Optimal turns at Indianapolis Motor Speedway with JR Hildebrand | Physics | Khan Academy

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  • Опубліковано 14 жов 2024
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КОМЕНТАРІ • 30

  • @overtheunder5
    @overtheunder5 11 років тому +23

    You should do more of these with other professional athletes in other sports. Maybe a video with a baseball pitcher discussing the mechanics of a pitch. Or maybe a video with a football player discussing energy and ATP used during a football game. How bout a soccer player and how topspin alters the direction of the soccer ball. Just some suggestions.

  • @ctcahoe
    @ctcahoe 11 років тому +4

    Nice job guys, keep them coming. You could probably do 20 of these on racing. Physics is a huge part of racing and there are so many laws that can be demonstrated with an IndyCar. Doppler Effect, Bernoulli's Principle, Newton's Third Law. For a race car driver, JR is as smart as they come.

  • @Akhiii_
    @Akhiii_ 4 роки тому

    This is soooo cool, how did I only come across this collab just now?!?

  • @benrice2793
    @benrice2793 11 років тому +1

    You have a soothing voice.

  • @mrbungle3310
    @mrbungle3310 Рік тому

    Not always the fastest drivers in normal circuits drive perfect geometrical lines...but its an interesting video,i know drivers that think this way every corner...but you have to take into consideration bumps and camber

  • @xStaticSumi
    @xStaticSumi 11 років тому

    Really cool! I liked the collaboration of minds :)

  • @iwillfoolu
    @iwillfoolu 11 років тому

    This is a good primer for the extremely complex issue of optimal racing turns. Look up "The Physics of Racing, Part 5: Introduction to the Racing Line by Brian Beckman" for a more accurate depiction of the arcs. Also check out the wiki page on this for info on early/late apexes and chicanes.

  • @alevelscientist
    @alevelscientist 11 років тому +4

    the apex

  • @EdgeRetro
    @EdgeRetro 11 років тому

    Seems like you're hitting on the right idea, but the way it's drawn is fairly misleading. The 3rd circle goes off the track, so initially entering that radius (right where you have the car drawn) would require a tighter turn than even the inner circle. Perhaps they start much further back and make an increasingly tightening turn?

    • @rsimister
      @rsimister 11 років тому +1

      for constant angular velocity the work done will equal any losses due to rolling friction or drag. you could call 'useful' work 0 here, but I don't think he is using 'work' in the mechanical sense, more as a synonym for 'effort'.

  • @emangi1
    @emangi1 11 років тому +2

    Could you also consider whether a drift is more effective than a straight apex turn.

    • @Fewawidood
      @Fewawidood 11 років тому +5

      No, it isn't. When you are drifting, you have broken the optimal "mew", the sticking friction force between the tire and the road, and lowered it dramatically. If you have less grip, you aren't going to be able to go around as fast, obviously. Also, with drifting, the back end if swinging outwards, and that makes for inefficient driving as well

    • @emangi1
      @emangi1 11 років тому

      Kinda knew the answer already lol. Still, it'd be nice to see it in a Khan-video descriptive illustration.

    • @kwltommy
      @kwltommy 11 років тому

      effortless35 When drifting the cars take corners way slower than if a car would take the corner not drifting. In few cases, such as rally racing, drivers will angle their cars or drift because the surface(usually dirt) they are racing on behaves differently than pavement. The front wheels have less friction in dirt than pavement so it will just slide when turned, this is called under steering.

    • @seattergj
      @seattergj 11 років тому

      kwltommy
      Also drifting those cars at that speed would wear out the tires in a couple laps, if they didn't put themselves into the wall.

  • @getbusy21
    @getbusy21 11 років тому

    yes

  • @ageofkz
    @ageofkz 11 років тому

    Hmmm but turning does not require any work done because centripetal force is perpendicular to direction of travel, so F.s=0?

    • @nikhilwardrobe
      @nikhilwardrobe 9 років тому

      +Lim Kaizhuo yeah you right but the engines need to do some work against the friction to make a turn that was he referring !

  • @GamingNinja07
    @GamingNinja07 11 років тому +1

    INDYCAR FTW

  • @kametrixomtikara8726
    @kametrixomtikara8726 11 років тому

    But what's the exact curve? How to calculate it? (Might do that myself in the next couple minutes xD)

    • @rsimister
      @rsimister 11 років тому

      I guess you construct 3 quadratics, one for the outside line, one for the inside then one 'unknown' line for the apex, then you solve such that the apex maximum is less than the outside but more than the inside? cool problem i'm sure there is a way more elegant solution than my lame idea :S

  • @kght222
    @kght222 11 років тому

    i've always thought about it as a gradient, but that is likely the wrong word.

  • @anoirtrabelsi8645
    @anoirtrabelsi8645 11 років тому +3

    We need IMO problems

  • @foghorn51rater
    @foghorn51rater 11 років тому

    Well J.R. Hildebrand wants go take the last corner in the whole race really high, have his car slip on the marbles and smash into the wall.

  • @MrTweetyhack
    @MrTweetyhack 11 років тому +1

    All I want to see are more crashes and more fatalities.

    • @teampenske17
      @teampenske17 11 років тому +8

      I understand crashes but fatalities? WTF is wrong with you?

  • @nhunt
    @nhunt 11 років тому +1

    First order and obvious. Inaccurate drawing - the "largest diameter circle" that he sketches doesn't even fit within the track boundaries.

  • @Hammelj
    @Hammelj 11 років тому

    1st