 inches." [Elektroda, Frank Bushnell, post #21667151]
Why it matters: This FAQ helps engineers, students, and hobbyists estimate real signal delays and edge‑case behaviors in ultra‑long conductors.
Quick Facts
- Typical copper propagation speed: 0.59c–0.77c (velocity factor range). [Elektroda, Frank Bushnell, post #21667151]
- Rule of thumb: ~1 ns delay per 6 in (15 cm) of wire. [Elektroda, Frank Bushnell, post #21667151]
- Time across 3×10^10 m at 0.5c ≈ 200 seconds. [Elektroda, Frank Bushnell, post #21667151]
- If velocity factor ≈0.9, time ≈111 seconds for 3×10^10 m. [Elektroda, Rodney Green, post #21667147]
- Assuming 0.1c gives about 1000 seconds across that length. [Elektroda, stephen Van Buskirk, post #21667145]
How long until DC appears at the far end of a 3×10^10 m wire?
Treat the DC step like a fast edge. Delay ≈ distance/propagation speed. Using 0.5c gives about 200 seconds for 3×10^10 m. Faster dielectrics (e.g., VF ≈0.9) yield ≈111 seconds; very slow assumptions (0.1c) give ≈1000 seconds. [Elektroda, Frank Bushnell, post #21667151]
What is propagation delay in a wire?
Propagation delay is the time a voltage or field disturbance takes to travel along the conductor. It depends on the medium’s velocity factor, typically expressed as a fraction of light speed, not on electron drift speed. [Elektroda, Frank Bushnell, post #21667151]
What is velocity factor (VF)?
Velocity factor is the signal speed divided by light speed. Many copper structures show VF around 0.59–0.77; some environments approach 0.9, changing your delay estimate dramatically. [Elektroda, Frank Bushnell, post #21667151]
Is 0.1c a realistic speed in wire?
It’s an extreme assumption. One poster used 0.1c to illustrate a slow case, yielding ≈1000 s across 3×10^10 m. Practical cables often propagate much faster. [Elektroda, stephen Van Buskirk, post #21667145]
Could it be as high as 0.9c?
Yes. With favorable geometry and dielectric, a velocity factor near 0.9 is plausible, giving roughly 111 s over 3×10^10 m. [Elektroda, Rodney Green, post #21667147]
How do I quickly estimate wire delay?
- Pick an approximate velocity factor (e.g., 0.6–0.77).
- Compute speed: v = VF × 3×10^8 m/s.
- Time = distance ÷ v.
A handy rule: ~1 ns per 15 cm. [Elektroda, Frank Bushnell, post #21667151]
Do electrons themselves travel that fast?
No. The energy and electromagnetic disturbance propagate near VF×c, while individual electrons drift slowly. Think fields, not particles, for timing. [Elektroda, Earl Albin, post #21667148]
What happens if the far end is open (no return path)?
An open end behaves like a capacitor. The launched step charges it, producing a voltage gradient and potential overshoot, then reflections that can ring and decay. [Elektroda, Steve Lawson, post #21667156]
Will I get the same result with a proper return path?
With a matched return path and source impedance, reflections reduce and the step settles faster. Mismatch or open ends create oscillations and slower settling. [Elektroda, Steve Lawson, post #21667156]
Can the wire radiate most of the energy?
Yes for extremely long, space‑exposed wires. Radiation resistance can dominate, so much of the launched energy radiates instead of reaching the end. [Elektroda, Rodney Green, post #21667146]
Is the delay noticeable in everyday electronics?
Usually no. For typical lengths, delay feels instantaneous, though engineers do create nanosecond delays with wire loops for timing tasks. [Elektroda, stephen Van Buskirk, post #21667145]
What practical rule should I remember?
“Wires have an approximate propagation delay of 1 ns for every 6 inches (15 cm) of length.” Use it for quick back‑of‑the‑envelope timing. [Elektroda, Frank Bushnell, post #21667151]
What is a step response in this context?
A step response is how the line reacts when you apply an abrupt DC voltage change. That edge travels at the line’s propagation speed. [Elektroda, Abdul Rawoof Shaik, post #21667150]
What is radiation resistance?
Radiation resistance models power lost as electromagnetic radiation from a conductor. On astronomical lengths, it can make the line look resistive. [Elektroda, Rodney Green, post #21667146]
Can the open end exceed the source voltage?
Yes. Inductive energy near the source can charge the line’s capacitance, briefly creating a higher potential at the open end before it rings down. [Elektroda, Steve Lawson, post #21667156]
What concrete example ties this together?
At 0.5c, 3×10^10 m takes ≈200 s; at 0.9c, ≈111 s; at 0.1c, ≈1000 s. One poster summarized: compute d ÷ (VF×c). [Elektroda, Frank Bushnell, post #21667151]
