Martian
Senior Member
Final estimate for J-20 canards' radar return energy is 3.276x10^-19
To my eyes, the only obvious weakness in China's J-20 front-aspect stealth is the canard. There is a total of three radar-signature contributors from the canards.
1. By far, the largest radar-reflecting source is the initial scatter from the canard. I will calculate this today.
2. There is a much smaller secondary scatter (which I'll calculate later to prove that it's inconsequential) as the radar waves bounce off the canard, bounce off the fuselage, and return to the enemy aircraft's receiver.
3. There is also another secondary scatter, which should equal the secondary scatter from the canard, as radar waves bounce off the fuselage, bounce off the canard, and return to the enemy aircraft's receiver.
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I have revised upward the size of the J-20's canards in the following calculations.
The J-20 has two canards. I'll estimate that each canard is 2m (or 6 feet) long. I'll further estimate that each canard is 0.75m (or 2.25 feet) high. The total canard area facing an enemy radar is 2m x 0.75m x 2 [canards] = 3 m^2.
I don't have the faintest idea of the illumination cone for a directed military radar. However, to be of any use, I'll estimate that the illumination cone is 1km in radius. If the illumination cone is significantly smaller than 1km, I don't see how you can find an enemy fighter within a reasonable amount of time.
Area of a circle = pi * r^2 = 3.14 * (1km*1km) = 3.14 * (1,000m*1,000m) = 3.14 x 10^6 m^2
Initial scatter ratio = (3 m^2) / (3.14 x 10^6 m^2) = 9.554 x 10^-7 = 0.0000009554. This is the ratio of the emitting radar energy that hits the J-20's canards.
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After hitting the canards, we know that 99.999% of the reflected energy is reduced by the military-grade RAM. (See ) This means that only 0.00001 (e.g. 1 - 0.99999 = 0.00001) of the impacting radar energy survives contact with the canard's RAM surface.
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However, only a tiny fraction of the radar energy that is scattered by the RAM-covered canards will make it back to the enemy aircraft's transmitter/receiver radar.
The total energy that is scattered from the canards' surface will radiate into a spherical shell that is determined by the following formula:
Surface area of a sphere = 4 * pi * r^2
For the radius, we will use 13.5 km from the detection range of an EADS DR 174 ground-based mobile 3D radar.
Total surface area of spherical shell = 4 * 3.14 * (13.5km x 13.5km) = 2289 km^2 = 2289 (1,000m x 1,000m) = 2289 x 10^6 m2 = 2.289 x 10^9 m2
Based on the picture shown at the bottom of the post, I will estimate the radius of the receiving radar to be 0.5m.
Total surface area of receiving radar = pi * r^2 = 3.14 * (0.5m x 0.5m) = 0.785 m^2
The fraction of the reflected radar energy from the RAM-coated canards that is seen by the receiver is:
Reflected radar energy from canards seen by receiver = 0.785 m^2/2.289 x 10^9 m2 = 3.429 x 10^-10
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Putting it all together, the amount of the emitting radar energy that returns to the receiver from two RAM-coated canards located 13.5 km away is:
Fraction of emitting radar energy returning to receiver = 9.554 x 10^-7 x 0.00001 x 3.429 x 10^-10 = 3.276 x 10^-19 or 0.0000000000000000003276.
QED: The J-20's canards have an insignificant impact on its stealthiness.
Figure 5 shows the phased array radar used on the F/A 22 Raptor. This radar employs approximately 2000 microwave transmitter/receiver pairs, each the size of a pack of chewing gum. (courtesy USAF)
To my eyes, the only obvious weakness in China's J-20 front-aspect stealth is the canard. There is a total of three radar-signature contributors from the canards.
1. By far, the largest radar-reflecting source is the initial scatter from the canard. I will calculate this today.
2. There is a much smaller secondary scatter (which I'll calculate later to prove that it's inconsequential) as the radar waves bounce off the canard, bounce off the fuselage, and return to the enemy aircraft's receiver.
3. There is also another secondary scatter, which should equal the secondary scatter from the canard, as radar waves bounce off the fuselage, bounce off the canard, and return to the enemy aircraft's receiver.
----------
I have revised upward the size of the J-20's canards in the following calculations.
The J-20 has two canards. I'll estimate that each canard is 2m (or 6 feet) long. I'll further estimate that each canard is 0.75m (or 2.25 feet) high. The total canard area facing an enemy radar is 2m x 0.75m x 2 [canards] = 3 m^2.
I don't have the faintest idea of the illumination cone for a directed military radar. However, to be of any use, I'll estimate that the illumination cone is 1km in radius. If the illumination cone is significantly smaller than 1km, I don't see how you can find an enemy fighter within a reasonable amount of time.
Area of a circle = pi * r^2 = 3.14 * (1km*1km) = 3.14 * (1,000m*1,000m) = 3.14 x 10^6 m^2
Initial scatter ratio = (3 m^2) / (3.14 x 10^6 m^2) = 9.554 x 10^-7 = 0.0000009554. This is the ratio of the emitting radar energy that hits the J-20's canards.
----------
After hitting the canards, we know that 99.999% of the reflected energy is reduced by the military-grade RAM. (See ) This means that only 0.00001 (e.g. 1 - 0.99999 = 0.00001) of the impacting radar energy survives contact with the canard's RAM surface.
----------
However, only a tiny fraction of the radar energy that is scattered by the RAM-covered canards will make it back to the enemy aircraft's transmitter/receiver radar.
The total energy that is scattered from the canards' surface will radiate into a spherical shell that is determined by the following formula:
Surface area of a sphere = 4 * pi * r^2
For the radius, we will use 13.5 km from the detection range of an EADS DR 174 ground-based mobile 3D radar.
Total surface area of spherical shell = 4 * 3.14 * (13.5km x 13.5km) = 2289 km^2 = 2289 (1,000m x 1,000m) = 2289 x 10^6 m2 = 2.289 x 10^9 m2
Based on the picture shown at the bottom of the post, I will estimate the radius of the receiving radar to be 0.5m.
Total surface area of receiving radar = pi * r^2 = 3.14 * (0.5m x 0.5m) = 0.785 m^2
The fraction of the reflected radar energy from the RAM-coated canards that is seen by the receiver is:
Reflected radar energy from canards seen by receiver = 0.785 m^2/2.289 x 10^9 m2 = 3.429 x 10^-10
----------
Putting it all together, the amount of the emitting radar energy that returns to the receiver from two RAM-coated canards located 13.5 km away is:
Fraction of emitting radar energy returning to receiver = 9.554 x 10^-7 x 0.00001 x 3.429 x 10^-10 = 3.276 x 10^-19 or 0.0000000000000000003276.
QED: The J-20's canards have an insignificant impact on its stealthiness.
Figure 5 shows the phased array radar used on the F/A 22 Raptor. This radar employs approximately 2000 microwave transmitter/receiver pairs, each the size of a pack of chewing gum. (courtesy USAF)
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