Fall with Air Drag - Which Value for Velocity to Use?

In summary, you should use vn in place of (vn+vn+1)/2 when simulating a fall with air drag. This will fix the error in the book's free fall solution.
  • #1
dracayr
2
0

Homework Statement


I want to numerically simulate a fall with air drag. In class, we used these equations to produce a table simulating free fall (using a ti calculator):
Code:
dt=0.1s
a=9.81m/s²
t[SUB]n+1[/SUB]=t[SUB]n[/SUB]+dt
v[SUB]n+1[/SUB]=v[SUB]n[/SUB]+a*dt
x[SUB]n+1[/SUB]=x[SUB]n[/SUB]+v[SUB]n[/SUB]*dt

When I compared that system with the curve produced by the direct equation, x=1/2at², I noticed a slight error. After looking at the equations again, I figured that the error came from the fact that I used vn, so the curve would always be a bit lower than 1/2at². So I fixed that by changing the Equation to
Code:
x[SUB]n+1[/SUB]=x[SUB]n[/SUB]+((v[SUB]n+1[/SUB]+v[SUB]n[/SUB])/2)*dt

Now I want to create a similar system but with air drag. As before, my physics book says that I should use vn and not (vn+vn+1)/2. But the book was wrong when it came to free fall, so I don't know which value to use for v. I think I should use (vn+vn+1)/2. But since I don't know the direct equation this time (if there is one), I can't compare to the correct solution..

2a. Relevant equations
v=x/t
a=v/t
F=ma

Fd=b|v|m
vE=(mg/b)1/m
a=g(1-vm/vEm)

Fd is the air drag
b and m are constants depending on the drag (I think in the case of air, m is about 2)
vE is the end velocity the falling body will achieve


The Attempt at a Solution



These are the spreadsheets with the 2 systems:

http://dracayr.awardspace.com/physik.ods"
http://dracayr.awardspace.com/physik.xls"


dracayr
 
Last edited by a moderator:
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  • #2
Ah, you've unknowingly ran into problems solving differential equations numerically. Essentially, what you're doing is time-marching an ordinary differential equation. You're comparing results using upwind and central differencing. For what you're doing, just do as the book says.

Numerically, upwinding (essentially using the value of the velocity at the current time) will damp waves, so you will typically have a lower magnitude than the analytic solution.
 
  • #3
OK, thanks for the information :)

dracayr
 
Last edited:

What is "fall with air drag"?

"Fall with air drag" refers to the motion of an object falling through the air while experiencing air resistance or drag. This phenomenon is commonly seen when objects are dropped or thrown from a height.

How does air drag affect the motion of a falling object?

Air drag acts in the opposite direction of the object's motion, slowing it down and changing its trajectory. The larger the surface area of the object, the greater the air drag it experiences.

What factors affect the amount of air drag experienced by a falling object?

The amount of air drag depends on the speed of the object, the density of the air, and the shape and surface area of the object. Objects with a larger surface area or moving at higher speeds experience more air drag.

How does air drag change as an object falls?

As an object falls, its speed and the force of air drag acting on it increase. However, at a certain point, the force of air drag becomes equal to the force of gravity, resulting in a constant speed known as terminal velocity.

Can air drag ever be completely eliminated during a fall?

No, air drag cannot be completely eliminated during a fall. However, it can be minimized by reducing the surface area and increasing the density of the object, and by decreasing the density and viscosity of the air through which it is falling.

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