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I've made a mpg improvement with an aftermarket chip. Surprised.

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Comments

  • Markyt
    Markyt Posts: 11,864 Forumite
    cepheus wrote: »
    Here we are referring to the deliberate reduction of speed on a Motorway from say 80 to 70 or 70 to 60 mph, not the reduction in average speed caused by congestion which results in more accelerations and braking which can increase mpg.

    So that only applies under ideal, non real world conditions then?
  • Conor_3
    Conor_3 Posts: 6,944 Forumite
    cepheus wrote: »
    Aerodynamic resistance increases as the square of the speed, and rolling resistance is approximately proportional.

    Is wrong. I've done a proper course. The Mercedes manufacturers one for the Actros. Complete with graphs, charts and all kinds of send you to sleep technical data.

    Even on a HGV what you say is completely wrong. On a graph of wind resistance versus speed for a 14ft high, 8ft wide artic with a front profile of a breezeblock, the wind resistance climbs extremely slowly until you reach 53MPH at which the resistance increases far faster than the speed. to the point at around 60MPH where the line is climbing almost vertically if you have speed on the horizontal axis and resistance on the vertical. This is for a truck. The graph is going to be a lot flatter for a car that has a far more aerodynamic shape. It certainly isn't anywhere near the square of the speed.

    And then there's engine load, engine speed and the powerband and torque characteristics of the engine - all of these play a part.

    If driving slower is more economical were correct, all the official figures for fuel consumption would show 30MPH being more economical than 56MPH....
  • tomstickland
    tomstickland Posts: 19,538 Forumite
    10,000 Posts Combo Breaker
    Well, at high speeds you would expect fuel consumption to scale with speed squared. A drag force proportional to speed cubed is certainly the basis that I've built all my mathematical acceleration models on previously.
    I do believe that travelling at 70mph versus 80mph does provide quite a good saving, around 5mpg for my car.

    The model would fall apart at lower speeds because there are certain fixed overheads in running the engine and ancillieries and keeping the engine at an efficient rpm.

    If you were to drive along in 1st gear with the engine at its most efficient rpm then you might expect that to be very efficient. However, you'd be making very little progress relative to the time that all the other power sinks were operating.


    However, like all things, it's a balance between cost and other factors.
    There's no way I'd choose to travel at 60mph on a motorway. I don't mind doing 70, but I much prefer to travel at 80 indicated. This provides a good balance between economy, rate of progress and fitting in with other traffic.

    Regarding inefficient driving, well yesterday I travelled along the motorway at 80 indicated for about 20 miles. In that time I was leapfrogged by another vehicle 3 times! They would pass me when I was in the left hand lane, stay in the middle lane then catch up slower traffic. As they slowed I'd pull into the outside lane, pass them and whatever was holding them up, and pull back into the left hand lane. Two minutes later they'd come past again.

    So I was travelling at 80 indicated with no more throttle than necessary to hold that constant. Meanwhile they'd slowed to 65, waited, then accelerated hard to 90 and done that 3 times.
    Happy chappy
  • movilogo
    movilogo Posts: 3,239 Forumite
    Part of the Furniture 1,000 Posts Name Dropper Photogenic
    We know, power = force * velocity

    The power required to move a car in velocity V is

    P = (Cd*r*A*(V^2)⁄2 + m*m*g + m*g*sin(q))*V

    where

    Cd = coefficient of drag (usually 0.2 to 0.35)
    r = air density (1.3 kg⁄m3)
    A = frontal projected area of the car (~1.5 m2)
    V = velocity of car (m⁄s)
    m = coefficient of friction (~0.015 for rubber tyre on concrete road)
    m = mass of car (kg)
    g = acceleration due to gravity (9.8 m⁄s2)
    q = gradient (or slope) of the road (radian) equals 0 on level ground
    P = power required to move the car (Watt)

    At low RPM, usually power is developed from torque multiplication (using gearbox). So, you start the car in lower gear. As speed rises, engine can gradually develop enough power to accelerate the car. So, you then shift to higher gears.

    PS: I personally would never fit any after market chip to modify ECU settings! If your engine RPM crosses red line, serious engine damage may happen. Let leave insurance thing alone, modifying ECU in this way even invalidates manufacturer's warranty!
    Happiness is buying an item and then not checking its price after a month to discover it was reduced further.
  • tomstickland
    tomstickland Posts: 19,538 Forumite
    10,000 Posts Combo Breaker
    In less words, power required is proportional to speed cubed (the rolling resistance term is usually small).
    Since distance travelled is proportional to 1/speed, and if we assume that petrol consumption rate is proportional to power requirement then, approximately mpg is inversely proportional to speed squared.

