Point between two points given time?

Question

Let's say there are two separate points in a 3 dimensional space. An object at point A can move to point B at any speed given (let 'S' be units per second). If I start moving the object from point A to point B at 'sT' (let 'sT' be the start time in milliseconds), and I want to get the point it would be at 'eT' (let 'eT' be the elapsed time in milliseconds) milliseconds later, how would I go about calculating it?

(Edit: Now I actually have the time it'll take for the object to point A to point B at 'S' speed. So now I just need to get the point it'll be at after some 'eT'.)

For example, here is an object at point A moving to point B:

[movement]1

If point A is at (1, 12, 4) (in X, Y, Z format), point B is at (3, 5, 5), the object is moving at 2 units per second, and it started moving at 12323 milliseconds; how can I get the position it would be at 17421 milliseconds?

Edit. I now have the following information to work with: start time (ms), time it will take for the entire movement (ms), speed (units/second), points A and B. I tried setting up an equation given the answer by Eli Rose, but it's failing. It is returning very off value, but I probably just did something wrong.

(Ps. Apparently I can't post images until I have 10 reputation, I hope leaving links to the image is allowed.)

Answer 1

Every second, your object moves $2$ units towards $(3, 5, 5)$. It moves some number of units $v_x$ in the $x$ direction, some number of units $v_y$ in the $y$ direction, etc. can be written as a list of components $(v_x, v_y, v_z)$.

If we can find $(v_x, v_y, v_z)$ then we can just multiply each $v$ by the difference in times $\Delta t$ (which is $5.098$ seconds in your case) to get the distance it will have moved in the $x$-direction, in the $y$-direction, etc.

Then the position it's at will be $(1 + v_x\Delta t, 12 + v_y\Delta t, 4 + v_z \Delta t)$.

So what are $(v_x, v_y, v_z)$ ? It might seem like they would be $(2, 2, 2)$ but that would actually make for an object that's traveling too fast. Think about it in two dimensions -- if you're traveling two units to the right and two units up every second, you're traveling faster than two units per second overall.

How much faster? The Pythagorean Theorem can tell us, since the distance you cover is the hypotenuse of a right triangle whose legs are length $2$ and $2$.

$$\sqrt{v_x^2 + v_y^2} = \sqrt{2^2 + 2^2} = \sqrt{8} \approx 2.8 \text{ units per second}$$

Suppose you made the whole journey in one second. Then your $(v_x, v_y, v_z)$ would equal $(3 - 1, 5 - 12, 5 - 4) = (2, -7, -1)$. But this is too fast. How fast?

$$\sqrt{2^2 + (-7)^2 + (-1)^2} = \sqrt{54} \text{ units per second}$$

If we divide every $v$ by $\sqrt{54}$

$$(\frac{2}{\sqrt{54}}, \frac{-7}{\sqrt{54}}, \frac{-1}{\sqrt{54}})$$

then we get what $(v_x, v_y, v_z)$ would be if you traveled at $1$ unit per second. But this is too slow for us -- we want twice that.

$ \begin{aligned} v_x &= \frac{4}{\sqrt{54}}\\ v_y &= \frac{-14}{\sqrt{54}}\\ v_z &= \frac{-2}{\sqrt{54}} \end{aligned} $

In math, this concept is called vectors. In particular you are trying to find the velocity vector of your object. What you get by dividing everything by $\sqrt{54}$ is called a unit vector. This is what all physics engines for games, animations, etc. are based on.