Harness Voltage Drop Is a Geometry Problem

The number everybody computes

Voltage drop on a DC feed is not difficult. Resistance per unit length comes from the conductor gauge, you double the length because current has to return, and Ohm's law does the rest:

drop_V = current_A × resistance_per_m × length_m × 2

Every term on the right is easy to look up except one. Gauge is a design decision. Current is a requirement. Length is geometry, and geometry is the term that models routinely get wrong, not by a little, and, where only the endpoints are stored, always in the same direction.

Straight lines are not wire

If a model stores a connection as a pair of endpoints, the only length available is the straight-line distance between them. No harness has ever taken that path. Wire follows structure: around the inside of a shell, along a chassis rail, through a grommet, down the side that has the tie-down points rather than the side that is shorter.

The straight-line figure is therefore not an estimate of the run. It is a strict lower bound on it, and a bound that gets looser exactly when the routing gets constrained, which is to say, on the runs that matter.

The fix is unglamorous: store the route. A connection carries an ordered list of waypoints in world coordinates, and its length is the polyline through its endpoints and those waypoints, plus a stated slack for service loops and strain relief.

length = Σ |pᵢ₊₁ − pᵢ|  over the routed polyline
       + slack_m

The point is not that the polyline is sophisticated. It is that it is recorded, so the length is a measurement of a decision somebody made rather than a consequence of the model's storage format.

A worked case

This article discusses an earlier model and its harness calculations. The current native SysML v2 workspace is described in the Studio guide.

An earlier JSON-based Studio aircraft study used an A320-class transport whose 28 V DC loads are fed from an electrical power distribution centre, and whose twenty runs carry power, data, fuel and hydraulics.

Its harness was first drawn round the airframe rather than through it, with runs leaving their ports and looping outside the fuselage skin before coming back in. Drawn that way, the twenty runs measured 252.3 m, and the flight deck’s 28 V feed alone took 10.76 m. Routed through the airframe, the way the loom would actually be built, the same runs measure 148.1 m and the flight deck feed 7.65 m. Not one part moved. The only thing that changed is the path the model records.

Drop is proportional to that length, so it moved in proportion. At the 35 A the feed then stated, on 16 AWG:

Route recorded	Flight deck feed	Drop, 5 % allowed	Loss
Round the airframe	10.76 m	35.4 %	347 W
Through the airframe	7.65 m	25.2 %	247 W

Here the recorded path was longer than the real one, the opposite of the straight-line error above, and it is the same mistake: a length that follows from how the model was drawn rather than a measurement of the harness. Either way the drop is wrong by exactly the ratio of the two lengths.

Routed correctly, the feed still failed, and that is the other half of the lesson: length is one of three terms. The run stated 35 A for a flight deck that draws 350 W, which is 12.5 A at 28 V, on a 16 AWG conductor too thin for either figure. Stated at the current it carries, on 12 AWG, the same run drops 3.6 % and passes. That earlier model kept the feed as stated, at 35 A on 16 AWG, because catching that is what the example is for: that version marked the run red without anyone having written a drop constraint, which is the argument for computing it where the geometry lives.

The part that surprises people

The drop is only half of what that routing decision costs. The same resistance that produces the voltage drop is also dissipating power, and it has to come from somewhere upstream:

loss_W = current_A² × resistance_per_m × length_m × 2

Across the airliner the harness dissipates 1,886.7 W. Loss goes as the square of current, and the three generator feeders carry 200 A each: the two integrated drive generators lose 403.6 W apiece over 15.65 m, and the APU’s feeder, the longest at 31.99 m, loses 825.1 W on its own. The long, heavy-current runs are where the watts go. The 28 V bus is the exception that proves the rule: its four feeds lose 254.3 W between them, and 246.9 W of that is the one under-gauged feed to the flight deck. Three of them together lose 7.4 W.

Now the power figure. The airliner’s modelled loads draw 3,155 W in its nominal mode. Add the harness and the sources have to supply 5,041.7 W, three fifths as much again. A power budget that counted only loads would be wrong by well over a third of its answer, and one that counts conductors is only as right as the lengths it is given, which is the whole of the worked case above.

Why this belongs in the geometry model

The reason to compute drop where the parts live, rather than in a separate electrical tool, is that every input is already there and every output feeds something else that is already there.

• Route length falls out of the same coordinates that drive clearance.
• Copper mass falls out of gauge and length, and lands in the mass roll-up and the centre of gravity.
• Dissipated watts land in the power budget and in the thermal screen, as heat at a location.
• Moving a part re-routes its feeds, so the drop updates while you drag it.

Split those across tools and each one is individually correct while the system is wrong. Keep them in one model and moving the electrical power distribution centre 200 mm visibly changes the drop, the loss, the budget and the CG at once, which is the actual trade being made.

What to take from it

If you record only the endpoints of a connection, you have not modelled the harness; you have modelled a wish about the harness. The gap between the two is where the drop lives.

And when a limit is exceeded, the finding is worth nothing unless it says by how much and over what. “Voltage drop exceeded” sends someone hunting. “Flight deck 28 V feed drops 16.7 % against a 5 % limit, over 5.09 m of AWG 16 at 35 A” tells them the run, the path length, the gauge, the current and the margin, which is enough to decide between a heavier conductor and a shorter route, or, as here, to notice the current is nearly three times the load, without opening anything else.