To estimate the takeoff distance of the F-35C without a catapult at 60% fuel and no weapons, we need to consider several aerodynamic and performance factors:
Step 1: Define Aircraft Parameters
Weight (W): 21,176 kg (from previous calculation)
Takeoff Speed (V_TO): Estimated 150 knots (77 m/s) for this configuration
Thrust (T): 191 kN (afterburner)
Wing Area (S): 62.1 m²
Lift Coefficient at Takeoff (C_L_TO): Estimated ~1.0 with flaps deployed
Drag Coefficient (C_D): Estimated ~0.03 in takeoff configuration
Air Density (ρ): 1.225 kg/m³ (at sea level)
Step 2: Calculate Lift (L) at Takeoff Speed
Lift equation:
L = \frac{1}{2} \rho V^2 S C_L
L = \frac{1}{2} (1.225) (77)^2 (62.1) (1.0) ]
L ≈ 226.5 \text{ kN}
Since weight (21,176 kg × 9.81 m/s² = 207.7 kN) is lower than lift (226.5 kN), the aircraft can take off at this speed.
Step 3: Calculate Acceleration (a)
Net force equation:
F_{\text{net}} = T - D - R
Drag (D) =
Rolling resistance (R) ≈ 0.02 W (estimated for carrier decks)
D = \frac{1}{2} (1.225) (77)^2 (62.1) (0.03) \approx 6.8 \text{ kN}
R = 0.02 \times 207.7 \approx 4.2 \text{ kN} ]
F_{\text{net}} = 191 - 6.8 - 4.2 = 180 \text{ kN}
a = \frac{F_{\text{net}}}{m} = \frac{180,000}{21,176} \approx 8.5 \text{ m/s}^2 ]
Step 4: Estimate Takeoff Distance
Using kinematic equation:
V^2 = 2 a d
d = \frac{V^2}{2a} = \frac{(77)^2}{2(8.5)} ]
d ≈ 350 \text{ m}
Conclusion: Estimated Takeoff Distance
Takeoff run ≈ 350 meters (1150 feet) on a flat surface.
Now for, F-35C Takeoff from INS Vikrant's 14° Ski Jump (200m Runway)
Given:
Ski-jump angle: 14°
Ski-jump length: ~30m
Total available takeoff run: ~200m
Weight (60% fuel, no weapons): 21,176 kg
Thrust (afterburner): 191 kN
Lift coefficient (C_L_TO): ~1.0
Drag coefficient (C_D): ~0.03
Takeoff speed (V_TO, estimated): ~77 m/s (150 knots)
Acceleration (flat deck, calculated earlier): 8.5 m/s²
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Step 1: Check Acceleration Over 200m
Using the kinematic equation:
V^2 = 2 a d
V = \sqrt{2 (8.5) (200)} ]
V \approx 58 m/s
At 200m, the F-35C reaches ~58 m/s (112 knots), which is below the estimated takeoff speed of ~77 m/s. This means it would not take off purely based on runway acceleration—it relies on the ski jump.
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Step 2: Effect of the 14° Ski Jump
The ski jump provides a vertical velocity component (V_y) upon exit:
V_y = V \sin(14^\circ)
V_y \approx 58 \times 0.241 ]
V_y \approx 14 m/s
The horizontal velocity (V_x) remains:
V_x = V \cos(14^\circ)
V_x \approx 58 \times 0.97 ]
V_x \approx 56 m/s
Total velocity after ski-jump:
V_{\text{total}} = \sqrt{V_x^2 + V_y^2}
V_{\text{total}} = \sqrt{56^2 + 14^2} \approx 58.7 \text{ m/s} ]
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Step 3: Check If F-35C Can Stay Airborne
The aircraft will stay airborne if lift force (L) is at least equal to weight (W).
L = \frac{1}{2} \rho V^2 S C_L
L = \frac{1}{2} (1.225) (58.7)^2 (62.1) (1.0) ]
L \approx 161 \text{ kN}
Required lift (weight force):
W = 21,176 \times 9.81 = 207.7 \text{ kN}
Since 161 kN < 207.7 kN, the F-35C does not generate enough lift immediately after leaving the ski jump and would start descending.
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Step 4: Can the F-35C Recover in Flight?
The aircraft still has significant thrust (191 kN) vs. drag (~6.8 kN at 58 m/s), meaning it can accelerate further after launch.
The F-35C's high angle of attack (AoA) capability allows it to pitch up aggressively, increasing C_L to ~1.5-1.7, generating more lift.
If the pilot manages a smooth AoA transition, the aircraft may sustain flight after a brief altitude loss.
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Final Verdict: Can the F-35C Take Off from INS Vikrant?
Barely possible, but extremely risky.
Will experience altitude drop after the ski jump before regaining lift.
Would require perfect AoA control and afterburner thrust.
Would severely limit takeoff payload (no weapons, possibly less fuel).
Practical Answer: No, the F-35C is not suited for Vikrant's STOBAR setup. A STOVL aircraft like the F-35B or MiG-29K/Rafale-M with better low-speed handling is required.
We ain't getting f35 "C"
There's a reason UK went with F 35B for its carriers not the C varient
and Opposition - 
and Opposition - 
