The Verification Venue · pointed at the number every transmission debate starts from
The Crossing Where Direct Current Wins
Ask at what distance direct current becomes cheaper than alternating current and someone will hand you a figure, spoken with the confidence of a physical constant. It is not one. The crossing is a ratio of two differences, not a fact about physics. It is the extra money DC spends at its two ends, divided by the money it saves on every kilometre in between. Drag the four costs below and watch the crossing slide along the distance axis. Then go to sea, where the argument stops being about money entirely.
Every long transmission link is two ledgers. One charges up front, at the ends: converter stations for direct current, plainer transformer substations for alternating. The other charges by the kilometre, along the route. Direct current is dear at the ends and cheap in the middle; alternating current is the reverse. Somewhere between, if the numbers cooperate, the two ledgers cross. The instrument below draws both lines and their intersection from four numbers you control. Drag any slider and watch the crossing move.
Four ways to spend the money: pick a starting point
The presets are shapes, not quotes: every cost figure on this page is a round representative number chosen for legibility, and none comes from any real project. ↓
Break-even distance d*
750 km
below: AC wins · above: DC wins
Ledger gap at your distance
-20.0 M$
at 700 km, AC total minus DC total
Transformer substations at both ends. Raise it and the crossing retreats from DC.
Two converter stations with power electronics. This is the premium DC pays to enter the game.
Three conductors, bigger towers, a wider corridor. Shrink the gap to DC and the crossing runs away.
Two conductors, narrower towers. This discount is what buys the crossing in the first place.
Slide your project along the axis and read off which ledger is cheaper there, and by how much.
Write each ledger as a straight line. Alternating current costs A_ac up front plus B_ac per kilometre; direct current costs A_dc plus B_dc. Set them equal and solve. The crossing sits at d* = (A_dc − A_ac) / (B_ac − B_dc). The formula is worth staring at. Voltage is not in it. Current is not in it. Losses, terrain, the price of copper: none of it appears. The crossing is a ratio of two differences, and it belongs to the accounting, not to the electrons.
Drag the terminal sliders and the crossing slides in direct proportion. Shrink the per-kilometre gap and it runs toward the horizon; close the gap to zero and the lines are parallel, there is no crossing at all, and one technology dominates or loses everywhere. Below the crossing AC is cheaper, above it DC is, and at it they tie exactly. Which one wins is not a discovery waiting to be made: it is your four numbers, divided.
A warning the page owes you: real terminal and line costs are commercially sensitive, they vary by orders of magnitude between projects and countries, and no honest single break-even distance exists. The sliders ship with round representative values picked to make the structure legible. Treat the output as the shape of the argument, never as a forecast, and note that no real project, contract or price is named anywhere on this page.
Two: the crossing at sea
Underwater, the comparison changes shape, and the reason is not money. An insulated cable lying in seawater is a capacitor built along its whole length: conducting core, dielectric insulation, conducting screen. A coaxial cylinder holding permittivity ε has, exactly, C′ = 2πε₀εᵣ / ln(b/a) farads per metre. On a 50 Hz system every metre of that capacitance draws a charging current i′ = ωC′V, continuously, whether or not a single watt is being delivered at the far end.
Zero-delivery length L0
80.2 km
uncompensated ideal; compensation pushes this out, it does not remove it
Deliverable at your length
202.3 MW
charging 599 A of 800 A; headroom 531 A
Higher voltage raises the charging current in direct proportion. The sea punishes ambition.
Set by insulation geometry. Real extruded submarine cables scatter roughly 0.15 to 0.35.
The thermal current limit of one buried conductor. Charging current spends this budget first.
Slide past L0 and watch the power readout reach zero while the charging current keeps flowing.
The cable's ampacity is fixed by heat: bury it, and there is a maximum current its conductor can carry, period. The charging current spends that budget first. What remains for useful load current is the phasor headroom √(I_max² − (i′L)²), and the deliverable power is three phases times voltage times that square root. At L0 = I_max / (ωC′V) the headroom is gone: the cable is full of its own charging current and delivers nothing, at any price. No cost argument was involved. One of the two options simply stops existing.
