Presentation on theme: "應用數學期中報告 Little things falling from the sky 通訊 3A B956C0031 張瑞典."— Presentation transcript:
應用數學期中報告 Little things falling from the sky 通訊 3A B956C0031 張瑞典
There are always things falling from up there somewhere ： smoke particles, dust, and volcanic ash;rain, snow, sweet, hail, and so on. Galileo Galilei dropped objects of various weights and sizes from the top of the Leaning Tower of Pisa and observed their times of fall. Isaac Newton‘s second law was specially helpful.This law states that the summation of all the forces acting on an object is equal to the mass of the object times its acceleration.
For an object falling through air,this relationship gives ： ρ s gV － ρ a gV － F D =ρ s Va ----------(1) ρ s ： the density of the material composing the falling object. ρ a ： the density of air. g ： the gravitational force per unit mass. V ： the volume. a ： the acceleration of the object. The first term on the left expresses the weight of the object, the second term the buoyant force, and the third term the drag force.
Case 1 ： Object Falling in a Vacuum a= dU/dt and U=dy/dt,where y is the distance of fall and U is the velocity. In this case, F D =0, ρ a =0.Equation(1) simolifies to ρ s gV-0=ρ s V U (dU/dy) => g = U (dU/dy) Which can be written in the form ∫ 0 U Udu=g∫ 0 y dy We also obtain y=(1/ 2)gt 2 ; U=gt ; a=g Case 2 ： Object Falling in Air F D =(1/ 2) ρ a C D AU 2 C D ： a dimensionless drag coefficient. A ： the shadow or projected area of the falling object. U ： the velocity.
y 0 =V * / 2g -----------(5).This quantity, y 0,might be called a “distance constant”. y = y 0 log e cosh 2 (gt/ U * ) ----------(6) U = U * tanh(gt/ U * ) ----------(7) a = g sech 2 ((gt/ U * )
For our parachute design we are interested only in the terminal velocity U *. Setting dU/dy=0 in equation(8) and solving for U * gives ----------(9) We use the English system of units instead of the International System(SI). W ： We slelect a design weight W=200lb. ρ a ： ρ a =0.0024 slugs/ft 3 at pressure and temperature ofρ o = 14.7lb/in 2 and I=60°F. C D ： Extensive information concerning aerodynamic drag coefficient is given by Hoerner(1965). An open parachute has the shape of a hollow hemisphere. For this shape, we have the following information ： C D =1.4 when the hollow open side faces the air flow, and C D =0.42 when the closed solid side faces the air flow. Given A=308ft 2, and so the diameter of the parachute is D=19.81ft.
Wherein We Go Skydiving With the parachute you have now designed and fabricated, properly packed and attached to your body, please step out of the airplane door. Do not open your parachute yet. Please assume the basic free-fall position ： face to earth, horizontal, knees slightly bent, arms angled at about 45° in the horizontal plane. In this position, according to data obtained in a wind tunnel and published by Hoerner(1965), the value of C D A=9.0ft 2. If you are in a kind of fetal free-fall position then C D A=2.5ft 2.If you are in a head-first, diving position then C D A=1.2ft 2.
Wherein Dreadful Mistakes Are Made Well, Joe Smog, a guy never gets anything right.On his first free- fall drop, he went into the head-first diving position instead of the required horizontal face-to-earth position. PROBLEM 1. Assuming that W=200lb and C D A=1.2ft 2,show that Joe’s terminal velocity was U * =373ft/s(mi/hr), that he reached this velocity in about 30 seconds, and that, by that time, he had fallen over 8,000 feet. Answers ： By using equation(9) => U * = 2W/ρ a C D A = [ (2×200)/ (0.0024×1.2) ] 1/ 2 ≒ 372.68 ft/s =373ft/s. From the figure below,we can see that when x ≒ 2.6, tanh(x) ≒ 0.99 ≒ 1
By using equation(7)=>U=373tanh[(32.2t/ 373)],when 32.2t/ 373 ≒ 2.6,we gets t ≒ 30(s). By using equation(5) and (6)=> y 0 =V * / 2g ≒ 2160.39 y = y 0 log e cosh 2 (gt/ U * ) ≒ 2160.39log e 44.9 ≒ 2160.39×3.8 ≒ 8209.48 feet. Then Joe pulls the ripcord on his chute. Would you believe that the guy had put his parachute on upside down ？ This is almost impossible to do!
PROBLEM 2. In this case,the closed solid side of the hemisphere faced the air flow, not the hollow open side. So the drag coefficient was C D =0.42 instead of C D =1.40(Again, this is not easy to do!) Show that Joe’s terminal velocity was U * =35.5ft/s(24mi/hr), and that this is like jumping from a 20-foot wall when you hit the ground. Answers ： By using equation(9) => U * = 2W/ρ a C D A = [(2×200)/(0.0024×0.42×308)] ≒ 35.89ft/s Because jumping from a 20-foot wall is not to high,we verify the answer with equation U= 2gy ≒ (2×32.2×20) 1/ 2 ≒ 35.89ft/s => The statement is perfectly match the answer !
Fall of an Object from a Very High Altitude For example, at an elevation of,say, 500meters, the density of air is about 95 ％ of the sea-level density.However, at an elevation of 5,000 meters, it is only50 ％ of the sea-level value. Accordingly, for objects falling from very high altitudes, it can no longer be assumed that the air density is constant and so the preceding analysis is no longer valid. Take a break ： Let’s watch short Skydiving films !