Four Ways to Calculate Sine Without Trig

Is it possible to sin without trig? That is a question that has plagued theologians for centuries. As evil as trigonometry is, modern science shows us that yes, it is possible to sin without trig. Here are some ways that I’ve come across.

1 – Slope Iteration Method

The above image uses 1024 samples from 0 to 2pi to aproximate sine iteratively using it’s slope. Red is true sin, green is the aproximation, and yellow is where they overlap.

This method comes from Game Programming Gems 8, which you can snag a copy of from amazon below if you are interested. It’s mentioned in chapter 6.1 A Practical DSP Radio Effect (which is a really cool read btw!).
Amazon: Game Programming Gems 8

This method uses calculus but is actually pretty straightforward and intuitive – and very surprising to me that it works so well!

The derivative of sin(x) is cos(x). That means, for any x you plug into sin, the slope of the function at that point is the cosine value of that same x.

In other words, sin(0) is 0, but it has a slope of cos(0) which is 1. Since slope is rise over run (change in y / change in x) that means that at sin(0), if you take an infinitely small step forward on x, you need to take the same sized step on y. That will get you to the next value of sine.

Let’s test that out!
sin(0) = 0 so we start at (0,0) on the graph.

If we try a step size of 0.1, our approximation is:
sin(0.1) = 0.1

The actual value according to google is 0.09983341664. so our error was about 0.0002. That is actually pretty close!

How about 0.25?
sin(0.25) = 0.25

The real value is 0.24740395925, so our error is about 0.003. We have about 10 times the error that we did at 0.1.

what if we try it with 0.5?
sin(0.5) = 0.5

The actual value is 0.4794255386, which leaves us with an error of about 0.02. Our error is 100 times as much as it was at 0.1. As you can see, the error increases as our step size gets larger.

If we wanted to, we could get the slope (cosine value) at the new x value and take another step. we could continue doing this to get our sine approximation, knowing that the smaller the step that we use, the more accurate our sine approximation would be.

We haven’t actually won anything at this point though, because we are just using cosine to take approximated steps through sine values. We are paying the cost of calculating cosine, so we might as well just calculate sine directly instead.

Well luckily, cosine has a similar relationship with sine; the derivative of cos(x) is -sin(x).

Now, we can use cosine values to step through sine values, and use those same sine values to step through cosine values.

Since we know that cos(0) = 1.0 and sin(0) = 0.0, we can start at an angle of 0 with those values and we can iteratively step through the values.

Here is a sample function in C++

// theta is the angle we want the sine value of.
// higher resolution means smaller step size AKA more accuracy but higher computational cost.
// I used a resolution value of 1024 in the image at the top of this section.
float SineApproximation (float theta, float resolution)
    // calculate the stepDelta for our angle.
    // resolution is the number of samples we calculate from 0 to 2pi radians
    const float TwoPi = 6.28318530718f;
    const float stepDelta = (TwoPi / resolution);

    // initialize our starting values
    float angle = 0.0;
    float vcos = 1.0;
    float vsin = 0.0;

    // while we are less than our desired angle
    while(angle < theta) {

        // calculate our step size on the y axis for our step size on the x axis.
        float vcosscaled = vcos * stepDelta;
        float vsinscaled = vsin * stepDelta;

        // take a step on the x axis
        angle += stepDelta;

        // take a step on the y axis
        vsin += vcosscaled;
        vcos -= vsinscaled;

    // return the value we calculated
    return vsin;

Note that the higher the angle you use, the more the error rate accumulates. One way to help this would be to make sure that theta was between 0 and 2pi, or you could even just calculate between 0 and pi/2 and mirror / reflect the values for the other quadrants.

This function is quite a bit of work to calculate a single value of sine but it’s real power comes in the fact that it’s iterative. If you ever find yourself in a situation where you need progressive values of sine, and have some fixed angle step size through the sine values, this algorithm just needs to do a couple multiplies and adds to get to the next value of sine.

One great use of this could be in DSP / audio synthesis, for sine wave generation. Another good use could be in efficiently evaluating trigonometry based splines (a future topic I plan to make a post about!).

You can see this in action in this shadertoy or look below at the screenshots:
Shadertoy: Sin without Trig

64 Samples – Red is true sine, green is our approximation, and yellow is where they are the same

128 Samples

256 Samples

1024 Samples

2 – Taylor Series Method

Another way to calculate sine is by using an infinite Taylor series. Thanks to my friend Yuval for showing me this method.

You can get the Taylor series for sine by typing “series sin(x)” into wolfram alpha. You can see that here: Wolfram Alpha: series sin(x).

Wolfram alpha says the series is: x-x^3/6+x^5/120-x^7/5040+x^9/362880-x^11/39916800 ….

what this means is that if you plug a value in for x, you will get an approximation of sine for that x value. It’s an infinite series, but you can do as few or as many terms as you want to be able to trade off speed for accuracy.

For instance check out these graphs.

Google: graph y = x, y = sin(x)

Google: graph y = x-x^3/6, y = sin(x)

Google: graph y = x-x^3/6+x^5/120, y = sin(x)

Google: graph y = x-x^3/6+x^5/120-x^7/5040, y = sin(x)

Google: graph y = x-x^3/6+x^5/120-x^7/5040+x^9/362880, y = sin(x)

Google: graph y = x-x^3/6+x^5/120-x^7/5040+x^9/362880-x^11/39916800, y = sin(x)
(Note that I had to zoom out a bit to show where it became inaccurate)

When looking at these graphs, you’ll probably notice that very early on, the approximation is pretty good between -Pi/2 and + Pi/2. I leveraged that by only using those values (with modulus) and mirroring them to be able to get a sine value of any angle with more accuracy.