With representative numbers the limit lands uncomfortably close. A 220 kV three-phase cable with C′ near 0.25 µF/km and an ampacity of 800 A draws about ten amperes of charging current per kilometre, so around eighty kilometres of cable consumes all eight hundred. The sliders let you move every one of those numbers; the zero-delivery length answers at once. One honesty note beside the readout: this is an idealised uncompensated cable. Real submarine AC links hang shunt reactors along the route that cancel much of the charging current, which pushes L0 outward. Compensation stretches the limit; it does not repeal it. What the curve shows is the ceiling that compensation has to fight, not the specification of any installed cable.
So the two halves of this page answer to different masters. On land, AC versus DC is bookkeeping with a crossing you can move by moving the numbers. At sea, past L0, the question of which is cheaper has no content, because one option delivers zero watts. Economics decides the first; electromagnetism settles the second.
The check · every number recomputed in front of you
Everything this page prints is recomputed in your browser from the constants declared in its source. Each table below computes its headline quantity twice, by methods that share no code: a closed-form formula, and a numerical root-find that only knows how to evaluate the underlying functions and bisect. They agree to within one part in 10⁹ or the row says MISMATCH in red. The last row of each table tracks your current sliders, live.
On land. Closed form is the ratio of differences. Root-find evaluates only the two total-cost functions and bisects their difference; it never sees the formula.
| case | A_dc − A_ac | B_ac − B_dc | d* closed form | d* root-find | verdict |
|---|
At sea. L0 from the closed form I_max / (ωC′V_ph), and again by bisecting the ampacity balance I_max − ωC′V_ph·L down to zero.
| case | V ll | C′ µF/km | I max | L0 closed | L0 bisect | verdict |
|---|
What is exact: the crossing formula is algebra, true wherever the two cost lines are straight; the coaxial capacitance is the closed-form solution of Laplace's equation for that geometry; the headroom square root is Pythagoras on orthogonal current phasors under the stated assumptions. Everything else is assumption, listed below. Run the offline verifier yourself: node research/the-crossing-where-direct-current-wins/verify-the-crossing-where-direct-current-wins.mjs (add --mutate to watch it corrupt its own engine and confirm the checks notice).
What's idealised here, and what's exactly true
Exactly true. Given linear costs A + B·d, the crossing d* = (A_dc − A_ac)/(B_ac − B_dc) is an identity: substitute it into both totals and they come out equal, for any values with B_ac > B_dc. The root-find column confirms this numerically without using the formula. The capacitance C′ = 2πε₀εᵣ/ln(b/a) is the exact solution of Laplace's equation between coaxial cylinders. Given a thermal limit I_max and a charging current in quadrature with the load current, the deliverable current √(I_max² − (i′L)²) follows from Pythagoras.
Idealised. Costs are linear and deterministic, with each technology's end costs lumped into one terminal number. The cable is lossless: no series resistance, no dielectric loss, no skin effect. The load is at unity power factor, three-phase balanced. Charging current is computed per metre as a lumped shunt, neglecting distributed-parameter wave corrections that matter only on cables approaching a wavelength. There is no reactive compensation on the sea layer. Vacuum permittivity is taken as ε₀ = 8.8541878128×10⁻¹² F/m (the CODATA 2018 recommended value) and the relative permittivity behind the default capacitance as 2.3, representative of extruded XLPE insulation but not a measurement of any particular cable.
Representative, not universal. Every cost constant is invented for shape. Published discussions tend to place the overland crossing in the hundreds of kilometres and the raw uncompensated submarine limit near a hundred, and the defaults were tuned to land there, but the page claims the structure, not the digits. Capacitance per kilometre for real extruded submarine cables scatters broadly with voltage class and insulation thickness; ampacity depends on burial depth, seabed temperature and neighbouring circuits. Across that scatter the ordering survives: charging current grows linearly with length, deliverable power dies at a finite L0, and the land crossing stays a ratio of two differences.