I investigated the relationship between the burned area and yield a little, but, as I said just above, I do not think it is that important whether the area scales with yield to the power of 2⁄3 or 1. Feel free to skip to the next section. In short, an exponent of:
2⁄3 makes sense if the energy released by the detonation is uniformly distributed in a spherical region (centred at the detonation point). This is apparently the case for blast/pressure energy, so an exponent of 2⁄3 is appropriate for the blasted area.
1 makes sense if the energy released by the detonation propagates outwards with negligible losses, like the Sun’s energy radiating outwards into space. This is seemingly the case for thermal energy, so an exponent of 1 is appropriate for the burned area.
Thanks to a chat with Stan Pinsent, I have realised the maximum burned area, which will arguably be seeked in order to maximise damage, is indeed proportional to yield. Given a burst height h, and a point P on the ground whose distance from the point on the ground directly below the explosion point is d, the energy flux at point P along the direction from the explosion point to P is I=kmh2+d2, where m is the yield, and k is the constant of proportionality. If the angle between the ground and the direction from the explosion point to P is θ, sinθ=h(h2+d2)1/2, and the energy flux orthogonal to the ground at point P is Iorth=Isinθ=kmh(h2+d2)3/2. If the burned area has an energy flux orthogonal to the ground of at least Ib, its radius will be db=max(0,((kmhIb)2/3−h2)1/2). So the burned area will be Ab=πdb2=max(0,π((kmhIb)2/3−h2)), which tends to 0 as the burst height goes to 0 or infinity. The burst height h1 which maximises the burned area is such thatdAb(Ab>0)dh=2π(13(kmIb)2/3h1−1/3−h1)=0, i.e. it is h1=3−3/4(kmIb)1/2. Consequently, the maximum burned area is Abmax=Ab(h=h1)=π(131/2−133/2)kmIb≈1.21kmIb, which is proportional to yield.
Thanks to a chat with Stan Pinsent, I have realised the maximum burned area, which will arguably be seeked in order to maximise damage, is indeed proportional to yield. Given a burst height h, and a point P on the ground whose distance from the point on the ground directly below the explosion point is d, the energy flux at point P along the direction from the explosion point to P is I=kmh2+d2, where m is the yield, and k is the constant of proportionality. If the angle between the ground and the direction from the explosion point to P is θ, sinθ=h(h2+d2)1/2, and the energy flux orthogonal to the ground at point P is Iorth=Isinθ=kmh(h2+d2)3/2. If the burned area has an energy flux orthogonal to the ground of at least Ib, its radius will be db=max(0,((kmhIb)2/3−h2)1/2). So the burned area will be Ab=πdb2=max(0,π((kmhIb)2/3−h2)), which tends to 0 as the burst height goes to 0 or infinity. The burst height h1 which maximises the burned area is such thatdAb(Ab>0)dh=2π(13(kmIb)2/3h1−1/3−h1)=0, i.e. it is h1=3−3/4(kmIb)1/2. Consequently, the maximum burned area is Abmax=Ab(h=h1)=π(131/2−133/2)kmIb≈1.21kmIb, which is proportional to yield.
However, I have noted Suh 2023, based on Richelson 1980, uses exponents of:
2⁄3 for yields smaller than 1 Mt (see Equation 1).
1⁄2 for yields larger than 1 Mt (see Equation 2).
I have asked the author to share his thoughts on this comment.