Re: Energy Dissipation Of A Sphere Passing Through a Medium
Posted: Thu Nov 22, 2012 7:58 pm
Hello,
Sorry if what I'm about to write is TMI to some people...
I agree with the approach presented, but I think I disagree with a couple
of the assumptions:
>So we'll use an approximate solution that is valid for spheres which
>move slowly through a medium.
>The drag force on such a sphere is: f = 1/2Cpav**2
2 comments:
1: I think this is the approximation for high velocity drag, not low velocity
drag, i.e. Reynolds number > ~1000 (from Wikipedia http://en.wikipedia.org/wiki/Drag_(physics) .
For low velocity the drag is proportional to the velocity,
not the velocity squared (same wiki link).
2: Regardless of comment 1, as the projectile slows down the drag force will
decrease. The drag force is not a constant. It
relates to the instantaneous velocity, not the initial velocity. Assuming constant drag
force might be a good assumption at high velocities, but at low velocities
this assumption makes a big difference.
If the high velocity drag term is used, I think this is the resulting differential
equation (dropping constants):
d2x/d2t = (dx/dt)^2 , which would yield a natural logarithm solution x(t) = ln(c1 + t) +c2
But anyway, I think the low velocity drag is better, which is proportional to velocity,
not velocity squared, and gives an exponential solution (dropping constants):
v(t) = 1 - exp(-t)
Happy Thanksgiving...
Sorry if what I'm about to write is TMI to some people...
I agree with the approach presented, but I think I disagree with a couple
of the assumptions:
>So we'll use an approximate solution that is valid for spheres which
>move slowly through a medium.
>The drag force on such a sphere is: f = 1/2Cpav**2
2 comments:
1: I think this is the approximation for high velocity drag, not low velocity
drag, i.e. Reynolds number > ~1000 (from Wikipedia http://en.wikipedia.org/wiki/Drag_(physics) .
For low velocity the drag is proportional to the velocity,
not the velocity squared (same wiki link).
2: Regardless of comment 1, as the projectile slows down the drag force will
decrease. The drag force is not a constant. It
relates to the instantaneous velocity, not the initial velocity. Assuming constant drag
force might be a good assumption at high velocities, but at low velocities
this assumption makes a big difference.
If the high velocity drag term is used, I think this is the resulting differential
equation (dropping constants):
d2x/d2t = (dx/dt)^2 , which would yield a natural logarithm solution x(t) = ln(c1 + t) +c2
But anyway, I think the low velocity drag is better, which is proportional to velocity,
not velocity squared, and gives an exponential solution (dropping constants):
v(t) = 1 - exp(-t)
Happy Thanksgiving...