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Main Question or Discussion Point
if an objects terminal velocity in air is 18m/second, when what would be an equation that describes its velocity from t0?
No. Its terminal velocity is still 18m/s. Regardless of how fast or slow it starts at, that is the velocity that it will approach, due to air resistance.yes, I fallowed that, so If I stick a cantaloupe out the window of the car and when I go 20m/s the angel is 45 degrees then the terminal velocity is 20m/s and the force of gravity is equal to the air resistance.
You are still up against the same basic problem. You have an unknown curve (the graph of drag versus velocity) for which you know only one parameter: f(18) = 1.The cantaloupe is on a rope, from a pole sticking out of the window and if the terminal velocity is 18m/s, when the car is going 18m/s, the anel of the rope will be 45 degrees. Vt= G.
i am getting closer, from this page:http://www.sciencebits.com/MR_Stokes_Drag
I get this:View attachment 238034
I need to figure out how to put this in an Arduino. If i look at the slope in 10ths of seconds along the curve I should be able to Vt
I got to get some college kid to do the real calculus. I had Calc 3 was in 1998 :) the application is, if my airplane is at 500 feet and doing 40MHP what is the lead angle to hit an abandoned car with said cantaloupe. Physics is fun
Just a heads up, that the drag for on an object near terminal velocity in atmospheric air is going to be dominated by quadratic drag. See for instance [1] for description of how linear and quadratic drag depends on the Reynolds number.The web page you are looking at addresses that problem by assuming linear drag.
For Reynolds number small that is not what this article says. It depends therefore on the density of object.Just a heads up, that the drag for on an object near terminal velocity in atmospheric air is going to be dominated by quadratic drag. See for instance [1] for description of how linear and quadratic drag depends on the Reynolds number.
I am not sure why you say that.For Reynolds number small that is not what this article says.
If by "it" you mean drag, then that is not correct since the drag force on an object do not depend on the density of that object. Object density do come into play for dynamical problems (like for instance a free falling object near terminal velocity) usually expressed via the so called ballistic coefficient.It depends therefore on the density of object.
My apologies. You were referring to the "cantaloupe on a rope" experiment where I was talking about the "falling object" which was the original question. I just wanted to clarify that for an arbitrary falling object the linear velocity drag regime can be appropriate. It is sometimes difficult to follow the thread ....I am not sure why you say that.
The paper I linked to describes ways to model linear and quadratic drag, i.e. drag force varying linearly and quadraticly with respect to speed, with linear drag mostly being associated with low Reynold numbers (Re << 1) and quadratic drag with high Reynold numbers (Re > 1000). The goal of the paper is to then go on describing how to model drag when 1 < Re < 1000. In context of this thread, my point of linking to the paper was really just to say that if you stick a sphere-like object out the window of a car moving around 18 m/s (Re >> 1000) then drag clearly is in the quadratic regime.
If by "it" you mean drag, then that is not correct since the drag force on an object do not depend on the density of that object. Object density do come into play for dynamical problems (like for instance a free falling object near terminal velocity) usually expressed via the so called ballistic coefficient.
Fair enough.My apologies. You were referring to the "cantaloupe on a rope" experiment where I was talking about the "falling object" which was the original question.
For free falling objects under one gravity in atmospheric air linear drag would perhaps be appropriate for dust flakes or similar microscopic objects, in which case the bouyancy force should also be considered as it becomes comparable to the drag force.I just wanted to clarify that for an arbitrary falling object the linear velocity drag regime can be appropriate.
Yes i now understand you are correct. Maybe very small insects and. for instance, Millikan oil drops but much larger goes too fast and any smaller the fluctuations take over. Thank you for taking the trouble to educate me!Fair enough.
For free falling objects under one gravity in atmospheric air linear drag would perhaps be appropriate for dust flakes or similar microscopic objects, in which case the bouyancy force should also be considered as it becomes comparable to the drag force.
