Implicit vs Parametric vs Explicit Surfaces

Implicit Surface

It’s always R = 0 where R is a function of one or more variables.

Like the unit circle equation:
x^2 + y^2 -1 = 0.

Parametric Surface

The components of the output are based on some parameter or parameters

Like the quadratic bezier curve (which A,B,C and CurvePoint are points in N dimensions):
CurvePoint = f(t) = A*(1-t)^2 + B*2t(1-t) + C*t^2

Or the unit circle:
x = cos(t)
y = sin(t)

Or surfaces like this:
SurfacePoint3D = f(u,v)

Explicit Surface

The more usual looking type functions where you have one variable on the left side (dependent variable), and another variable on the right side (independent variable).

Like lines:
y = mx + b

or height fields:
height = f(x,y)

More Info

Here’s a cool set of slides that explain this stuff in more detail (and beyond), and the pros and cons of using various forms.

Representing Smooth Surfaces

Bezier Surface Properties

Here’s a couple pretty cool properties of Bezier surfaces that I learned recently.

The first one is that if you consider a “convex hull” being made up of the control points (connect all the control points into a convex shape), the curve will lie entirely inside that shape. That means you can use the shape of the control points as a “quick test” for rendering or collision detection. Note though, you could also just make a sphere that enclosed all the control points and do a sphere test instead, if you would rather have a simpler/quicker test at the cost of some wasted space (more false positives).

The second interesting property is that you can do back face culling of a Bezier surface if all the control points face away from the camera. while it’s true this isn’t EXACTLY proper back face culling, the odds are good it’s good enough for your needs, especially given how quick a test it is.

The third interesting property is that if you want to transform a bezier surface with something like a translation, rotation, or scale, you can apply the transform to the control points, and the curve will be transformed by the same transformation.”A Bézier surface will transform in the same way as its control points under all linear transformations and translations.” (from Wikipedia: Bezier Surface)

… but unfortunately, as promising as these properties are, it still seems infeasible to render a decent number of bezier surfaces via real time raytracing (something i was planning on) and it seems to only get worse when moving to b-splines and nurbs surfaces, so it seems like this may not be the way to go. It’s still possible though that raymarching these surfaces could be doable, but I haven’t explored too much in that direction yet.

Bezier Curves

Bezier curves are pretty cool. They were invented in the 1950s by Pierre Bezier while he was working at the car company Renault. He created them as a succinct way of describing curves mathematically that could be shared easily with other people, or programmed into machines to make curves that matched the ones created by human designers.

I’m only going to go over bezier curves at the very high level, and give some links to html5 demos I’ve made to let you play around with them and understand how they work, so you too can implement them easily in your own software.

If you want more detailed information, I strongly recommend this book: Focus on Curves and Surfaces

Quadratic Bezier Curves

Quadratic bezier curves have 3 control points. The first control point is where the curve begins, the second control point is a true control point to influence the curve, and the third control point is where the curve ends. Click the image below to be taken to my quadratic bezier curve demo.


A quadratic bezier curve has the following parameters:

  • t – the “time” parameter, this parameter goes from 0 to 1 to get the points of the curve.
  • A – the first control point, which is also where the curve begins.
  • B – the second control point.
  • C – the third control point, which is also where the curve ends.

To calculate a point on the curve given those parameters, you just sum up the result of these 3 functions:

  1. A * (1-t)^2
  2. B * 2t(1-t)
  3. C * t^2

In otherwords, the equation looks like this:

CurvePoint = A*(1-t)^2 + B*2t(1-t) + C*t^2

To make an entire curve, you would start with t=0 to get the starting point, t=1 to get the end point, and a bunch of values in between to get the points on the curve itself.

Cubic Bezier Curves

Cubic bezier curves have 4 control points. The first control point is where the curve begins, the second and third control points are true control point to influence the curve, and the fourth control point is where the curve ends. Click the image below to be taken to my cubic bezier curve demo.


A cubic bezier curve has the following parameters:

  • t – the “time” parameter, this parameter goes from 0 to 1 to get the points of the curve.
  • A – the first control point, which is also where the curve begins.
  • B – the second control point.
  • C – the second control point.
  • D – the fourth control point, which is also where the curve ends.