    Torque multiplication doesn't change the engine power. It changes the torque available at the wheels by reducing the rotational speed. So it does provide more acceleration force, but the power is the same.

    Engine damage can result from high rpm use. The manufacturer's redline is an arbitrarily selected "safe" rpm. Quite often the rod bolts will break, or piston damage will occur. However that doesn't mean that a particular engine cannot withstand occasional high rpm use. For example, my old 1.3 Astra 8v had no rev limiter andI took it off the clocks several times without incident.

    With a 20 year old 210K engine do you think I car about a manufacturer's warranty?
    Happy chappy
  • raskazz
    raskazz Posts: 2,877 Forumite
    On the insurance note, if you make any sort of changes in this vein without telling your insurer then you need your head examining. Engine chipping is usually rated amongst the riskiest modifications. Hence you are highly likely to find your policy is void if you get caught.
  • cepheus
    cepheus Posts: 20,053 Forumite
    Markyt wrote: »
    So that only applies under ideal, non real world conditions then?

    Yes, but motorway driving without congestion is what the OP was originally referring to, not the stop start driving characteristic of urban conditions. Actually, most cars would be most efficient in mpg terms at a constant 30-40mph in top gear. Mpgs would then be in the 70s and 80s especially for small engines. The reason why we rarely see this quoted is because it is rarely possible to drive this way, and when average speed decreases to this sort of level you will find mpg actually decreases due to the amount of stop start driving.


    PS The power (not resistance) to overcome aerodynamics increases as the cube of speed.

    The power to overcome air resistance increases roughly with the cube of the speed, and thus the energy required per unit distance is roughly proportional to the square of speed. Because air resistance increases so rapidly with speed, above about 30 mph (48 km/h), it becomes a dominant limiting factor. Driving at 45 rather than 65 mph (72 rather than 105 km/h), results in about one-third the power to overcome wind resistance, or about one half the energy per unit distance, and much greater fuel economy can be achieved. Increasing speed to 90 mph (145 km/h) from 65 mph (105 km/h) increases the power requirement by 2.6 times, the energy by 1.9 times, and drastically decreases fuel economy. In practice, rather than doubling or halving the fuel economy, the difference is actually closer to 40-50%, because engine efficiency varies greatly with the torque/speed operating point. Rolling resistance, which is broadly proportional to speed, is also a factor particularly at lower speeds.

    http://en.wikipedia.org/wiki/Fuel_economy_in_automobiles
  • moonrakerz
    moonrakerz Posts: 8,650 Forumite
    Part of the Furniture 1,000 Posts Combo Breaker
    movilogo wrote: »
    We know, power = force * velocity

    The power required to move a car in velocity V is

    P = (Cd*r*A*(V^2)⁄2 + m*m*g + m*g*sin(q))*V

    where

    Cd = coefficient of drag (usually 0.2 to 0.35)
    r = air density (1.3 kg⁄m3)
    A = frontal projected area of the car (~1.5 m2)
    V = velocity of car (m⁄s)
    m = coefficient of friction (~0.015 for rubber tyre on concrete road)
    m = mass of car (kg)
    g = acceleration due to gravity (9.8 m⁄s2)
    q = gradient (or slope) of the road (radian) equals 0 on level ground
    P = power required to move the car (Watt)

    At low RPM, usually power is developed from torque multiplication (using gearbox). So, you start the car in lower gear. As speed rises, engine can gradually develop enough power to accelerate the car. So, you then shift to higher gears.

    PS: I personally would never fit any after market chip to modify ECU settings! If your engine RPM crosses red line, serious engine damage may happen. Let leave insurance thing alone, modifying ECU in this way even invalidates manufacturer's warranty!


    Bulls**t baffles brains ! ;)
  • tomstickland
    tomstickland Posts: 19,538 Forumite
    10,000 Posts Combo Breaker
    It's not !!!!!!!! though; it's the textbook formula for rolling resistance force due to aerodynamic drag and rolling resistance, converted to power by multiplication by velocity.
    Which is a long way of saying drag power is mainly proportional to speed cubed.
    Happy chappy
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