### Eddie Sez:

Descent VVI — If you've ever flown an airplane without a flight director, or a PAR approach in just about any airplane, computing a target VVI is the first prerequisite along the way to being able to land at minimums. But even if you are flying a very high tech cockpit, having a target VVI can save you if the electrons or winds misbehave. The AFIIC 60-to-1 rule for this has nothing to do with 60-to-1 or trigonometry, but it works.

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.

### Definition

Nautical miles per minute times descent angle times 100 gives vertical velocity in feet per minute.

### Example

Figure: PAR Controller, from the movie "Airplane."

Let's say your are flying an ILS without a flight director (or find yourself flying a PAR approach), the glide path is based on a 3° angle, and you need to know what your target VVI should be on glide path. If your approach speed is 130 and you have a 10 knot headwind, you assume you will have a ground speed of 120 knots. Of course this may not be true at higher elevations where your KTAS will be higher, impacting your ground speed and you may want to adjust. If you have a ground speed readout, use that. But the math is pretty easy:

$\mathrm{Descent VVI \left(ft/min\right)}=\left(\mathrm{Velocity \left(nm/min\right)}\right)\left(\mathrm{Descent angle \left(degrees\right)}\right)100$

Doing the math:

If you want to know more about PAR approaches: Instrument Procedures / Precision Approach Radar (PAR).

### Proof

Figure: Descent VVI, from Eddie's notes.

The AFIIC rule for computing a target vertical velocity on final approach can be written thusly:

$\mathrm{Descent VVI \left(ft/min\right)}=\left(\mathrm{Velocity \left(nm/min\right)}\right)\left(\mathrm{Descent angle \left(degrees\right)}\right)100$

The mathematically pure answer is fairly easy, as long as we keep the units consistent. An engineer refers to a change in a quantity with the Greek symbol delta, Δ. A change in altitude, A, for example, can be found by subtracting the first altitude, A1, from the second altitude, A2. The engineer cites this as ΔA = A1 - A2. The airplane's position change along its route of flight, ΔS, is nothing more than the velocity. The math for our Descent VVI is fairly easy, as long as we keep the units straight:

$\mathrm{\Delta A}=\left(\mathrm{\Delta S}\right)sin\left(\theta \right)$

For example, let's say you are flying down a 3° glide path at 120 knots, which comes to 2 nm/min. We want the end result in ft/min; that comes to 120 (nm/hr) (6076) (ft/nm) (1/60) (hr/min) = 12,152 ft/min. The rate of altitude change is:

Now we certainly can't fine tune our vertical descent to that degree, 634 ft/min. If we permit ourselves some approximation, we can derive an easier to use rule of thumb.

$\mathrm{V \left(nm/min\right)}=\frac{\mathrm{\Delta S \left(nm\right)}}{\mathrm{\Delta t \left(min\right)}}$

Therefore . . .

$\mathrm{\Delta t \left(min\right)}=\frac{\mathrm{\Delta S \left(nm\right)}}{\mathrm{V \left(nm/min\right)}}$

Also . . .

$\mathrm{\Delta S \left(nm\right)}=\frac{\mathrm{\Delta A \left(nm\right)}}{sin\theta }$

Combining . . .

$\mathrm{\Delta t \left(min\right)}=\frac{\frac{\mathrm{\Delta A \left(nm\right)}}{sin\theta }}{\mathrm{V \left(nm/min\right)}}$

And . . .

$\mathrm{\Delta t \left(min\right)}=\frac{\mathrm{\Delta A \left(nm\right)}}{\left(sin\theta \right)\mathrm{V \left(nm/min\right)}}$

Now looking at the VVI side of things, we need to be careful about the units. Since VVI is presented in feet per minute, we have to convert the change in altitude, ΔA, into feet. There are 6,076 feet in a nautical mile, but we are in the business of approximating for these rules of thumb, so we'll use 6,000:

$\mathrm{VVI \left(feet/min\right)}=\left(\frac{\mathrm{\Delta A \left(nm\right)}}{\mathrm{\Delta t \left(min\right)}}\right)6000\mathrm{ft/nm}$

We can substitute our earlier value for Δt . . .

$\mathrm{VVI \left(feet/min\right)}=\left(\frac{\mathrm{\Delta A \left(nm\right)}}{\frac{\mathrm{\Delta A \left(nm\right)}}{\left(sin\theta \right)\mathrm{V \left(nm/min\right)}}}\right)6000\mathrm{ft/nm}$

And through the magic of algebra . . .

$\mathrm{VVI \left(feet/min\right)}=6000\mathrm{\left(feet/nm\right)}\left(sin\theta \right)\mathrm{V \left(nm/min\right)}$

Now, for small angles, the math wizards tell us that the sin( θ ) = θ expressed in radians. We can convert θ to radians knowing 2π radians = 360 degrees.

$\mathrm{VVI \left(feet/min\right)}=6000\mathrm{\left(feet/nm\right)}\left(\theta \right)\left(\frac{2\pi }{360}\right)\mathrm{V \left(nm/min\right)}$

Simplifying:

$\mathrm{VVI \left(feet/min\right)}=100\left(\theta \right)\left(\frac{\pi }{3}\right)\mathrm{V \left(nm/min\right)}$

It is true that π = 3.1416... and so on. But let's call it 3:

$\mathrm{VVI \left(feet/min\right)}=100\left(\theta \right)\left(\frac{3}{3}\right)\mathrm{V \left(nm/min\right)}$

And now we have a rule of thumb worthy of AFIIC:

$\mathrm{VVI \left(feet/min\right)}=100\left(\theta \right)\mathrm{V \left(nm/min\right)}$

The AFIIC rule of thumb comes to (2 nm/min) (3 degrees) (100) = 600 fpm (just 5% in error).

### Book Notes

Portions of this page can be found in the book Flight Lessons 2: Advanced Flight, Chapter 11.

### Bottom Line

So what about the claims this rule of thumb is based on 60-to-1? My conclusion: It is more about trigonometry.

 Rule of Thumb 60-to-1? Trigonometry π Descent Math ✓

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.