5 ms·

I call bullshit on the claim that you can transfer as much energy through optical fiber as through a copper wire of the same width. I'm pretty sure they calcula

I call bullshit on the claim that you can transfer as much energy through optical fiber as through a copper wire of the same width. I'm pretty sure they calculate the power transmitted taking into account only the 'active' part of the fiber, its core which diameter is in micrometers. So it's akin to pretending that to transfer electricity you just need copper core without the insulators

beambot 3y ago
I suspect their statement is true if you consider the fully-loaded diameter too -- i.e. fiber + cladding can carry more power than copper + insulation, especially for small gauges. Edit: Quick google... This paper [1] shows a 180um fiber + cladding carrying 150W over 1km. According to [2], that diameter equates to a 33awg wire, which has 678 ohm/km resistance and a maximum current of 0.072A. At 1km, the wire would be 678ohms. For maximum power transfer, you want the load to equal the resistance of the wire. Thus you could deliver 0.072A into a 678ohm load - i.e. your maximum power would be 3.5 Watts. The fiber is better by 40x. [1] https://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=8908271&casa_token=W_8YBHm8hssAAAAA:86Dy1EppIMO4Inbg7u6ZAEciApnsS5-DbJ0wAWPSiOQdHDhbksO7BE-8j8BWQHCr5CXjW1CyODh5yJQ&tag=1 https://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=8908271... [2] https://www.badgerwire.com/tech-info/wire-gauge-table/ https://www.badgerwire.com/tech-info/wire-gauge-table/
- pi-e-sigma 3y ago
  Nah, just google 'high power fiber cable', these are used to transfer light in laser steel cutters. As these are already in practical, industrial use we can assume that their spec is in line with what is currently realistically achievable for an optical fiber power transmission over a distance of a few meters max. For instance a SMA905 Fiber Cable can deliver meager 50 Watts and its external diameter is 5 millimeters which is laughable low compared to how much power you could push through a copper cable with the same external diameter, that is including electrical insulator.
  - beambot 3y ago
    I don't think it's a reasonable assumption to look at industrial use cases over small distances compared to long-distance operations -- the design criteria are different.
    - pi-e-sigma 3y ago
      The design criteria is exactly the same. Have an optical medium with as little attenuation as you can. If you can't have it in a medium that is only a few meters long how could you possibly achieve it in a medium of a few kilometers? Apply your logic to copper cables. Can you have a copper cable able to deliver power over a long distance but unable to deliver the same power over a shorter distance?
      - quailfarmer 3y ago
        The analogy doesn’t hold because one of the hardest parts of this technology is building a device that can extract the optical energy as electrical energy, and dissipate whatever heat it generates. The power density of the fiber itself is easy by comparison.
        pi-e-sigma 3y ago
        The analogy holds perfectly. Remember, we are discussing a claim that optical fiber with all required cladding/protection can carry as much power as a copper wire with all insulation having the same external dimension as the fiber cable. It is simply not possible, fiber will always have less power carrying capacity and your claim that fiber power density is 'easy' has no basis in the reality.
      - mlyle 3y ago
