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Speeding up before a Hill to Save Gas

  1. Nov 15, 2008 #1
    Is there any way to prove that speeding up before a hill in a car will save you gas compared to maintaining your speed on the hill?

    Here is the situation:

    [​IMG]

    Notice, the one car maintains 10m/s throughout while the second car accelerates to 20m/s on the flat part. Both cars will have the same potential and kinetic energy at the end.

    My original approach was to convert MPG of my car (35 mi/gal) to J/m (aka Netwons) by using the conversion factor of 34.8 x 10^6 J/L, which I did and got 2338.95 Netwons of force. Since MPG is presumably measured on a flat surface with a constant speed, that means my car requires 2338.95 Netwons of force to maintain speed so the force of friction is also 2338.95 N. Therefore, since I had the force of friction total of the internal parts of my car, I thought the rest of the problem would be easy. However, when I did all the math, the result showed that the energy required was the same, but that goes against the saying that you should speed up before hills--is that unfounded?

    Thanks for any help and if you need any clarification, I'd be willing to explain anything. This really interests me so I'd love to know what you guys think.

    -Paul
     
  2. jcsd
  3. Nov 15, 2008 #2
    Although it's fairly insignificant, if you wait until you are on the hill to accelerate (simply maintaining speed on the hill) gravity will be pulling down on the gas you would've burned speeding up before hand, resulting in a greater gravitational force and more energy lost from the system.
     
  4. Nov 15, 2008 #3
    Well, even though I am not accounting for drag, the higher speed of the second car would make it have to work harder so things like the amount of gas in the tank are not significant either. Although it's the little things about physics, isn't it?
     
  5. Nov 15, 2008 #4

    rbj

    User Avatar

    the raw physical energy needed to climb the hill is the same whether you speed up or not.

    the "proof" you need requires knowledge about how internal combustion engines work and how they optimally work. there is a reason for why we have different gears with different gear ratios. it is not optimal for you to start your car out from a dead stop in 5th gear.

    driving at high speed in a lower gear is worse gas mileage than in a higher gear. "lugging the engine" by a significant amount is worse gas mileage than not lugging it. the only way to be operating your car in the high gear without lugging the engine is to be going fast. at a given speed, "lugging the engine" increases as the load increases. the load increase when you're climbing as opposed to descending or level travel.
     
  6. Nov 15, 2008 #5
    It depends on the efficiency curve for each transmission gear under the two load conditions, level ground and slope.

    For a steep enough hill that the car could not maintain a velocity in the lowest gear, but will stall, hitting the slop at high speed may be the only way to make it.
     
  7. Nov 15, 2008 #6
    Is it possible to find energy curves for the gears for ground and slopes? I suppose it probably isn't and would be company "secrets." Hmm.
     
  8. Nov 16, 2008 #7
    I dont' know if you are going to find efficiency curves, or energy curves. Googling "power curves" will give you hits.
     
  9. Nov 16, 2008 #8
    Well, after finding a few racing power curves (sorry I don't use a drag racer to climb up hills), I got a hold of a 2001 Dodge minivan engine power curve. Not being totally familiar, does anyone have any idea how to use it?

    [​IMG]

    I assume I'd find the RPM on a flat surface while accelerating at 5 m/s2 and then find the RPM on a hill while keeping constant velocity of 10m/s then find the corresponding horsepower, which would equal Hp X 745.69987 = Watts. Then use Watts to find the rest? It's a shame I don't have a 2001 minivan. I can't find power curves for a Ford Focus.

    Edit:
    Some digging gave me something:
    [​IMG]
     
    Last edited: Nov 16, 2008
  10. Nov 16, 2008 #9

    OmCheeto

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    Gold Member

    Your calculations are correct. The energy is the same. In the textbook example you've given anyways.

    If you are looking for real world number about the two scenario's, it get really complicated.

    One number that people don't often quote is the fuel consumed by the vehicle at idle.
    This plays a big part in the fuel efficiency of a vehicle at slow speeds.

    My guess is that you should speed up to the optimal mpg based velocity of the vehicle and maintain that speed up the hill.
     
    Last edited: Nov 16, 2008
  11. Nov 16, 2008 #10
    Well, my idle fuel consumption is the green circle since my car basically idles at 1000 rpm.

    [​IMG]

    Does that mean idling, it is just as efficient as it is at 6000rpm (aka, not really that efficient?)

    At 2000rpm, that's my prime efficiency, correct?
     
  12. Nov 16, 2008 #11
    As you climb the hill, the potential energy is going to increase, meaning the kinetic energy will decrease. Gassing uphill just means you're adding more energy to the system, meaning you will have a larger kinetic energy at the top of the hill than hitting the hill with a constant velocity. The only possible implication I can think of would be the ability to climb over a series of hills, but even then it'd be better go gas through the downslope because there is a greater engine efficiency and far less resistance.
     
  13. Nov 16, 2008 #12
    The curve at the bottom serves for efficiency. At 1950 rmp, or so, the engine puts out the most power per gallon of fuel.

    To turn rpm into miles per hour, you need to know about the transmission and the rest of the drive train.
     
  14. Nov 16, 2008 #13
    Couldn't you turn the KwH or the Joules into Kinetic Energy without it? (1/2 mv^2)?
     
  15. Nov 16, 2008 #14

    rcgldr

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    Homework Helper

    That's at full throttle. The graph would need a 3rd dimension to include the efficiency versus rpm versus throttle position. Then you'd need to compare the efficicency versus power for the power required for speed and rate of climb.
     
  16. Nov 16, 2008 #15
    Okay, I think I am going to not pursue this any further. I appreciate all the help but it's now beyond the scope of my abilities to handle. Thanks for all the replies.

    -Paul
     
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