# Fighter Pilot Math

To be sure,designinga fighter requires aerodynamictheories and principles involving advanced mathematics, but flying one requires mainly simple arithmetic. And the most common concept pilots need to master is ‘rate.’ Most of us are already familiar with various rates used in everyday life—like the speed of an automobile or the $3.499 price of gasoline (I hate that trailing 9!). Mathematically, rate is defined as the ratio between two related quantities in different units, like miles per hour or dollars per gallon. Common rates used in flying include speed (kts, knots, or nautical miles per hour), rate of climb or descent (fpm, or feet per minute), and fuel burn (pph, or pounds per hour). I'm not sure if it is considered being “good at math” to know that flying along at 360 kts is the equivalent to 6 nm per minute, or covering a mile every 10 seconds1; or that when air traffic control directs you to climb from 20,000 feet to 29,000 feet and to be level within the next six minutes that you need to maintain a minimum climb rate of 1,500 fpm2, but those are examples of the types of rate calculations that need to be second nature to fighter pilots. Fuel burn awareness is also important, especially for Navy folks around the ship where fuel is always at a premium—as is a place to land. If I’m in my trusty F/A-18C, for example, burning 2,400 pph per engine at high holding, it’s 20 minutes to land, and I need to cross the ramp with at least a 5.0 (5,000 pounds of fuel on board), how much fuel should I have right now? Easy, 6.63. Other math I routinely used during my flying career involved angular relationships. Without delving into all the reasons why (let’s leave that to the mathematicians), it is accepted that 1° equates to 100 feet at 1 nm. That is, if an aircraft is flying down a 3° glideslope, say, then the pilot would expect to be 1,200 feet above the touchdown elevation when 4 miles away, 900 feet at 3, 600 at 2,and300 at 1.

360 kts = 360 nm / 60 min = 6 nm / min = 6 nm / 60 sec = 1 nm / 10 sec.

29,000 – 20,000 = 9,000 feet. 9,000 feet ÷ 6 minutes = 1,500 fpm.

2,400 pph * 2 engines = 4,800 pph total. 20 minutes = 1/3 of an hour. 1/3 of 4,800 lbs = 1,600 lbs. 1,600 lbs + 5,000 lbs = 6,600 lbs. Never mind the fuel saved in the idle descent, that’s gravy.

If x° = x*100 ft / nm then the number of nm = x*100 ft / x°. In our example 2,000 / 2.5 = 8 (ignoring units, which is usually a mathematical no-no).

If 1° = 100 ft / nm then 1° = 6,000 feet / 60 nm, and 6,000 feet is 1 nm (not exactly, but close enough).

Arcing from the 240 to 270 radial is 30°, and 30° = 60,000 feet / 20 nm, and 60,000 feet is 10 nm.

240 kts = 4 nm / min, and 10 nm ÷ 4 nm / min = 2.5 minutes.

Given footnote 5, 1° = 1/3 nm at 20 nm or 3° = 1 nm there. So a 3 nm turn requires 9°, thus the turn should begin 9° before the 270 while arcing clockwise, which is the 261 radial.