Flight levels divided by nautical miles equals gradient.

I first learned about this rule of thumb at the Air Force Instrument Instructor's Course (AFIIC) where we were told it is derived from the 60-to-1 concept. But is that true? Well let's find out. But first, we'll define the rule, provide an example of its application, and then we'll provide a mathematical proof of the rule.

Thanks as always to Al Klayton, retired Air Force electrical engineer and civilian pilot for help with the trickier math.

Flight levels divided by nautical miles equals gradient.

Figure: Gradient example, from Eddie's notes.

You are at FL410, flying to Boston, and ATC gives you direct ALB with 435 nm to go and you have absolutely nothing to do for the next 48 minutes. But when you get to 180 DME they throw in a curve ball: "Cross 100 miles west of Albany at 10,000 feet."We know the descent gradient is equal to the flight levels to lose divided by the nautical miles to go, but the math

$$\frac{310}{80}=?$$isn't so straight forward.

We want to do this without a calculator, slide rule, or Stephen Hawking in the jump seat. Well, 8 is a factor of 32 — it divides easily — and if we can make the gradient with 320 flight levels to lose we can certainly do it with only 310 flight levels. And that makes life for us as pilots easier:

$$\frac{320}{80}=4$$If we achieve a 4° descent gradient, we will make it to our restriction with room to spare. So leave the boards stowed for now and see what kind of VVI we've got with the power levers at idle.

Figure: 60-to-1 Descent Gradient, from Eddie's notes.

The basic 60-to-1 theory tells us that 1 nautical mile horizontally requires 100 feet vertically at a 1° gradient. A flight level is equal to 100 feet, so the rule follows naturally. If you are at FL450, for example, and center directs you to descend to and maintain 15,000 feet before crossing a fix that you are 75 nm away from, your descent gradient will be:

$\mathrm{Gradient}=\frac{300}{75}=4\xb0$But there is some poetic license here, since an airplane climbing or descending isn't behaving as if on an arc of a circle. The real answer has more to do with trigonometry:

$\mathrm{Vertical\; change}=6076\mathrm{feet}tan\left(1\right)=106.06\mathrm{feet}$In the case of our example, the actual gradient is:

$\mathrm{Gradient}=atan\left(\frac{300}{\left(75\right)\left(6076\right)}\right)=3.77\xb0$So the theory is off by 6 percent, not too bad.

For more about this rule of thumb, see: Procedures & Techniques / Descent.

Portions of this page can be found in the book Flight Lessons 1: Basic Flight, Chapter 24.

Portions of this page can be found in the book Flight Lessons 2: Advanced Flight, Chapter 3.

So what about the claims this rule of thumb is based on 60-to-1? My conclusion: Not really.

Rule of Thumb | 60-to-1? | Trigonometry | π |

Gradient | ✓ |

60-to-1 — Rule of thumb is based on the mathematical relationship of a 360° circle and/or 6076' to 1 nm.

Trigonometry — Rule of thumb is based on the relationship to a right angle and the derived trigonometric functions.

π — Rule of thumb is based on the relationship of a 360° circle, the number π, and/or 6076' to 1 nm.