Week 5: Track Position#
The stop also chooses a gap
The question#
Why can the stop lap that is fastest for one car lose once we ask where that car rejoins among everyone else?
This week’s boundary:
- one dry, uninterrupted synthetic race
- fixed traces for three other cars
- no rival response and no randomness
Week 4’s race loop is supplied. We add the field, not another loop.
The words for place#
- running order — the sequence of cars at one point in the race
- track position — a car’s place in that sequence
- rejoin — where a car enters the running order after leaving the pits
- traffic packet — the cars just ahead inside this scenario’s stated time threshold
Running order is a snapshot. Finishing position is the classification after the final lap.
The words for cost#
- traffic cost — synthetic time added for rejoining close behind cars that must be followed or passed
- finishing position — ordinal place in the classified field: P1, P2, P3
A fixed trace is a supplied car path and clock that never changes when SlopU changes its call. It can block SlopU. It is not a rival, because it does not respond.
Keep both clocks#
Clean-air finishing time is Week 4’s isolated total.
Adjusted finishing time is that total plus this scenario’s traffic cost.
adjusted_finish = clean_air_finish + traffic_costNever overwrite the first value. The difference is evidence about what the one-car model omitted.
Compare at one line#
For each fixed car: gap = SlopU rejoin clock − car crossing clock.
The packet contains cars with 0 < gap ≤ 1.000 s.
Price the packet once#
Synthetic rule:
packet empty → traffic_cost = 0
packet not empty → traffic_cost = 0.4 + 0.8 × packet_size0.4 sprices reaching a supplied passing window0.8 sprices each car in the packet1.000 sis the supplied packet threshold
These are SlopU scenario values, not Formula 1 facts. Apply the result once to the final total.
Six laps: price both candidates#
All values are synthetic.
| Candidate | Clean-air finish | Cars in packet | Calculation | Adjusted finish |
|---|---|---|---|---|
| early stop | 599.600 | 2 | 599.600 + 0.4 + 0.8×2 | 601.600 |
| late stop | 599.800 | 0 | 599.800 + 0 | 599.800 |
Fixed cars: Lark finishes at 600.000; Wren at 600.800.
The time-only comparison chooses early by 599.800 − 599.600 = 0.200 s.
Early stop: sort the adjusted field#
Packet calculation:
0.4 + 0.8 × 2 = 2.0 s
| Position | Car | Adjusted finish |
|---|---|---|
| P1 | Lark | 600.000 |
| P2 | Wren | 600.800 |
| P3 | SlopU early | 601.600 |
The isolated winner rejoins behind both fixed cars and finishes third.
Late stop: sort again#
The packet is empty, so the traffic cost is zero.
| Position | Car | Adjusted finish |
|---|---|---|
| P1 | SlopU late | 599.800 |
| P2 | Lark | 600.000 |
| P3 | Wren | 600.800 |
The call flips: give away 0.200 s in clean air, gain two finishing places.
Places are ordinal. We report them beside the time margin, never add them to it.
Back to SlopU’s 30 laps#
Week 4 supplies the candidate totals. The fixed-trace ledger supplies the cars inside one second ahead at the common line.
| Stop after lap | Clean-air finish | Rejoin clock | Packet ahead |
|---|---|---|---|
| 11 | 2804.620 | 1044.175 | Moss |
| 12 | 2804.500 | 1137.430 | Moss, Tern |
| 13 | 2804.515 | 1230.740 | none |
Week 4 chooses lap 12 by isolated finishing time. What does the field choose?
Predict before the reveal#
Fixed finishing clocks:
Oriole 2803.900 · Moss 2805.000 · Tern 2805.700
Apply the packet rule once to each SlopU candidate. Then sort each four-car field.
Which stop gives SlopU its best finishing position?
Price all three candidates#
| Stop | Packet calculation | Traffic cost | Adjusted finish |
|---|---|---|---|
| 11 | 0.4 + 0.8×1 | 1.200 | 2804.620 + 1.200 = 2805.820 |
| 12 | 0.4 + 0.8×2 | 2.000 | 2804.500 + 2.000 = 2806.500 |
| 13 | empty packet | 0.000 | 2804.515 + 0.000 = 2804.515 |
Clean-air time remains in the first ledger. The adjusted clock is a new field, not a replacement.
Classify each complete field#
All four cars complete the same 30 laps.
| Stop 11 order | Stop 12 order | Stop 13 order |
|---|---|---|
| P1 Oriole 2803.900 | P1 Oriole 2803.900 | P1 Oriole 2803.900 |
| P2 Moss 2805.000 | P2 Moss 2805.000 | P2 SlopU 2804.515 |
| P3 Tern 2805.700 | P3 Tern 2805.700 | P3 Moss 2805.000 |
| P4 SlopU 2805.820 | P4 SlopU 2806.500 | P4 Tern 2805.700 |
Best position first, adjusted time only as a tie-break: stop after lap 13.
State what changed#
Compare Week 4’s stop after lap 12 with the position-first stop after lap 13.
| Quantity | Calculation | Result |
|---|---|---|
| clean-air time given up | 2804.515 − 2804.500 | 0.015 s |
| adjusted time gained | 2806.500 − 2804.515 | 1.985 s |
| finishing places gained | P4 → P2 | 2 places |
Oriole remains ahead at 2803.900. SlopU improves to second; it does not win.
What the regulation supports#
Issue 08 Article B2.5.5 classifies cars first by complete laps covered and, for cars on the same lap, by the order in which they cross the Line.
Every supplied car completes 30 laps, so sorting adjusted finishing times represents that same-lap crossing order in this bounded scenario.
Retirements, lapped cars and the 90% classification rule sit outside the fixture. The FIA supplies none of the traffic equation, packet threshold or coefficients; those remain synthetic.
Two failures worth catching#
Double charge
adjusted = clean_air + packet_cost + slower_traffic_lapsThe supplied packet cost already prices the effect. Slowing laps as well charges it twice unless the whole model is rewritten consistently.
Wrong sort
assert classify(stop_12).slopu_position == P4Sorting its clean-air 2804.500 would put SlopU P2. The assertion catches a
classifier that ignored the adjusted 2806.500.
What passes forward#
Assignment 1’s contest criterion asks for a rival treated as responding rather than as a fixture. Week 5 is the smaller failure case: adding fixed cars already changes the call.
You now have a comparison that:
- preserves clean-air and adjusted clocks
- prices a rejoin packet exactly once
- ranks complete same-lap fields by adjusted time
- reports seconds and positions separately
Your lab: implement it, reproduce both flips and state why the fixed traces are not yet a rival. Week 6 lets the rival move.