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  1. Join Date
    Oct 2002
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    10,820
    #11
    In ideal circumstances, like in a vaccuum and frictionfree environment, the force required to maintain a speed of 30kph & 100kph should be the same.
    pare, the force required to maintain a steady velocity (meaning it is now a vector, velocity = speed and direction) in space where it is almost a perfect vacuum and thus no friction/drag will be zero. kaya off ang engines ng space shuttle after it attains it's orbital velocity.

    outer space, btw, is not a perfect vacuum. free electrons, neutrons, "space dust", leftover particles from comets, solar wind material, not to mention the theoretical "dark matter" permeates outer space.

    clarification lang, di ako kumokontra ha. :mrgreen:

    peace.

  2. Join Date
    Oct 2002
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    29,354
    #12
    well, kaya nga di space yung example ko... i did say "vaccuum & frictionfree environment" lang. Ideal circumstance lang. Just like some physics problem solving we got during high school & college. :mrgreen:

  3. Join Date
    Oct 2002
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    2,716
    #13
    just have an idea on this topic, gumawa ako calculation:

    given a car consumes 10km/liter on highway speed 100km/hr, how much fuel is injected on each chamber?

    assuming typical data for my civic esi, the engine is at 3000rpm (approx) when running 100 kph at 5th gear, and noting there are 2 injections per revolution for a 4-cylinder engine,

    the car is consuming 1 liter in 6 minutes (10km/liter)/(100km/hr)=(1/10)hr
    in 6 minutes, the engine made 18,000 revolutions (6 x 3000rpm)
    the car consumed 1 liter in 36,000 injections
    hence, the amount of fuel per injection is 1/36,000 liters or 0.02777 ml.

    yun lang ...

  4. Join Date
    Oct 2002
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    29,354
    #14
    the problem is your assumptions are wrong. It nearly impossible to measure fuel consumption based on time unless your car is on a threadmill (trying to get close to ideal conditions). It doesn't take into considerations that air drag, road inclines, acceleration rate, car weight and other friction inducing forces play a significant role in a car's fuel consuption.

    A better measure would be km/L.

  5. Join Date
    Oct 2002
    Posts
    10,820
    #15
    korek. drag alone may take up to 90% of an engine's power output when moving at a steady speed (depending on speed, ok). transmission losses, traction loss, road friction, alternator, power steering and and a/c takes up the rest.

  6. Join Date
    Oct 2002
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    2,716
    #16
    Quote Originally Posted by ghosthunter

    A better measure would be km/L.
    i used 10km/L in my calculations ... purposely to take care of every imaginable losses ... your insight is correct that a better measure is km/L that's why i used it ... i'm not interested in calculating losses, rather just the amount of fuel injected on every injection and starting with km/L value is what i thought to be most sensible

  7. Join Date
    Oct 2002
    Posts
    377
    #17
    nabasa ko rin somewhere that each car uses only approximately 15% of its total energy output to turn the wheels. dami kasing moving parts and other variables kaya daming energy lost sa isang auto due to friction and heat.

    On another topic, nabasa ko rin pala na ang drag is exponential. forgot the exact formula but its something like <some variable> x velocity squared. So ibig sabihin the drag that a vehicle encounters at 80kph is four times the drag at 40kph. I guess to achieve maximum fuel efficiency, one must balance it between drag and travel time (since the car travelling at 80kph experiences 4x as much drag but for only half the time as a car travelling at 40kph for the same distance).

  8. Join Date
    Oct 2002
    Posts
    10,820
    #18
    F(drag) = A x Cd x V x V x m,

    where A = cross sectional area of the object perpendicular to the direction of motion
    Cd = coefficient of drag
    V = velocity
    m = density of the fluid through which the object is moving. in the case of a car, that will be the density or air. for a submarine it will be the density of sea water.

    you will find this formula in basic fluid mechanics.

  9. Join Date
    Oct 2002
    Posts
    10,820
    #19
    oh **** ngayon ko lang napansin, tricycle drayber na ako! funyeta kaya pala parang magaan ang paa ko di na ko padyak drayber!

    are you taking up engineering kevinp? looks like it eh. well, welcome to the club. :mrgreen:

  10. Join Date
    Oct 2002
    Posts
    377
    #20
    are you taking up engineering kevinp? looks like it eh. well, welcome to the club.
    actually, nope. i'm more of a computer guy. hobby ko lang yung interest ko sa cars. di ako nag fluid mechanics nung college e (or di ako nakinig :mrgreen: ). i wouldn't mind though combining my interest in cars and computers, like making traction control software for F1 cars (in my dreams siguro) :D

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