Top Gun

Launch worksheet

Launch solution 

The Movie

Top Gun is a blockbuster 1986 military film starring Tom Cruise in one of his earliest major roles. It probably did more for the military in recruiting pilots than any other movie or advertisement campaign they have ever done. With fighter jets and action there are plenty of mathematical scenarios including carrier launches and landings and buzzing the control tower.

You can view the trailer here:

Math Scenario 1: Carrier Launch

To take off from the carrier in your fighter, you get the assistance of a catapult launch system. The catapult helps accelerate the plane over the 60 or 70 metres of the deck so that the jet leaves the carrier at a high enough speed to fly.

We can work out the average acceleration of the plane during the catapult launch. An F-14 (the Top Gun movies had F-14s) leaves the carrier at about 310 km/hr, after being accelerated by the catapult for about 2.8 seconds. However, that final speed includes the carrier’s speed which is already travelling at about 60 km/hr.

Acceleration = change in speed / time

Acceleration = (310 km/hr – 60 km/hr) / 2.8s

Acceleration = 89.29 km/hr/s

People often like to express accelerations as the “number of gs” – how strong the acceleration is relative to gravity, which is what keeps us on the surface of the earth. gravity is 9.81 m/s/s:

Number gs = acceleration / gravity

Number gs = 89.29 km/hr/s / 9.81 m/s/s

Need to do some conversion to get into consistent units of metres and seconds:

Number gs = 89.29 km/hr/s / 9.81 m/s/s

Number gs = 89.29 km/hr/s * 1000 m/km / 3600 s/hr / 9.81 m/s/s

Number gs = 2.528 gs

So that’s an average acceleration felt by the pilot of about 2.5 gs. For an 80 kilogram pilot, they would feel like they were being pressed back into the seat and weighed about 200 kg during launch!

Math Scenario 2: Landing 

[] Public domain photo prepared by an officer or employee of the United States Government as part of that person’s official duties under the terms of Title 17, Chapter 1, Section 105 of the US Code.

Early in the movie, Maverick’s (Tom Cruise) wingman Cougar is low on fuel and shaken by an engagement. He has to land in a hurry, but he’s lost his confidence and his jet is maneuvering wildly.

We can examine the mathematical challenges of landing on a carrier.

The initial stages of landing – from about 20 km to 2 km – is done using the assistance of a Long Range Laser Lineup System (LLS) [1].

The final stage is done visually, with the pilot lining up visual cues in the aircraft with lights or markings on the carrier deck. The pilot typically aims for the middle arrestor wire (out of four arrestor wires), which helps rapidly slow the jet down once it lands.

Landing typically involves a deceleration from about  240 km/hr to zero in about 2 seconds. If the carrier is moving in the same direction as the aircraft at 60 km/hr, the deceleration is:

Deceleration = change in speed / time

Deceleration = (240 – 60) / 2

Deceleration = 90 km/hr/s

Deceleration = 90 km/hr/s * 1000 m/km / 3600 s/hr

Deceleration = 25 m/s/s

Deceleration = 2.55 gs

If Cougar had missed the initial landing attempt, he would have had to execute a go-around, which in commercial operations occurs about 1-3 times in every 1000 landing approaches.

Math Scenario 3: Buzzing the Tower

One of the cheeky moves pulled off by Maverick is called “buzzing the tower” – flying very close the to the air control tower without permission:

Such a move requires expert control of the plane, and any momentary distraction could prove fatal. We can examine how much of a navigation error would result in a catastrophic collision with the tower or the ground, rather than an impressive fly by.

Let’s say Maverick plans to clear the tower by 20 metres. If his plane is flying at 700 km/hr, and he starts his line up 1000 metres out, how big a heading error (in degrees) would result in him smashing into the control tower?

We can use trigonometry – in this case tan – to work it out:

tan(heading error) = opposite / adjacent

Now take the inverse tan of both sides:

heading error = atan(opposite / adjacent)

heading error = atan(20 / 1000)

heading error = 1.146 degrees

Math Scenario 4: High G Turns

Fighter pilots are famous for going through “high G” turns where they are subjected to feeling high G forces many times more than you normally do during your typical daily activities.

The exact amount of g force a human can withstand without blacking out varies, depending on the orientation of the body and whether the human is trained to withstand the force. Experienced pilots wearing special suits may be able to withstand up to 9 gs in turns, while a normal person may black out at 5 gs or less.

Plane airframes are also only rated to a certain g force, after which they start to lose their structural integrity.

The formula for centripetal force (the force felt on the object due to circular motion) is:

F = mv^2/r

If we’re just interested in calculating G forces, we can remove the “m” for mass and replace this formula with:

G force = v^2 / r

Let’s take an example of a fighter jet travelling at 800 km/hr, trying to turn in a circle of radius 10000 metres (we’ll ignore gravity for this calculation). We’ll first need to convert to standard units of m/s:

800 km/hr = 800 km/hr * 1000 m/km / 3600 s/hr

800 km/hr = 222.22 m/s

G force = v^2 / r

G force = 222.22^2 / 10000

G force = 4.938

A force that might be too much for a civilian, but probably withstandable by a trained fighter pilot.

Real Life Example: Testing Carrier Aircraft

Carrier landings are tough on the aircraft, and even tougher if the carrier deck is pitching upwards, increasing the impact of the landing.

Carrier aircraft have been tested for this stress by being repeatedly lifted up above the ground and dropped to simulate a certain impact speed [2].

Let’s say designers though that their aircraft would need to be robust to an impact of 6 m/s.

Ignoring air resistance, we can work out how high the aircraft should be dropped from. First we can work out the total drop time required:

impact velocity = drop time * acceleration 

drop time = impact velocity / acceleration

drop time = 6 m/s / 9.81 m/s/s

drop time = 0.6116 seconds

Then we can workout how far the plane will fall during this time (it starts at zero speed and then falls):

drop height = 0.5 * a * t^2

drop height = 0.5 * 9.81 * 0.6116^2

drop height = 1.835 m

The new F-35C is rated to a maximum impact of 26.4 feet per second during carrier landings, which is 8.05 m/s.