226 The UMAP Journal 24.3(2003) Analysis Our model allows us to predict a stunt persons fall given parameters for the box catcher. Now we would like to find analytic results to guide box-catcher Using the equations of motion( 8), we determine the force that the stunt person feels falling through the box catcher: 1 IS, As F=m dJ2△yA,.|2-al<△ 42-m39,|2-2opl≥△H We want to make this force large enough to stop but not ha arm the performer Therefore, we demand that F≤(1-6)Fmax However, we wish to find solutions that both minimize cost(fewest boxes) and conform to spatial constraints(we don't want the box catcher to be taller than the obstacle that the stunt person is jumping over). Thus, it is in our best interest to maximize the force applied to the performer subject to(9) We are faced with two independent equations with three unknowns; we solve for two of them in terms of the third. With the simplifying assumption that the top face of each box is a square(A=w), we can solve for the optimal dimensions of the box given the stunt persons impact velocity Y? w As 12g2A Fmax(1-8)=Thresh pA._m where i2 is the stunt persons velocity after causing the box to buckle(but before expelling all the air). So we have Define ?, the maximum number of gees of acceleration felt by the stunt person, thresh (1-6) max 2(-23+1)2m) (10) (11) 12m39C2226 The UMAP Journal 24.3 (2003) Analysis Our model allows us to predict a stunt person’s fall given parameters for the box catcher. Now we would like to find analytic results to guide box-catcher design. Using the equations of motion (8), we determine the force that the stunt person feels falling through the box catcher: F = ms dus dt =    1 2∆H T2 S Y V As A , |z − ztop| < |∆H|; 1 2 ρAs u2 s α2 − msg, |z − ztop|≥|∆H|. We want to make this force large enough to stop but not harm the performer. Therefore, we demand that F ≤ (1 − δ)Fmax. (9) However, we wish to find solutions that both minimize cost (fewest boxes) and conform to spatial constraints (we don’t want the box catcher to be taller than the obstacle that the stunt person is jumping over). Thus, it is in our best interest to maximize the force applied to the performer subject to (9). We are faced with two independent equations with three unknowns; we solve for two of them in terms of the third. With the simplifying assumption that the top face of each box is a square (A = w2), we can solve for the optimal dimensions of the box given the stunt person’s impact velocity: π2 12 Y τ 3 w 2 As A = Fmax(1 − δ) = Fthresh, 1 2 ρ α2 Asu˜2 − msg = Fthresh, where u˜2 is the stunt person’s velocity after causing the box to buckle (but before expelling all the air). So we have u˜2 = u2 o − 4τ T2 S msY w As A . Define γ, the maximum number of gees of acceleration felt by the stunt person, by Fthresh = (1 − δ)Fmax = γmsg. We get 3 = π2 48γ Y TS 2 τ 2 g u2 0 − 2(γ + 1)α2 msg ρAs (10) and A(w) w = w = π2 12γ AsY msg τ 3 2 . (11)