To calculate a point on the curve given those parameters, you just sum up the result of these 4 functions:

  1. A * (1-t)^3
  2. B * 3t(1-t)^2
  3. C * 3t^2(1-t)
  4. D * t^3

In otherwords, the equation looks like this:

CurvePoint = A*(1-t)^3 + B*3t(1-t)^2 + C*3t^2(1-t) + D*t^3


You might think the math behind these curves has to be pretty complex and non intuitive but that is not the case at all – seriously! The curves are based entirely on linear interpolation.

Here are 2 ways you may have seen linear interpolation before.

  1. value = min + percent * (max – min)
  2. value = percent * max + (1 – percent) * min

We are going to use the 2nd form and replace “percent” with “t” but they have the same meaning.

Ok so considering quadratic bezier curves, we have 3 control points: A, B and C.

The formula for linearly interpolating between point A and B is this:
point = t * B + (1-t) * A

The formula for linearly interpolating between point B and C is this:
point = t * C + (1-t) * B

Now, here’s where the magic comes in. What’s the formula for interpolating between the AB formula and the BC formulas above? Well, let’s use the AB formula as min, and the BC formula as max. If you plug the formulas into the linear interpolation formula you get this:

point = t * (t * C + (1-t) * B) + (1-t) * (t * B + (1-t) * A)

if you expand that and simplify it you will end up with this equation:
point = A*(1-t)^2 + B*2t(1-t) + C*t^2

which as you may remember is the formula for a quadratic bezier curve. There you have it… a quadratic bezier curve is just a linear interpolation between 2 other linear interpolations.

Cubic bezier curves work in a similar way, there is just a 4th point to deal with.

Next Up

The demos above are in 2d, but you could easily move to 3d (or higher dimensions!) and use the same equations. Also, there are higher order bezier curves (more control points), but as you add control points, the computational complexity increases, so people usually stick to quadratic or cubic bezier curves, and just string them together. When you put curves end to end like that, they call it a spline.

Next up, be on the look out for posts and demos for b-splines and nurbs!

Soft Maximum vs Hard Maximum

The other day i stumbled on an interesting concept called a “Soft Maximum”.

If you think of the normal maximum, you might have something like this:

float maxValue = max(valueA, valueB);

if valueA and valueB come from functions, there’s usually going to be a sharp bend in the graph of the above where the maximum value changes from valueA to valueB or vice versa.

Sometimes, instead of a sharp bend, you would like a smooth transition between the two values – like when using this for graphics or advanced mathematics.

Here’s the formula for soft max:

double SoftMaximum(double x, double y)
	double maximum = max(x, y);
	double minimum = min(x, y);
	return maximum + log( 1.0 + exp(minimum - maximum) );

Here are 2 really interesting links on computing and using soft max:

Soft Maximum

How to Compute the Soft Maximum

Check out the images below for an example of when you might use this. This is from a shadertoy shader The Popular Shader. The first image is with using normal max, and the second image uses soft max.



Converting RGB to Grayscale

If you were converting an RGB pixel to grayscale, you might be like me and be tempted to just add the red, green and blue components together and divide by 3 to get the grayscale equivalent of the color.

That’s close, but not quite correct!

Red, green and blue are not equal brightness, so doing a straight average gives you biased results.

There’s a wikipedia page on this topic here, but the equation to use is below:
grayScale = red * 0.3f + green * 0.59f + blue * 0.11f;

Here are some sample images to show you the difference.



Weighted Average Equation:


You might be wondering “why the heck would i want to convert RGB to grayscale?”

Well… if you render a scene once, convert it to grayscale and shove it into the red channel, then render the scene again slightly offset to the side, convert that to grayscale and shove it into the blue channel, you can get some neat images like the below. Red/Blue 3d glasses required, click the images to view the full size versions (;



Transformation Matrix Basics

Here are some interesting tidbits of info that I’ve found useful for being able to think about matrix math in a more intuitive way. We start off with 2d matrix math but extend it to 3d at the end.

