F-14 Tomcat Carrier Launch Calculations Explained

The calculations of an F-14 Tomcat Carrier Launch, add to Tom Cruise’s Top Gun movie mystique. We’ll do the calculations to understand why.

The ocean’s airports

the aircraft carrier USS Theodore Roosevelt
The aircraft carrier USS Theodore Roosevelt (Photo Credit: Dvidshub)

The USS Theodore Roosevelt is a nuclear-powered aircraft carrier. It stands 20 stories above the ocean and is 1,092 feet long with a 196,000 square-foot flight deck.

This herculean ship houses 80-plus airplanes and around 6,000 people for several months. But, how do fighter jets take off from these floating airports?…

To answer this question, we need to first understand how planes take off.

The science behind airplane takeoffs

Bernoulli’s principle says fluids, which move faster than their surroundings have less pressure. Our fluid is air, and the goal is to create less pressure on top of the wings than below, by going fast. This pressure differential creates an upwards force on the wings, lifting the airplane.

On a limited runway though, planes use two unique elements not found in land airports to go fast.

#1 Catapult system

The typical airport runway is 2,300 feet long. While an aircraft carrier has a runway of only around 300 feet. This means planes have 2,000 fewer feet of runway to gain speed to take off.

To compensate, aircraft carriers use steam-powered catapults to create lift. This machinery creates enough force to launch planes from 0 to 170 miles per hour in 2 seconds. To point out, the catapult force differs, based on the following airplane attributes:

  • Weight
  • Aerodynamic shape
  • Engine output

If the catapult pressure is incorrectly set though, the following can happen:

  • Pressure set too low: a plane will fall into the ocean from moving too slowly.
  • Pressure set too high: the sudden jerk could break an airplane’s nose gear.

#2 Ocean assistance

Aircraft carriers move fast in oceans while cutting through the air. This can create added airflow on the flight deck when traveling in the direction of a jet’s takeoff. In return, a jet’s required thrust decreases.

Catapult system speed calculation

f-14 tomcat carrier launch
F-14-tomcat-carrier-launch (Photo Credit: Robert Sullivan)

With each launch, operators calculate the catapult’s settings by the specific plane type. But, operators also consider these other following variables:

  • Wind speed across the aircraft carrier deck
  • Air density
  • Airplane carrier speed

Important Note: launching into the wind, the catapult needs to create greater force. Because oncoming wind reduces a jet’s speed. 

To simplify our F-14 Tomcat carrier launch calculation, we’ll disregard these other variables. We’ll just focus on the F-14 Tomcat, Maverick’s (Tom Cruise’s) fighter jet of choice. This incredible jet can sweep its wings backward and forward to do the following:

  • In flight to go high speed, the wings swing back creating a more aerodynamic shape.
  • For takeoff and landing, the wings swing open. This maximizes the surface area of the wings, which we’ll learn the importance of later.

Important Note: the F-14 wings look amazing, but they were a maintenance nightmare. The high number of moving parts created endless points of failure. Even more, the added hardware increased the jet’s weight.

How fast will the catapult shuttle need to travel to launch an F-14 Tomcat?

According to the Aerospace Web, the max takeoff weight of an F-14 Tomcat is 74,350 pounds (33,725 kilograms). We’ll input this weight into the lift formula to do our calculation.

L = C_{L}\times(\dfrac{1}{2}\rho v^{2}) \times s

Where,
L = lift force in pounds or newtons (i.e. the F-14 Tomcat weight)
C_{L} = lift coefficient (i.e. the max takeoff lift coefficient)
\rho = density of air in the takeoff region – at sea level, it’s \rho is 0.0023769 slug/ft^{3}
v = true airspeed (i.e. how fast the airplane travels)
s = surface area of the wings (the F-14 Tomcat wing area when fully forward is 565 ft^{2} (52.49 m^{2}))

Important Note: the C_{L} value is normally determined through experimentation. It considers the shape, inclinations, material density, and flow conditions of a wing. This value gives the performance of airfoils and wings when it comes to lift. In other words, how well a plane can take off. 

Using Haw Hamburg, the takeoff lift coefficient for a fighter jet is between 1.4 to 2.0. This value is from Dr. Jan Roskam’s amazing Airplane Design books. In our calculation, we’ll assume a C_{L} of 1.4.

For the F-14 Tomcat to takeoff, the lift must overcome gravity and equal the weight of the jet. So let’s plug in our values, and solve for v.

v = \sqrt{\dfrac{2L}{\rho s C_{L}}}

v = \sqrt{\dfrac{2\times 74,350}{0.0023769\times565 \times 1.4}}=281.23 feet/second \: (85.72 meters/second)

The lift equation tells us the higher the lift coefficient, the slower a plane needs to travel to take off. Not surprisingly, a slower-traveling airplane is safer on takeoff. This is why a high aspect ratio wing is desirable, even more so, on an aircraft carrier, as we’ll soon learn.

Catapult system speed mistake calculation

What would happen if the catapult system was set to the wrong speed? Say 0.9v, where ‘v’ is the desired catapult speed.

