6 ms·
This is excellent. I don't understand why there's an arc at the end when the lineman disconnects and pulls away the safety rod. That's there to keep the helico
by scrumper 8y ago
This is excellent.
I don't understand why there's an arc at the end when the lineman disconnects and pulls away the safety rod. That's there to keep the helicopter at the same potential as the transmission line, right? So why's there a potential difference when the safety rod is withdrawn?
- tlb 8y agoIt's a low-current arc. The circuit is completed by the capacitance between the helicopter and the ground. Assuming a spherical helicopter 1M in radius, the capacitance to ground is >>> 4 * pi * 8.8e-12 1.1058406140636071e-10 The current, at 220 kV and 60 Hz can be calculated >>> 60 * 2 * pi * 220e3 * 1.1058406140636071e-10 0.009171625977844317 or 9 milliamps. Enough to feel a good tingle. The power in the arc is >>> 220e3 * 0.009171625977844317 2017.7577151257497 or 2 kilowatts, so you can see the arc in daylight.
- scrumper 8y agoThank you for this.
- kragen 8y agoThank you! I wanted to calculate this, so I'm glad you did.
- tedsanders 8y agoFYI, and not that it matters, I think your current is off by a factor of 2/pi. There's no 2*pi in the current formula because the 60 Hz is the full frequency not the angular frequency. And there's an extra factor of 4 because in each full cycle sine wave there are four 220 kV steps: 0-220, 220-0, 0-(-220), (-200)-0. (and all of this of course assumes that the metal helicopter has low enough resistance to charge up to the full voltage)
- tlb 8y agoThe current through a capacitor is I = ωC, where ω is in radians per second (ie, 2𝜋 60 Hz). I think 220 kV is the RMS voltage, so you get RMS current from the formula above. Peak voltage would be √2 * 220 kV. Indeed, both the capacitor and the arc are in series, so actual current and power is somewhat less. As the arc grows longer, there should be an optimum length where you get 1/√2 of the maximum power.
- p1necone 8y agoI chuckled at "Assuming a spherical helicopter".
- deelowe 8y agoThis is how you know the parent is an engineer.
- function_seven 8y agoNah, physicist. If parent was an engineer, he's just reference the row and column from some table of helicopter capacitances. ("See p.34, table 7 for the MD-500. The one in the video is a 530, but close enough")
- deelowe 8y agoGood point.
- SmellyGeekBoy 8y agoAt least it's not flying in a vacuum.
- baybal2 8y agoAssuming it is an AC line, it is inductance
- Casseres 8y agoMy understanding is that it's caused from the rotorblades moving through the air. Apparently there are other factors as well. This StackExchange answer can address it better than I can: https://aviation.stackexchange.com/questions/9357/can-rotorcraft-pick-up-a-dangerous-electric-charge https://aviation.stackexchange.com/questions/9357/can-rotorc...
- deleted 8y ago[deleted]
- widforss 8y agoIt's an AC line, so when the helicopter retracts it will get a static charge instead of fluctuating in sync with the line, giving an immediate potential difference. I have no source for this, just amateur reasoning.
- zaarn 8y agoIt's very likely capacitance. At 200kV even a couple dozen meters above ground the capacitance would be enough to get a couple microamps over an arch (it would likely tickle a bit). AC makes isolating rather complicated since increasing your resistance to ground naturally turns you into a capacitor which acts like a resistor in AC (which is why thin rubber soles on your shoes don't protect you, you want think ones to increase resistance and decrease capacitance) so a small current will flow. At high voltage this small current can still be transmitted over an plasmaarc.