It’s been fifteen years since I had math in school, and even then trigonometry was barely just touched on. Every time I try to do game development though, I’m hitting a brick wall because people throw around concepts like tangents, radians and square roots of things like it’s nothing and I completely forgot about all of that. Wikipedia isn’t really that helpful because it explains the concepts from a math perspective which means it’s dry, complicated, and doesn’t answer the question “What would I use this mathematical concept for?”. So for my own sake, I’m blogging about a game problem and how to solve it with math in a way that I can understand.

A note on Vectors: I use “vector” for a collection of 2 points (since I’m dealing with 2D here). In Mathematics, vector has very well defined meanings, but here it’s just a X and Y value and a vector might both define a Point in space or an adjustment that’s done over time.

**Movement from A to B over time**

Possibly the earliest (and easiest) problem is movement of something from A to B over time. I know that my object can move at 3 units per second (this is something I made up for the game, you’ll have to figure out what speed you use in whatever game and coordinate system you use). I know where it is right now (A at 4,4) and where I want it to eventually be (B at 9,6), but how do I get there?

What we need is a movement vector, that is a set of X/Y values to add to A over time. What does this mean? Let’s say that our movement vector is 0.6,0.4 just for the sake of the example. That means that after 1 second, A should be at 4.6,4.4. Then, a second later it would be at 5.2,4.8 and so on. The movement vector is called **velocity** because it doesn’t describe a point in space but a magnitude of which a point gets shifted over time.

How do we find out the velocity? First, we need to know the difference between points A and B so that we know what we have to cross. Let’s call this the difference vector, and it’s calculated by B.X-A.X,B.Y-A.Y or in the example, 9-4,6-4 which results in 5,2. That’s neat, but that’s the complete distance I have to cross it doesn’t tell me how much to move over time.

Here’s the point I struggled with for a long time: **Normalized Vectors**. People call this the **unit vector** usually without going to deep into it. The textbook definition: A unit vector is a vector of length 1. What does that mean? For a long time, I thought that means that X + Y have to equal 1. The difference vector above seems to be 5+2 = 7. Why do I need to have a vector of length 1 and how to I turn my vector of 7 into a vector of 1? Just subtract X/Y by 7 to get 0.71,0.29?

First off, the length of the difference vector is **not** 7. This was a mistake I made for a long time. Rather, the length of the vector is the hypotenuse of a right triangle with origin 0,0 and points at the X and Y coordinates. Let’s just quickly go over the parts of a right triangle (image courtesy of Wikipedia):

The Hypotenuse *c* is the longest side and is the one opposing the right angle. Side *a* is the *adjacent* while side *b* is the *opposite*. Angle B seems to be the source of the naming since side *b* sits adjacent to it while side *a* is on the opposite.

Point A is 0,0. Point B is my difference vector, 5,2. Point C is chosen by us to sit at 5,0 so that we get a right triangle like this:

This is where the pythagorean theorem comes in, which defines that a^{2} + b^{2} = c^{2}. We want *c*, and we have *a* (2) and *b* (5). Let us calculate *c*^{2}:

- a
^{2} = 5 * 5 = 25
- b
^{2} = 2 * 2 = 4
- c
^{2} = 25 + 4 = 29

Getting *c* is now simply the square root of *c*^{2}, which is 5.385164807134504 or 5.39 for short.

The **length** of the vector 5,2 is 5.39, not 7. Because it takes 5.39 units to cross the distance from 0,0 to 5,2 or from 4,4 to 9,6. Because if the length were 7 it means you would first walk the X-Axis for 5 units and then the Y-Axis for 2 Units, when you could just cross diagonally.

Quick recap:

- Source (A): 4,4
- Destination (B): 9,6
- Desired Speed: 3 units per second
- Difference: 5,2
- Length of Difference: 5.39
- Velocity: ???

So how do we get the velocity and what does this stuff have to do with normalized vectors? As said, a normalized vector is a vector of length 1, and we can get the normalized difference vector by dividing both 5 and 2 by 5.39 which gets us a vector of 0.93,0.37 (which, ignoring a rounding error for the sake of the blog post, has a length of 1 = sqrt(0.93^{2} + 0.37^{2})). We can now multiply this with the desired speed – 3 – to get the Velocity: 2.79,1.11 per second.

- Source (A): 4,4
- Destination (B): 9,6
- Desired Speed: 3 units per second
- Difference: 5,2
- Length of Difference: 5.39
- Normalized Difference: 0.93,0.37
- Velocity (=Change applied to A every second): 2.79,1.11

This would then be multiplied with the frame delta time to get the amount of movement in a frame (so in a 60 fps scenario you would multiply it by 1/60 = 0.016666…).

I still haven’t explained the use of a normalized vector. I think that Gamedev.net has done a good job explaining it, but in my own understanding: The difference vector has a magnitude of it’s own – with a length of 5.39 it would mean that my object would move 5.39 units per second. So normalizing a vector to me is an application of the rule of three. How do I get from 5.39 units per second to the desired 3 units per second? By going down to 1 unit per second:

- 5.39 = 5,2
- 1 = 5/5.39 , 2/5.39 = 0.93 , 0.37
- 3 = 3*0.93 , 3*0.37 = 2.79 , 1,11