When using just x-x^3/6, there was an obvious problem at the top and bottom of the sine waves:

When i boosted the equation with another term, bringing it to x-x^3/6+x^5/120, my sine approximation was much better:

You can see this method in action in this shadertoy:
Shadertoy: Sin without Trig II

3 – Smoothstep Method

The third method may be my favorite, due to it’s sheer elegance and simplicity. Thanks to P_Malin on for sharing this one with me.

There’s a function in graphics called “smoothstep” that is used to take the hard linear edge off of things, and give it a smoother, more organic feel. You can read more about it here: Wikipedia: Smoothstep.

BTW if you haven’t read the last post, I talk about how smooth step is really just a 1d bezier curve with specific control points (One Dimensional Bezier Curves). Also, smoothstep is just this function: y = (3x^2 – 2x^3).

Anyhow, if you have a triangle wave that has values from 0 to 1 on the y axis, and put it through a smoothstep, then scale it to -1 to +1 on the y axis, you get a pretty darn good sine approximation out.

Here is a step by step recipe:

Step 1 – Make a triangle wave that has values on the y axis from 0 to 1

Step 2 – Put that triangle wave through smoothstep (3x^2 – 2x^3)

Step 3 – Scale the result to having values from -1 to +1 on the axis.

That is pretty good isn’t it?

You can see this in action in this shadertoy (thanks to shadertoy’s Dave_Hoskins for some help with improving the code):
Shadertoy: Sin without Trig III

After I made that shadertoy, IQ, the creator of shadertoy who is an amazing programmer and an impressive “demoscene” guy, said that he experimented with removing the error from that technique to try to get a better sin/cos aproximation.

You can see that here: Shadertoy: Sincos approximation

Also, I recommend checking out IQ’s website. He has a lot of interesting topics on there: IQ’s Website

4 – CORDIC Mathematics

This fourth way is a method that cheaper calculators use to calculate trig functions, and other things as well.

I haven’t taken a whole lot of time to understand the details, but it looks like it’s using imaginary numbers to rotate vectors iteratively, doing a binary search across the search space to find the desired values.

The benefit of this technique is that it can be implemented with VERY little hardware support.

The number of logic gates for the implementation of a CORDIC is roughly comparable to the number required for a multiplier as both require combinations of shifts and additions.
Wikipedia: Coordinate Rotation Digital Computer

Did I miss any?

If you know of something that I’ve left out, post a comment, I’d love to hear about it!

One Dimensional Bezier Curves

I was recently looking at the formula for bezier curves:

Quadratic Bezier curve:
A * (1-T)^2 + B * 2 * (1-T) * T + C * T ^2

Cubic Bezier curve:

(more info available at Bezier Curves Part 2 (and Bezier Surfaces))

And I wondered… since you can have a Bezier curve in any dimension, what would happen if you made the control points (A,B,C,D) scalar? In other words, what would happen if you made bezier curves 1 dimensional?

Well it turns out something pretty interesting happens. If you replace T with x, you end up with f(x) functions, like the below:

Quadratic Bezier curve:
y = A * (1-x)^2 + B * 2 * (1-x) * x + C * x ^2

Cubic Bezier curve:
y = A*(1-x)^3+3*B*(1-x)^2*x+3*C*(1-x)*x^2+D*x^3

What makes that so interesting is that most math operations you may want to do on a bezier curve are a lot easier using y=f(x), instead of the parameterized formula Point = F(S,T).

For instance, try to calculate where a line segment intersects with a parameterized bezier curve, and then try it again with a quadratic equation. Or, try calculating the indefinite integral of the parameterized curve so that you can find the area underneath it (like, for Analytic Fog Density). Or, try to even find the given Y location that the curve has for a specific X value. These things are pretty difficult with a parameterized function, but a lot easier with the y=f(x) style function.

This ease of use does come at a price though. Specifically, you can’t move control points freely, you can only move them up and down and cannot move them left and right. If you are ok with that trade off, these 1 dimensional curves can be pretty powerful.

Below is an image of a 1 dimensional cubic bezier curve that has control points A = 0.5 B = 0.25 C = 0.75 D = 0.5. The function of this curve is y = 0.5 * (1-x)^3 + 0.25 * 3x(1-x)^2 + 0.75 * 3x^2(1-x) + 0.5 * x^3.

You can ask google to graph that for you to see that it is in fact the same: Google: graph y = 0.5 * (1-x)^3 + 0.25 * 3x(1-x)^2 + 0.75 * 3x^2(1-x) + 0.5 * x^3


Another benefit to these one dimensional bezier curves is that you could kind of use them as a “curve fitting” method. If you have some data that you wanted to approximate with a quadratic function, you could adjust the control points of a one dimensional quadratic Bezier curve to match your data set. If you need more control points to get a better aproximation of the data, you can just increase the degree of the bezier curve (check this out for more info on how to do that: Bezier Curves Part 2 (and Bezier Surfaces)).

Smoothstep as a Cubic 1d Bezier Curve

BIG THANKS to for telling me about this, this is pretty rad.

It’s kind of out of the scope of this post to talk about why smoothstep is awesome, but to give you strong evidence that it is, it’s a GLSL built in function. You may have also seen it used in the post I made about distance fields (Distance Field Textures), because one of it’s common uses is to make the edges of things look smoother. Here’s a wikipedia page on it as well if you want more info: Wikipedia: Smoothstep

Anyhow, I had no idea, but apparently the smoothstep equation is the same as if you take a 1d cubic bezier curve and make the first two control points 0, and the last two control points 1.