Why Use A Matrix?

You might ask why you might go through all the trouble of using a matrix for doing transformations like translation, rotation, scaling and shearing / skewing.

Why not just manually modify the points, putting them through equations to get the results. Well, there are two main reasons.

The first one is for performance. The function for rotating a point in 2d looks like this:

x’ = x * cos(theta) – y * sin(theta)
y’ = x * sin(theta) + y * cos(theta)

If you have 1000 points, that means you are calculating sin twice and cos twice for each point which is 4000 trig operations. If you are smart (or your compiler is!), you’ll only do sin and cos once for each point, but that’s still 2000 trig operations.

If you are super duper smart (or your compiler is…), you’ll notice that theta is the same for all 1000 points, and perhaps you’ll calculate sin(theta) and cos(theta) once ahead of time and use those values for each point.

That last step is basically what matrix math does for you. A 2d rotation matrix looks like the below:

[ cos(theta), sin(theta) ]
[-sin(theta), cos(theta) ]

That means that once you have calculated your rotation matrix, you don’t need to keep performing trig operations. You have your values and can use them over and over very cheaply.

This especially saves processor time when you combine multiple transformations together. If you needed to perform an operation that did some crazy combination of 2000 rotations, 2000 translations and 2000 scale adjustments, instead of needing to do those 6000 operations on each point, you can just calculate the final matrix (by combining the matrix of each of those 6000 operations into one matrix) and then use that single matrix to your hearts content.

Another reason why you might want to use matrices instead of doing transforms by hand is that it’s a lot simpler writing code that does general transformations instead of deciding after the fact “hey i want to add scaling now and i need to touch all my transformation related code to implement it”.

Using a matrix, you don’t have to know or care what the transform is, it will just do it and you can move on with your life.

Matrix Vector Multiplication Is Just Dot Products!

As a quick refresher, in 2d, a dot product is just 2 multiplies and an add (A.X * B.X + A.Y * B.Y) and in 3d, a dot product is just 3 multiplies and 2 adds (A.X * B.X + A.Y * B.Y + A.Z * B.Z).

A lot of modern hardware (both CPU and GPU) has varying amounts of support built in for vector and matrix math to make it faster, but matrix vs vector multiplies are really not that bad to begin with. In 2d, a matrix * vector operation is just 2 dot products! In 3d, it’s 3 dot products. Look at the below to see what I mean:

VX’ = AX * VX + BX * VY
VY’ = AY * VX + BY * VY

and in 3d:

VX’ = AX * VX + BX * VY + CX * VZ
VY’ = AY * VX + BY * VY + CY * VZ
VZ’ = AZ * VX + BZ * VY + CZ * VZ

You Can Make a Matrix From Basis Vectors

Let’s say that you are working in 2d and you want to rotate a point. Let’s say that for some reason you know what the rotated X and Y axis are supposed to be. You can actually create a rotation matrix from that knowledge alone without having to do any trig or geometry type math.

Like for instance, if you wanted an object’s x axis to point parallel to the vector of a moving object, and you knew the object’s normalized velocity vector (the direction it is moving, normalized to have a vector length of 1) was (0.34, 0.93).

That normalized velocity vector would be your X axis, and you could use the “faked 2d cross product” of flipping x and y and negating one to get a vector perpendicular to that X axis (aka your Y axis) . So…

X axis = ( 0.34, 0.93)
Y axis = (-0.93, 0.34)

(note, which one you negate on the Y axis matters only as much as you care which of the 2 directions you want the Y axis to point. Basically if you don’t care if it points up or down, it doesn’t matter which one you flip. If you do care, you need to pick the right one for the direction you want. This may not be the best way, but i do it by visual inspection, or by evaluating the math and seeing if it’s pointing in the way that i want or not. In other words… try it one way, and if it’s backwards, do it the other way.)

Now that we have an X and a Y axis, we just use the X axis as the first row, and the Y axis as the second row and get our rotation matrix:

[ 0.34, 0.93]
[-0.93, 0.34]

To see that it really works, try multiplying the vector (1,0) by that matrix to see if we get the right number out (it should be the same vector as the velocity of the object we are orienting to). We are basically verifying here that our X axis comes out to what it should.

