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Thread: New Bristol Fighter T Veryon killer 270mph 1016bhp car

  1. #33
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    I think the veyron's a very good looking beast

  2. #34
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    Quote Originally Posted by Zak33 View Post
    It looks.......tall! I'd test most stuff at speed, but you'd not get me in that. Way too much glass too.... my brain is shouting "what is rigidity like in it?"
    I know what you mean, it "feels" like it should be more like this:
    (excuse the dodgy photochop)



    Much slinkier!


    Original for comparison:

    (\__/)
    (='.'=)
    (")_(")

  3. #35
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    The chopped roof looks cool, dont think it'd fit in though.

  4. #36
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    Quote Originally Posted by chuckskull View Post
    The chopped roof looks cool, dont think it'd fit in though.
    Umm GT40?
    (\__/)
    (='.'=)
    (")_(")

  5. #37
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    Ugh typos got me again. Why am I still hungover at nearly 3pm

    Lets try again...

    The chopped roof looks cool, don't think I'd fit in though.

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    Blimey!

    Shorter legs, or sit in the back?
    (\__/)
    (='.'=)
    (")_(")

  7. #39
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    Quote Originally Posted by Stoo View Post
    Blimey!

    Shorter legs, or sit in the back?
    I've driven a mini from the back seat. That was a scary and very uncomfortable experience

  8. #40
    Now with added sobriety Rave's Avatar
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    Quote Originally Posted by badass View Post
    Yes it was the roadster, but remember theoretical power required to reach top speed is proportional to the top speed cubed.
    Wind resistance is proportional to the square of the speed as you say, but in addition to that power is proportional to force times velocity.

    I get a top speed of 229 MPH based on that so I was out in the first case
    That does however assume that in both cases the car hits peak power at its top speed.
    Meant to reply to this earlier, never got round to it. Thought I might as well do it now in the light of the interminable "plane on a treadmill" thread, which I read up until the point Nicho saw the light, and then gave up on.

    Power is proportional to force x velocity as you say- and consequently as a car accelerates, the accelerative force the engine is able to exert upon it decreases. In the hypothetical example of putting a rocket in deep space and then firing off the engines, the astronauts would experiece hard acceleration to start with which would gradually diminish as their speed increased. Kinetic energy is mass times velocity squared- which all matches up neatly, although I found it a bit counter-intuitive when I started A-Level physics.

    In this instance (I.E. top speed of a car) we are talking about the power of the engine versus the frictional resistance of the wind, which increases with the square of the speed (as does, AFAIK, the mechanical resistance via the tyres and bearings, which in any case is far less important). Consequently to double your speed you need four times the power. I fail to see where any calculation involving the cube of the speed comes in?

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    Quote Originally Posted by Rave View Post
    In this instance (I.E. top speed of a car) we are talking about the power of the engine versus the frictional resistance of the wind, which increases with the square of the speed (as does, AFAIK, the mechanical resistance via the tyres and bearings, which in any case is far less important). Consequently to double your speed you need four times the power. I fail to see where any calculation involving the cube of the speed comes in?
    Wind resistance increases as the square of the speed, correct.
    The result of that is the force (not power) to overcome that resistance is squared. Bare in mind you are travelling at twice the speed, and at twice the speed to generate the same force, you need twice the power - as in power=force times velocity.
    so you have power is proportional by the speed squared and multiplied by the top speed again - i.e. the top speed cubed.
    power=force times velocity.
    p=fv
    force of resistance~velocity^2
    f~V^2
    substitute f in the top equation with v^2 and you get p~v^2v
    p~v^3

    Where ~ means proportional to.
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  10. #42
    Now with added sobriety Rave's Avatar
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    Oh hellfire= I've just spent literally over any hour trying to pick your equations apart- in an alcohol befuddled state- before realising that my previous post was wrong. Power must equal force x velocity x mass- otherwise the weight of the object would have no bearing on its acceleration. Obviously, it does.

    Anyway, terminal velocity of any object (in our earthbound frame of reference), be it a car with the throttle pinned, or a skydiver jumping out of a plane with only gravity accelerating them, occurs when the accelerative force reaches equilibrium with the decelerative force due to air resistance.

    Air resistance (and AFAIK rolling resistance from the tyres etc.) is proportional to the square of the speed. If you have a car whose air and rolling resistance increases with the square of its speed (which is more or less any car), then if you quadruple the power you double the terminal velocity. The mass of the car only affects the time it will take to reach terminal velocity, as the engine has to impart more kinetic energy to it to accelerate it.

    If that doesn't convince you, I'll bust out the equations tomorrow,

  11. #43
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    Quote Originally Posted by Rave View Post
    Oh hellfire= I've just spent literally over any hour trying to pick your equations apart- in an alcohol befuddled state- before realising that my previous post was wrong. Power must equal force x velocity x mass- otherwise the weight of the object would have no bearing on its acceleration. Obviously, it does.
    Power is just Force x Velocity.

    The force required is affected by the mass.

  12. #44
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    Quote Originally Posted by Rave View Post
    Oh hellfire= I've just spent literally over any hour trying to pick your equations apart- in an alcohol befuddled state- before realising that my previous post was wrong. Power must equal force x velocity x mass- otherwise the weight of the object would have no bearing on its acceleration. Obviously, it does.
    power=force x velocity
    There's no acceleration in there.
    acceleration= force/mass
    However at top speed no acceleration is occuring so it doesn't matter.
    Anyway, terminal velocity of any object (in our earthbound frame of reference), be it a car with the throttle pinned, or a skydiver jumping out of a plane with only gravity accelerating them, occurs when the accelerative force reaches equilibrium with the decelerative force due to air resistance.
    Yep
    Air resistance (and AFAIK rolling resistance from the tyres etc.) is proportional to the square of the speed. If you have a car whose air and rolling resistance increases with the square of its speed (which is more or less any car), then if you quadruple the power you double the terminal velocity. The mass of the car only affects the time it will take to reach terminal velocity, as the engine has to impart more kinetic energy to it to accelerate it.

    If that doesn't convince you, I'll bust out the equations tomorrow,
    Substitute the power in bold there with force. That is where you have gone wrong.
    To generate the same force at double the speed requires double the power.
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