The equation for smoothstep is: y = 3*x^2 – 2*x^3

The equation for the bezier curve i mentioned is: y = 0*(1-x)^3+3*0*(1-x)^2*x+3*1*(1-x)*x^2+1*x^3

otherwise known as: y = 3*1*(1-x)*x^2+1*x^3

If you work it out, those are the same equations! Wolfram alpha can verify that for us even: Wolfram Alpha: does 3*x^2 – 2*x^3 = 3*1*(1-x)*x^2+1*x^3.

Kinda neat 😛

Moving Control Points on the X Axis

There’s a way you could fake moving control points on the X axis (left and right) if you really needed to. What I’m thinking is basically that you could scale X before you plug it into the equation.

For instance, if you moved the last control point to the left so that it was at 0.9 instead of 1.0, the space between the 3rd and 4th control point is now .23 instead of .33 on the x axis. You could simulate that by having some code like this:

if (x > 0.66)
  x = 0.66 + (x - 0.66) / 0.33 * 0.23

Basically, we are squishing the X values that are between 0.66 and 1.0 into 0.66 to 0.9. This is the x value we want to use, but we’d still plug the raw, unscaled x value into the function to get the y value for that x.

As another example, let’s say you moved the 3rd control point left from 0.66 to 0.5. In this situation, you would squish the X values that were between 0.33 and 0.66 into 0.33 to 0.5. HOWEVER, you would also need to EXPAND the values that were between 0.66 and 1.0 to be from 0.5 and 1.0. Since you only moved the 3rd control point left, you’d have to squish the section to the left of the control point, and expand the section to the right to make up the difference to keep the 4th control point in the same place. The code for converting X values is a little more complex, but not too bad.

What happens if you move the first control point left or right? Well, basically you have to expand or squish the first section, but you also need to add or subtract an offset for the x as well.

I’ll leave the last 2 conversions as an exercise for whoever might want to give this a shot 😛

Another complication that comes up with the above is, what if you tried to move the 3rd control point to the left, past the 2nd control point? Here are a couple ways I can think of off hand to deal with it, but there are probably other ways too, and the right way to deal with it depends on your needs.

  1. Don’t let them do it – If a control point tries to move past another control point, just prevent it from happening
  2. Switch the control points – If you move control point 3 to the left past control point 2, secretly have them start controling control point 2 as they drag it left. As far as the user is concerned, it’s doing what they want, but as far as we are concerned, the control points never actually crossed
  3. Move both – if you move control point 3 to the left past control point 2, take control point 2 along for the ride, keeping it to the left of control point 3

When allowing this fake x axis movement, it does complicate the math a bit, which might bite you depending on what you are doing with the curve. Intersecting a line segment with it for example is going to be more complex.

An alternative to this would be letting the control points move on the X axis by letting a user control a curve that controls the X axis location of the control points – hopefully this would happen behind the scenes and they would just move points in X & Y, not directly editing the curve that controls X position of control points. This is a step towards making the math simpler again, by modifying the bezier curve function, instead of requiring if statements and loops, but as far as all the possibly functions I can think of, moving one control point on the X axis is probably going to move other control points around. Also, it will probably change the shape of the graph. It might change in a reasonable way, or it might be totally unreasonable and not be a viable alternative. I’m not really sure, but maybe someday I’ll play around with it – or if you do, please post a comment back and let us know how it went for you!


Here are some links to experiment with these guys and see them in action:
HTML5 Interactive 1D Quadratic Bezier Demo
HTML5 Interactive 1D Cubic Bezier Demo
Shadertoy: Interactive 1D Quadratic Bezier Demo
Shadertoy: Interactive 1D Cubic Bezier Demo

What if My Equation DOESN’T Equal Zero??

Take a simple equation such as y = 2x. You can transform that into the equation 2x – y = 0, and then could write it as f(x,y) = 2x – y = 0.

Now we have some function of x and y that equals zero. If we plug in numbers that are valid points in the original equation y = 2x (in other words, coordinates where y is double x), we’ll get zero out as a result:

f(x,y) = 2x - y

f(1,2) = 2 * 1 - 2 = 0
f(2,4) = 2 * 2 - 4 = 0
f(0.5, 1.0) = 2 * 0.5 - 1.0 = 0

If we plug numbers in that don’t fit that pattern, we get non zero answers as a result:

f(x,y) = 2x - y

f(2,1) = 2 * 2 - 1 = 3
f(5,5) = 2 * 5 - 5 = 5
f(-10, 20) = 2 * -10 - 20 = -40

What do these numbers mean? Are they useful for anything? As it turns out, they do have meaning and they are useful! They are part of what is needed to calculate the distance from the point (x,y) to the closest point on the curve. The other part of what we need is called the “gradient vector”.

Tangent Vector & Gradient Vector

As you probably know, there is a tangent vector at every point on a curve (let’s ignore things like asymptotes for now), and the tangent vector basically points from one point on the curve to the next point on the curve. The tangent vector can also be seen as a slope of that point on the curve.

A gradient vector is another property of a point, but it is different than a tangent vector in a couple ways. First off, it is perpendicular to the tangent vector, so sticks out away from the curve, and could be seen as a normal vector. Secondly, every point in the graph has a gradient vector, even if that point isn’t part of the curve and is just floating by itself in empty space.