I still maintain that a quadratic drag model is most appropriate for cases involving air, gravity and objects you can see.
I am a bit curious about how buoyancy enters in. One would expect small metallic flakes (for instance) to have the same density as larger metallic ingots. If buoyancy is negligible for one, it should be negligible for the other as well.dust flakes or similar microscopic objects, in which case the bouyancy force should also be considered as it becomes comparable to the drag force.
But the terminal velocity scales with massI am a bit curious about how buoyancy enters in. One would expect small metallic flakes (for instance) to have the same density as larger metallic ingots. If buoyancy is negligible for one, it should be negligible for the other as well.
Are you, perhaps, considering "fluffy" dust tufts where the behavior is well approximated as a combination of the fibrous tuft itself plus the air that is effectively entrained within?
But the the downforce from gravity and the buoyancy from air scale identically. If downforce from gravity dominates at one scale, it dominates at all scales.But the terminal velocity scales with mass
But the drag force goes up slower (maybe like an area not a volume). We don't worry about the buoyancy of an F-15 fighter!But the the downforce from gravity and the buoyancy from air scale identically. If downforce from gravity dominates at one scale, it dominates at all scales.
Viscous resistance does not scale identically, of course. But viscous resistance and buoyancy are different effects.
I admit I have very little specific knowledge about the aerodynamics of dust flakes. I was thinking that the "effective density" of a flake, if compared to a sphere of same mass and characteristic size (e.g. diameter, projected area or similar), would be low, but I agree that this is not relevant for whether or not buoyancy is significant enough to be included. Thinking about it I would say that if someone said that the seemingly long "fall time" for dust has nothing do to with buoyancy but can alone be explained as a consequence of its terminal velocity being significant lower than normal circulation air speeds in a room (e.g. due to heating, people moving about, etc) then I would think that sounded quite plausible. I guess it would be easy to test, but I am also sure my wife would not approve if I start to conduct home experiments with falling dust, so I am unfortunately left speculating.I am a bit curious about how buoyancy enters in. One would expect small metallic flakes (for instance) to have the same density as larger metallic ingots. If buoyancy is negligible for one, it should be negligible for the other as well.
Are you, perhaps, considering "fluffy" dust tufts where the behavior is well approximated as a combination of the fibrous tuft itself plus the air that is effectively entrained within?
For the sanity of this discussion (when it involves cantaloupes and F-15 fighters, at least) I suggest you guys better forget I mentioned buoyancy. I mentioned it once and thought I got away with it.But the drag force goes up slower (maybe like an area not a volume). We don't worry about the buoyancy of an F-15 fighter!
I remember doing the Millikan experiment (>40 years ago!) and using linear air drag with a buoyancy correction to ascertain the drop size by driving it up and down in the "telescope". So you are absolved of responsibility.For the sanity of this discussion (when it involves cantaloupes and F-15 fighters, at least) I suggest you guys better forget I mentioned buoyancy. I mentioned it once and thought I got away with it.
Buoyancy correction should be around one part in one thousand for oil in air. It's an easy correction -- just subtract the density of air from the density of oil. See https://www.physics.uci.edu/~advanlab/millikan.pdf where the effective density of the oil, ##\rho'## is defined in that manner.I remember doing the Millikan experiment (>40 years ago!) and using linear air drag with a buoyancy correction to ascertain the drop size by driving it up and down in the "telescope". So you are absolved of responsibility.
I deal with this for clean rooms, where larger particles settle out of the air faster than smaller ones. My understanding was that as the particles get smaller they behave more like gas molecules. I guess I never went deeper into what that means, but I guess that random molecular motion and turbulence become a larger contributor to a particle's movement as it gets smaller. This is why gasses don't stratify by density much in a room.I am a bit curious about how buoyancy enters in. One would expect small metallic flakes (for instance) to have the same density as larger metallic ingots. If buoyancy is negligible for one, it should be negligible for the other as well.