[1 0]
[ 0.34, 0.93]
[-0.93, 0.34]
X’ = 0.34 * 1 – 0.93 * 0 = 0.34
Y’ = 0.93 * 1 + 0.34 * 0 = 0.93

now, let’s check our y axis

[0 1]
[ 0.34, 0.93]
[-0.93, 0.34]
X’ = 0.34 * 0 – 0.93 * 1 = -0.93
Y’ = 0.93 * 0 + 0.34 * 1 = 0.34

Note that when you put your X and Y axis basis vectors into the matrix, that they should be normalized, otherwise they will do strange things to your point – like introduce scaling and skewing.

You Can Get Basis Vectors From a Matrix

The process works backwards too which is really handy. If you have some matrix of unknown rotation, you can get the basis vectors out the same way you put them in.

You might have already seen this when looking at the matrix from the last example

[ 0.34, 0.93]
[-0.93, 0.34]

The first row (0.34, 0.93) is the X axis, and the second row (-0.93, 0.34) is the Y axis.

One caveat to be aware of though is that if you are working with a “matrix from the wild” where you don’t know if it’s a rotation matrix only, or if it might have some other transforms in it (scaling or skewing), the rows may not be normalized.

If you know for sure that it’s just a rotation matrix, you can take the basis vectors right out of the matrix. If you don’t know for sure, you need to normalize the basis vectors after you pull them out.

Why is this useful?

If in 3d, you had the matrix representing the camera transform, you could grab the 3rd row to get the forward vector. You could use this vector when launching a projectile from the player’s position so that it would go where they were aiming.

Again, in 3d if you had the camera matrix, you could grab the first row to get the “left vector” and you could add or subtract that from the player’s position to do strafing left and right.

No complex math required (:

2d Translation Matrix

To be able to have a matrix that can do translation, you need to go to a 3×3 matrix.

Below is what a translation matrix looks like. (TX,TY) is the translation.

[1 0 0]
[0 1 0]
[TX TY 1]

When you want to transform a 2d point by a 3×3 matrix like the above, you need to use a 1 for the Z component. Let’s see what happens when we transform a 2d point by this 3×3 translation matrix.

[X Y 1]
[1 0 0]
[0 1 0]
[TX TY 1]
X’ = X * 1 + Y * 0 + 1 * TX
Y’ = Y * 0 + Y * 1 + 1 * TY
Z’ = 1 * 0 + 1 * 0 + 1 * 1
X’ = X + TX
Y’ = Y * TY
Z’ = 1

If you want to transform a 2d VECTOR (something that represents a direction, not a location) by a 3×3 matrix, you need to use a zero in the Z component instead of a 1. You may have heard this before, but let’s see why:

[X Y 0]
[1 0 0]
[0 1 0]
[TX TY 1]
X’ = X * 1 + Y * 0 + 0 * TX
Y’ = Y * 0 + Y * 1 + 0 * TY
Z’ = 1 * 0 + 1 * 0 + 0 * 1
X’ = X
Y’ = Y
Z’ = 0

As you can see, the vector was not affected by the translation of the matrix. If the 3×3 matrix had scaling and rotation in it as well as the translation, the vector WOULD be affected by those things as it should be. The translation is the only thing that doesn’t apply when you use a Z value of zero.

Note that if you want to get the 2d basis vectors from a 3×3 matrix, you just ignore the 3rd row and the 3rd column and do the same thing you would do with a 2×2 matrix. Again, making sure to normalize the basis vectors when it’s appropriate.

Combining Matrices

To combine transforms together (whether 2×2 or 3×3 matrices) you just multiply them together.

The order of matrix multiplication matters though. A * B is not the same as B * A.

To make things confusing, OpenGL and DirectX use different representations of matrices (one is “column major” the other is “row major”) which means that the matrices in each API are transposes of the other.

To make things even more confusing, if AT and BT are the transpose of A and B, then A * B = BT * BA. This means that premultiplication and postmultiplication (aka is A on the left or the right in A * B) swap meanings when going from one API to the other.