        There's absolutely no reason for the application you cited to optimize diameter and isn't approaching physical limits. There's no reason you can't shove many optical fibers into that 5mm diameter. People don't, in short distance applications, to have simpler systems and connectors. There's big reasons why we don't use power over fiber (the endpoints are very expensive, and the overall system efficiencies are low). None of them have to do with the highest optical power density you could hit.
- heisenzombie 3y ago
  You’re right, though I suspect you could increase that “max” current figure quite substantially if we’re assuming this is an underwater ROV and the wire has water cooling for free. You’d need to be pumping a whopping 300A/mm^2 through that wire to match the 150W fiber, though, so your conclusion isn’t affected. Also, your calculation is for equivalent diameters. If mass was more important then, well, the fiber is 1/3 the density of copper…
- aidenn0 3y ago
  It's been a long time since I did any physics, but explain to me why you can't run the 0.072A into e.g. a 10kohm load and have 51W of power dissipated in the load? Sure you'd be running at several hundred volts, but high voltage is how you get more power through less copper. [edit] Upon further reflection your claim about maximum power transfer being when the load and cable are dissipating the same total power doesn't pass the sniff test, because it would imply that a load connected by the same 33awg cable, but only 1m long could only drive a load of 3.5mW since the load would then only be 678 mohm.
  - aidenn0 3y ago
    For other's curious: the "load resistance matches source resistance" is true for maximum power transfer of fixed voltage sources. In the case of current capacity of a wire, the voltage is variable, but the maximum current is fixed. In that case we are selecting the input voltage for maximum power so (Rs, I are fixed, Rl can vary, and for a purely resistive load, Vin is a function of Rl): Vin=I(Rs+Rl) Vin = Vs + Vl Vl/Vs=Rl/Rs Pl = (I^2*Rl) Clearly we can always select a Rl (and thus a Vin) that gets the desired Pl at a fixed current. Obviously at some point we are limited by shielding of the wire, and DC/DC conversion at the end-point. We eventually may also be limited by interactions between the medium surrounding the wire and EM fields generated by turning the circuit on and off. But, when you can control the voltage, there's no simple calculation from wire impedance to X maximum watts of load.
- coryrc 3y ago
  Your maximum current is too conservative, power transmission calculations are conservative enough to run wire through fiberglass insulation in a hot attic. Closer is "chassis wiring" which rates that wire for 0.47A: https://www.powerstream.com/Wire_Size.htm https://www.powerstream.com/Wire_Size.htm To deliver 150W with output voltage of 300V (0.5A) would require an input voltage of 640V at an efficiency of ~45%. Not great. The water could handle a lot higher heat distribution but efficiency is already terrible. > For maximum power transfer at a fixed input voltage.
- onetimeuse92304 3y ago
  Remember, that power is voltage times current. You can increase voltage to push as much power as you want through any wire. This is the reason why overhead lines use such high voltages. Now, the limitation is the insulation of the wire. The insulation will determine how high voltage you can use and when you increase voltage, the insulation requirements grow very quickly. The fiberoptic can transfer more power, just not for the reason you think it can.
  - TaylorAlexander 3y ago
    That's right and since we are talking about ROVs, I recall that it is common to power the ROV with relatively high voltages specifically to allow for smaller and lighter tethers.
  - deleted 3y ago
    [deleted]
resters 3y ago
how is this different than the skin effect in wire?

