Numerical Method Terminal Velocity
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Parachutist of mass(m) = 68.1 kg
drag coefficient(c) = 12.5 kg/s
g= gravitational acceleration = 9.81
ti=0
vi=0
ti+1 - ti = 0.1
here is the eqn
v(ti+1)=v(ti)+(g-(c/m)*v(ti))*(ti+1 - ti)
I just want to plot within such a point that v(ti+1)-v(t)<0.001
thanks
댓글 수: 3
James Tursa
2017년 2월 22일
What have you done so far? What specific problems are you having with your code?
Torsten
2017년 2월 23일
If you set the initial condition for v to be zero, the solution is v=0 for all times. I guess this is not what you want.
Best wishes
Torsten.
@Torsten: Why?
v(2) = v(1) + (g - (c/m)*v(1)) * 0.1 =
= 0 + (g - 0) * 0.1
This is an acceleration.
@Ali Enes Yildirim: What have you tried so far? The only pitfall if not to confuse the index of the times and the value.
답변 (2개)
The terminal velocity is reached, when there is no further acceleration. This means that g-(c/m)*v(ti) must be 0.0 and you can calculate the result without any iterations or rough limits.
If you really want to calculate this by a loop:
v(1) = 0;
ti = 1;
tStep = 0.1;
vStep = inf; % Arbitrary large value to allow entering the loop
while vStep > 0.001
... increase ti by 1 (not by 0.1)
... calculate new speed and store it in v(ti) using tStep (not ti)
... calculate the step in the velocity vStep
end
Ali Enes Yildirim
2017년 2월 23일
0 개 추천
댓글 수: 1
Jan
2017년 2월 27일
As said already: you can solve it manually: v(final) = g*m/c
카테고리
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