I’m not sure who is to blame for that one, but here’s an interesting link on the subject:

Note that row vs column major matrices also change which direction the basis vectors are stored in… so instead of X axis being the 3 top numbers, it would be the 3 left numbers! It should be easy enough to tell which is which by inspection, but keep it in mind!

Anyways, I’ll continue to use the conventions set in this article above in the vector / matrix multiplication.

If you multiply a translation matrix by a rotation matrix, you’ll get a matrix that rotates a point, and then translates it.

[1 0 3]
[0 1 5]
[0 0 1]
[ 0.34 0.93 0]
[-0.93 0.34 0]
[ 0 0 1]
[ 0.34 0.93 3]
[-0.93 0.34 5]
[ 0 0 1]

If, however, you multiply a rotation matrix by a translation matrix, you’ll get a matrix that translates a point, then rotates it. Going that direction, the translated point is rotated.

[ 0.34 0.93 0]
[-0.93 0.34 0]
[ 0 0 1]
[1 0 3]
[0 1 5]
[0 0 1]
[ 0.34 0.93 5]
[-0.93 0.34 1.4]
[ 0 0 1 ]

Which way you multiply entirely depends on what it is you are trying to achieve. And, well… it also depends on whether you are dealing with row major or column major matrices!

Multiplying a 3×3 matrix by another 3×3 matrix is the same as doing nine 3d dot products.

Inverting Matrices

Taking the transpose of a matrix doesn’t have any intuitive geometrical (or other) meaning that I’m aware of. I’ve looked on the net and all I could find was some “simple” explanations involving general relativity. Awesome right? LOL.

On the other hand, inverse matrices have a very intuitive and very useful meaning. Inverse matrices do the reverse of whatever the matrix does.

That means if you have a matrix that translates by (7,5) and then rotates by (45, 30) degrees, the inverse matrix will rotate by (-45, -30) degrees and then translate by (-7,-5).

This is super useful sometimes (:

Inverting a 2×2 matrix is actually really easy. I could explain it but you really ought to check out this page to see how. I recommend doing the exercises at the bottom to make sure you firmly understand how to do it!

Inverting a 3×3 matrix is fairly easy too, but kind of tedious. Here’s a page that explains how:

After you are done with that, here are some problems to run through to make sure you really do understand it:

Not all matrices are invertible. If you read the links and walk through the exercises, you’ll see why. Basically, an uninvertable matrix will cause a divide by zero in the inversion process. I believe this comes up when you have a matrix with parallel basis vectors, or if you have a zero scaling matrix (a multiply by zero can’t be reversed!) but i could be mistaken or not have the full picture there. If I’ve mis-spoken, post a comment!

Extending to 3d

Extending the above to 3d matrices is pretty simple and there aren’t really many suprises.

One difference is that in the above where we use a 2×2 matrix in 2d, we would use a 3×3 matrix in 3d because of the extra Z coordinate.

If you want to do translation in 3d, you have to use a 4v4 matrix instead of a 3×3 (just like in 2d how you had to move from a 2×2 to a 3v3, in 3d you have to move from a 3×3 to a 4×4). A 3d translation matrix looks like this:

[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
[TX TY TZ 1]

Inverting a 4×4 matrix is pretty tedious, but follows the same patterns as inverting a 3×3 and a 2×2.

You can find info on a 3d rotation matrix here, if you don’t want to build it up with basis vectors:

In our example above where we had the X axis and used a “fake 2d cross product” to get the perpendicular vector, when you move into 3d you’ll probably want to use the cross product to get perpendicular vectors.

Like for instance if you know 2 of the basis vectors, you can use cross product of those 2 to get the third.

Z = X x Y

If, however, you only have one basic vector (say “Z” because maybe you have a “camera direction”), you can use cross product to get 2 other vectors so long as you can make certain assumptions about the orientations involved. Like for instance, you might do this:

Fwd = normalize(Camera.Forward)
Left = normalize(Fwd x (0, 1, 0))
Up = normalize(Fwd x Left)

rotation matrix =
[Left.X Left.Y Left.Z]
[Up.X Up.Y Up.Z]
[Fwd.X Fwd.Y Fwd.Z]

The above only works if the camera can never look straight up, and it also assumes that your camera doesn’t have any roll – but it is a useful technique if those assumptions are ok.

