Relative Speed and Striking Speed ```Name: Myra S. Status: Educator Age: 40s Location: N/A Country: N/A Date: July 2003 ``` Question: I do not understand how the foam insulation that broke off the fuel tank could strike the shuttle wing at a speed of 500 mph as reported. The debris would have essentially the same speed as the shuttle when it broke off, and I know the shuttle was accelerating, but how can the difference over such a short distance amount to a speed difference of 500 mph? What was the acceleration at the moment? Replies: To understand the damage which foam insulation falling from the external fuel tank and striking the shuttle wing could do, a test was done in which a 1.67 lb block of foam moving at either 779 ft/s of 768 ft/s (about 530 miles per hour) was made to strike a spare wing. It opened a large hole in the wing, approximately 16 inches in diameter. The speed seems quite large, but for the test to be realistic it must approximate the actual speed of the foam when it struck the wing. I have attempted to calculate the speed to be expected using the parameters described below and obtain approximately the quoted speed, most of which is due to the air flow. One web site kindly provided a polynomial fit to the speed of the shuttle during the boost phase which lasts 126 seconds before the boosters are dropped. The piece of foam struck the wing between 81 and 82 seconds after liftoff, so it is appropriate to use this equation. The equation is (velocity, v, in ft/s, time, t, in seconds): v = 1.302E-3 t^3 - 0.09029 t^2 + 21.61 t - 3.083 ft/s This is obviously an approximate fit and is not perfect. For, example, I am sure that the speed of the shuttle is not -3.083 ft/s at t=0. I differentiated this equation to find the acceleration and plugged in some numbers. The speed and acceleration vary from -3 ft/s and 21.61 ft/s^2 at t=0 s to 3891 ft/s and 60.89 ft/s^2 at t = 126 s according to this equation. At t = 81.5 s, the speed is 1814 ft/s and the acceleration is 32.85 ft/s^2. When the foam breaks off, it accelerates downward like any freely falling body, ignoring air resistance, at g ~ 32 ft/s^2, so the relative acceleration of the shuttle and the foam is close to 65 ft/s^2, ignoring air resistance. However, the force of the air on the foam, traveling at 1800 ft/s relative to the air immediately after breaking off, cannot be ignored. This force is given by the equation R = C r A v^2/2, where C is a drag factor, typically in the range 0.4 to 1.0, r is the density of the air, A is the cross sectional area of the object, and v is its speed relative to the air. The difficult part here is the air density. I will use the atmospheric equation, which assumes (incorrectly) that the temperature is independent of altitude. Let us continue in metric units to make our life bearable. The atmospheric equation is: P = Po exp (-mgz/kT). P is the atmospheric pressure at an altitude of z when Po is the pressure at z=0, m is the mass of the molecules the atmosphere is composed of, k = 1.38E-23 J/K is Boltzman's constant, g = 9.8 m/s^2, and T = 300K (?) is the temperature. Taking nitrogen (N2) as the sole constituent of the atmosphere, m = 28 * 1.66E-27 kg. Since the shuttle was destroyed at approximately 200,000 ft altitude, z = 60 km = 6E4 m. Then P = Po exp (-6.6) = 0.014 Po, so r = 0.014 * 1.3 kg/m^3. Finally, taking C = 0.5 (a guess), A = 0.1 m^2 (11 in by 19 in) and v = 540 m/s (1800 ft/s), I get R = 133 N. Taking the mass of the foam to be 0.75 kg (1.67 lb), its acceleration due to the air velocity is a = F/m = 177 m/s^2 = 590 ft/s^2. Adding this to the 65 ft/s^2 found above, the total acceleration is 655 ft/s^2. The external fuel tanks are 154 ft tall and the boosters are 150 ft tall. Since the foam did not come from the very top and the shuttle wing is not at the bottom, I guess the foam fell 100 ft before striking the wing. Its speed is then given by: v^2 = 2ad = 2 655 ft/s^2 100ft, which gives v = 360 ft/s Given the inaccuracy of some of the estimates, I consider this to be in reasonable agreement with the 779 ft/s quoted by NASA. If C were taken as 1.0 instead of 0.5, I would calculate 500 ft/s. Using a temperature well below 300K, which is certainly reasonable, would increase it further. You are welcome to further refine the calculation. Best, Dick Plano.... Click here to return to the Engineering Archives

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