text: Consider a slender body with pointed edges at the front and back. The supersonic flow past this body will be nearly parallel to the -axis everywhere since the shock waves formed (one at the leading edege and one at the trailign edge) will be weak; as a consequence, the flow will be potential everywhere, which can be described using the velocity potential , where  is the incoming uniform velocity and  characteising the small deviation from the uniform flow. In the linearized theory,  satisfies
  where ,  is the sound speed in the incoming flow and  is the Mach number of the incoming flow. This is just the two-dimensional wave equation and  is a disturbance propagated with an apparent time  and with an apparent velocity .
  Let the origin  be located at the leading end of the pointed body. Further, let  be the cross-sectional area (perpendicular to the -axis) and  be the length of the slender body, so that  for  and for . Of course, in supersonic flows, disturbances (i.e., ) can be propagated only into the region behind the Mach cone. The weak Mach cone for the leading-edge is given by , whereas the weak Mach cone for the trailing edge is given by , where  is the squared radial distance from the -axis. 
  The disturbance far away from the body is just like a cylindrical wave propagation. In front of the cone , the solution is simply given by . Between the cones  and , the solution is given by
  whereas the behind the cone , the solution is given by
  The solution described above is exact for all  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  and a fcator depending on the shape function .
  The drag force  is just the -component of the momentum per unti time. To calculate this, consider a cylindircal surface with a large radius and with an axis along the -axis. The momentum flux density crossing through this surface is simply given by . Integrating  over the cylindrical surface gives the drag force. Due to symmetry, the first term in  upon integration gives zero since the net mass flux  is zero on the cylindrical surface considered. The second term gives the non-zero contribution,
  At large distances, the values  (the wave region) are the most important in the solution for ; this is because, as mentioned earlier,  is a like disturbance propating with a speed   with an apparent time . This means that we can approximate the expression in the denominator as  Then we can write, for example,
  From this expression, we can calculate , which is also equal to  since we are in the wave region. The factor  appearing in front of the integral need not to be differentiated since this gives rise to the small correction proportional to . Effecting the differentiation and returning to the original variables, we find
  Substituting this in the drag force formula gives us
  This can be simplified by carrying out the integration over . When the integration order is changed, the limit for  ranges from the  to . Upon integration, we have
  The integral containing the term  is zero because  (of course, in addition to ).
  The final formula for the wave drag force may be written as 
  or
  The drag coefficient is then given by
  Since  that follows from the formula derived above, , 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 agiven volume  and length  can be obtained from the wave drag force formula. This shape is known as the Sears–Haack body.