The interesting part here is that a gradient vector of a point not on the curve will actually lead you straight to the closest point on the curve to the point. This is because of a handy property where the line connecting a point to the closest point on the curve will be perpendicular to the curve (perpendicular to the tangent).

So, if we can calculate a gradient vector for a point, we have the vector that points towards the closest point on the curve. Technically it might be pointing away from the closest point on the curve, instead of towards it but that gets cleared up in the math.

Calculating Gradient Vector

The gradient vector is just a partial derivative for each of the variables of the equation. If you have no idea what I’m talking about, here are 2 options:

1) You can watch some videos and learn about it here: Khan Academy: Multivariable calculus
2) You can have wolfram alpha calculate it for you, by typing “gradient” before the function. Click this link to see what I mean: Wolfram Alpha: gradient x^2-y

Here are some example equations, along with their gradient vector, to make sure that you either understand what’s going on, or are properly entering your equations into wolfram alpha. Note that a gradient vector may not always be constant, in which case, it changes value for each point! If it is constant, it’s the same gradient vector across the entire graph space.

f(x,y) = 2x – y
gradient(x,y) = (2, -1)

cubic function:
f(x,y) = x^3 – 10y
gradient(x,y) = (3x^2, -10)

f(x,y) = sin(x) – y
gradient(x,y) = (cos(x), -1)

f(x,y) = x^2 + y^2 – 5
gradient(x,y) = (2x, 2y)

Calculating Distance

Ok, so now that we have the value of our equation at some point (x,y) and we have the ability to find our gradient vector at that point, we have everything we need to calculate the distance from the point to the closest point on the curve.

All you need to do is divide the absolute value of the equation value at that point by the length of the gradient vector at that point.

Example 1 – Line

Let’s start with something super simple… y = x, aka x – y = 0 aka f(x,y) = x – y.

the gradient of that function is (1, -1) at every point on the graph. The magnitude of the gradient vector is the square root of 2.

What is the distance from the line to point P(2,0)?

Well.. the value of the equation at that point f(2,0) is 2, and the absolute value of that is still 2.

Next we divide that value by the magnitude of the gradient to get 2 / square root(2), which is just square root (2). So, the distance is square root of 2.

Since the value of the function at that point was positive before taking an absolute value, that means that the gradient points from the line to the point, so you can flip the gradient around to get (-1,1), and that is the direction that the closest point on the line is from the point P(2,0). If we travel square root of 2 units from P(2,0) down the vector (-1,1) we will get to the point (1,1) which is on the graph. That is the closest point on the graph to our point.

Note that it works out in this case, but usually the gradient vector won’t be the right magnitude since it’s not normalized. It happened that in this case it is, by dumb luck, but it usually isn’t.

Example 2 – Another Line

Let’s try a more complex line. y = 2x aka 2x – y = 0 aka f(x,y) = 2x – y.

the gradient of that function is (2, -1) with a magnitude of square root of 5 at all points on the graph.

What is the distance from that line to point P(1,5)?

Well, the value of the function at (1,5) is -3. When you take the absolute value of that, it becomes 3.

So, the distance is 3 / square root (5).

In this case, since the value of the function at the point was negative before we took the absolute value, it means that the gradient points from the point to the graph, so the line y=2x is 3/square root(5) units of length away from the point P(1,5) and the direction to travel from the point to get to the line is the same direction as the gradient vector at that point aka (2,1).

Example 3 – Quadratic Function

For the last example lets do y=x^2 aka x^2 – y = 0 aka f(x,y) = x^2 – y.

The gradient of that function is (2x,-1) and since it isn’t constant at all points, we can’t get the magnitude of the gradient yet.

What is the distance from that function to point P(2,0)?

Well, the value of f(2,0) is 4, which is still 4 if you take the absolute value.

The gradient at P(2,0) is (4,-1) with a magnitude of square root of 5.

So, the distance from point P(2,0) to the curve f(x,y) = x^2 – y is 4/square root(5) and to get from the point to the curve you should travel that distance down the same direction as the vector (4,-1).


Ok ok… the other shoe dropping here is that this distance calculation is only a distance ESTIMATE, and not always the right answer.

This is because if the gradient is a constant or linear function, it will be correct, but higher order gradients don’t always follow straight lines from the point to the curve.

The good news though is that this estimation will be less than or equal to the actual distance. Maybe not as useful as a for sure distance calculation, but an upper bound estimate still has it’s uses (like in ray marching, or drawing vector graphics).

Basically, the straight line tells you the closest it could possibly be, but the actual gradient may take a longer, curving path, from the point to the curve.

More Info

Here’s the article that got me started down this path, trying to find the answer to this question, while trying to understand the main topic:
IQ: Distance Estimation

And here’s a link that really helped me understand what a gradient was, and what it was all about:
Vector Calculus: Understanding the Gradient

Khan academy videos about these topics:
Khan Academy: Multivariable calculus

Analytic Fog Density


There are a number of ways to implement the effect of fog with modern real time rendered graphics. This blog post will explain how to render fog that has varying density, based on a function of X,Y,Z location in space, like in the picture above.

Faked Fog

One way is to “fake it” and do something like set the color of a pixel on an object to be based on it’s height. For instance you might say that pixels with a y axis value above 15 are unfogged, pixels with y axis values between 15 and 10 progressively get more fogged as they get closer to 10, and pixels with y axis values less than 10 are completely fogged. That can make some fog that looks like this:


A strange side effect of doing that though, is if you go down “into” the fog, and look out of the fog, things that should be fogged won’t. For instance, looking up at a mountain from inside the fog, the mountain won’t be fogged at all, even though it should be because you are inside of the fog.

