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Car speed tests

  1. Mar 11, 2005 #1
    Hi all. Newbie here who has a simple question.

    In car magazines, you see these speed tests they do. For example, they'll list a 0-60 time, 0-100 time, and a 0-130 time. My question is, would the times be different if you just made one run from 0-130 and mark down the times when you hit 60, 100, and 130 as opposed to doing three separate runs of 0-60, 0-100, and 0-130. We assume that on the three individual runs, the driver doesnt slow down until he passes the intended mark. Also, lets say you do 0-60 in 5.0 seconds, 0-100 in 10 seconds, and 0-130 in 15 seconds, can we use these numbers and make the following conclusions or is there more to it?

    from 60-100, it takes 5.0 seconds
    from 60-130, it takes 10.0 seconds


    Thanks for your help.
     
    Last edited: Mar 11, 2005
  2. jcsd
  3. Mar 11, 2005 #2

    ZapperZ

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    There are two different aspects of your question.

    First is the IDEAL case where in each test, the acceleration is a constant no matter what test is being conducted. If this is the case or the assumption that one is making, then having different tests for 0-60, 0-100, etc. doesn't make sense since all you need is just the 0-130, then you know everything. This is becaue the time and the speed are proportional to each other. So if you know the time it takes to go from 0 to 130, you will also know the time it will take to go from 0-60, 0-100, etc.

    [tex]v = at[/tex] since initial velocity is zero.

    However, the fact that the times you listed are not linear with the velocity range, i.e. 0-60 takes 5 secs, while 0-100 takes 10 secs, means that one cannot assume a constant acceleration (which is usually the case in real life). The performance of the car may be different at different speeds (which is what happens when you shift gears or transmission). That is why you do test this over different velocity range. So if this is the case, then no, you simply cannot infer from just one test what the time would be over a fraction of the speed.

    Zz.
     
  4. Mar 11, 2005 #3
    that is if you dont run out of petrol on the way...
     
  5. Mar 11, 2005 #4
    Thank you for your help. The bottom numbers I used were just random. The actual numbers are below. I'm a car enthusiast who is comparing the performance of two cars. This helps.

    Here are the actual numbers

    0-60 mph 4.8
    0-100 mph 12.6
    0-130 mph 25.0
     
    Last edited: Mar 11, 2005
  6. Mar 11, 2005 #5

    krab

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    No. The times would not be different. Why should they be? Different would mean that the 0-60mph time would depend upon what you did after you hit 60mph. That contradicts causality.
    The numbers are assuredly NOT linear. Linear would mean that if 0-60mph is 4.8 sec, 0-130mph would be 4.8*130/60=10.4 sec. It would not be linear even if your wheels exerted a constant force to the road. This is because of aerodynamic drag. That's also the reason you top out at some speed. So if 135 mph is top speed, the time needed to reach it (or just higher than it) is infinite.

    In fact, your tires do not exert a constant force, because vehicles are power-limited. So the force to drive the car is power divided by speed. Power varies a bit because you cannot stay at peak power except with a continuously-variable transmission, but this is not a large effect. So a good description of an accelerating vehicle is given by
    [tex]m{dv\over dt}={P\over v}-bv^2[/tex]
    where b is an aero drag coefficient, m is mass, P is power (to the road). You will notice acceleration is zero when [itex]P=bv^3[/itex]. The v that satisfies this formula is of course top speed. So you can rewrite the equation of motion as
    [tex]m{dv\over dt}=P\left({1\over v}-{v^2\over v_\infty^3}\right)[/tex]
    Which is nicer because you don't need to worry about how to find b. Solve this equation and you have v as a function of time. No road test needed. I've done this and it agrees with car magazine test reports. (Well actually it's a little more complicated than this because the force does not go to infinity at zero velocity as P/v would indicate, so you also need the torque in first gear.)
     
    Last edited: Mar 11, 2005
  7. Mar 13, 2005 #6

    vcc

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    Also impacting the times is discrepancies in how a driver shifts. I, also, am a car enthusiast, and I know that for seasoned drivers, shifting is different for different speeds if one were to optimally shift through the car's powerband. For a 0-60 run, a driver would want to launch into the next gear's powerband as he/she approaches 60mph, due to the fact that 60mph is probably attained on the 2nd gear of the car.