beambot 3y ago

I suspect their statement is true if you consider the fully-loaded diameter too -- i.e. fiber + cladding can carry more power than copper + insulation, especially for small gauges. Edit: Quick google... This paper [1] shows a 180um fiber + cladding carrying 150W over 1km. According to [2], that diameter equates to a 33awg wire, which has 678 ohm/km resistance and a maximum current of 0.072A. At 1km, the wire would be 678ohms. For maximum power transfer, you want the load to equal the resistance of the wire. Thus you could deliver 0.072A into a 678ohm load - i.e. your maximum power would be 3.5 Watts. The fiber is better by 40x. [1] https://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=8908271&casa_token=W_8YBHm8hssAAAAA:86Dy1EppIMO4Inbg7u6ZAEciApnsS5-DbJ0wAWPSiOQdHDhbksO7BE-8j8BWQHCr5CXjW1CyODh5yJQ&tag=1 https://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=8908271... [2] https://www.badgerwire.com/tech-info/wire-gauge-table/ https://www.badgerwire.com/tech-info/wire-gauge-table/

pi-e-sigma 3y ago
Nah, just google 'high power fiber cable', these are used to transfer light in laser steel cutters. As these are already in practical, industrial use we can assume that their spec is in line with what is currently realistically achievable for an optical fiber power transmission over a distance of a few meters max. For instance a SMA905 Fiber Cable can deliver meager 50 Watts and its external diameter is 5 millimeters which is laughable low compared to how much power you could push through a copper cable with the same external diameter, that is including electrical insulator.
- beambot 3y ago
  I don't think it's a reasonable assumption to look at industrial use cases over small distances compared to long-distance operations -- the design criteria are different.
  - pi-e-sigma 3y ago
    The design criteria is exactly the same. Have an optical medium with as little attenuation as you can. If you can't have it in a medium that is only a few meters long how could you possibly achieve it in a medium of a few kilometers? Apply your logic to copper cables. Can you have a copper cable able to deliver power over a long distance but unable to deliver the same power over a shorter distance?
    - quailfarmer 3y ago
      The analogy doesn’t hold because one of the hardest parts of this technology is building a device that can extract the optical energy as electrical energy, and dissipate whatever heat it generates. The power density of the fiber itself is easy by comparison.
      - pi-e-sigma 3y ago
        The analogy holds perfectly. Remember, we are discussing a claim that optical fiber with all required cladding/protection can carry as much power as a copper wire with all insulation having the same external dimension as the fiber cable. It is simply not possible, fiber will always have less power carrying capacity and your claim that fiber power density is 'easy' has no basis in the reality.
    - mlyle 3y ago
      There's absolutely no reason for the application you cited to optimize diameter and isn't approaching physical limits. There's no reason you can't shove many optical fibers into that 5mm diameter. People don't, in short distance applications, to have simpler systems and connectors. There's big reasons why we don't use power over fiber (the endpoints are very expensive, and the overall system efficiencies are low). None of them have to do with the highest optical power density you could hit.
heisenzombie 3y ago
You’re right, though I suspect you could increase that “max” current figure quite substantially if we’re assuming this is an underwater ROV and the wire has water cooling for free. You’d need to be pumping a whopping 300A/mm^2 through that wire to match the 150W fiber, though, so your conclusion isn’t affected. Also, your calculation is for equivalent diameters. If mass was more important then, well, the fiber is 1/3 the density of copper…
aidenn0 3y ago
It's been a long time since I did any physics, but explain to me why you can't run the 0.072A into e.g. a 10kohm load and have 51W of power dissipated in the load? Sure you'd be running at several hundred volts, but high voltage is how you get more power through less copper. [edit] Upon further reflection your claim about maximum power transfer being when the load and cable are dissipating the same total power doesn't pass the sniff test, because it would imply that a load connected by the same 33awg cable, but only 1m long could only drive a load of 3.5mW since the load would then only be 678 mohm.
- aidenn0 3y ago
  For other's curious: the "load resistance matches source resistance" is true for maximum power transfer of fixed voltage sources. In the case of current capacity of a wire, the voltage is variable, but the maximum current is fixed. In that case we are selecting the input voltage for maximum power so (Rs, I are fixed, Rl can vary, and for a purely resistive load, Vin is a function of Rl): Vin=I(Rs+Rl) Vin = Vs + Vl Vl/Vs=Rl/Rs Pl = (I^2*Rl) Clearly we can always select a Rl (and thus a Vin) that gets the desired Pl at a fixed current. Obviously at some point we are limited by shielding of the wire, and DC/DC conversion at the end-point. We eventually may also be limited by interactions between the medium surrounding the wire and EM fields generated by turning the circuit on and off. But, when you can control the voltage, there's no simple calculation from wire impedance to X maximum watts of load.
coryrc 3y ago
Your maximum current is too conservative, power transmission calculations are conservative enough to run wire through fiberglass insulation in a hot attic. Closer is "chassis wiring" which rates that wire for 0.47A: https://www.powerstream.com/Wire_Size.htm https://www.powerstream.com/Wire_Size.htm To deliver 150W with output voltage of 300V (0.5A) would require an input voltage of 640V at an efficiency of ~45%. Not great. The water could handle a lot higher heat distribution but efficiency is already terrible. > For maximum power transfer at a fixed input voltage.
onetimeuse92304 3y ago
Remember, that power is voltage times current. You can increase voltage to push as much power as you want through any wire. This is the reason why overhead lines use such high voltages. Now, the limitation is the insulation of the wire. The insulation will determine how high voltage you can use and when you increase voltage, the insulation requirements grow very quickly. The fiberoptic can transfer more power, just not for the reason you think it can.
- TaylorAlexander 3y ago
  That's right and since we are talking about ROVs, I recall that it is common to power the ROV with relatively high voltages specifically to allow for smaller and lighter tethers.
- deleted 3y ago
  [deleted]

resters 3y ago

how is this different than the skin effect in wire?