A better way to do it, if you intend for the camera to be able to go into the fog, is to calculate a fogging amount for a pixel based on how far away it is from the view point, and how dense the fog is between the view point and the destination point.

If you are doing ray based rendering, like ray tracing or ray marching, you might find yourself trying to find how much fog is between points that don’t involve the view point – like if you are calculating the reflection portion of a ray. In this case, you are just finding out how much fog there is between the point where the reflection happened and the closest intersection. You can consider the point of reflection as the “view point” for the purpose of fogging.

Sometimes, the entire scene might not be in fog. In this case, you have to find where the fog begins and ends, instead of the total distance between the view point and the destination point.

In any case, the first thing you need to do when fogging is figure out the point where the fog begins, and the point where the fog ends. Then, you can figure out how much fog there is based on how the fog density works.

Constant Density Fog


The simplest sort of fog is fog that has the same density all throughout it.

What you do in this case is just multiply the fog density by the distance spent in the fog to come up with a final fog value.

As an example, your fog density might be “0.04” and if you are fogging a pixel 10 units away, you multiply density by distance. FogAmount = 0.04 * 10.0 = 0.4.

Doing this, you know the pixel should be 40% fogged, so you interpolate the pixel’s color 40% towards the fog color. You should make sure to clamp the fog amount to be between 0 and 1 to avoid strange visual anomolies.

The image below shows a constant fog density of 0.04.


Here’s an image of the same constant density fog as viewed from inside the fog:


A problem with constant fog density though, is that if you view it from edge on, you’ll get a very noticeable hard edge where the fog begins, like you can see in the image below:


Linear Density Fog


With linear fog density, the fog gets denser linearly, the farther you go into the fog.

With a fog plane, you can get the density of the fog for a specified point by doing a standard “distance from plane to point” calculation and multiplying that by how much the fog density grows per unit of distance. If your plane is defined by A*x+B*y+C*y+D = 0, and your point is defined as X,Y,Z, you just do a dot product between the plane and the point, giving the point a W component of one.

In other words…

FogDensity(Point, Plane) = (Plane.NormX * Point.X + Plane.NormY * Point.Y + Plane.NormZ * Point.Z + Plane.D * 1.0) * FogGrowthFactor

Here’s a picture of linear fog with a fog growth factor of 0.01:


The same fog viewed from the inside:


And lastly, the fog viewed edge on to show that the “hard line” problem of linear fog is gone (dramatic difference isn’t it?!):


Analytic Fog Density – Integrals


Taking a couple steps further, you might want to use equations to define fog density with some function FogDensity = f(x,y,z,).

How could you possibly figure out how much fog there is between two given points when the density between them varies based on some random function?

One way would be to take multiple samples along the line segment between the view point and the destination point, and either calculate the fog amount in each section, or maybe average the densities you calculate and multiply the result by the total distance. You might have to take a lot of samples to make this look correct, causing low frame rate, or accepting low visual quality as a compromise.

If you look at the graphs for the previous fog types, you might notice that we are trying to find the area under the graphs between points A and B. For constant density fog, the shape is a rectangle, so we just multiply width (time in fog) by height (the constant fog density) to get the fog amount. For linear density fog, the shape is a trapezoid, so we use the trapezoid area formula which is height (in this case, the distance in the fog) times the sum of the base lengths (the fog densities at points A and B) divided by 2.

How can we get the area under the graph between A and B for an arbitrary formula though? Well, a way exists luckily, using integrals (thanks to my buddy “Danny The Physicist” for educating me on the basics of integrals!).

There’s a way to transform a formula to get an “indefinite integral”, which itself is also a formula. I won’t go into the details of how to do that, but you can easily get the indefinite integral of a function by typing it into Wolfram Alpha.

Once you have the indefinite integral (let’s call it G(x)) of the fog density formula (let’s call it F(x)), if you calculate G(B) – G(A), that will give you the area under the graph in F(X) between A and B. Yes, seriously, that gives us the area under the graph between our points, thus giving us the amount of fog that exists between the two points for an arbitrary fog density function!

Note that when you plug a value into the indefinite integral and get a number out, that number is called the definite integral.

Analytic Fog Density – Implementation Details

Now that the theory is worked out let’s talk about implementation details.

First off, coming from an additive audio synthesis type of angle, I figured I might have some good luck adding together sine waves of various frequencies and amplitudes, so I started with this:

sin(x*F) * A

F is a frequency multiplier that controls how long the sine wave is. A is an amplitude multiplier that controls how dense the fog gets max.

Next, I knew that I needed a fog density function that never goes below zero, because that would mean if you looked through a patch of negative fog density, it would make the other fog you were looking through be less dense. That is just weird, and doesn’t exist in reality (but maybe there is some interesting visual effect hiding in there somewhere??), so the formula evolved to this, making sure the function never went below zero:

(1 + sin(x*F)) * A

Plugging that equation into wolfram alpha, it says the indefinite integral is:

(x – (cos(x*F)) / F) * A

You can check that out here:
Wolfram Alpha: (1 + sin(x*F)) * A.

It’s also kind of fun to ask google to graph these functions so you can see what they do to help understand how they work. Here are the graphs for A = 0.01 and F = 0.6:
Fog Density: graph (1 + sin(x*0.6)) * 0.01
Indefinite Integral: graph (x – (cos(x*0.6)) / 0.6) * 0.01

So, if you have point A and B where the fogging begins and ends, you might think you can do this to get the right answer:
FogAmount = G(B.x) – G(A.x)

Nope! There’s a catch. That would work if A and B had no difference on the y or z axis, but since they probably do, you need to jump through some hoops. In essence, you need to stretch your answer across the entire length of the line segment between A and B.