That’s about it! I hope you found at least some of this information useful (:

if I missed anything you think belongs here, post a comment and share with the rest of us!

Converting To and From Polar / Spherical Coordinates Made Easy

As a game developer there is just too much darn stuff to learn. You could spend your entire life learning things and never know it all.

It’s a blessing in that you are seldom bored, but also sometimes a curse in that there almost always is a better way to do something, and that you would know about it if you had spent your time learning X instead of Y 😛

I find that you sort of have to triage what you learn and what you choose to keep fresh in your mind, which can be a challenge sometimes. If you can find the commonalities between things that can help some – like understanding how encryption, hashing, pseudo random number generators and chaos theory all overlap – or how skeletal animation blending and audio synthesis are both trying to be continuous waves above all else. Also, if you put in the time investment to learn something to where it becomes intuitive, that frees up neurons to make room for other stuff. But, of course, we have a finite amount of time, so can’t always spend the time needed to get to that level on every single topic.

How about you… do you have to do a juggling act like this to keep sharp and stay effective as a game (or non-game) programmer? I’d be interested to hear how others deal with this sort of thing with such a large knowledge space that we work in.

In any case, I usually work with spherical or polar coordinates only on rare occasions, so whenever i do, the process usually is to google the equations, drop them in, and move on with my life. I was recently implementing an orbit camera for a raytracer on (Raytraced Refraction) and when my copy/pasting wasn’t working, I was forced to take a deeper look at why it wasn’t working. Amazingly, this time around, it finally clicked and is now an intuitive thing so I figured I’d share the explanation that makes most sense to me in case it helps anyone else.

Converting Polar Coordinates to Cartesian (2D)

Polar coordinates have two components – a distance and an angle – and represent a point in 2d space.

The distance is called the radial coordinate, or the radius and the angle is called the angular coordinate or polar angle.

Just like you probably expect, the angle defines what direction the point is in, and the radius defines how far away it is. Super simple. Below is a picture of a polar coordinate point at (3, 45) where 3 is the distance and 45 is the angle.


So how do we convert that to rectangular coordinates? well, first thing to do is to convert the angle to rectangular coordinates on a unit circle to get a direction vector. Then, you multiply that direction vector by the radius to get the final coordinate.

To convert the angle to a point on a unit circle and get the direction vector it’s super simple…

X = cos(angle)
Y = sin(angle)

For every point on the unit circle, it’s X coordinate is the cosine of the angle, and it’s Y coordinate is the sine of the angle.

Looking at the diagram below, see if you can figure out why arccosine only returns an angle between 0 and 180, and why arcsine only returns an angle between -90 and 90 (hint, what if i asked you to tell me what angle gives 0.7 in the x component). Also see if you can understand why sin(x)^2 + cos(x)^2 = 1 (hint: distance formula).


Ok so now that we can get our direction vector, we just need to multiply it by the radius. So… to convert from polar to rectangular (cartesian) coordinates, you do this:

X = cos(angle) * radius
Y = sin(angle) * radius

Converting Cartesian to Polar Coordinates (2D)

So how do we convert from rectangular coordinates to polar?

Well, since we have the X and the Y coordinates, and we know that tangent(angle) = Y / X, we can use arctangent to get our angle. Unfortunately atan has a similar problem to asin and acos in that it doesn’t know which quadrant you are talking about. For instance, look at the diagram above again and tell me “which angle gives me a value of 1 when i divide the Y component by the X component?”. The answer is 45 degrees and 225 degrees both. This is because with a positive value of 1, we don’t know if X and Y were both negative, or if they were both positive. Similarly, if i asked which angle gave an answer of -1, we wouldn’t know if it was the X or Y that was negative.

Instead of using atan, we want to use atan2, which takes 2 parameters – Y and X – so that it can figure out the correct angle for you.

