what gauge wire will I need for my matter replicator?

[The_Tao_s_Revenge]

So I was doing my electrical home when just for shits n giggles I decided to figure out what gauge wire the enterprise uses incase I want to future proof my house by installing a replicator hook up.

Assumptions:
it has the same voltage limits as standard house wiring. 600V
it’s copper wire based.
it can only replicate 10 kilograms of mass at a time.
the replicator has a 100% efficiency at converting electrical energy to mass.

What I know
10 kilograms converted to to amps at 600v = 1500000000 amps or 1.5 giga-amps

10*300,000²\600

i = mc²\v

break down of my work. e=mc². Energy in joules (e) is mass in kilograms *lighti = mc² . one joule = watt of energy which divided by the voltage to get amps .

What I don’t know. the formula for how many amps a wire of particular width can carry.

What is it?

[The_Tao_s_Revenge]

oh and a couple more assumptions.

the wire only runs 40 feet from th branch terminal and won’t get any hotter then 70 degrees C

[Sunspace]

I used to work around electric resistance-welders. The welding tips carried 5V at 20000A (20 kA). They were connected to busbars that were about a foot wide and two inches thick. I think that for giga-amp currents, your equipment would require connection to apartment-building-sized copper busbars. Time to get out the superconductors…

[Squink]

Linkypoo (pdf)

Wire gauge is the resistance per unit length in decibels, where 10 AWG wire has 1 ohm per thousand

I had no idea!

However, on to the relevent bits:

A reasonable value for the current carrying capacity of a wire can be estimated from the rule that the cross-sectional area should be 700 “circular mils” per amp. The (rather perverse) definition of a circular area measured in “circular mils” is that it is the area of the square whose side has the same length as the diameter of the circle, i.e. just the diameter (in mils or 0.001 inch) squared. The current carrying capacity in amps is therefore (with D measured in inches)
I = 10^6 D^2/700

1500000000 = 10^6 D^2/700 = 1428.57143 D^2
1050001.05 = D^2
D =1 024.7 inches.

So by the conservative estimate of the 700 circular mils rule, you’d need a wire 85.4 feet, or 26 meters in diameter.

[The_Tao_s_Revenge]

Wow that’s a big wire. I hope home depot carries it.

[Sunspace]

[QUOTE=The Tao’s Revenge]
Wow that’s a big wire. I hope home depot carries it.
[/QUOTE]
I want to see the spool it comes on. Also, where will you plug it in?

[The_Tao_s_Revenge]

hehe it’s my new toy.

I worked out some easy formulas for mass to amp conversion
(i is amps, x is units of mass, v is volts.)

for pounds
i = x8164662660(5/v)
for stones
i = x583190190(5/v)
for kilograms
i = x18000000000(5/v)

Basically 8164662660 for example is one pound at 5 volts.

when converting amps between volt levels:

(i is new amp amount, h is old amp amount, v is new volt amount, t is old volt amount)

i = h(t/v)

Thanks for satisfying my curiousity Squink
And Sunspace I’ll have to get an adaptor!

[gotpasswords]

So how much would a 40 foot long piece of 86 foot diameter wire cost?

And where would you find the 900 foot long screwdriver to connect it to the circuit breaker’s screw terminals?

To be slightly more realistic, you’d need to change your replicator from running on 600 volts to long distance transmission voltages. The Pacific Intertie runs at 800 KV DC and has a capacity of 2 gigawatts, (normal operation is closer to 1.4 GW) using conductors that are about 1,200 mm[sup]2[/sup] in cross-sectional area. In terms of standard wire gauge, this is somewhere between 10 and 11 #0000 wires. A “four-aught” wire is almost a half-inch in diameter, so 11 of them would be a bundle of wire about as big around as your wrist.

Of course, insulating 800,000 volts will be your challenge.

[gotpasswords]

Screwed that up and missed the edit window. The invisible cat walked across the desk and sent the post too soon…

To be slightly more realistic, you’d need to change your replicator from running on 600 volts to long distance transmission voltages. The Pacific Intertie runs at 800 KV DC and has a capacity of 3.1 gigawatts, using conductors that are about 1,200 mm[sup]2[/sup] in cross-sectional area. In terms of standard wire gauge, this is somewhere between 10 and 11 #0000 wires. A “four-aught” wire is almost a half-inch in diameter, so 11 of them would be a bundle of wire about as big around as your wrist.

Of course, insulating 800,000 volts will be your challenge. FWIW, the Intertie actually runs at +/- 400 KV, which makes insulation a bit easier. Either conductor is 400 KV to ground, but 800 KV to each other.

[Sunspace]

So how do they insulate in the megavolt range? It can’t all be air…

[The_Tao_s_Revenge]

[QUOTE=gotpasswords]
So how much would a 40 foot long piece of 86 foot diameter wire cost?