To do that, firstly you need to get that fog amount down to unit length. You do that by modifying the formula like so:
FogAmount = (G(B.x) – G(A.x)) / (B.x – A.x)

This also has a secondary benefit of making it so that your fog amount is always positive (so long as your fog density formula F(X) can’t ever go negative!), which saves an abs() call. Making it always positive ensures that this works when viewing fog both from the left and the right.

Now that we have the fog amount down to unit length, we need to scale it to be the length of the line segment, which makes the formula into this:
FogAmount = (G(B.x) – G(A.x)) * Length(B-A)/(B.x – A.x)

That formula will now give you the correct fog amount.

But, one axis of fog wasn’t enough to look very good, so I wanted to make sure and do one sine wave on each axis. I used 0.01 amplitude for each axis, but for the X axis i used a frequency of 0.6, for the Y axis i used a frequency of 1.2 and for the Z axis i used a frequency of 0.9.

Also, I wanted to give a little bit of baseline fog, so I added some constant density fog in as well, with a constant density of 0.1.

As a bonus, I also gave each axis a “movement factor” that made the sine waves move over time. X axis had a factor of 2.0, Y axis had a factor of 1.4 and Z axis had a factor of 2.2.

Putting all of this together, here is the final fog equation (GLSL pixel shader code) for finding the fog amount between any two points at a specific point in time:

float DefiniteIntegral (in float x, in float amplitude, in float frequency, in float motionFactor)
	// Fog density on an axis:
	// (1 + sin(x*F)) * A
	// indefinite integral:
	// (x - cos(F * x)/F) * A
	// ... plus a constant (but when subtracting, the constant disappears)
	x += iGlobalTime * motionFactor;
	return (x - cos(frequency * x)/ frequency) * amplitude;

float AreaUnderCurveUnitLength (in float a, in float b, in float amplitude, in float frequency, in float motionFactor)
	// we calculate the definite integral at a and b and get the area under the curve
	// but we are only doing it on one axis, so the "width" of our area bounding shape is
	// not correct.  So, we divide it by the length from a to b so that the area is as
	// if the length is 1 (normalized... also this has the effect of making sure it's positive
	// so it works from left OR right viewing).  The caller can then multiply the shape
	// by the actual length of the ray in the fog to "stretch" it across the ray like it
	// really is.
	return (DefiniteIntegral(a, amplitude, frequency, motionFactor) - DefiniteIntegral(b, amplitude, frequency, motionFactor)) / (a - b);

float FogAmount (in vec3 src, in vec3 dest)
	float len = length(dest - src);
	// calculate base fog amount (constant density over distance)	
	float amount = len * 0.1;
	// calculate definite integrals across axes to get moving fog adjustments
	float adjust = 0.0;
	adjust += AreaUnderCurveUnitLength(dest.x, src.x, 0.01, 0.6, 2.0);
	adjust += AreaUnderCurveUnitLength(dest.y, src.y, 0.01, 1.2, 1.4);
	adjust += AreaUnderCurveUnitLength(dest.z, src.z, 0.01, 0.9, 2.2);
	adjust *= len;
	// make sure and not go over 1 for fog amount!
	return min(amount+adjust, 1.0);

More Info

I ended up only using one sine wave per axis, but I think with more sine waves, or perhaps different functions entirely, you could get some more convincing looking fog.

At some point in the future, I’d like to play around with exponential fog density (instead of linear) where the exponential power is a parameter.

I also think that maybe squaring the sine waves could make them have sharper density changes perhaps…

One thing that bugs me in the above screenshots is the obvious “hard line” in both constant and linear fog where it seems fog crosses a threshold and gets a lot denser. I’m not really sure how to fix that yet. In traditional rasterized graphics you could put the fog amount on a curve, to give it a smoother transition, but in ray based rendering, that could make things a bit odd – like you could end up with an exponential curve butting up against the start of a different exponential curve (due to reflection or refraction or similar). The fog density would end up looking like log graph paper which would probably not look so great – although honestly I haven’t tried it to see yet!

If you have any questions, or feedback about improvements you know about or have discovered in any of the above, post a comment and let me know!

Here’s a good read on fog defined by a plane, that also gets into how to make branchless calculations for the fog amounts.
Unified Distance Formulas for Halfspace Fog

Interactive demo with GLSL pixel shader source code that you can also edit in real time with WebGL:

Feistel Networks – Do They Have to use XOR?

If you have no idea what a Feistel network is, but like cryptography and/or random number generation algorithms, read this link first:
Fast & Lightweight Random “Shuffle” Functionality – FIXED!

As a quick refresher, to encrypt data with a Feistel network, you break the plain text data into a left and a right side and do N rounds of this operation:

Left[i+1]  = Right[i];
Right[i+1] = Left[i] ^ RoundFunction(Right[i], key);

Where RoundFunction is ideally some chaotic function that returns some pseudo-random-esque number based on the inputs. For instance, RoundFunction could be MD5 so that it returned the MD5 hash of the data and the key, where the key could be considered the salt of the hash. The better the round function, the better your encryption algorithm will be.

To decrypt data with a Feistel network, you break the data into a left and right side and do the same number of rounds of this operation:

Right[i] = Left[i+1];
Left[i] = Right[i+1] ^ RoundFunction(Left[i+1], key);

Ok, onto the question….

Does it Have to use XOR?

