For an estimate, we can use the Brequet range equation:

$range = V\cdot t_f = \frac{L}{D} \times I_{sp} \times \ln\left(\frac{W_i}{W_f}\right) $

with:

• $V$: optimal velocity
• $t_f$: flight time
• $L$: lift
• $D$: drag
• $I_{sp}$: propulsion efficiency
• $W_i$: initial weight
• $W_f$: final weight

Assuming the ratio of lift to drag ratio $\left(\frac{L}{D}\right)$ and the propulsion efficiency will stay constant, the range will depend on the ratio of initial and final weight.

$range = C \times \ln\left(\frac{W_i}{W_f}\right)$

In case the aircraft departs with the maximum payload, the weight is:

$W_i = W_{oe} + W_{payload} + W_{fuel} = W_{mto}$

• $W_{oe}$: operating empty weight: 127 500 kgf
• $W_{payload}$: payload weight: 51 000 kgf
• $W_{mto}$: maximum take-off weight: 227 000 kgf
• $W_{fuel}$: fuel weight = $W_{mto} - W_{payload}-W_{oe}$ = 48 500 kgf

The final weight at maximum range is then: $W_f = W_{oe} + W_{payload}$

Filling in the numbers we find:

4000 km = $C \times \ln\left(\frac{\textrm{227 000}}{\textrm{127 500 + 51 000}}\right)$

That means $C \approx \textrm{16 640}$ for the Beluga XL.

If we now remove the payload and take-off with the same amount of fuel, we find:

• $W_i$ = 176 000 kgf
• $W_f$ = 127 500 kgf
• $range \approx \textrm{16 640} \times \ln\left(\frac{\textrm{176 000}}{\textrm{127 500}}\right) \approx 5365$km

But that is just removing the payload without adding fuel. Of course we can now bring more fuel on board without exceeding the maximum take-off weight.

The fuel capacity of the Beluga XL is 73000 kg (source:Aircraft Characteristics Airport And Maintenance Planning Beluga XL manual). This changes the numbers to:

• $W_i$ = 200 500 kgf
• $W_f$ = 127 500 kgf
• $range \approx \textrm{16 640} \times \ln\left(\frac{\textrm{200 500}}{\textrm{127 500}}\right) \approx 7530$km

So the Beluga can fly approximately 7500 km when empty, enough to cross the Atlantic from Toulouse, even if there is normal headwind.

For an estimate, we can use the Brequet range equation:

$range = V\cdot t_f = \frac{L}{D} \times I_{sp} \times \ln\left(\frac{W_i}{W_f}\right) $

with:

• $V$: optimal velocity
• $t_f$: flight time
• $L$: lift
• $D$: drag
• $I_{sp}$: propulsion efficiency
• $W_i$: initial weight
• $W_f$: final weight

Assuming the lift to drag ratio $\left(\frac{L}{D}\right)$ and the propulsion efficiency will stay constant, the range will depend on the ratio of initial and final weight.

$range = C \times \ln\left(\frac{W_i}{W_f}\right)$

In case the aircraft departs with the maximum payload, the weight is:

$W_i = W_{oe} + W_{payload} + W_{fuel} = W_{mto}$

• $W_{oe}$: operating empty weight: 127 500 kgf
• $W_{payload}$: payload weight: 51 000 kgf
• $W_{mto}$: maximum take-off weight: 227 000 kgf
• $W_{fuel}$: fuel weight = $W_{mto} - W_{payload}-W_{oe}$ = 48 500 kgf

The final weight at maximum range is then: $W_f = W_{oe} + W_{payload}$

Filling in the numbers we find:

4000 km = $C \times \ln\left(\frac{\textrm{227 000}}{\textrm{127 500 + 51 000}}\right)$

That means $C \approx \textrm{16 640}$ for the Beluga XL.

If we now remove the payload and take-off with the same amount of fuel, we find:

• $W_i$ = 176 000 kgf
• $W_f$ = 127 500 kgf
• $range \approx \textrm{16 640} \times \ln\left(\frac{\textrm{176 000}}{\textrm{127 500}}\right) \approx 5365$km

But that is just removing the payload without adding fuel. Of course we can now bring more fuel on board without exceeding the maximum take-off weight.

The fuel capacity of the Beluga XL is 73000 kg (source:Aircraft Characteristics Airport And Maintenance Planning Beluga XL manual). This changes the numbers to:

• $W_i$ = 200 500 kgf
• $W_f$ = 127 500 kgf
• $range \approx \textrm{16 640} \times \ln\left(\frac{\textrm{200 500}}{\textrm{127 500}}\right) \approx 7530$km

So the Beluga can fly approximately 7500 km when empty, enough to cross the Atlantic from Toulouse, even if there is normal headwind.