Re: Chinese Engine Development
The problem with satellite pictures is that with such poor resolution it's adequate enough to get length + wingspan, but it's a lot harder to directly get area for measurements. It is hard to tell where shadow ends and wing begins, so you're going to have a much higher error rate.
With regards to what I'm doing, however, is that if you see the wing area calculation for the F-22, it's based on calculating the triangle formed by the forward sweep of the wing, then using that to form a hexagon including the rearward sweep of the wing.
Doing something equivalent for the J-20, the shape formed is roughly a hexagon and you can solve it trigonometrically.
Split the shape into roughly 3 sections, upper triangle, lower triangle, and middle rectangle. The area of the upper triangle is wingspan * length of upper triangle * .5, the area of the lower triangle is wingspan * length of lower triangle * .5, while the area of the middle rectangle is length of the middle rectangle * wingspan.
Added together, it's roughly (wingspan * length of upper triangle * .5) + (wingspan * length of lower triangle * .5) + (length of the middle rectangle * wingspan), factor out and you get ( (length of upper triangle + length of lower triangle) * .5 + length of middle rectangle) * wingspan. Multiply and divide each term by 2 and you get roughly (length of upper triangle + length of lower triangle + 2*length of middle rectangle) * wingspan * .5. If you consider that length of upper triangle + length of lower triangle + length of middle rectangle = length of wing, then you can simply the equation to (length of wing + length of middle rectangle) * wingspan * .5.
Now, to get length of wing + length of middle rectangle, you can simplify by finding these two figures as a proportion of total aircraft length. In my case, using
(if you insist on using satellite pictures), I get 56% as the percent of total length formed by the length of the wing + the length of the rectangular section.
Finally, multiply total length by .56, then multiply it by wingspan, then divide by 2, and you get 77m^2, which is less than that of the F-22.
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Also, with regards to using the J-35 as a maneuverable front-line fighter, it makes sense, considering that larger aircraft tend to be more optimized for BVR than their smaller siblings. The trick is that the effectiveness of your radar scales linearly with a one-dimensional scaling factor, while the detection range of enemy radars against you only scales as the square root of such a scaling factor.
As an example, let's say you have an aircraft with an RCS of 1 m^2, width of 5, and a radar aperture of 1 meter. Bear with me. You scale the aircraft up by 10%, so its wingspan increases by 10%, its height increases by 10%, and its length increases by 10%. That increases the RCS by 21%, so now your RCS is 1.21 m^2. Your radar's size also increases by 21%, so now you transmit 21% more energy per radar pulse, as well as having 21% more surface area with which to receive incoming radar signals, so the total signal received on return has gone up by 46.41%. That translates to having 10% more detection range versus enemy targets.
On the other hand, your RCS has also gone up, but the way RCS works is that having a 21% larger RCS only results in an increase of 4.88% detection range against your aircraft.
So put together, your aircraft is now more capable as a BVR platfom, because your increased size isn't wholly balanced by increased detection range versus your plane.
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The other part is that in WVR, it is a lot harder to have a disproportionate technological advantage over your opponent, because it is a lot easier to make a missile ultra-maneuverable, so kill-death ratios are going to be steadily approaching 1. In this sort of case, if you are aiming to specialize in WVR warfare, smaller becomes better, simply because smaller means cheaper and that you'll win in a war of attrition.