Tornado Wind Speed 4106Dd

1. **Problem Statement:** Calculate the wind speed required from a tornado to lift and throw three train cars, each weighing 3 tons, a distance of 5 yards. 2. **Assumptions and Known Values:** - Mass of each train car $m = 3$ tons = $3 \times 2000 = 6000$ pounds (assuming US tons). - Number of cars $n = 3$. - Total mass $M = n \times m = 3 \times 6000 = 18000$ pounds. - Distance thrown $d = 5$ yards = $15$ feet. 3. **Physics Principles and Formulas:** - To loft and throw the cars, the tornado must exert a force overcoming gravity and provide horizontal acceleration. - Weight force $W = M \times g$, where $g = 32.2$ ft/s$^2$ (acceleration due to gravity). - The wind force $F$ can be approximated by drag force formula: $$F = \frac{1}{2} \rho C_d A v^2$$ where: - $\rho$ is air density (about 0.0765 lb/ft$^3$ at sea level), - $C_d$ is drag coefficient (approx 1.2 for a boxy object), - $A$ is cross-sectional area, - $v$ is wind speed in ft/s. 4. **Estimating Cross-sectional Area $A$:** - Assume each train car has a cross-sectional area about $10$ ft wide by $15$ ft tall = $150$ ft$^2$. - For 3 cars, total $A = 3 \times 150 = 450$ ft$^2$. 5. **Calculate the force needed to lift the cars:** - Weight force $W = 18000 \times 32.2 = 579600$ lb*ft/s$^2$ (note: pounds already a force unit, so weight in pounds force is 18000 lb$_f$). - Actually, 1 pound force (lbf) is the force to accelerate 1 lb mass at $g$, so weight $W = 18000$ lbf. 6. **Set wind force equal to weight to just lift:** $$\frac{1}{2} \rho C_d A v^2 = W$$ $$v^2 = \frac{2W}{\rho C_d A}$$ 7. **Plug in values:** $$v^2 = \frac{2 \times 18000}{0.0765 \times 1.2 \times 450} = \frac{36000}{41.31} \approx 871.3$$ $$v = \sqrt{871.3} \approx 29.52 \text{ ft/s}$$ 8. **Convert ft/s to mph:** $$1 \text{ ft/s} = 0.6818 \text{ mph}$$ $$v = 29.52 \times 0.6818 \approx 20.13 \text{ mph}$$ 9. **Consider horizontal throw:** - To throw 5 yards (15 ft), the wind must impart horizontal velocity. - Assuming a simple projectile motion and neglecting air resistance, horizontal velocity $v_x$ needed to cover $d=15$ ft before landing. - If lofted vertically with initial vertical velocity $v_y$ to reach height $h$, time in air $t = \frac{2 v_y}{g}$. - Horizontal velocity $v_x = \frac{d}{t}$. 10. **Estimate vertical velocity to loft:** - To just lift, vertical velocity $v_y$ must overcome gravity. - Assume $v_y = 10$ ft/s (a rough estimate). - Time in air $t = \frac{2 \times 10}{32.2} = 0.62$ s. - Horizontal velocity $v_x = \frac{15}{0.62} = 24.19$ ft/s. 11. **Total wind speed needed:** - Combine vertical and horizontal components: $$v_{total} = \sqrt{v_y^2 + v_x^2} = \sqrt{10^2 + 24.19^2} = \sqrt{100 + 585.2} = \sqrt{685.2} \approx 26.18 \text{ ft/s}$$ - Convert to mph: $$26.18 \times 0.6818 = 17.86 \text{ mph}$$ 12. **Conclusion:** - The tornado wind speed needed to lift and throw the three 3-ton train cars about 5 yards is approximately **20 mph**. - This is a simplified estimate; actual tornado winds are much higher (often over 100 mph) to cause such damage due to complex dynamics and additional forces.