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Ballistic Entry Calculator

This model represents an object that enters the Earth’s atmosphere. It calculates peak deceleration, peak heating, and the altitudes and times at which that happens using the first order ballistic entry model published by C. Gazley Jr of RAND Corporation in 1960, which itself traces back to the Allen-Eggars model published in 1958.

For this model, the atmosphere’s density is approximated to fall off exponentially with increasing altitude, gravity is approximated as constant regardless of altitude, and the entering body is approximated to have zero lift and a constant coefficient of drag. These approximations simplify the underlying trajectory equations and allow them to be solved in closed form.

A consequence of these approximations is that the flight path angle at which the body enters the atmosphere is constant throughout. Real trajectories reportedly have constant flight path angles provided that the entry angle is 5 degrees or more below horizontal.

Press to see the model. Try your own inputs or the examples below. Refresh your browser to clear results.

Allen and Eggers’ Iron Sphere

On page 2 of their seminal 1958 paper, Allen and Eggers consider the vertical descent of a 1-foot diameter solid iron sphere.

1.  First, let’s use the atmospheric parameters they do by specifying sea-level atmospheric density as 0.0034 slugs/ft³ (?!) «ctrl+g» rs.slug/ft3 = 0.0034 and atmospheric scale height as 22,000 ft «ctrl+g» b.ft = 22000 .
2. The body’s ballistic coefficient is implied by the information that it is a 1-foot diameter BC.Sphere.Dh.ft = 1 iron BC.Sphere.«ctrl+g»r = 7874 sphere with unity drag coefficient BC.Cd = 1 . Given these inputs, the model infers a ballistic coefficient of 1,600 kg/m² and related quantities grouped within the ballistic coefficient sub-model BC .
3. Turning to page 3, let’s take the case where the sphere enters the atmosphere at 40 miles altitude he.mi = 40  and 30,000 ft/s Ve.ft/s = 30000 , vertically (γₑ=-90°) ge = -90 . The model infers that the sphere decelerates at a peak of 234 g’s (peak_decn.-dV/dt ) at 43,854 ft altitude (peak_decn.h.ft ), which corresponds to the peak in figure 3 of the paper.
4. To compare with figure 2 of the paper as well, let’s choose 50,000 ft altitude h.f = 50000 . The model infers that velocity at this altitude (V.ft/s ) is 20,555 ft/s, which visually matches the figure.
5. To estimate heating, the model needs to know that the radius of curvature is 0.5 ft Rn.ft = 0.5 . The model infers peak heating (peak_heat.qw ) of 66 MW/m² (6,600 W/cm² or 45E5 ft-lbf/s/ft²) at 68,000 ft altitude (peak_heat.h.ft ). This enormous heat input would occur if the walls of the sphere were cold, and it would take a blackbody at 5,840 K (peak_heat.Wall.T ) to produce the same radiant exitance. Note that this value will not exactly match the values in the paper because this model deviates from the paper and instead uses the stagnation point heating correlation of Kemp & Riddell (1957) to estimate radiant exitance.

So what about at ground level? Setting altitude to zero h.ft = 0 , the radiant sphere of melting iron hits the ground at 233 m/s () only 17 seconds after atmospheric entry (t.s )! Note that when the ballistic coefficient is lower, setting altitude too low may result in a consistency error. This is a sign that the model is out of its regime, and instead the body has slowed to terminal velocity. This model does not yet calculate terminal velocity or detect when it is reached, so it is the user’s responsibility to notice when the velocity gets unrealistically low.

One can also find the altitude at which a velocity (such as terminal velocity) occurs: Clearing the altitude h = and setting a velocity, for example 300 m/s V = 0.3 , the model infers that the iron sphere decelerates through 300 m/s at 482 meters altitude.