Recently a friend of mine was using Feistel networks for something pretty amazing (so amazing, I can’t even talk about it), but in doing so, he asked me an interesting question. He asked “do you think this HAS to be XOR here, where we combine the round functions result back into the data?”. Well, it turns out, it doesn’t!

The operation has to be a reversible operation though, and you have to do the reverse operation when decrypting that you did while encrypting.

For instance, when encrypting you could add the round function result in, but then when decrypting, you would have to subtract the round function result out.

Or, you could do bitwise rotation left when encrypting, and right when decrypting perhaps.

Basically, anything that has a reverse operation can be used.

You have to be careful though because you might be lured into the trap of thinking that this includes something like multiplication and division.

If you multiply when you encrypt, you might get an integer overflow and lose data that can’t be corrected by doing a divide. For instance, if you multiply 255*2 in an unsigned 8 bit number you get 254 as a result. If you divide 254 by 2 to “undo” the multiplication, you get 127 which is obviously not 255, so we’ve lost some data. In an unsigned 8 bit number, ((255*2)/2) = 127.

If you go the other way and divide on encryption, and multiply on decryption, that doesn’t work either. For instance, when you divide 3 by 2, you get 1 with integer math, and when you multiply by 2, you get 2. So, with integers… ((3/2)*2) = 2.

Confusing note: you ARE able to do irreversible operations within the round function though. Feel free to do a divide or whatever you want in there. If that is difficult to understand how that could possibly work, you aren’t alone. Step through the code a bit by hand with a simple round function and a low number of rounds and you might be able to understand better how it does what it does.

I’m really not sure if anyone else out there does this variation on the traditional Feistel networks or not, but it is pretty interesting to combine the RoundFunction result back into the data with something other than XOR.

Source Code

Here’s some simple C++ code below to play with if you want to mess around with this stuff.


static const unsigned int c_numRounds = 4;

void PrimeRandomNumberPump ()
	// if you are curious about this, check out:
	for (unsigned int index = 0; index < 20; ++index)

unsigned char RoundFunction (unsigned char value, unsigned char key)
	// Not a particularly effective round function, but the round function
	// isn't the point of this code.
	// If you want a better round function, try plugging in a hash function
	// or another chaotic function that has big changes in output for
	// small changes in input.  Also, you could change c_numRounds to a
	// higher number if you want better encryption.
	return value + key | (value * key) + 3;

void Encrypt (unsigned char &left, unsigned char &right, unsigned char key)
	for (unsigned int index = 0; index < c_numRounds; ++index)
		// Feistel Network Encryption:
		//  Left[i+1]  = Right[i];
		//  Right[i+1] = Left[i] ^ RoundFunction(Right[i], key);

		// let's do addition to combine the value of the round function on 
		// encryption, instead of doing xor.  Xor is used in feistel networks
		// because xor is it's own inverse operation.
		unsigned char oldLeft = left;
		left = right;
		right = oldLeft + RoundFunction(right, key);

void Decrypt (unsigned char &left, unsigned char &right, unsigned char key)
	for (unsigned int index = 0; index < c_numRounds; ++index)
		// Feistel Network Decryption:
		//  Right[i] = Left[i+1];
		//  Left[i] = Right[i+1] ^ RoundFunction(Left[i+1], key);

		// let's do subtraction to combine the value of the round function on 
		// decryption, instead of doing xor.  Xor is used in feistel networks
		// because xor is it's own inverse operation.
		unsigned char oldRight = right;
		right = left;
		left = oldRight - RoundFunction(left, key);

void DoTest (unsigned char plainText1, unsigned char plainText2, unsigned char key, int &tests, int &errors)
	// encrypt the plaintext
	unsigned char cipherText1 = plainText1;
	unsigned char cipherText2 = plainText2;
	Encrypt(cipherText1, cipherText2, key);

	// decrypt the cipher text
	unsigned char decryptedData1 = cipherText1;
	unsigned char decryptedData2 = cipherText2;
	Decrypt(decryptedData1, decryptedData2, key);

	// if the decrypted data doesn't match the plaintext data, count it as an error
	// and show the details
	if (decryptedData1 != plainText1 || decryptedData2 != plainText2)
		printf("plaintext = 0x%02X%02Xrn", (unsigned int)plainText1, (unsigned int)plainText2);
		printf("ciphertext = 0x%02X%02Xrn", (unsigned int)cipherText1, (unsigned int)cipherText2);
		printf("decrypteddata = 0x%02X%02Xrnrn", (unsigned int)decryptedData1, (unsigned int)decryptedData2);

void main (void)
	// generate a key
	unsigned char key = (unsigned char)rand();

	// run tests with the key
	int errors = 0;
	int tests = 0;
	for (unsigned int y = 0; y < 256; ++y)
		for (unsigned int x = 0; x < 256; ++x)
			DoTest((unsigned char)y, (unsigned char)x, key, tests, errors);
	// display the test results
	printf("%i tests ran, %i errors encountered. key = 0x%02Xrn", tests, errors, key);

Bezier Curves Part 2 (and Bezier Surfaces)

This is a follow up post to Bezier Curves. My plan was to write a post about b-splines and nurbs next, but after looking into them deeper, I found out they aren’t going to work for my needs so I’m scratching that.

Here’s some basic info on b-splines and nurbs though before diving deeper into Bezier curves and surfaces.

B-Splines (Basis Splines)

Bezier curves are nice, but the more control points you add, the more complex the math gets because the degree of the curve function increases with each control point added. You can put multiple Bezier curves end to end to be able to have more intricate curves, but another option is to use B-Splines.

