# Easy Mental Arithmetic for Pilots

How to quickly and accurately do maths as a pilot

## Easy Mental Arithmetic for Pilots

When flying an aircraft, whether it’s a Cessna 152 or an A380, pilots need to be able to do fairly basic but quite quick mental arithmetic. Of course, there are occasions where complete accuracy is critical, but the vast majority of the time, you don’t need to be ‘bang on’, but rather work to rough ‘ballpark’ figures. Whether it be calculating required descent rates or speed/distance/time calculations, general rules of thumb can help you make these calculations quickly and reasonably accurately.

These rules of thumb also work very well when completing pilot numerical reasoning aptitude tests as part of airline pilot assessments. Such tests often require you to work quickly, but with a number of multiple-choice answers, you often just need to reach a rough figure rather than an exact one. Quickly being able to calculate a good estimate in test conditions can prove to be invaluable in passing an airline pilot selection process.

Some of the most important rules of thumb involve using the 3 times table, the 1 in 60 rule and being able to divide or multiply by 10. Let’s have a look at some examples:

### Distance Required to Descend for Aircraft Calculations

Most aircraft plan to descend at an angle of approximately 3 degrees. To calculate how much distance an aircraft needs to fly to achieve a given reduction in altitude, based on a 3-degree angle of descent, a basic rule of thumb can be used:

Distance Required to Reduce Altitude = Total Altitude to Lose / 1,000 x 3.

For example, if you are at 40,000ft and you need to be at 10,000ft at 30NM before the airfield, you need to lose a total of 30,000ft. Divide 30,000ft by 1,000 (simply take away the last 3 numbers when dividing by 1,000), which gives you 30 and multiply this by 3 gives you 90. It will therefore take you about 90NM to reduce altitude by 30,000ft. If you need to be at 10,000ft by 30NM before the airfield, then add this to the 90NM which gives you a start of descent point of 120NM.

This assumes still air conditions at a constant speed. Whilst it depends on aircraft types, for commercial aircraft, adding an extra mile for every 10kts of airspeed you need to lose is a good ballpark figure. So, in the above example, if you start the descent at 300kts IAS, and need to be at 200kts IAS by the time you reach 10,000ft, you would add 10NM to the distance required to lose the altitude (so 90NM becomes 100NM).

Having a headwind or tailwind also needs to be factored when calculating the distance required to descend. As a rough guide, add 1 NM to the distance required to descend for every 10kts of tailwind and reduce the distance by 1 NM for every 10kts of headwind.

### Descent Rate Required to Achieve a 3 Degree Descent Angle when Flying

So, you’ve calculated the distance required to descend to a given altitude using the above method. Using the above example, you will be descending at a 3-degree angle over 90NM. But if you are descending at 3 degrees, what descent rate do you need to achieve? An easy way to calculate this is using this basic formula.

3 Degree Descent Rate = 5 x Ground Speed

For example, if you are flying at a ground speed of 300kts, multiply 300 by 5 and this tells you that you would need to descend at 1,500fpm to achieve a 3-degree descent profile. Some people prefer to multiply the ground speed by 10 then divide by 2. Clearly your ground speed will change with altitude as the True Airspeed and Head/Tailwind changes so you will need to periodically review your rate of descent throughout the manoeuvre.

### Descend to an Altitude within a Fixed Time Period

ATC will sometimes require an aircraft to descend to a given altitude within a specific time period. For example, “FDF123 descend to Flight Level 320 to be level within 4 minutes”. In this type of scenario, you need to calculate how many feet per minute you need to descend in order to achieve this restriction. This can be calculated using the following method:

Feet Per Minute Required = Total Altitude to Lose / Number of Minutes

For example, if flight FDF123 is maintaining FL360 (36,000ft) and has been told to descend to FL320 (32,000ft) within 4 minutes, the total altitude required to lose is 4,000ft. 4,000 divided by 4 is 1,000, so the aircraft needs to descend at 1,000ft per minute to meet the restriction.

In such a scenario, you don’t necessarily need to be exact, sometimes you can simplify and be conservative with your calculations since the request is usually ‘within 4 minutes’ not ‘exactly 4 minutes’. For example, if you are flying at 25,000ft and are told to descent to 17,000ft to be level within 9 minutes, we know the calculation is 8,000 / 9 (which equals 888 fpm). However, we can turn these into round numbers to make the calculations easier, just remember to do it in a conservative way to ensure the restriction can be made. For example, we can hopefully quite quickly work out that if we descended at 1,000 fpm, we would descend 8,000ft in 8 minutes. Yes, we’d be levelling off one minute earlier than the restriction required, but we have achieved ATCs request.

### Speed, Distance and Time Calculations for Pilots

We are probably all aware of the relationship between variables from school and have heard of the Speed, Distance & Time triangle. When flying, we should always be aware of our speed so calculating distance and time is more relevant.