Next is the easy part of finding the radius. treating your point as a vector (or continuing to treat it like a vector if it IS a vector), the radius is just the magnitude of the vector (or distance from the origin if you want to think of it in “point” terms instead of vectors).

So, converting rectangular to polar coordinates is done like this:

radius = sqrt(X * X + Y * Y)
angle = atan2(Y, X)

Converting Spherical Coordinates to Cartesian (3D)

Spherical coordinates have the same components as polar coordinates, but then an added component: an angle which determines pitch / vertical rotation (think: looking up and looking down, instead of the polar angle which is in charge of looking left and right).

In math, they usually call the radius rho, the polar angle theta, and the azimuth angle phi, so a formal polar coordinate looks like this:

(rho, theta, phi)

For our examples let’s assume that X and Y make up the horizontal plane and that Z is the vertical (3d) axis.

If you are scared to make the jump from 2D polar coordinates to 3D spherical coordinates don’t be! The way to deal with these guys is to break the 3d problem into two 2d problems, using the exact same stuff as described above.

So, the first thing we want to do is completely ignore this new 3rd component phi and think back to our 2d case. We are also going to ignore the radius for now.

XTheta = cos(theta)
YTheta = sin(theta)

This is our direction vector on the horizontal plane (same as the 2d case, not accounting for radius yet).

Next we want to pretend like we are looking at our 3d world from the side and use our phi angle to convert from polar to rectangular coordinates:

XPhi = cos(phi)
YPhi = sin(phi)

One way to think of what this other angle phi means, is that it is controlling where in the unit sphere the theta circle sits. As the theta circle gets higher or lower on the sphere, it shrinks or grows. It’s only at zero angle that it has a radius of 1.0. So, calculating these values, YPhi represents how high on the sphere the theta circle should sit, and XPhi is how large the circle should be.

So, to combine the X,Y theta values and the X,Y Phi values, we use YPhi as the vertical component, and XPhi as a radius for the theta circle, which we can do like this:

X = XTheta * XPhi
Y = YTheta * XPhi
Z = YPhi

The above equation will give us a point on a unit sphere, so from here, we need to multiply in the radius and our equation becomes:

X = XTheta * XPhi * radius
Y = YTheta * XPhi * radius
Z = YPhi * radius

If we put sin and cos back in, instead of xtheta (etc), the equation becomes that familiar, and previously complex equation:

X = cos(theta) * cos(phi) * radius
Y = sin(theta) * cos(phi) * radius
Z = sin(phi) * radius

Hopefully the equation makes more sense now, and hopefully you can look at that and intuitively understand why those values are what they are.

Converting Cartesian to Spherical Coordinates (3D)

To convert from spherical coordinates to rectangular, the first thing to do is to get the radius, which is done in the exact same way as in 2d. We just take the magnitude of the vector (aka the distance of the point from the origion) and we are done.

To get theta and phi, we do the same thing of separating this 3d problem into two 2d problems.

In fact, to get theta, we do the exact same thing as we do for polar coordinates! We use atan2(Y,X) to get our angle… THAT’S ALL!

So, we have this so far:

radius = sqrt(X * X + Y * Y + Z * Z)
theta = atan2(Y, X)

How do we figure out phi? Well, if you said that we should do atan2(Z,Y) or atan2(Z,X) you were pretty close but it’s actually arccos(Z / radius).

The reason for this is because neither X, nor Y is the “X” component of the “phi 2d polar coordinate”. you’d have to take the length of the (X,Y) vector and use that if you wanted to use atan2 to calculate phi. Instead of calculating that vector length, we can instead use a value we already have. cosine is Y / hypotenuse length, and hypotenuse length is the radius (length of our vector), so we might as well use that radius we already have to be able to use arccos.

The final equations for converting rectangular to spherical are:

radius = sqrt(X * X + Y * Y + Z * Z)
theta = atan2(Y, X)
phi = acos(Z / radius)

More info / alternate forms available on wikipedia here:
From Cartesian to Spherical Coordinates
Spherical Coordinate System