Martian
Senior Member
My calculation for J-20's edge-diffraction radar return is 0.000016
From a frontal perspective, the obvious weakness in the J-20's stealth design in comparison to the F-22 is the presence of the J-20's canards. I have made an attempt to quantify the effect on the radar return from those canards to determine whether it is significant or not. Here is my post from another forum.
I made a mistake in calculating the edge-diffraction radar return. It should be lower. I was looking at Gambit's diagram for reflection and I realized that the J-20's RAM coating would absorb the incoming radar signal twice, not just once.
Therefore, the correct calculation is as follows:
1% x 1% x 40% x 40% = 0.01 x 0.01 x 0.4 x 0.4 = 0.000016 of the radar energy is reflected back due to edge diffraction. This is extremely small and it should be ignored.
Basically, I'm saying that curve-shaped canards, which are made from composite material and RAM-coated, should have no effect on the J-20's overall RCS.
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Let's do some math to get a feel for the insignificance of edge diffraction.
A. Pretend that you're shining a flashlight (e.g. an electromagnetic source in the visible-light range) at the edge of a piece of paper. How much light is hitting that paper's edge? Let's just say 1% of the total light being emitted by the flashlight. (Initial scatter)
B. A significant portion of the radar waves that impact upon the RAM-coated canard of the J-20 will be absorbed. Let's say 60% is absorbed and 40% is reflected. (Initial absorption)
C. Of the 1% of light that scatters due to edge diffraction, how much of it could theoretically bounce back toward the light source? Given the continuous curved surface of the J-20's body and shaped nose, let's say 1% makes it back. (Secondary scatter)
D. A significant portion of the radar waves that impact upon the RAM-coated body of the J-20 will be absorbed. Let's say 60% is absorbed and 40% is reflected. (Secondary absorption)
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"In electromagnetic wave propagation, the knife-edge effect or edge diffraction is a redirection by diffraction of a portion of the incident radiation that strikes a well-defined obstacle such as a mountain range or the edge of a building.
The knife-edge effect is explained by Huygens-Fresnel principle, which states that a well-defined obstruction to an electromagnetic wave acts as a secondary source, and creates a new wavefront. This new wavefront propagates into the geometric shadow area of the obstacle."
"2. A second way of stopping radar reflections is by coating the plane with material that soaks up radar energy. These typically consist of carbon or a magnetic ferrite-based substance. The result is that the B-2 is reported to have the same RCS as a child's tricycle! Radar-absorbing paints tend to consist of liquid chemical bases mixed with carbon or iron oxide powders. Still other coatings come in the form of sheets impregnated with iron oxides. These sheets, which are used on the Lockheed F-117 and Northrop B-2, can be cut or shaped to fit the outside surface panels of the host aircraft."
From a frontal perspective, the obvious weakness in the J-20's stealth design in comparison to the F-22 is the presence of the J-20's canards. I have made an attempt to quantify the effect on the radar return from those canards to determine whether it is significant or not. Here is my post from another forum.
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I made a mistake in calculating the edge-diffraction radar return. It should be lower. I was looking at Gambit's diagram for reflection and I realized that the J-20's RAM coating would absorb the incoming radar signal twice, not just once.
Therefore, the correct calculation is as follows:
1% x 1% x 40% x 40% = 0.01 x 0.01 x 0.4 x 0.4 = 0.000016 of the radar energy is reflected back due to edge diffraction. This is extremely small and it should be ignored.
Basically, I'm saying that curve-shaped canards, which are made from composite material and RAM-coated, should have no effect on the J-20's overall RCS.
----------
Let's do some math to get a feel for the insignificance of edge diffraction.
A. Pretend that you're shining a flashlight (e.g. an electromagnetic source in the visible-light range) at the edge of a piece of paper. How much light is hitting that paper's edge? Let's just say 1% of the total light being emitted by the flashlight. (Initial scatter)
B. A significant portion of the radar waves that impact upon the RAM-coated canard of the J-20 will be absorbed. Let's say 60% is absorbed and 40% is reflected. (Initial absorption)
C. Of the 1% of light that scatters due to edge diffraction, how much of it could theoretically bounce back toward the light source? Given the continuous curved surface of the J-20's body and shaped nose, let's say 1% makes it back. (Secondary scatter)
D. A significant portion of the radar waves that impact upon the RAM-coated body of the J-20 will be absorbed. Let's say 60% is absorbed and 40% is reflected. (Secondary absorption)
----------
"In electromagnetic wave propagation, the knife-edge effect or edge diffraction is a redirection by diffraction of a portion of the incident radiation that strikes a well-defined obstacle such as a mountain range or the edge of a building.
The knife-edge effect is explained by Huygens-Fresnel principle, which states that a well-defined obstruction to an electromagnetic wave acts as a secondary source, and creates a new wavefront. This new wavefront propagates into the geometric shadow area of the obstacle."
"2. A second way of stopping radar reflections is by coating the plane with material that soaks up radar energy. These typically consist of carbon or a magnetic ferrite-based substance. The result is that the B-2 is reported to have the same RCS as a child's tricycle! Radar-absorbing paints tend to consist of liquid chemical bases mixed with carbon or iron oxide powders. Still other coatings come in the form of sheets impregnated with iron oxides. These sheets, which are used on the Lockheed F-117 and Northrop B-2, can be cut or shaped to fit the outside surface panels of the host aircraft."
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