    On a 0-130 run, however, shift points are generally lower at lower gears.
    By increasing the shift-frequencies at lower gears, the driver is actually cutting down the link time between gears (time for clutch depression and gear changing to allow revs to drop into the optimal range for the next shift).
    In an 9000rpm (peak) vehicle, with the powerband at 7000, making the shift at 7,000 to drop to 6000 on the next gear will be shorter than making the shift at 8500 rpm and allowing it to drop to 7000. Although landing on 7000rpm is optimal for the vehicle, the time to drop from 8500 to 7000 is significantly longer than the time it takes to drop from 7000 to 6000 due to the increased circular momentum in the flywheel.

    Depending on the torque specs on the car, as the driver increases speed, and approaches the 130mph mark, engine speed at shifts begin to increase. It would, in this case, be optimal to hit 130mph right after the powerband of the highest gear that the driver used to attain this speed. Therefore, to launch the latest gear into 7000rpm, and preventing any shifts onto gears higher that gear would be optimal.
    As a result, there is a way to approach these later gears that may not be suitable for a 0-60 run. Nor would the way to approach 60 be suitable for this 130mph run.

    Let's assume that 130mph utilizes the car's 5th gear, and that reaching 60 only utilizes the car's 2rd gear.
    For a driver to reach 60mph, he would likely make the shift from gear 1-2 at 8500 rpm to hit the second gear at 7000rpm, and fully utilize the powerband by attaining 60mph right after it.
    On the other hand, for a river to reach 130mph, he would likely make a shift form gear 1-2 at 7000rpm, drop to 6000rpm, shift from 2-3 at the same range, 3-4 at a slightly higher range, and ease his way to 8500->7000rpm, from 4-5. If the driver shifted like this for the 0-60 run, the time to reach 60mph would be slower. However, it would be optimal for the sake of reaching 130mph.
     
    Last edited: Mar 13, 2005
  8. Mar 14, 2005 #7
    Your explanation of shifting techniques make very good sense. The two cars I'm interested in have different gearing. One is geared to reach 62mph in second gear before redlining (or fuel cutoff) while the other car has to make a third shift to reach 60mph accounting for a slower 0 to 60 time. The second car however has more aggressive gearing as you reach higher speeds and catches up and passes the first car when you hit 100mph.
     
  9. Mar 14, 2005 #8

    Danger

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    Hi folks;
    Although not as important over all as the factors mentioned (and very nicely detailed) by the responding gentlemen, there are a few other tricky bits that I remember from my old Porsche-hunting days. The following all increase with speed, and therefore affect performance: rolling friction; electromagnetic drag on the alternator; increased air resistance of the fan (if belt-driven); oil windage in the crankcase; gear and tranny lube windage; bearing friction in axles, gearboxes, etc.; breathing resistance in the air cleaner; and a few others that elude me at present.
     
  10. Mar 14, 2005 #9

    brewnog

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    But those don't affect any difference there may be between the 0-60 sections of 0-60 and 0-130mph tests.
     
  11. Mar 14, 2005 #10

    Danger

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    Hi again;
    Correct. It was just a bit of trivia to reinforce the fact that acceleration isn't constant. I've only had net access for a couple of weeks (I actually joined the site before getting it). This whole thing is absolutely fascinating, and I can't quite help myself from poking my nose in wherever I find the discussion within some range of my experience. (You'll see that I've done a bunch of posts today.) I just realized that this site is some sort of official help system rather than just for discussion, and sincerely hope that my participation isn't a bother.
     
  12. Mar 14, 2005 #11

    brewnog

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    No worry! You'll find that there's often a fine line between providing extra information, and confusing people (a line most of us cross occasionally). But your participation was welcome!

    Anyway, welcome. Be prepared never to leave, you're trapped here now.
     
  13. Mar 14, 2005 #12

    Danger

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    Hi yet again;
    Sadly true. I've been sitting at this stupid machine ever since I got home from work 4 1/2 hours ago. It's like when I look something up in a dictionary and can't stop reading. Your the first person I've had any actual interaction with on line, so your welcome is most appreciated. I don't know exactly what that buddy list thing is, but you're going to be the first one on it.
     
  14. Mar 14, 2005 #13

    brewnog

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    Head down to general discussion and you can make some more friends :smile:
     
  15. Mar 14, 2005 #14

    Danger

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    I don't know how, but it sounds great. Please advise.
     
  16. Mar 14, 2005 #15

    brewnog

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    All the way back up to https://www.physicsforums.com, then go down to the General Discussion section at the bottom, and carry on as you did here! Watch out for that Moonbear though, I've heard she bites. :smile:
     
  17. Mar 16, 2005 #16

    vcc

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    I'll do the same thing.
    The environment here seems very friendly.
     
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