So when we put loads under the wings we modify the aerodynamics and that is a source of difficulty for the fan boys because you have to understand a little bit of what is happening and in the end you have to at least know how to make rules of three.
To start, I suggest you read this post:
View topic - Drag index calculation • F-16.net
The example being treated concerns the F-16 and the reference surface (300 ft^2) in question is the wing surface as is traditionally the case.
For the Rafale it is 492 ft^2.
To calculate the drag index of the Rafale we can put ourselves in the same conditions as the example, we therefore need the drag of the Rafale smooth at M 0.8 at sea level. This drag will be balanced by the thrust of the engines or these consume 44 kg of fuel per minute in these conditions, or 2640 kg per hour.
As at this speed we do not use post-combustion, the specific consumption is 0.80 kg/daN·h which gives a thrust of 3300 daN or 7425 lbf.
Calculation of the Rafale Drag index:
Reference surface S of the Rafale: 492 sq. ft^2.
Drag D of the smooth Rafale at 0.8 mach, at sea level: 7425 pounds.
D = Cd * q * S
Cd = D / q*S
q at .8 sea level is 947 lb/ sq ft
S = 492 sq ft^2
n/a Cd = 0.0159
Basic airplane drag index: 159
The index for a 2000 l tank is 20.5 and for a 1150 l tank it is 14.2.
We will now calculate the index of conformal tanks using the same principle as that which allowed me to calculate that of tanks but I only give this calculation once: the difference is that for tanks I suppose that Half of the fuel added by the tanks is used to compensate for the extra drag whereas for conformal tanks I guess it's only 1/4.
We first consider a smooth Rafale
Its range with a full tank of fuel is noted as X in km and a full tank of fuel is noted as Y in liters.
The consumption to travel 1km is therefore Z=Y/X
We now consider a Rafale with two conformal tanks, each containing y liters of fuel.
According to the rule it takes half a tank to compensate for the drag. The autonomy will be (Y+1.5y)/Z and the consumption for 1 km will be
Z1 = (Y+2y)/((Y+1.5y)/Z)
The consumption being proportional to the drag index, the increase in drag index for the plane is 159*((Z1/Z)-1)
Or Z1/Z= (Y+2y)/(Y+1.5y) = 1 +0.5 y/(Y+1.5y) and therefore the increase in the drag index is 159*0.5y/( Y+1.5y).
As the Rafale can carry 5750 l internally this gives:
For a 1150 l tank: 159*575/(5750 + 1725) = 12.23 and as it is for the plane this is half per tank or 6.12.
How are drag indexes used?
Usually we say that the drag index of a Rafale carrying a GBU 12 is 159+4.27= 163.27 and if the range of a smooth Rafale is 1850 km then that of a Rafale carrying a GBU 12 will be 1850*159/163.27 = 1801.61.
In fact, it is the drag that increases and the Rafale is forced to increase its thrust to compensate. It is because we only have only one specific consumption that the rule of three works (but it is an approximation), if the specific consumption varied, it would have to be taken into account.
With this method we calculate the consumption for each segment when loads are released.
This data allows us to calculate interesting things for example we know that the Ferry Range is 3700 Km with 3 tanks of 2000 l :
We therefore have an index of 159 +3*20.5 =220.5 and a fuel volume of 6000 +5750 = 11750 l
At index 159 we could do 3700km *220.5/159 =X but with only 5750l we would do
5750*X/11650 =5750*3700*220.5/(159*11650)=2532.5km which means that the F-35 could cover 1.31 times this distance or 3318 km with our previous calculation which did not take into account the hold penalty.
We therefore have a theoretical F-35 which does not have a hold for which we will try to calculate the index, knowing that we will then calculate the index of the hold which will be added to this theoretical index.
The F-35 will consume 8278 kg to fly 31% further than a Rafale which consumes 4700 kg. To fly the same distance it will therefore consume 8278/1.31 = 6319 kg. Furthermore, the specific consumptions are not the same and as we only want to measure aerodynamic effects we must still correct the mass of fuel required by assuming that we have installed engines of the same efficiency in both aircraft:
the new mass is therefore 6319 * 80/88.6 = 5706
In the Rafale frame of reference the Index of the F-35 is therefore 159*5706/4700 = 193, but we must put ourselves in the frame of reference of the F-35 which has a wing area of 460 sq. ft^2. and so the Index will be: 193 * 460/492 = 180.5
To calculate the hold index we will consider the combat range of the air-air configuration which is 1410 km or 2820 km round trip.
We have 3318*180.5/X = 2820 where X is the drag index with the hold.
We have X = 212.38 and the hold index is 212.38 - 180.5 = 31.88