Predicting Change in multiple dimensions

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Claimed by rbose7 Fall 2015

Edited by Richard Udall Summer 2019

Here we analyze how to apply the same iterative methods we have hitherto used in one dimension to predict motion in multiple dimensions. In essence the situation may be broken down into three one dimensional problems, but when we are dealing with varying forces, it is necessary to generalize the probability that motion in one dimension will alter the forces in another dimension.

The Main Idea

Just as in one dimension, the linear (also known as translational) motion of an object is the vector quantity equal to the product of the mass and velocity of an object. Unlike in one dimension, we must now consider the vectors components in each direction (see 3-Dimensional Position and Motion for more detail). By utilizing Newton's second law in each dimension, it is possible to predict the motion of the object in each dimension, making use of the same modelling techniques we have applied before.[1]

A Mathematical Model

The change in momentum due to a force is expressed in mathematical terms as:

[math]\displaystyle{ \Delta \vec{p} = \vec{p}_{f}-\vec{p}_{i} = m\vec{v}_{f}-m\vec{v}_{i} }[/math]

Or by relating it to force:

[math]\displaystyle{ \Delta \vec{p} = \vec{F} \Delta t }[/math]

Relation to Velocity

Given an object with velocity [math]\displaystyle{ \vec{v} = (v_x,v_y,v_z) }[/math] and mass [math]\displaystyle{ m }[/math], the object's momentum will be

[math]\displaystyle{ \vec{p} = m\vec{v} = m(v_x,v_y,v_z) = (m v_x,m v_y,m v_z) }[/math]

Relation to Force

Given the force [math]\displaystyle{ \vec{F} = (F_x,F_y,F_z) }[/math], and change in time [math]\displaystyle{ \Delta t }[/math], we may express the change in momentum as

[math]\displaystyle{ \Delta p = \vec{F} \Delta t = (F_x,F_y,F_z) \Delta t= ( F_x \Delta t,F_y \Delta t ,F_z \Delta t) }[/math]

[math]\displaystyle{ \vec{p}_f = \vec{p}_{i} + \Delta p = \vec{p}_i + (F_x \Delta t,F_y \Delta t,F_z \Delta t) }[/math]

This can also be expressed as:

[math]\displaystyle{ \vec{F}_{net} = \frac {\text{d}\vec{p}} {\text{d}t} }[/math]


[math]\displaystyle{ \Delta p = \int_{t_1}^{t_2} F(t)\, dt\,. }[/math]

A Computational Model

The principles of iterative prediction in multiple dimensions are exactly the same as the Fundamentals of Iterative Prediction with Varying Force, except that as per the above the process must be performed with 3 times as much information. The most important difference is that the force in one direction may depend on the position and velocity in another - Two Dimensional Harmonic Motion will consider such cases - which requires the system to update each coordinate simultaneously, rather than treating it like three separate one dimensional problems.

A Generic 3D Simulator

This is an example of a generic three dimensional motion simulator constructed using numpy.

Specific Examples Using VPython

This is an example of an object moving at a constant velocity in multiple dimensions, with no net force acting upon it. This is an example of object being acted upon by the force of gravity. Because gravity acts in the 'y' direction, the object's y component for velocity decreases. This is a simulation of an object launched from a cliff. It accelerates in the -y direction due to gravity until it hits the ground.



A ball of mass [math]\displaystyle{ 1 \; kg }[/math] has initial velocity [math]\displaystyle{ (3,4,0) \; m/s }[/math] in a pool of water, where the z-axis is the axis of gravity. The ball has exactly the density of water, so it buoyancy exactly cancels out gravity, and it only feels the force of drag due to the water. It feels linear drag, described by the equation [math]\displaystyle{ \vec{F}_d = -b\vec{v} }[/math], with [math]\displaystyle{ b = 0.1 \; kg/s }[/math]. Find the vector form of the drag force on the ball, and state its magnitude.


All we have to do is plug our values into the given equation, but we do need to get the vector right:

[math]\displaystyle{ \vec{F}_d = -b \vec{v} \rightarrow (F_x,F_y,F_z) = -b(v_x,v_y,v_z) }[/math]

The scalar term (the coefficient) will multiply out through the vector, so it results in

[math]\displaystyle{ (F_x,F_y,F_z) = -(0.1 \; kg/s)\cdot(3,4,0) \; m/s = (-0.3,-0.4,0) \; N }[/math]

Now, to find the magnitude we take a pythagorean sum:

[math]\displaystyle{ |\vec{F}_d| = \sqrt{F_x^2 +F_y^2+F_z^2} = \sqrt{0.25 \; N^2} = 0.5 \; N }[/math]

Note that the magnitude is strictly positive, and that the negatives are therefore coming from the direction of the vector. Although it's not asked for, note as well that it may be useful to determine the angle of the force with the x-axis, which we do by computing

[math]\displaystyle{ \theta = \arctan\biggr{(} \frac{F_y}{F_x}\biggr{)} = 53.13^\circ = 0.927 \; \text{rad} }[/math]


In the same situation as described above, compute the position (assuming the ball starts at the origin) after two time steps of [math]\displaystyle{ 0.1 \; s }[/math].


