# Car parked on top of an electric cord. **URL:** \ **Category:** Factual Questions\ **Created:** [February 9, 2013, 1:07am UTC](https://boards.straightdope.com/t/car-parked-on-top-of-an-electric-cord/649749 "2013-02-09T01:07:23Z")\ **Posts on this page:** 5\ **Page:** 3
**Author:** ![DSeid](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dseid/32/20194_2.png) [@DSeid](https://boards.straightdope.com/u/DSeid)\ **Post date:** [February 9, 2013, 7:41pm UTC](https://boards.straightdope.com/t/car-parked-on-top-of-an-electric-cord/649749/41 "2013-02-09T19:41:54Z")
That was to snailboy - mc, no it is math. ---
**Author:** ![snailboy](https://avatars.discourse-cdn.com/v4/letter/s/97f17d/32.png) [@snailboy](https://boards.straightdope.com/u/snailboy)\ **Post date:** [February 9, 2013, 7:46pm UTC](https://boards.straightdope.com/t/car-parked-on-top-of-an-electric-cord/649749/42 "2013-02-09T19:46:37Z")
> [@Musicat](#): > > Without verifying your numbers, and assuming my calculator is operating normally, 900 lbs / 24 inches is approximately 38 lbs/inch. > > _Which has nothing to do with the internal tire pressure._ > > The numbers are coincidental, but the coincidence is meaningless. If I increase the internal tire pressure to 48 PSI, does that mean that the pressure per square inch on the ground has doubled? Assuming you mean 76 psi, which is double the original 38 psi, then yes, there would be twice as much pressure on the ground (assuming it’s flat ground). The tire would have a smaller contact patch because the extra pressure makes it more rigid. It’s the same weight but less area to transfer it to. ---
**Author:** ![Musicat](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/musicat/32/20189_2.png) [@Musicat](https://boards.straightdope.com/u/Musicat)\ **Post date:** [February 9, 2013, 7:50pm UTC](https://boards.straightdope.com/t/car-parked-on-top-of-an-electric-cord/649749/43 "2013-02-09T19:50:17Z")
> [@snailboy](#): > > Assuming you mean 76 psi, which is double the original 38 psi, then yes, there would be twice as much pressure on the ground (assuming it’s flat ground). The tire would have a smaller contact patch because the extra pressure makes it more rigid. It’s the same weight but less area to transfer it to. So doubling the pressure reduces the footprint of a tire (cylinder) by half? Interesting trigonometry. How much pressure could I have before the tire (and vehicle) was weightless? ---
**Author:** ![snailboy](https://avatars.discourse-cdn.com/v4/letter/s/97f17d/32.png) [@snailboy](https://boards.straightdope.com/u/snailboy)\ **Post date:** [February 9, 2013, 8:02pm UTC](https://boards.straightdope.com/t/car-parked-on-top-of-an-electric-cord/649749/44 "2013-02-09T20:02:46Z")
> [@Musicat](#): > > So doubling the pressure reduces the footprint of a tire (cylinder) by half? Interesting trigonometry. How much pressure could I have before the tire (and vehicle) was weightless? I already said the weight will remain the same, only the pressure will change, that is unless you rip a hole in the bottom of the tire and turn it into a hovercraft. ---
**Author:** ![scr4](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@scr4](https://boards.straightdope.com/u/scr4)\ **Post date:** [February 9, 2013, 8:05pm UTC](https://boards.straightdope.com/t/car-parked-on-top-of-an-electric-cord/649749/45 "2013-02-09T20:05:20Z")
> [@Musicat](#): > > So doubling the pressure reduces the footprint of a tire (cylinder) by half? Yes, if the tire didn’t have any structural strength by itself. An actual car tire will deviate from this formula slightly. > [@](#): > > Interesting trigonometry. How much pressure could I have before the tire (and vehicle) was weightless? It’s not trigonometry, just arithmetic: Weight = pressure \* area If you increase the pressure to infinity, you’d end up with an infinitesimal contact patch. The weight is the same. [Previous page](https://boards.straightdope.com/t/car-parked-on-top-of-an-electric-cord/649749.md?page=2)