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Dynamic Events parachute simulations

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free fall in vacuum – exact v(t) = -gt + v 0 y(t) = -½gt 2 + v 0 t + y 0 2000 meters, zero velocity 0 = -½gt 2 + 2000 (g = 9.81 ms -2 ) t 2 = 2000*2/9.81 t = 20.2

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free fall – Euler’s method see http://spiff.rit.edu/classes/phys317/lectures/heun/heun.html t i, t i+1 v i, v i+1 a i so, v i+1 = v i + a i *( t i+1 - t i ) y i+1 = y i + v i *( t i+1 - t i )

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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

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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

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results

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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, A 1 a measure of viscosity term proportional to v 2, A 2 drag In air, the viscosity can be ignored ( A 1 << A 2 ). 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. F D = ½CρAv 2

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Drag coefficients: C Parachutist – spread eagle.49 Can of soup.88 Flat dinner plate1.11 Parachute1.34 “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)

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acceleration with drag ma = ½CρAv 2 - mg a = ½CρAv 2 /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

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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

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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

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Free fall in air

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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-the- size-of-skydiving-parachutes/)

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With parachute

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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

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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/sas- british-air-troop.html)

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4.3 Free Fall and the Acceleration due to Gravity

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