We have the first iterations force already worked out, so we compute the change in momentum:

[math]\displaystyle{ \Delta \vec{p} = \vec{F}_d \Delta t }[/math]

[math]\displaystyle{ \Delta \vec{p} = ((-0.3,-0.4,0) \; N)\cdot (0.1 \; s) = (-0.03,-0.04,0) \; \frac{kg\cdot m}{s} }[/math]

The adding this to our initial momentum gives

[math]\displaystyle{ \vec{p}_1 = (2.97,3.96,0) \frac{kg \cdot m}{s} }[/math]

[math]\displaystyle{ \vec{v}_1 = (2.97,3.96,0) \frac{m}{s} }[/math]

Then we compute the average velocity over the time step

[math]\displaystyle{ \vec{v}_{01} = ((2.97,3.96,0)+(3,4,0))/2 \; \frac{m}{s} = (2.985,3.98,0) \; \frac{m}{s} }[/math]

And so find displacement

[math]\displaystyle{ \vec{r}_1 = ((2.985,3.98,0)\; \frac{m}{s}) (0.1 \; s) = (0.2985,0.398,0) \; m }[/math]

Now we repeat the entire process:

[math]\displaystyle{ \vec{F}_d = -(0.1 \; \frac{kg}{s})((2.97,3.96,0) \; \frac{m}{s}) = (-0.297,-0.396,0) \; N }[/math]

[math]\displaystyle{ \Delta \vec{p} = ((-0.297,-0.396,0) \; N)\cdot(0.1 \; s) = (-0.0297,-0.0396,0) \frac{kg\cdot m}{s} }[/math]

[math]\displaystyle{ \vec{p}_2 = (2.9403,3.9204,0) \; \frac{kg\cdot m}{s} }[/math]

[math]\displaystyle{ \vec{v}_2 = (2.9403,3.9204,0) \; \frac{m}{s} }[/math]

[math]\displaystyle{ \vec{v}_{12} = (2.95515,3.9402,0) \; \frac{m}{s} }[/math]

[math]\displaystyle{ \vec{r}_2 = (0.594015,0.79202,0) \; m }[/math]

This is our final answer.


In the situation given above, approximate the time it takes the ball to come to a rest (note that in this ideal case it will never fully stop, but Xeno's paradox is not our problem) and its position once it has stopped. It is recommended to do this with an iterative prediction program, and use a time step of width [math]\displaystyle{ 0.01 \; s }[/math] or smaller. Now, set [math]\displaystyle{ b = 0.2 \; \frac{kg}{s} }[/math] and give the new results. Finally, set the mass to [math]\displaystyle{ m = 0.3 \; kg }[/math] (keeping [math]\displaystyle{ b = 0.2 \frac{kg}{s} }[/math]), and give these results. Deduce the formula for final position from these relationships.


I used the program linked in the Computational Method section above. Plugging in the force equation and other parameters, we get a result for convergence time of [math]\displaystyle{ \Delta t \approx 40 s }[/math], and a final position [math]\displaystyle{ \vec{r}_f = (30,40,0) \; m }[/math]. Doubling the value of [math]\displaystyle{ b }[/math] gives a time of convergence of [math]\displaystyle{ \Delta t \approx 20 \; s }[/math] and [math]\displaystyle{ \vec{r}_f = (15,20,0) \; m }[/math]. Tripling the mass, we now have a convergence time of [math]\displaystyle{ \Delta t \approx 60 \; s }[/math] and [math]\displaystyle{ \vec{r}_f = (45,60,0) \; m }[/math]. The formula is therefore fairly simple: [math]\displaystyle{ \vec{r}_f = \frac{v_0\cdot m}{b} }[/math]. This result may be found analytically from a straightforward application of the separation of variables technique, sketched out in 1 dimension (since there is no dimensional dependence, it extends trivially to multiple dimensions) below:

[math]\displaystyle{ m\frac{\text{d}v}{\text{d}t} = -bv \rightarrow \ln (v) = -\frac{bt}{m} + C }[/math]

[math]\displaystyle{ v = A \exp(-\frac{bt}{m}) }[/math]

[math]\displaystyle{ v(0) = v_0 = A }[/math]

[math]\displaystyle{ x = \int_0^t v(t') \text{d} t' = -\frac{v_0 m}{b} (\exp(\frac{bt}{m}) - 1) }[/math]

For large [math]\displaystyle{ t }[/math], the above solution has the exponential goint to zero, cancelling out the negative and giving the relationship we found.


[This should be extended by a student following the Template]

For my major, computer science, there are a few applications. For example, every computational model in the section above is coded using Python. A big part of computer science is modeling real life examples on a computer to mimic situations that would be otherwise impractical or impossible to set up. For example, the model we created of the electron and proton moving would take several expensive instruments to set up and visualize the two particles. For several other instances, using the principle of momentum and analyzing the change in momentum we can model several situations using coding that would otherwise not be possible.


The first school of thought in dynamics was the Aristotelian school, which was completely incorrect, in that it claimed that motion had to be continuously sustained by a force being imparted, leading to claims such as an arrow being driven by the air.[2] The first to move away from this was John Philoponus, who proposed the idea of impetus, which held that objects were imparted with some essence of motion, which is then depleted over the course of the objects motion. This is substantially closer to reality, but still not accurate. Galileo noted from experiment the existence of inertia, and the separability of dimensions. [3] Descartes stated what we now view as momentum more explicitly, constructing it based on the principles of Cartesian metaphysics [4], and in doing so critiqued Galileo's experimentalism (although the balance between theory and experiment is still a matter of discussion today, modern physics is constructed primarily with Galileo's experimental method rather than Descartes' metaphysical justifications). As with the corresponding developments in mathematics (in which Descartes' coordinate geometry laid the basis for calculus), momentum as a concept, and the methods of analyzing motion, were developed in greater detail by Newton in the Principia Mathematica. [5]

See also

External Links

Further Reading

  • Matter and Interactions, 4th Edition