## The 60:1 Rule and related derivations

At some point you've probably run into the 60:1 rule and wondered how it came to be and what all we can do with it. At its core, the 60:1 rule uses two simple geometric approximations to give us a host of valuable derivations to fly with. To start, grab a piece of paper and sketch out an equilateral triangle. Label each internal angle with 60 degrees, and each face with the letter "r." Next, draw a circle that's centered on one corner of the triangle that touches the other corners and has the same radius r. Label the arc between the vertices "a." It should look something like this:

You'll notice that arc a and length r are almost the same length. If you remember back to geometry class, the circumference of a circle is 2πr, and that π≈3.14. Thus, 2πr≈6.28r. If you stack 6 triangles in the circle (6 * 60 = 360 degrees), and each has edge length r, then the circumference of the resulting hexagon is 6r. Thus, relationship is 6.28r/6r=1.0467. In other words, the length of the arc a is within 5% of the length of r. This means that

**for small angles where we plan to round to the nearest degree, a 5% error is trivial**. That's our first little approximation. Next, we want to look at what happens when we chop length r and arc a into small chunks. Let's say we chop each length up into 60 chunks radially from the opposite corners. Realistically, we could chop each r or a into any n chunks, but let's use n=60 because we use degrees when we fly (if we navigated by radians, the same r≈a approximation would apply, and we'd just call it "the rule of 1."). Blame the Babylonians for the 60 degrees. Ok, so now each face of our triangle is divided into 60 roughly equal chunks. I've sketched out the bottom right corner and left the rest open for de-cluttering:One relationship that jumps right out at us is that if r≈a, then for each degree, we have r/60 units of length on the opposite edge or arc. In other words,

**for every degree of deflection, we have 1/60th of****r as distance****displaced from the bottom line**. So if r is 60 nm, then a degree puts us a mile off course. Our second hand-wave approximation, closely related to the first, is that the length of an arc opposite θ and the height of a right triangle with angle θ are the same. Let's zoom in on that bottom right corner of the arc that we chopped into 60 pieces:Looking at that 3-degree slice, we can approximate the height of the vertical line forming the right triangle as the arc length around the outside. Combining that with our earlier discovery, we can say that

One useful coincidence for us is that a nautical mile is 6076 feet. 6076≈6000. If we take our 1/60 rule and add two zeros on the end, we get 100/6000. Combine that with our earlier findings, we derive that GPA*100 = ft/nm.

Now let's look at what we can derive from this.

**a gradient of θ degrees gives us θ units of vertical displacement for every 60 units of lateral displacement**. For those engineers and trigonometricians out there screaming foul, tan(1)=.017455, and 1/60 is .016667. That's still only 4.73% off. Close enough for us pilots.One useful coincidence for us is that a nautical mile is 6076 feet. 6076≈6000. If we take our 1/60 rule and add two zeros on the end, we get 100/6000. Combine that with our earlier findings, we derive that GPA*100 = ft/nm.

Now let's look at what we can derive from this.