Kármán–Moore theory

Kármán–Moore theory is a linearized theory for supersonic flows over a slender body, named after Theodore von Kármán and Norton B. Moore, who developed the theory in 1932.^[1][2] The theory, in particular, provides an explicit formula for the wave drag, which converts the kinetic energy of the moving body into outgoing sound waves behind the body.^[3]

Consider a slender body with pointed edges at the front and back. The supersonic flow past this body will be nearly parallel to the ${\displaystyle x}$-axis everywhere since the shock waves formed (one at the leading edge and one at the trailing edge) will be weak; as a consequence, the flow will be potential everywhere, which can be described using the velocity potential ${\displaystyle \varphi =xv_{1}+\phi }$, where ${\displaystyle v_{1}}$ is the incoming uniform velocity and ${\displaystyle \phi }$ characterising the small deviation from the uniform flow. In the linearized theory, ${\displaystyle \phi }$ satisfies

${\displaystyle {\frac {\partial ^{2}\phi }{\partial y^{2}}}+{\frac {\partial ^{2}\phi }{\partial z^{2}}}-\beta ^{2}{\frac {\partial ^{2}\phi }{\partial x^{2}}}=0,}$

where ${\displaystyle \beta ^{2}=(v_{1}^{2}-c_{1}^{2})/c_{1}^{2}=M_{1}^{2}-1}$, ${\displaystyle c_{1}}$ is the sound speed in the incoming flow and ${\displaystyle M_{1}}$ is the Mach number of the incoming flow. This is just the two-dimensional wave equation and ${\displaystyle \phi }$ is a disturbance propagated with an apparent time ${\displaystyle x/v_{1}}$ and with an apparent velocity ${\displaystyle v_{1}/\beta }$.

Let the origin ${\displaystyle (x,y,z)=(0,0,0)}$ be located at the leading end of the pointed body. Further, let ${\displaystyle S(x)}$ be the cross-sectional area (perpendicular to the ${\displaystyle x}$-axis) and ${\displaystyle l}$ be the length of the slender body, so that ${\displaystyle S(x)=0}$ for ${\displaystyle x<0}$ and for ${\displaystyle x>1}$. Of course, in supersonic flows, disturbances (i.e., ${\displaystyle \phi }$) can be propagated only into the region behind the Mach cone. The weak Mach cone for the leading-edge is given by ${\displaystyle x-\beta r=0}$, whereas the weak Mach cone for the trailing edge is given by ${\displaystyle x-\beta r=l}$, where ${\displaystyle r^{2}=y^{2}+z^{2}}$ is the squared radial distance from the ${\displaystyle x}$-axis.

The disturbance far away from the body is just like a cylindrical wave propagation. In front of the cone ${\displaystyle x-\beta r=0}$, the solution is simply given by ${\displaystyle \phi =0}$. Between the cones ${\displaystyle x-\beta r=0}$ and ${\displaystyle x-\beta r=l}$, the solution is given by^[3]

${\displaystyle \phi (x,r)=-{\frac {v_{1}}{2\pi }}\int _{0}^{x-\beta r}{\frac {S'(\xi )d\xi }{\sqrt {(x-\xi )^{2}-\beta ^{2}r^{2}}}}}$

whereas the behind the cone ${\displaystyle x-\beta r=l}$, the solution is given by

${\displaystyle \phi (x,r)=-{\frac {v_{1}}{2\pi }}\int _{0}^{l}{\frac {S'(\xi )d\xi }{\sqrt {(x-\xi )^{2}-\beta ^{2}r^{2}}}}.}$

The solution described above is exact for all ${\displaystyle r}$ when the slender body is a solid of revolution. If this is not the case, the solution is valid at large distances will have correction associated with the non-linear distortion of the shock profile, whose strength is proportional to ${\displaystyle (M_{1}-1)^{1/8}r^{-3/4}}$ and a factor depending on the shape function ${\displaystyle S(x)}$.^[4]

The drag force ${\displaystyle F}$ is just the ${\displaystyle x}$-component of the momentum per unit time. To calculate this, consider a cylindrical surface with a large radius and with an axis along the ${\displaystyle x}$-axis. The momentum flux density crossing through this surface is simply given by ${\displaystyle \Pi _{xr}=\rho v_{r}(v_{1}+v_{x})\approx \rho _{1}(\partial \phi /\partial r)(v_{1}+\partial \phi /\partial x)}$. Integrating ${\displaystyle \Pi _{xr}}$ over the cylindrical surface gives the drag force. Due to symmetry, the first term in ${\displaystyle \Pi _{xr}}$ upon integration gives zero since the net mass flux ${\displaystyle \rho v_{r}}$ is zero on the cylindrical surface considered. The second term gives the non-zero contribution,

