Numerical Method Terminal Velocity

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Ali Enes Yildirim
Ali Enes Yildirim am 22 Feb. 2017
Kommentiert: Jan am 27 Feb. 2017
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 Kommentare
Torsten
Torsten am 23 Feb. 2017
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.
Jan
Jan am 23 Feb. 2017
Bearbeitet: Jan am 23 Feb. 2017
@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.

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Antworten (2)

Jan
Jan am 23 Feb. 2017
Bearbeitet: Jan am 23 Feb. 2017
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
Ali Enes Yildirim am 23 Feb. 2017
this is about falling parachutist problem. I solved it with EXCEL by using a lots of rows and columns. ı just wanted make sure whether there is a simple way to solve Numerical Methods with MATLAB. Thank you MS Excel and you guys trying to help me :)
  1 Kommentar
Jan
Jan am 27 Feb. 2017
As said already: you can solve it manually: v(final) = g*m/c

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