We’ll use our previously calculated F-14 Tomcat values to answer this question. But, we’ll also use the following two jet engine thrust values:

Next, we’ll assume the following:

  • F-14 Tomcat will not use its afterburners
  • The aircraft carrier deck for liftoff is 60 feet (18.29 meters) above the water
  • F-14 Tomcat will leave the edge of the runway traveling horizontally

To start, we calculate how much lift the F-14 produces, by substituting 0.9v into our lift equation.

L = C_{L}\times(\dfrac{1}{2}\rho v^{2}) \times s

L = 1.4\times(\dfrac{1}{2}\times 0.0023769 \times 0.9 \times 281.23^{2}) \times 565 = 66,914.96 lbs \: (297,652.57 \: Newtons)

Once the F-14 Tomcat takes off, the lift on the jet is the following:

\dfrac{66,914.96}{74,350} \times 100 = 90.00% gravity

The 90% signifies the jet falls with an acceleration of 10.00% of gravity.

\Rightarrow 0.1 \times gravity = 0.1 \times 32.2 feet/sec^{2}
= 3.22 feet/sec^{2}\: (0.98 meter/sec^{2})

Where acceleration due to gravity is = 32.2 feet/sec^{2}\: (9.8 meter/sec^{2})

Calculating the time for the jet to fall 60 feet, the height of the aircraft carrier, we have the following:

h = \dfrac{1}{2}at^{2}

t = \sqrt{\dfrac{2h}{a}} = \sqrt{\dfrac{2\times 60}{3.22}} = 6.10 \: seconds

Important Note: we ignore the increase in horizontal velocity for a few seconds after launch. This increase in velocity does create more lift. But, our calculation would then become a hairy calculus problem. 

Calculating thrust

Finally, we calculate the required jet trust to accelerate from 0.9v to v in 6.10 seconds.

The jet’s acceleration: \bigtriangleup v = at, where 281.23 \times 0.9 = 253.11 feet/sec

a = \dfrac{\bigtraingleup v}{t} = \dfrac{281.23-253.11}{6.10} = 4.61 feet/sec^{2}

The jet’s thrust: F=ma

Mass of a F-14 Tomcat = \dfrac{74,350 lbs}{32.2 feet/sec^{2}}=2,309 \: slugs

F = 2,309 \times 4.61 = 10,644.49 \: pounds

The calculated required engine thrust is less than the F-14 Tomcat’s full thrust. So, even with the catapult speed mistake, the jet will still safely take off. This wiggle room is important, given human errors are unavoidable.

Lift factors: manufacturer versus pilot controlled

Having gone over the lift equation, let’s now look more closely at the lift variables. I’ll cram a lot of complex aerodynamic concepts into simple words. So bare with me, as I show how much control a pilot has in creating lift.

s: the airplane manufacturer determines the surface area of the wings. The pilot has no control over this.

\rho: air density is dependent on an airplane’s altitude. The higher an airplane travels, the less dense the air becomes. Then at a set mission altitude, \rho becomes constant, and the pilot can’t change the air density.

V: true airspeed the pilot controls, better called, the Indicating Air Speed (IAS). The pilot views the dynamic pressure, \dfrac{1}{2}\rho v^{2}, as IAS.

C_{L}: the airplane manufacturer controls most parts of this dimensionless figure. This figure is a function of the following:

  • Camber of the airfoil
  • The shape of the planform
  • The Angle of Attack (AOA)

The graphic below shows these 3 factors.

main factors that determine coefficient of lift

Which lift coefficient factors can pilots control?

The pilot can only control the AOA. The AOA is the angle between oncoming air or relative wind and a set assumed line on the airplane. By controlling the pitch of an airplane, AOA, the lift changes.

But if AOA changes with a certain speed adjustment, the lift may not change. Rather, the airplane will maintain level flight, as the lift coefficient changes only.

Lift Coefficient vs Angle of Attack

The relationship between the lift coefficient and AOA is best seen in the below graph.

To increase C_{L}, you increase AOA. But, increase AOA too much and the airplane will stall, and lift decreases. This is where you see the curve start to go downwards in the graph.

Every airplane has its own unique lift coefficient versus the AOA graph. The graph depends on the following:

  • The shape and size of the wings
  • The curvature of the lead edge of the airfoil

lift coefficient versus angle of attack

What’s the takeaway with all these lift variables?

The manufacturer controls most parts of an airplane’s lift. While many interdependent complex takeoff variables exist, as we discussed.

For example, if I increase the pitch of an airplane by 5°, you won’t know how AOA impacts lift. Maybe I lowered my airspeed as well, to maintain a constant altitude with the increased AOA.

Conclusion

The F-14 Tomcat forever lives in pop culture because of the Hollywood movie Top Gun. I remember wanting to become a fighter pilot as a kid because of this movie alone.

Many years later, I learned the F-14 Tomcat has many shortcomings. For one, the jet design was before computer-aided design assistance existed. This led to many aerodynamic flaws. Regardless though, it was an amazing feat of engineering for its time, and it looks badass to this day. I’m still in awe, by how machinery this heavy took off on such short 300-foot runways.

What do you find most fascinating about an F-14 tomcat carrier launch? What part of the engineering with an F-14 Tomcat carrier launch do you think is most critical?


Featured Image Photo Credit: U.S. Navy via Photographer’s Mate 3rd Class Todd Frantom (image cropped)

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