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Math Help - Predicting Tartget Speed

  1. #1
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    Predicting Tartget Speed

    Hey
    - I have a Ball moving at a Speed S.
    - The Ball has an InitialSpeed iS
    - The Ball has a TargetSpeed tS

    Every second:
    - Ball is displaced S number of pixels
    - S is multiplied by deacceleration D ( D is a number between 0 - 1 )

    ...
    I need an expression, returning how many pixels the Ball moves before reaching TargetSpeed tS ( tS is allways lower or equal to iS )
    ( Time is integer only )

    Hope this makes sense.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Predicting Tartget Speed

    Forgive me but I'm going to change your notation a bit, so that it's less confusing.

    Let:
    v = velocity,
    V_i = initial velocity
    V_t = target velocity
    T = time to reach target velocity
    X = distance traveled to reach target velocity

    The equation of motion that you are describing is this:

    v = V_i(D^t)

    From this we have  \frac {V_t}{V_i}= D^T = e^{T \ln( D)}. Hence the time to reach the target velocity is  T = \frac {\ln(V_T/V_i)}{\ln D}

    The distance traveled in time T is:

     X = \int_{t=0} ^{t=T} v dt = \int_{t=0} ^{t=T} V_i e^{(ln (D))t} dt = \frac {V_i}{ln(D)}(D^T-1)

    Substituting the value for T we get:

     X = \frac {V_i}{ln(D)}(D^{\frac {\ln(V_T/V_i)}{\ln D}} - 1)

    Hope this helps!
    Last edited by ebaines; May 1st 2013 at 11:53 AM.
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  3. #3
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    Re: Predicting Tartget Speed

    Thanks Ebaines

    ...
    I made a simple computer program:

    V_i = 4
    V_t = 3
    D = 0.99

    This expression is copied directly from the program ( after implementing it ):
    ( ( 0.0 + value( "VelocityInitial" ) ) / Ln( 0.99 ) ) * ( 0.99 * ( Ln( ( 0.0 + value( "VelocityTarget" ) ) / value( "VelocityInitial" ) ) / Ln( 0.99 ) ) - 1 )

    Ill simplefy it so it is easier to read:
    ( ( 0.0 + V_i ) / Ln( D ) ) * ( D * ( Ln( ( 0.0 + V_t ) / V_i ) / Ln( D ) ) - 1 )

    ...
    By doing a Step by Step test in the programm, i find:
    - It takes 29 turns/ steps/ seconds for V_i to reach V_t
    - As V_i reaches V_t, the Ball has traveled very close to 100 pixels

    However, the expression provided ( see below )
    ( ( 0.0 + V_i ) / Ln( D ) ) * ( D * ( Ln( ( 0.0 + V_t ) / V_i ) / Ln( D ) ) - 1 )

    outputs: -10880.4

    ...
    Am i doing someting wrong in the expression, does this expression not fit my situation, or does the problem lie somewhere else ( in the program maybe )?
    Last edited by CakeSpear; May 1st 2013 at 01:17 PM.
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: Predicting Tartget Speed

    Quote Originally Posted by CakeSpear View Post
    Ill simplefy it so it is easier to read:
    ( ( 0.0 + V_i ) / Ln( D ) ) * ( D * ( Ln( ( 0.0 + V_t ) / V_i ) / Ln( D ) ) - 1 )
    You misread the forumula - it should be:
    ( ( 0.0 + V_i ) / Ln( D ) ) * ( D ^ ( Ln( ( 0.0 + V_t ) / V_i ) / Ln( D ) ) - 1 )

    Note thatin this formula the number D is raised to the power of (ln(V_t/V_i))/lnD, not multiplied by it.

    Quote Originally Posted by CakeSpear View Post
    By doing a Step by Step test in the programm, i find:
    - It takes 29 turns/ steps/ seconds for V_i to reach V_t
    - As V_i reaches V_t, the Ball has traveled very close to 100 pixels

    However, the expression provided ( see below )
    ( ( 0.0 + V_i ) / Ln( D ) ) * ( D * ( Ln( ( 0.0 + V_t ) / V_i ) / Ln( D ) ) - 1 )

    outputs: -10880.4
    Make the correction and you'll find the formula gives an answer of 99.5 pixels, and T=28.6 steps.
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  5. #5
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    Re: Predicting Tartget Speed

    Aha, didnt see that.

    Excellent it works!
    Thanks again Ebaines!
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