4 ms·

It'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 ca

by tlb 8y ago

It'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 ago
Thank you for this.
kragen 8y ago
Thank you! I wanted to calculate this, so I'm glad you did.
tedsanders 8y ago
FYI, 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 ago
  The 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 ago
I chuckled at "Assuming a spherical helicopter".
- deelowe 8y ago
  This is how you know the parent is an engineer.
  - function_seven 8y ago
    Nah, 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 ago
      Good point.
- SmellyGeekBoy 8y ago
  At least it's not flying in a vacuum.

by tlb 8y ago

scrumper 8y ago

Thank you for this.

kragen 8y ago

Thank you! I wanted to calculate this, so I'm glad you did.

tedsanders 8y ago

FYI, 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 ago
The 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 ago

I chuckled at "Assuming a spherical helicopter".

deelowe 8y ago
This is how you know the parent is an engineer.
- function_seven 8y ago
  Nah, 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 ago
    Good point.
SmellyGeekBoy 8y ago
At least it's not flying in a vacuum.