${\displaystyle F=-2\pi r\rho _{1}\int _{-\infty }^{\infty }{\frac {\partial \phi }{\partial r}}{\frac {\partial \phi }{\partial x}}dx.}$

At large distances, the values ${\displaystyle x-\xi \sim \beta r}$ (the wave region) are the most important in the solution for ${\displaystyle \phi }$; this is because, as mentioned earlier, ${\displaystyle \phi }$ is a like disturbance propating with a speed ${\displaystyle v_{1}/\beta }$ with an apparent time ${\displaystyle x/v_{1}}$. This means that we can approximate the expression in the denominator as ${\displaystyle (x-\xi )^{2}-\beta ^{2}r^{2}\approx 2\beta r(x-\xi -\beta r).}$ Then we can write, for example,

${\displaystyle \phi (x,r)=-{\frac {v_{1}}{2\pi {\sqrt {2\beta r}}}}\int _{0}^{x-\beta r}{\frac {S'(\xi )d\xi }{\sqrt {x-\xi -\beta r}}}=-{\frac {v_{1}}{2\pi {\sqrt {2\beta r}}}}\int _{0}^{\infty }{\frac {S'(x-\beta r-s)ds}{\sqrt {s}}},\quad s=x-\xi -\beta r,\,\,r\gg 1.}$

From this expression, we can calculate ${\displaystyle \partial \phi /\partial r}$, which is also equal to ${\displaystyle -\beta \partial \phi /\partial x}$ since we are in the wave region. The factor ${\displaystyle 1/{\sqrt {r}}}$ appearing in front of the integral need not to be differentiated since this gives rise to the small correction proportional to ${\displaystyle 1/r}$. Effecting the differentiation and returning to the original variables, we find

${\displaystyle {\frac {\partial \phi }{\partial r}}=-\beta {\frac {\partial \phi }{\partial x}}={\frac {v_{1}}{2\pi }}{\sqrt {\frac {\beta }{2r}}}\int _{0}^{x-\beta r}{\frac {S''(\xi )d\xi }{\sqrt {x-\xi -\beta r}}}.}$

Substituting this in the drag force formula gives us

${\displaystyle F={\frac {\rho _{1}v_{1}^{2}}{4\pi }}\int _{-\infty }^{\infty }\int _{0}^{X}\int _{0}^{X}{\frac {S''(\xi _{1})S''(\xi _{2})d\xi _{1}d\xi _{2}dX}{\sqrt {(X-\xi _{1})(X-\xi _{2})}}},\quad X=x-\beta r.}$

This can be simplified by carrying out the integration over ${\displaystyle X}$. When the integration order is changed, the limit for ${\displaystyle X}$ ranges from the ${\displaystyle \mathrm {max} (\xi _{1},\xi _{2})}$ to ${\displaystyle L\to \infty }$. Upon integration, we have

${\displaystyle F=-{\frac {\rho _{1}v_{1}^{2}}{2\pi }}\int _{0}^{l}\int _{0}^{\xi _{2}}S''(\xi _{1})S''(\xi _{2})[\ln(\xi _{2}-\xi _{1})-\ln 4L]d\xi _{1}d\xi _{2}.}$

The integral containing the term ${\displaystyle L}$ is zero because ${\displaystyle S'(0)=S'(l)=0}$ (of course, in addition to ${\displaystyle S(0)=S(l)=0}$).

The final formula for the wave drag force may be written as

${\displaystyle F=-{\frac {\rho _{1}v_{1}^{2}}{2\pi }}\int _{0}^{l}\int _{0}^{\xi _{2}}S''(\xi _{1})S''(\xi _{2})\ln(\xi _{2}-\xi _{1})d\xi _{1}d\xi _{2},}$

or

${\displaystyle F=-{\frac {\rho _{1}v_{1}^{2}}{2\pi }}\int _{0}^{l}\int _{0}^{l}S''(\xi _{1})S''(\xi _{2})\ln |\xi _{2}-\xi _{1}|d\xi _{1}d\xi _{2}.}$

The drag coefficient is then given by

${\displaystyle C_{d}={\frac {F}{\rho _{1}^{2}v_{1}^{2}l^{2}/2}}.}$

Since ${\displaystyle F\sim \rho _{1}v_{1}^{2}S^{2}/l^{2}}$ that follows from the formula derived above, ${\displaystyle C_{d}\sim S^{2}/l^{4}}$, indicating that the drag coefficient is proportional to the square of the cross-sectional area and inversely proportional to the fourth power of the body length.

The shape with smallest wave drag for a given volume ${\displaystyle V}$ and length ${\displaystyle l}$ can be obtained from the wave drag force formula. This shape is known as the Sears–Haack body.^[5][6]

• Taylor–Maccoll flow