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What is integrated with the Euler integration?

IMHO: integration to obtain the velocity at time t to a place. Acceleration a is constant over time interval t. Right?

S(t0 -> delta_t) = S0(t) + ∫ s'(t) dt

s'(t) is derivated S

Please see Tim van Beek Answer. It's the right one :)

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    @anon that was what i had :( i can't give mire information then i have ... but Tim understood my partial Informtaion....2011-08-02

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Without further information I would guess that you are talking about the Euler integration scheme which is a method for a numerical approximation of an ordinary differential equation.

If you are further talking about Newtonian mechanics, you mean that there is a point mass of mass $m$ whose behaviour is described by the equation $ F = m * a $ Force = mass times accelaration. This is an ordinary differential equation of second order for the position of the point mass as a function of time $t$, $S(t)$, because we have S''(t) = a(t) that is the first derivative of $S$ is the velocity $v(t)$, and the second is the acceleration $a(t)$. In order to get a unique solution, we need to specify initial conditions for both $S(t)$, $ S(t =0) = S_0 $ which is the initial position and for $v(t)$, $ v(t = 0) = v_0 $ which is the initial velocity of the point mass. Now you can prescribe the force $F(t)$ as a differentiable function of time and calculate the position $S(t)$ numerically using the Euler method. If you set the force to zero, then you have zero acceleration and the unique solution in closed form is $ S(t) = S_0 + v_0 t $ In this case the numerical approximation via the Euler method will coincide with the solution obtained in closed form. But in general there will be an approximation error.

HTH

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    thats what i was looking for :) now i makes sense it's fuxxing physic.. i got the quarter information about this equation and couldn't go on... So the Euler integration returns the location/place to the given time...2011-08-02