And where would you find the 900 foot long screwdriver to connect it to the circuit breaker’s screw terminals?

To be slightly more realistic, you’d need to change your replicator from running on 600 volts to long distance transmission voltages. The Pacific Intertie runs at 800 KV DC and has a capacity of 2 gigawatts, (normal operation is closer to 1.4 GW) using conductors that are about 1,200 mm[sup]2[/sup] in cross-sectional area. In terms of standard wire gauge, this is somewhere between 10 and 11 #0000 wires. A “four-aught” wire is almost a half-inch in diameter, so 11 of them would be a bundle of wire about as big around as your wrist.

Of course, insulating 800,000 volts will be your challenge.
[/QUOTE]

That just blows my mind. If my math is right that thing is carrying enough energy to transmute into physical objects. I must have did my math wrong. That can’t be right can it? There’s a power line with enough juice to make physical objects? We really use that much energy?

[QUOTE=gotpasswords]
Screwed that up and missed the edit window. The invisible cat walked across the desk and sent the post too soon…

To be slightly more realistic, you’d need to change your replicator from running on 600 volts to long distance transmission voltages. The Pacific Intertie runs at 800 KV DC and has a capacity of 3.1 gigawatts, using conductors that are about 1,200 mm[sup]2[/sup] in cross-sectional area. In terms of standard wire gauge, this is somewhere between 10 and 11 #0000 wires. A “four-aught” wire is almost a half-inch in diameter, so 11 of them would be a bundle of wire about as big around as your wrist.

Of course, insulating 800,000 volts will be your challenge. FWIW, the Intertie actually runs at +/- 400 KV, which makes insulation a bit easier. Either conductor is 400 KV to ground, but 800 KV to each other.
[/QUOTE]

I’ll have to figure that out later, I’m still recovering from the Intertie. That’s alot more doable though!

[matt]

[QUOTE=The Tao’s Revenge]

10*300,000²\600

i = mc²\v

break down of my work. e=mc². Energy in joules (e) is mass in kilograms *lighti = mc² . one joule = watt of energy which divided by the voltage to get amps .

[/QUOTE]
'Fraid you messed up at the first calc. C is not 300,000 m/s, it’s 300,000,000 m/s! So, assuming you replicate 10kg in one second, your current is 1500,000 giga-amps! Or 1.5 peta-amps.

I’m not even going to touch the calc for the wire diameter, other than to state it will be TOO BIG.

[HongKongFooey]

[QUOTE=Sunspace]
So how do they insulate in the megavolt range? It can’t all be air…
[/QUOTE]
Air is the cheapest and lightest insulation for extra high voltage lines but requires a lot of space between the lines and of course they have to be insulated from the supporting structure. Underground high-voltage cables (and, rarely, some overhead lines) are insulated, I can’t find a decent link to any of them but there is picture of various types of extra high voltage cables on the main page here. Click on the image to see the cross-sections of various cables that are used.

[The_Tao_s_Revenge]

[QUOTE=matt]
'Fraid you messed up at the first calc. C is not 300,000 m/s, it’s 300,000,000 m/s! So, assuming you replicate 10kg in one second, your current is 1500,000 giga-amps! Or 1.5 peta-amps.

I’m not even going to touch the calc for the wire diameter, other than to state it will be TOO BIG.
[/QUOTE]

Oh thank god. That means we’re not using enough energy to transduce into physical objects. I don’t why but that just screwed with my ideas of a coherent reality.

I was using kilometers. Friggen metric system, now I know how Nasa felt.

[snailboy]

You make a mistake (or an unannounced assumption) in your first post. Joules do not equal watts. Joules equal watts times seconds. If your machine replicates this matter in one second at a consistent rate, this works out. However, even the ones on Star Trek take a few seconds. The longer you allow the machine to take, the thinner the wire can be.

[Q.E.D]

[QUOTE=gotpasswords]
In terms of standard wire gauge, this is somewhere between 10 and 11 #0000 wires. A “four-aught” wire is almost a half-inch in diameter, so 11 of them would be a bundle of wire about as big around as your wrist.
[/QUOTE]

Oh, they can get bigger than that, although you’re right that in terms of standard B & S gauge sizes, 4/0 is the largest. After that, wire sizes are specified by their size in units of MCM (thousands of circular mils). The very largest we deal with where I work is 500 MCM, which is a very fat wire indeed, having a conductor over .8 inches in diameter and capable of conservatively carrying 380 amps (@ 75[sup]o[/sup] C ampacity). Not fun to bend, I tell you! Fortunately, that’s not my job.

[Steve_Wright]

Just for the sake of geekiness … The Enterprise doesn’t have wires; it uses an electro-plasma distribution system to conduct power from the matter-antimatter reactor to the ship’s systems.

And now, I suppose, we know why - wires would be too big to be practical. (Good grief, does this mean Star Trek’s scientific advisors actually got something right for once?)