There is a ‘magic triangle’ which can help us quickly remember how to calculate speed, distance, and time. You simply cover up the entity you are trying to find and the reveals how to calculate it. For example, if you cover the ‘S’ you can see that the calculation for speed is distance divided by time.

• Speed = Distance / Time
• Distance = Speed x Time
• Time = Distance / Speed ### Easy Speed / Distance / Time Calculations for Pilots

Speed is the distance you travel over a specific time period, so they are intrinsically related. It’s worth understanding some rules of thumb which can help you make quick calculations about distance and time calculations.

• 30kts = 0.5NM per minute
• 60kts = 1NM per minute
• 120kts = 2NM per minute
• 180kts = 3NM per minute
• 240kts = 4NM per minute
• 300kts = 5NM per minute
• 480kts = 8NM per minute
• 540kts = 9NM per minute
• 600kts = 10NM per minute

Remember that there are 60 minutes in an hour. Well therefore, if we divide any speed by 10, this will tell us what distance the aircraft is travelling in 6 minutes at its current speed.

For example, if an aircraft is flying at 150kts, this tells us that it is travelling 15NMs every 6 minutes (150 divided by 10 = 15). Another example is that if the aircraft is travelling at 370kts it is covering 37NM every 6 minutes. You could then halve this number to see how far the aircraft travel in 3 minutes, 1.5 minutes etc.

Put another way, if asked ‘how many miles will you travel in 20 minutes at a speed of 180kts?’. 180 divided by 10 is 18, so 18 miles every 6 minutes. So, if we multiply this number by 3, we know how many miles are covered in 18 minutes (3 x 18 miles = 54NMs). If we cover 18 miles every 6 minutes, we know we cover 9 miles every 3 minutes (it’s then easy to see that it’s actually a mile a minute in this example!). So therefore, we are covering 63 miles every 21 minutes. Knock 3 miles off and we get to 60 NM.

#### Example Distance to Descend Questions for Pilots

Here’s a few example questions. We’ve got lots more pilot numerical reasoning test example questions over on our dedicated page. The BBC GCSE Bitesize website is also a great resource to help you practice your mental arithmetic.

If an aircraft is flying at an intermediate altitude of 25,000ft and is instructed by ATC to achieve an altitude of 13,000ft by a fix on the arrival, what distance before the fix should the pilots initiate the descent, assuming a planned 3-degree descent profile at a constant speed and still wind?

• A) 40 NM
• B) 38 NM
• C) 36 NM
• D) 28 NM

25,000ft minus 13,000ft = 12,000ft to lose. 12,000 divided by 1000 = 12, multiplied by 3 = 36.

ATC have told you to self-position to a 10NM extended centreline from the landing runway. There are no restrictions other than needing to achieve an altitude of 3,000ft and at a speed of 180kts at the 10NM point. You are currently level at 8,000ft at 250kts and anticipate an average tailwind of 10kts. In order to achieve the restriction, at what point before the 10NM fix should you commence the descent?

• A) 23 NM
• B) 15 NM
• C) 22NM
• D) 16NM

8,000ft – 3,000ft = 5,000ft. 5,000 divided by 1,000 = 5, multiplied by 3 = 15 NM. 250kts – 180kts = 70kts = add on an extra 7NM (1NM per 10kts of airspeed to lose). Add 1NM per knot of tailwind. 15 NM (distance required) + 7 NM (to account for deceleration) + 1 NM (to allow for tailwind) = 23NM.

#### Example Descent Rate Required Questions

With a 280kts ground speed, what rate of descent do you need to achieve in order to maintain a 3-degree descent angle?

• A) 2,800 fpm
• B) 1,400 fpm
• C) 700 fpm
• D) 2,000 fpm

5 x 280 = 1,400. Or 280 x 10 = 2,800 / 10 = 1,400.

With a 420kts ground speed, what rate of descent do you need to achieve in order to maintain a 3-degree descent angle?

• A) 2,000 fpm
• B) 1,800 fpm
• C) 2,800 fpm
• D) 2,100 fpm

5 x 420 = 2,100 fpm. Or 420 x 10 = 4,200 / 10 = 2,100.

#### Descend to an Altitude within a Fixed Time Period Questions

If an aircraft is maintaining 27,000ft and has been told by ATC to descend to 13,000ft within 10 minutes, what rate of descent is required?

• A) 1,400 fpm
• B) 1,200 fpm
• C) 1,300 fpm
• D) 1,500 fpm

27,000 – 13,000 = 14,000. 14,000 divided by 10 minutes = 1,400 fpm

If an aircraft is maintaining 35,000ft and has been told by ATC to descend to 30,000ft within 3 minutes, what rate of descent is required to the nearest 100fpm?

• A) 1,600 fpm
• B) 1,700 fpm
• C) 1,500 fpm
• D) 1,800 fpm

35,000 – 30,000 = 5,000 fpm divided by 3 minutes = 1,666 fpm.