Recent content by wvengineer

Sorry it's been a while. Just had time to thnk about this. Here's what I was doing. Force of drag = k*(v^2), k=(1/2)*(rho)*(Cd)*(Aref) m*a=k*(v^2) m*(dv/dt)=k*(v^2) dv/(v^2)=(k/m)*dt -(1/v)=[(k*t)/m]+C v=-m/[(k*t)+(C*m)] Now I havne't gone any further because it seems to me this isn't...

From what Nasu wrote I took it to mean b = 1/2*Cd*A*p and v=v^2

I believe I see it Nasu. I appreciate the help. I'll have to try it later. It's funny how this started as a break-room discussion, and then how to prove it got stuck in my head.

I see what your saying about as t approaches infinity then the particles velocity will at best approach zero. As far as the solution can you elaborate on that for me. The drag coefficient is constant (and the same for both particles), but how are you showing the change in the force of drag...

My problem is that I have a change of force with respect to velocity, and a change of deceleration with respect to the change in force. I'm not sure how to set this equal to each other for the two particles and calculate at what time the two will be traveling at the same velocity. This is with...

Now I would like to show it mathematically. The problem I'm having is we don't have a constant deceleration. I can get the force of drag for each particle. I can get the change in deceleration for each particle with respect to a certain velocity. I would now like to get this with respect to...

The faster one will "approach" the the velocity of the slower one given enough time (at t=∞). That was just for simplification to separate the difference of deceleration due to the different magnitudes of the force of drag from the difference in deceleration due to the different masses of the...

I think you're agreeing with me. My second post wasn't what I though, I was just wanting to make sure I understood Jeff's response. Here's my side of the discussion. As a disclaimer I understand this is not a practical scenario. I separated this into two things. First is with all thing...

Well if I give no consideration to the mass of the particles then given enough time the two particles will eventually be going the same velocity. After that they will decelerate at the same rate. Is that a true statement? I understand drag is independent of the particle's mass. I just keep...

Was having a breakroom discussion today and I figured I would get an outside opinion. Anywas I have to projectiles/particles flying through the air and are unaffected by gravity. Both have the same drg coefficient, cross-sectional area, air densityetc. The only differences are particle A has...