Does waverider have much lower drag than Sears Haack body? I tried to calculate drag for hyper velocity project using this paper for Sears Haack revolution body. Some parameters:

Diameter: 160mm (round number for simulating 155mm projectile)

Fineness ratio: 13.2, which means projectile's length is 2112mm

From the paper, the assumed wing area $S$ is 

$$0.077 \cdot l^2 = 0.077 \cdot (2.112)^2 = 0.3435 \ [m^2]$$

From the Figure 15 in the paper, the total Cd at 2km/s is about 0.004

So total drag at sea level 
$$D = 0.5 \cdot v^2 \cdot \rho \cdot S \cdot C_D$$

$$D = 0.5 \cdot 2000^2 \cdot 1.23 \cdot 0.3435 \cdot 0.004 = 3380 \ [N]$$

From the paper, the volume $V$ of the projectile is

$$V = 0.002655 \cdot l^3 = 0.025 \ [m^3]$$
Assume it is filled with aluminum, then the mass $m$ is about 
$$ 0.025 \cdot 3000 = 75 \ [kg]$$

So the drag is about 4.5g.

On the other hand, there is an empirical formula for hypersonic lift to drag ratio is:

$$\frac{L}{D} = \frac{4 \cdot (M + 3)}{M} = \frac{4 \cdot (6 + 3)}{6} = 6$$

It seems that if the projectile is shaped as a good waverider then drag can be as low as 1/6 of its weight, assuming the waverider is the best shape for hypersonic lift to drag ratio.

There is of course the problem about scaling where mass is increased at cubed speed, while area only increases at squared speed. While I don't have an complete computational model of waverider to compare directly with SH body, but it seems at the first glance, the waverider can get much lower drag than even the most aerodynamic shape, 1/6g vs 4.5g.

So is it true that the waverider has much lower drag than a SH body?
A waverider will prevent the compressed air under the airframe from escaping sideways. An equivalent view is that the sideways component of the air displaced by the body is used for lift creation by straightening it and pushing it down at the wingtips. This allows it to fly at a lower angle of attack for the same lift, which reduces supersonic lift-related drag linearly with the reduction in angle of attack.

The drag of a Sears-Haack body is the lowest for a given volume and a given fineness ratio. Increasing the fineness (essentially, the local inclination times the circumference of each body section) will reduce drag. This increased fineness, however, will now cause less sideways displacement of airflow which could be used for the waverider effect.

In the end, the question as asked is not answerable. What can be said is:


A blunt Sears-Haack body will profit the most from the waverider effect.
A waverider is still a Sears-Haack body if it has the right cross section distribution over length.
A waverider will maybe lower drag by 20% or 30% in a realistic configuration and at Mach numbers between Mach 3 to 5.


The empirical equation for the L/D of a waverider looks suspicious. Without knowing the constraints of its validity, I would not trust its results. Also, the drag coefficient of your Sears-Haack body looks like the zero-lift drag. You might need to add the lift-related drag to arrive at the complete drag.

And the biggest mistake is to operate it at ground level - 2000 m/s would be appropriate in 30 km altitude. Get the drag figure there and you will arrive at a much lower drag just by operating it in the appropriate density.