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:

$\displaystyle v = V_i(D^t)$

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

The distance traveled in time T is:

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

Substituting the value for T we get:

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

Hope this helps!

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 )?

Re: Predicting Tartget Speed

Quote:

Originally Posted by

**CakeSpear** 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** 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.

Re: Predicting Tartget Speed

Aha, didnt see that.

Excellent it works!

Thanks again Ebaines!

:D