Numerical Method Terminal Velocity
显示 更早的评论
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
3 个评论
James Tursa
2017-2-22
What have you done so far? What specific problems are you having with your code?
Torsten
2017-2-23
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.
回答（2 个）
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
2017-2-23
0 个投票
1 个评论
Jan
2017-2-27
As said already: you can solve it manually: v(final) = g*m/c
类别
在 帮助中心 和 File Exchange 中查找有关 Mathematics 的更多信息
2017-2-22
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
