CV-18 Fujian/003 CATOBAR carrier thread

[SinoaTerrenum]

Force is a push or pull that is exerted on an object, while kinetic energy is the energy of an object in motion. Force is what causes an object to move and accelerate, while kinetic energy is the energy the object has due to its motion. So wouldn't the catapult calculation be how much force it needs to generate, and not the kinetic energy of the plane already in motion (kinetic energy more relevant for kamikaze/ramming missions)?

Energy needed does not depend on acceleration? E = 1/2 m V^2 --> 0.5*32500*70 (250km/h) ^2 gets you 80MJ, 290km/h gets you 110MJ so 122MJ seems pretty reasonable?

 

[JebKerman]

122 MJ mentioned previously is energy not force.

A J-15 sitting stationary has 0 kinetic energy, a J-15 at 70m/s has ~80MJ of kinetic energy. So to accelerate from 0-70m/s you need to add at least 80MJ to the aircraft, doesn't matter how you do it. If your calculation gives you less then 80MJ you made a mistake somewhere.

To correct your calculation:
Take your example: Mass M = 40000, final speed V = 100m/s, acceleration a = 50m/s^2
So force required for the acceleration F = Ma = 40000*50 = 2MN (this Force in N, not Energy in J)
To find work done due to the force (Energy added = force x distance): W = F*d = 2MN * 100m = 200MJ

But if we simply calculate the kinetic energy for this example you will find: E = 1/2 * 40000kg * (100m/s)^2 = 200MJ. We get the same number without having to make any assumptions about acceleration and take off distance.

 

[taxiya]

Pardon my basic question but why does the launch require so much energy? Basic physics dictate Force = mass x acceleration.

Maximum loaded weight of J-15 is 32,500 kg, let's round up to 40,000 kgs for argument's sake, so mass = 40,000.
Take-off velocity on US carriers is ~170 knots/hr, or 87.22 m/s. Let's round that up to 100 m/s to be conservative.
It takes ~2 seconds to catapult aircraft off, so acceleration needs to be 50 m/s.
Taking integral of 50 m/s accelerating over 2 seconds from initial velocity of 0 is (1/2 * 50) * 2^2 = 100 meters, which is roughly the runway distance on a carrier (300 ft/90 meters but close enough)

50 * 40,000 = 2 million newtons, 2 million joules, or 2 MJ.
Even assuming a 10% conversion/efficiency ratio, only 20 MJ energy needed. Even reducing time to 1.5 seconds and acceleration to 67 m/s2 gets you 2,666,667 newtons/joules, or 2.67 MJ. Unless team catapults only 2-3% efficient?

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calculation based on your input gives 123MJ.

 

[taxiya]

delete as @JebKerman has made the explanation.

 

[SinoaTerrenum]

122 MJ mentioned previously is energy not force.

A J-15 sitting stationary has 0 kinetic energy, a J-15 at 70m/s has ~80MJ of kinetic energy. So to accelerate from 0-70m/s you need to add at least 80MJ to the aircraft, doesn't matter how you do it. If your calculation gives you less then 80MJ you made a mistake somewhere.

To correct your calculation:
Take your example: Mass M = 40000, final speed V = 100m/s, acceleration a = 50m/s^2
So force required for the acceleration F = Ma = 40000*50 = 2MN (this Force in N, not Energy in J)
To find work done due to the force (Energy added = force x distance): W = F*d = 2MN * 100m = 200MJ

But if we simply calculate the kinetic energy for this example you will find: E = 1/2 * 40000kg * (100m/s)^2 = 200MJ. We get the same number without having to make any assumptions about acceleration and take off distance.

Thanks! Seems like I was missing the distance and Force to Work conversion, appreciate it.

 

[asif iqbal]

122 MJ mentioned previously is energy not force.

A J-15 sitting stationary has 0 kinetic energy, a J-15 at 70m/s has ~80MJ of kinetic energy. So to accelerate from 0-70m/s you need to add at least 80MJ to the aircraft, doesn't matter how you do it. If your calculation gives you less then 80MJ you made a mistake somewhere.

To correct your calculation:
Take your example: Mass M = 40000, final speed V = 100m/s, acceleration a = 50m/s^2
So force required for the acceleration F = Ma = 40000*50 = 2MN (this Force in N, not Energy in J)
To find work done due to the force (Energy added = force x distance): W = F*d = 2MN * 100m = 200MJ

But if we simply calculate the kinetic energy for this example you will find: E = 1/2 * 40000kg * (100m/s)^2 = 200MJ. We get the same number without having to make any assumptions about acceleration and take off distance.

these generic calculations do not take into account real life parameters like wind, air and sea state

so are meaningless

 

[by78]

Not sure when this was taken, but sharing it nonetheless because it provides a nice comparison between Fujian and the Type 075.

 

[DaTang cavalry]

These are the latest pictures

 

[Quickie]

Thanks! Seems like I was missing the distance and Force to Work conversion, appreciate it.

The big missing part is the plane's aerodynamic drag force. The higher the takeoff weight of the aircraft, the higher the aerodynamic drag force. The catapult itself would not be 100% efficient and would present some resistive force.

 

[DaTang cavalry]

One more photo of the 003