Presets
Playback
Running
Speed of the simulation clock.
Line and substations
Layout, feeder resistance and rectifier supply.
Trains
Number, mass, power, speed, braking and dwell.
Regeneration
Voltage limits that decide how much braking energy can be returned.
Camera
Layers
Quality
Model checks
The science
A DC metro is a resistive ladder with one-way sources
Rectifier substations hold the line near their no-load voltage U0 (for a 750 V system about 790 V) behind an internal resistance Rss. The third rail and running rails form a feeder of loop resistance r (ohm/km) between the substations. A train is not a resistor but a constant-power load: it draws a current I = P / V, so when the voltage sags it needs more current, which sags the voltage further. For one train and one substation this gives the closed form
V = ( U0 + sqrt( U0^2 - 4 R P ) ) / 2, R = Rss + r d
with a maximum transferable power U0^2 / (4R) beyond which no solution exists (voltage collapse). Several trains and substations form a chain network solved here with Newton iteration (a tridiagonal linear system per step); below about 600 V a train must reduce its power (linearly to zero at 500 V), as the supply-voltage standards for DC railways require.
Regenerative braking
A braking train turns kinetic energy back into electrical power P_reg = eta m a v. Diode rectifier substations cannot take that power back, so it is usable only by other trains drawing current at that moment. If nobody absorbs it the line voltage rises; at the regeneration limit the converter ramps its output down and the rest is burnt in the brake resistor. The trade-off shown on the board: more trains and denser service raise the share of regenerated energy that is used; more substations reduce losses and voltage sag but cost money.
Train motion
Trains run the loop terminal to terminal, stopping at every station for the dwell time. Acceleration is limited by the traction force m a and by the power limit F = P / v; the running resistance is 1500 + 12 v + 6 v^2 newtons.
Numerical checks on this page
(1) The solver is compared with the closed form for one train and one substation. (2) On a standard scene (one braking, one accelerating, one cruising train) Kirchhoff's current law at every node and the power balance (substations = trains + feeder losses + rectifier losses) are verified. (3) A station-to-station run with unlimited traction power is compared with the trapezoid formula for the run time.
Honest simplifications
Both tracks are lumped into one feeder, substations are ideal Thevenin sources with a diode, there is no catenary or return-rail insulation model, no stray current, no substation transformer ratings or protection; the drive is represented by a constant efficiency; the regeneration limit is a linear ramp over 60 V; one tiny numerical leak to U0 (about 1 A at 200 V excess) keeps a network with all diodes blocked solvable. Numbers are plausible for teaching, not the data of any real metro.
What to try
- Load Weak supply: only 2 substations and watch the lowest train voltage fall towards the 600 V power-limit zone; graph 4 shows how it recovers with more substations.
- Switch Regenerative braking off: the energy drawn from the substations rises and graph 2 loses its green line.
- Load Few trains: regeneration has nowhere to go, the red wasted-power line grows and the regeneration used share drops.
- Load Rush hour: 12 trains: more simultaneous accelerations and braking, so the regenerated power is used almost entirely.
- Raise the feeder resistance to 0.06 ohm/km: the same trains now make the voltage sag far more.
- Lower the regeneration voltage limit to 800 V: converters cut back earlier and more energy is wasted.