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posted by martyb on Tuesday February 19 2019, @11:41PM   Printer-friendly
from the Wir-fahr'n-fahr'n-fahr'n-auf-der-Autobahn dept.

Brought to the floor by Senator John Moorlach of Orange County, SB-319 would direct the state's Department of Transportation to build two unlimited speed lanes on each side of Interstate 5 and State Route 99, the main north-south arteries linking cities like Los Angeles, San Francisco, and Sacramento. The sections of the roadways in question run straight through the supremely flat Central Valley, making for ideal high-speed driving conditions.

Perhaps paradoxically, California's answer to the German autobahn would be paid for by the state's Greenhouse Gas Reduction Fund. The text of SB-319 points out that the recent collapse of California's ambitious plan for a bullet train between Los Angeles and San Francisco, which was originally intended to trace the same route as the proposed unlimited speed lanes, has left residents without "access to high-speed, unabated transportation across the state."

http://www.thedrive.com/news/26554/california-might-add-lanes-with-no-speed-limits-to-major-highways


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  • (Score: 0) by Anonymous Coward on Wednesday February 20 2019, @02:18PM (4 children)

    by Anonymous Coward on Wednesday February 20 2019, @02:18PM (#803973)

    Nope, under ideal conditions, the dependence is quadratic.
    1. a car accelerating to 120mph will use at least 4 times as much fuel as the same car accelerating to only 60mph (hint: kinetic energy scaling with the square of the speed)
    2. the (aerodynamic) drag scales at least quadratic with the speed, the drag equation [wikipedia.org] guarantees it. It can scale even more if the Reynolds number also depends on the speed.

    Ok, let's talk about *speed* not *accelerating*. Don't confuse the two. KE is about speed, not accelerating. You want accelerating, you integrate. Also, don't talk about Raynold's number. That's from fluid mechanics and doesn't really apply here (think, speed of sound).

    Also, you are wrong, it's 2x. It's 4x the energy (due to drag) to travel at 2x the speed. Since you get there 2x faster, 4/2=2.... ok? So, naively speaking, traveling at 2x the speed will burn 2x the gas, roughly speaking, for same journey.. Traveling at 4x the speed will burn 4x the gas...

    ratio is v^2 (from drag) / v (from distance) so energy used to travel some distance is linear with speed.

    Now, if you want to talk about accelerations, it's completely different game that is quite independent of the actual speed. ;)

    QED.

  • (Score: 2) by c0lo on Wednesday February 20 2019, @03:32PM (3 children)

    by c0lo (156) Subscriber Badge on Wednesday February 20 2019, @03:32PM (#803999) Journal

    Fuel consumption proportional with energy you spend to achieve the work.

    Let's talk about energy:
    - on the accelerating side of the travel to reach the cruising speed from zero, you'll consume an energy proportional with the final kinetic energy, thus the square of the speed. Integrate all you want you aren't getting anything else (and this expenditure is relevant driving in heavy traffic, when you have cycles of acceleration/braking).

    - travelling at constant speed - for the same vehicle, the aerodynamic drag makes the only difference when it comes to travelling the same distance at different speeds. And the aerodynamic drag force scales with the square of the speed (in both laminar and turbulent cases, only the scaling factor is different between the two). Now, the energy expenditure against the aerodynamic drag is drag_force × distance (this is what your integration reduces to at constant speed). For the same distance travelled, will get to an energy expenditure going quadratic with the speed (plus the same energy in both cases lost due to friction with the road, which is independent of the speed)

    QED.

    Ummm... what exactly do you think you demonstrated?

    --
    https://www.youtube.com/watch?v=aoFiw2jMy-0 https://soylentnews.org/~MichaelDavidCrawford
    • (Score: 2) by Immerman on Wednesday February 20 2019, @04:09PM (2 children)

      by Immerman (3985) on Wednesday February 20 2019, @04:09PM (#804020)

      Absolutely correct. Worth mentioning as well that by the time you hit around 60mph, air resistance is generally the primary source of energy loss. Quadruple that by doubling your speed, and while you won't have quite 4x the total energy consumption per mile, you will almost certainly be well over 2x.

      • (Score: 2) by c0lo on Wednesday February 20 2019, @04:28PM (1 child)

        by c0lo (156) Subscriber Badge on Wednesday February 20 2019, @04:28PM (#804029) Journal

        and while you won't have quite 4x the total energy consumption per mile

        Yeah, switching between algo complexity and physics equations becomes harder with age - that friction term isn't quite ignorable in physics.
        A more refined model will take into consideration the mass of the fuel burnt by ICE over the travel - the crews in F1 races certainly don't ignore it.

        --
        https://www.youtube.com/watch?v=aoFiw2jMy-0 https://soylentnews.org/~MichaelDavidCrawford
        • (Score: 2) by Immerman on Wednesday February 20 2019, @04:54PM

          by Immerman (3985) on Wednesday February 20 2019, @04:54PM (#804043)

          Quite - without regenerative braking, acceleration can become a significant source of energy loss as well, especially if traffic speed fluctuates frequently.