B-Splines are basically Bezier curves which let you specify more control points without raising the degree of the Bezier curve. They do this by having control points only affect part of the total curve.

This way, you could make a quadratic b-spline which had 10 control points. Only a few control points control any given point on the curve, so the curve stays quadratic (and so does the math), but you get a lot more control points. A “Knot Vector” is what controls which parts of the curve the control points control.

A Bezier curve is actually a special case of B-Spline where all control points affect the entire curve.

Nurbs (Non Uniform Rational B-Spline)

Sometimes when working with curves, you want some control points to be stronger that others. You can accomplish this in Bezier curves and B-splines by doubling up or trippling up control points in the same location to make that control point twice, or three times as strong respectively.

What if you want a control point to be 1.3 times stronger though? That gets a lot more complicated.

Nurbs solve that problem by letting you specify a weight per control point.

Just like Bezier curves are a special case of B-Splines, B-Splines are a special case of nurbs. A B-Spline could be thought of as nurbs that has the same weighting for all control points.

Back to Bezier!

My end goal is to find a curve / surface type that is flexible enough to be used to make a variety of shapes by artists, but is efficient at doing line segment tests against on the GPU. To this end, B-Splines and Nurbs add algorithmic and mathematical complexity over Bezier curves, and seem to be out of the running unless I can’t find anything more promising.

My best bet right now looks like a Bezier Triangle. Specifically, a quadratic Bezier triangle, where each side of the triangle is a quadratic Bezier curve that has 3 control points. When I get those details fully worked out, I’ll report back, but for now, here’s some interesting info I found about generalizing bezier curves both in order (linear, quadratic, cubic, quartic, etc) as well as in the number of dimensions (line, curve, triangle, tetrahedron, etc).

Bezier Generalized

I found the generalized equation on the wikipedia page for Bezier triangles and am super glad i found it, it is very cool!

I want to show you some specifics to explain the generalization by example.

Quadratic Curve:
(A * S + B * T) ^ 2

Expanding that gives you:
A^2 * S^2 + A * B * 2 * S * T + B^2 * T^2

In the above, S and T are Barycentric Coordinates in a 1 dimensional Simplex. Since we know that barycentric coordinates always add up to 1, we can replace S with (1-T) to get the below:

A^2 * (1-T)^2 + A * B * 2 * (1-T) * T + B^2 * T^2

Now, ignoring T and the constants, and only looking at A and B, we have 3 forms: A^2, AB and B^2. Those are our 3 control points! Let’s replace them with A,B and C to get the below:

A * (1-T)^2 + B * 2 * (1-T) * T + C * T ^2

And there we go, there’s the quadratic Bezier curve formula seen in the previous post.

Cubic Curve:
(A * S + B * T) ^ 3

To make a cubic curve, you just change the power from 2 to 3, that’s all! If you expand that equation, you get:

We can swap S with (1-T) to get:


Looking at A/B terms we see that there is more this time: A^3, A^2B, AB^2 and B^3. Those are our 4 control points that we can replace with A,B,C,D to get:

There is the cubic Bezier curve equation from the previous chapter.

Linear Curve:
(A * S + B * T) ^ 1

To expand that, we just throw away the exponent. After we replace S with (1-T) we get:
A * (1-T) + B * T

That is the formula for linear interpolation between 2 points – which you could think of as the 2 control points of the curve.

One more example before we can generalize.

Quadratic Bezier Triangle:
(A * S + B * T + C * U) ^ 2

If you expand that you get this:

Looking at combinations of A,B & C you have: A^2, AB, AC, B^2, BC, C^2. Once again, these are your control points, and their names tell you where they lie on the triangle. A Bezier triangle is a triangle where the 3 sides of the triangle are bezier curves. A quadratic bezier triangle has quadratic bezier curves for it’s edges which mean that each side has 3 control points. Those 3 control points are made up of the 3 corners of the triangle, and then 3 more control points, each one being between end points. A^2, B^2 and C^2 represent the 3 corners of the triangle. AB is the third control point for the bezier curve on the edge AB. BC and AC follow that pattern as well! Super easy to remember.

In a cubic Bezier triangle, you get a lot more control points, but a new class of control point too: ABC. This control point is in the middle of the triangle like the name would imply.

Anyways, in the expanded quadratic bezier triangle equation above, when you replace the control points with A,B,C for the triangle corner control points (the squares) and D,E,F for the inbetween control points, you get the bezier triangle equation below:


Note that we are dealing with a simplex in 3d now, so once again, instead of needing ALL Barycentric coordinates (S,T,U) we could pick one and replace it. For instance, we could replace U with (1-S-T) to have one less variable floating around.

All Done for Now

You can use this pattern to expand either in “surface dimension”, or in the dimension of adding more control points (and increasing the order of the equation). I love it because it’s super simple to remember that simple equation, and then just re-calculate the equation you need for whatever your specific usage case is.

If this stuff is confusing, check out the wiki page for Bezier Triangles, it has a great graphic that really shows you what I’m trying to explain:
Bezier Triangle

Next up I either want to make an HTML5 interactive app for messing around with Bezier triangles, or if I can figure out how to intersect a line segment with a quadratic Bezier triangle, i’ll probably just have some real cool looking screenshots to post along w/ the equation I ended up using (;

Special thanks to wolfram alpha for crunching some of these equations. Check it out, it’s really cool!
Wolfram Alpha – Cubic Bezier Curve Expansion

For more bezier fun check out my next Bezier post: One Dimensional Bezier Curves.

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 (;