Reversible Dynamics, Asymmetric Actions: Parking as a Trajectory-Level Affordance
Reversible Dynamics, Asymmetric Actions: Parking as a Trajectory-Level Affordance
This research paper investigates the paradoxical advice of backing into parking spaces versus driving in nose first. The authors define three clearance metrics: static fit (F), eventual access (A_infinity), and resource-bounded access (A_K). Using certified arguments, they prove that while a car is dynamically reversible, forward-in and reverse-in tasks differ in the number of corrections required to maintain specific clearances under finite reversal budgets. The study proposes experiments to test if observers can distinguish these clearance types from simpler visual cues like body sweep.
This preprint proposes a formal distinction between static fit, eventual access, and resource-bounded access for vehicle parking maneuvers, using certified reachability arguments to demonstrate that forward-reverse asymmetry arises from finite correction budgets rather than dynamic reversibility.
Source: arXiv Robotics — research abstracts · Read original article ↗
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What the source reports
Publisher-reported claims, with original evidence. These results have not been independently verified by RoboSignal.
Reported numbers
clearance
0.5
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orward and reverse. They satisfy $F\ge A_\infty\ge A_K$. In one fully specified scene (one car, bay, aisle and start), a certified argument proves $A_\infty=F=0.5000$ m for both tasks. In this scene, all certified forward/reverse asymmetry therefore arises at finite $K$: it lies in how many corrections each task needs, not in the clearance either can eventually keep. With one reversal, an explicit path gives reverse-
Open source S4clearance
0.441
View original evidence
in at least $0.4410$ m clearance. A certified reachability argument caps forward-in's one-reversal clearance at $d_0<0.1720002$ m. With two reversals, forward-in gets at least $0.2833$ m. To keep a clearance strictly above any $d$ with $d_0\le d<0.2833$ m, reverse-in thus needs exactly one reversal and forward-in exactly two when both tasks move forward first. If both move in reverse first, the counts are two and thr
Open source S5clearance
0.2833
View original evidence
in at least $0.4410$ m clearance. A certified reachability argument caps forward-in's one-reversal clearance at $d_0<0.1720002$ m. With two reversals, forward-in gets at least $0.2833$ m. To keep a clearance strictly above any $d$ with $d_0\le d<0.2833$ m, reverse-in thus needs exactly one reversal and forward-in exactly two when both tasks move forward first. If both move in reverse first, the counts are two and thr
Open source S5reversals
1
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in at least $0.4410$ m clearance. A certified reachability argument caps forward-in's one-reversal clearance at $d_0<0.1720002$ m. With two reversals, forward-in gets at least $0.2833$ m. To keep a clearance strictly above any $d$ with $d_0\le d<0.2833$ m, reverse-in thus needs exactly one reversal and forward-in exactly two when both tasks move forward first. If both move in reverse first, the counts are two and thr
Open source S5reversals
2
View original evidence
in at least $0.4410$ m clearance. A certified reachability argument caps forward-in's one-reversal clearance at $d_0<0.1720002$ m. With two reversals, forward-in gets at least $0.2833$ m. To keep a clearance strictly above any $d$ with $d_0\le d<0.2833$ m, reverse-in thus needs exactly one reversal and forward-in exactly two when both tasks move forward first. If both move in reverse first, the counts are two and thr
Open source S5
- paper: not_reported
Source excerpts and review record
Automatically extracted; no manual editorial approval recorded.
orward and reverse. They satisfy $F\ge A_\infty\ge A_K$. In one fully specified scene (one car, bay, aisle and start), a certified argument proves $A_\infty=F=0.5000$ m for both tasks. In this scene, all certified forward/reverse asymmetry therefore arises at finite $K$: it lies in how many corrections each task needs, not in the clearance either can eventually keep. With one reversal, an explicit path gives reverse-
Open source S4
in at least $0.4410$ m clearance. A certified reachability argument caps forward-in's one-reversal clearance at $d_0<0.1720002$ m. With two reversals, forward-in gets at least $0.2833$ m. To keep a clearance strictly above any $d$ with $d_0\le d<0.2833$ m, reverse-in thus needs exactly one reversal and forward-in exactly two when both tasks move forward first. If both move in reverse first, the counts are two and thr
Open source S5
Source：arXiv Robotics — research abstracts · arxiv.org