Hypervelocity Drag

Drag on an accelerating launch loop vehicle with a hemispherical nose

Based on Trajectory Optimization for an Apollo-type Vehicle under Entry Conditions Encountered During Lunar Returm by John W. Young (famous astronaut) and Robert E. Smith Jr., May 1967, NASA TR-R-258, Langley Research Center.

Equations on Page 5 in Foot-second-slug-BTU :

• 1a convective power: ~ ~ \dot Q_c = 20 \rho^{1/2} \left( V \over 1000 \right)^3 Btu/ft²-s

• 1b radiative power: ~ ~ \dot Q_r = 6.1 \rho^{3/2} \left( V \over { 10 000 } \right)^{20} Btu/ft²-s

• Equations assume an effective nose radius of 1 foot

• Equations from Shock Layer Radiation During Hypervelocity Re-Entry by Robert M. Nerem and George H. Stickford, AIAA Entry Technology Conference, CP-9, American Institute of Aeronautics and Astronautics, Oct. 1964, pp 158-169. (not downloaded yet)

Density in slugs/ft³: multiply kg/m³ by 1.9403203e-3

Power in Btu/ft²-s: multiply by 11350.54 to get W/m²

Velocity in ft/s: divide m/s by 0.3048

Metric equations:

• Metric 1a convective power: ~ ~ \dot Q_c = 3.53e-4 \rho^{1/2} ~ V^3 Watts

• Metric 1b radiative power: ~ ~ \dot Q_r = 1.24e-69 \rho^{3/2} ~ V^{20} Watts

These are for a 1 foot diameter nose, and scale by {r_n}^{-1/2} according to equation 4B-4 on page 520 of Part 4B (Entry Heat Transfer) of the SAE Aerospace Applied Thermodynamics Manual. That sites reference 1, A study of the motion and aerodynamic heating of missiles entering the earth's atmosphere at high supersonic speeds, H. Julian Allen and A. J. Eggers, Jr, NACA TN 4047, 1957. If r_n is in meters, scale by 0.552 {r_n}^{-1/2} .

If we scale these for a half-spherical nose, area \pi {r_n}^2 , we get:

• Total nose convective power: ~ ~ \dot Q_c = 6.1e-4 ( {r_n}^3 ~ \rho )^{1/2} ~ V^3 Watts

• Total nose radiative power: ~ ~ \dot Q_r = 2.2e-69 ( {r_n} ~ \rho )^{3/2} ~ V^{20} Watts

Effective time:

Assume constant acceleration for the vehicle, v = a t , to a maximum velocity V = a T .

define t_{eff} = {\Large { T \over { n+1 } } } = { \Large { V \over { a ( n+1) } } }

If the drag power \dot Q = k v^n = k a^n t^n , then the time integrated power:

Q=k a^n{\Large {T^{n+1}\over {n+1}}}=k a^n T^n{\Large {T\over{n+1}}} = k V^n t_{eff} = \dot Q_{max} t_{eff}

There will also be additional exit or climb-out time for the launch loop added to t_{eff} , TBD. This additional time will be proportionally larger for the radiation fraction, but that will remain small, especially in thinner, higher altitude atmosphere.

The drag losses are much higher; most of the lost energy ends up heating the upper atmosphere (where it radiates efficiently into space, not to the ground). The drag power is P = C_D \rho Area V^3 and the drag loss is P = C_D \rho Area V^3 T/4

Examples:

For a 1 meter diameter nose, V=11 km/s, a=3*9.8m/s, T=374 s, C_D = 2.0 and density at 80, 100, and 120 km: