4 D. B. MEADE

Mathematically, the jerk is the time derivative of the acceleration,

j(t) := \frac{da}{dt} = −\frac{k′(t)}{m}v(t) − \frac{k(t)}{m}a(t). (2.5)

When k is piecewise constant, as in (2.3), the snatch force has a jump discontinuity at the instant of deployment: [j(t_{d})] := j(t_{d}^{+}) − j(t_{d}^{−}) = −\frac{[k(t_{d})a(t_{d})]}{m}. Using the Nagle/Saff parameters, a(60^{−}) ≈ 0 G, a(60^{+}) ≈ 6 G, and [j(60)] ≈ −8.4 G/s. These jumps seem to be fairly significant; are they realistic?

3. Real-World Considerations. The preceding discussion illustrates techniques used to answer a wide variety of mathematical and physical questions about ODE models. The analysis raised several additional questions about the applicability of the results to a real-life parachute jump. To address these issues it is necessary to obtain real-life data about skydiving; two accessible references for this material include [8] and [9].

Training jumps for the Parachute Team at the United States Air Force Academy begin 4,000′ (1,219 m) above ground level (AGL). The free-fall portion of the jump lasts about 10 seconds; free-falls longer than 13 seconds are grounds for removal from the team. The terminal velocity for free-fall is 120 miles/hr (176 ft/sec or 53.6 m/sec). The parachute requires approximately 3.2 seconds to fully deploy from the time the ripcord is pulled — at an altitude of at least 2,500′ (762 m) AGL. The snatch force felt when the lines and canopy are fully elongated is a heavy, but smooth, tug that is not particularly uncomfortable. While this force depends on the weight of the skydiver, it should not exceed 500 lbs ( ≈3 G for a 165 lb (75 kg) person). The harness and parachute are designed to withstand a force of 5000 lb (30 G). The landing velocity should be no worse than a free-fall from a 5′ (1.52 m) wall — between 15 and 17 ft/sec (4.6 and 5.2 m/sec). A reserve chute is required on all intentional jumps. If a malfunction occurs with the main chute at 3000′ (912 m) AGL, almost 6 seconds will be required to recognize and react to the problem and to deploy the reserve chute. The reserve chute must be opened no lower than 1000′ (304 m) AGL; deployment requires only 1.5 seconds and the landing velocity should not exceed 17.5 ft/sec (5.3 m/sec).

A free-fall terminal velocity of 120 miles/hr implies \frac{k_{1}}{m} = \frac{2}{11} and a landing velocity of 16 ft/sec corresponds to \frac{k_{2}}{m} = \frac{32}{16}. Note that the values of \frac{k_{1}}{m} found in Table 2.1 are relatively realistic, the values of \frac{k_{2}}{m} are uniformly low (so that the landing velocities are too high). The jump height and deployment time can be more varied; constraints on these parameters arise from e.g., the minimum height needed for the parachute to deploy and the applicability of the model at higher altitudes.^{3}

The statement that the snatch force should be smooth, and not exceed 3 G, suggests that (2.3) is not appropriate. A smooth snatch force is assured only when k ∈ C^{1}(0, ∞). The fact that the parachute does not open instantaneously provides the necessary flexibility to connect the two constant states in a continuous manner. The resulting form for k is

k(t) = \{ \begin{matrix}k_{1}, & t < t_{d} \\ k_{d}, & t_{d} ≤ t < t_{d} + τ_{d} \\ k_{2}, & t ≥ t_{d} + τ_{d} \end{matrix} (3.1)

where the time required for the parachute to open is denoted by τ_{d}. Note that an explicit solution to (2.1) with k as in (3.1) can be obtained by solving three IVPs.

3It should be noted that some of the textbook problems become more realistic simply by changing the units.