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Dynamic Events parachute simulations free fall in vacuum – exact v(t) = -gt + v0 y(t) = -½gt2 + v0t + y0 2000 meters, zero velocity 0 = -½gt2 + 2000 (g = 9.81 ms-2) t2 = 2000*2/9.81 t = 20.2 free fall – Euler’s method see http://spiff.rit.edu/classes/phys317/lectures/heun/heun.html ti, ti+1 vi, vi+1 ai so, vi+1 = vi + ai*( ti+1 - ti) yi+1 = yi + vi*( ti+1 - ti) euler’s method %free fall in vacuum dt = 1; tmax = 30; clear y1 t t1 v y; t = 0:dt:tmax; y0 = 2000; v(1:length(t)) = 0; y(1:length(t)) = 0; v(1) = 0; y(1) = y0; for i = 2:length(t) y(i) = y(i-1)+v(i-1)*dt; v(i) = v(i-1) + -9.81*dt; end comparison with exact % formula follows below texact = 0:tmax/100:tmax; yexact = y0 +.5*(-9.81)*texact.^2; vexact = -9.81*texact; above = yexact > 0; yexact = yexact.*above; subplot(1,2,1) plot(texact,yexact,'r',t,y,'b'),grid vexact = vexact.*above; subplot(1,2,2); plot(texact,vexact,'r',t,v,'b'),grid results air resistance If you do a free fall through air the acceleration is not constant. There are terms which change as your velocity changes. term proportional to v , A1 a measure of viscosity term proportional to v2 , A2 drag In air, the viscosity can be ignored ( A1 << A2). The drag means in mathematical terms is that we do not have an analytical solution to our problem. What it means in practical terms is that sooner or later we will hit a maximum velocity, or terminal velocity at the point when the deceleration due to drag just equals the acceleration of gravity. FD = ½CρAv2 FD = ½CρAv2 Drag coefficients: C Parachutist – spread eagle Can of soup Flat dinner plate Parachute .49 .88 1.11 1.34 “The drag coefficient C is 0.5 for a spherical object and can reach 2 for irregularly shaped objects …” (Physics for Scientists and Engineers, Volume 1 By Raymond A. Serway, John W. Jewett) acceleration with drag ma = ½CρAv2 - mg a = ½CρAv2/m - mg You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom it gets a slight shock and walks away, provided that the ground is fairly soft. A rat is killed, a man is broken, a horse splashes. On Being the Right Size by J. B. S. Haldane Air density and altitude Michael Richmond, Rochester Institute of Technology: density = (1.21 kg/m^3) * exp(-height/8000 m) Ignore temperature dependence. For much more detail, see: http://spiff.rit.edu/classes/phys317/lectures/mu ltiple_funcs/temp_profile.html free fall in air Cd = .5; Area = .8; mass = 85; v(1:length(t)) = 0; y(1:length(t)) = 0; v(1) = 0; y(1) = y0; for i = 2:length(t) acc = -9.81 + .5*Cd*Area*v(i-1)^2*airDensity(y(i-1))/mass; y(i) = y(i-1)+v(i-1)*dt; v(i) = v(i-1) + acc*dt; end Free fall in air 2000 0 -20 1500 -40 -60 1000 -80 -100 500 -120 -140 0 -160 -180 -500 0 10 20 30 40 -200 0 10 20 30 40 parachutes Expert skydivers use parachutes that range in size from 80 square feet to 200 square feet [ about 7.5 to 18.5 square meters] (http://www.livestrong.com/article/417149-thesize-of-skydiving-parachutes/) With parachute 2000 0 1800 -20 1600 -40 1400 -60 1200 -80 1000 -100 800 -120 600 -140 400 -160 200 -180 0 0 50 100 150 200 -200 0 50 100 150 200 Heun’s method while y(i) > 0 % current acceleration a = calc_a(y(i),area,v(i),m); %future velocity bad guess vfuture = v(i) + a*dt; %future acc based on future v afuture = calc_a(y(i),area,vfuture,m); %average acc a = (a+afuture)/2; % future velocity based on average acc v(i+1) = v(i) + a*dt; % future position based on average velocity y(i+1) = y(i)+(v(i+1)+v(i))*dt/2; t(i+1) =t(i) + dt; i = i+1; end HALO High Altitude, Low Opening: the trooper exits the aircraft at 10,000 metres, except the chute is deployed at around 760 metres. (http://www.sasspecialairservice.com/sasbritish-air-troop.html)