# Integrating Newton's equations of motion

1. Jul 19, 2008

### graffy76

Hello,

I have a little project I'm playing with that involves calculating a series of forces and summing them to define the motion of an object (in this case, a walking pedestrian).

The force equation is modeled on time-dependent vectors and scalars and is solved using Gear's predictor-corrector numerical method.

The Gear predictor portion of the method is described thus:

x[n+1] = x + hv + ((h^2/2)a + ((h^3)/6)A + ((h^4)/24)B

with v[n+1], a[n+1], A[n+1], and B[n+1] defined as the first through fourth derivatives of the above.

The force equation (= dv/dt) has time-dependent displacement and velocity vectors and scalars, but is not explicitly defined in terms of time. So, I obviously need to express the force equation in terms of time (or in this case, the time step, h).

My only thought would be to re-write the displacement and velocity components explicitly in terms of Newton's equations for curvlinear motion.

Thus:

if F(t) = v * x(t),

then F(t) = v * (x[n] + v[n]h + 1/2*a[n](t^2) ).

Such a rewriting would allow two derivatives of F(t) for the Gear model. By doing this, I think I'm just assuming that the motion of the pedestrians in time step h ( = 0.1 seconds in this case) is curvlinear - the same assumption that is taken to establish the Gear equations in the first place.

In any case, I haven't dealt with this level of math or physics since college (going on ten years now), and I'm a bit rusty.

Am I on track, or is this the wrong way to go about getting the two extra derivatives I need?

http://public.rz.fh-wolfenbuettel.de/~apel/files/thesis.pdf" [Broken]

see pages 21 - 29. The force equation is summarized on page 26 and the Gear predictor portion is summarized on page 28.

Any help is much appreciated.

Thanks.

Joel

Last edited by a moderator: May 3, 2017
2. Jul 21, 2008

### Angelos K

For me the link you provide doesn't work. I have not understood the question, but I can imagine that some of the very skilled people the PF hosts could give you valuable advice, if they had the information you intended to equip them with.

Good luck,
Angelos

3. Jul 22, 2008

### graffy76

Ok,

I realize my original post was a bit obscure, and I've had a few thoughts since then, so let me restate it...

I have a force equation which is used to describe the motion of pedestrians. According to Newton's 2nd and assumunig unit mass,

F(t) = ma = a = dv\dt

Where:

F(t)=v * e(t) \ tau

Where:

V = ped velocity (fixed scalar)
Tau = time constant
e(t) = vector pointing in the ped direction of travel.

Solving this is done using Gear's predictor corrector to the fifth order, which requires a first and second derivative of F(t).

It's finding those derivatives that is giving me problems, but I think I may have worked it out.

The problem is, how do I convert a time dependent function from an instantaneous definition to a definition with respect to t?

Look at e(t);

e(t) = (p - x)\(||p-x||)

e(t) is a unit vector where:
p=goal coordinate of pedestrian
x=current coordinate of pedestrian

Given that I need first and second derivatives of F(t) w.r.t t, and the only time dependent variable in F(t) is e(t), then I need to differentiate e(t) w.r.t. t. Looking at the definition of e(t), it's obvious that the pedestrian current coordinate, x, is the only time dependent variable.

Thus it seems to me, I must redefine x as:

x(t) = x + v * t + (1\2) * a * t^2

(Where i is initial or previous value)

which assumes that the pedestrians path is always curvlinear.

Now substitute x(t) in for both occurences of x in e(t).

Having done that, I'm now left with a rather ugly little equation that I need to derive twice w.r.t. t to get the derivatives I need to implement a fifth-order Gear predictor-corrector solution.

That's about as far as I can take it. Does it make sense? Is there a better way to convert a force equation defined instantaneously as a function of time t?