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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....
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