Originally Posted by
ANCPER
Well. A Squared, you'd be wrong.
Well, ANCPERS, no, I'm not.
It is interesting how you have changed your claim as you've been unable to defend it and you've had to walk it back, so to speak, to avoid admitting you were wrong.
You began by claiming:
Originally Posted by
ANCPER
the resistance thus terminal velocity will be the same and both will hit the ground at the same time regardless of mass.
and
Originally Posted by
ANCPER
When you are considering an object falling in the atmosphere the only determinate when it comes to how fast an object will fall is the area of the object and therefore the drag(resistance) it will produce, mass does not come into it.
But now you're claiming that:
Originally Posted by
ANCPER
so in a fall where the objects fail to reach their Vt the above equation is it and they will hit the ground at the same time.
Originally Posted by
ANCPER
So as to the effect of mass depends on whether the object hits the ground prior to reaching Vt.
First, you claim that mass has *nothing* to do with it, now suddenly, mass *does* have something to do with it.
Which means you were wrong. QED.
Of course in order to avoid merely saying “yeah, I was wrong, I was talking out of my nether regions” you've concocted some new fallacy. Now you're claiming that the objects fall at precisely the same rate until the less dense one reaches it's terminal velocity and the more dense one leaves it behind.
It's a more complex fallacy, but every bit as incorrect as your initial fallacy, which you've now abandoned.
A word about your “physics expert” She may well be a physics grad student, but that doesn't mean that she necessarily understands ballistics very well. That would be a subject that physics and engineering students learn in the first semester physics class in their freshman year. Second semester physics has moved on to other subjects. The thing is, Newtonian physics is pretty basic stuff and a physics student, even an undergrad moves on to other more advanced physics pretty quickly. By the time one is in in a physics graduate program, the motion of falling objects is something that they may not have considered in any depth since the first semester of their freshman year. (The astute reader will at this point be wondering: Hmmm, I wonder how A Squared knows about studying physics at a university?)
Of course it is entirely possible that she understands it perfectly well, but you just didn't understand her explanation of it, perhaps willfully on a subconscious level because of your demonstrated desire to not admit you're mistaken.
Regardless, whether it's because she doesn't understand the physics or because you didn’t understand her explanation, either way, the end result is the same; what you've taken away from that conversation is wrong.
Originally Posted by
ANCPER
She laughed at your helium balloon comparison
Uhhh, nobody mentioned helium balloons. The comparison was air filled balloons (which will fall down). You made up the part about helium. FWIW, the force of buoyancy on an air filled balloon and one filled with concrete is exactly the same. The only difference is that with air filled balloon, the buoyancy is large compared to the weight of the balloon, and small compared to the weight of the concrete filled one, but in either case the force of buoyancy that the air is exerting on the balloons is the same. And as far as the your link goes, you apparently missed the fact that your link is the same one I linked in my previous response to you, and discussed the information therein. Go back and look, it's still there in my post, the second link.
Now, here's why the current iteration of your theory is wrong. You claim that drag makes absolutely no difference until one object reaches it's terminal velocity, at which point it stops accelerating and the other continues accelerating, thus pulling “ahead” and falling faster.
In order to believe that, you'd have to believe that drag is completely non-existent until one object reaches terminal velocity.
OK, let's work through a simple example: Assume a 1 kg object and a 10kg object. The weights of the object (the force of gravity pulling on them) will be 9.8 newtons and 98 newtons respectively.
Let's further assume that the terminal velocity of the heavier object will be 50 m/s. If that is true, and the objects are externally identical (same shape, drag coefficient and frontal area) then the lighter object will reach terminal velocity at about 15.8 m/s
So at 15.8 meters/sec the force of drag (on both objects) will be 9.8 newtons. The lighter one will stop accelerating because that is equal to the force of gravity on the object and the forces cancel each other out, there is no extra force to accelerate the object. That's the definition of terminal velocity, where the force of gravity is exactly balanced by the force of drag. The heavier object will continue to accelerate because the force of drag is 9.8 newtons, but the force of gravity is 98 newtons, so there is still 88.2 newtons accelerating the object.
OK, that's just an explanation of what terminal velocity is. Now your claim is that up till the point that the lighter object reaches terminal velocity, the objects will fall with identical acceleration, thus identical speed, and will reach any point at the identical same time (short of the lighter one reaching terminal velocity)
You're wrong. Here's why: Using our example, at the instant they are released, both objects will begin accelerating at the same rate. There is no motion yet, so drag is zero, non-existent. The only force on the objects is the force of gravity. There is 9.8 newtons of gravitational force acting on the 1 kg mass which causes it to accelerate at 9.8 meters/sec/sec and there is 98 newtons of gravitational force acting on the 10 kg. mass, causing it to also accelerate at 9.8 meters/sec/sec.
That is at the instant of release, when velocity is zero, thus drag is zero, acceleration will be the same.
The thing is, drag does not remain at zero. As the speed increases, so does the drag. What is the drag when the objects are falling at half the lighter object's terminal velocity? Well, it will not be half, because drag is a function of the square of the speed. Half the lighter object's terminal velocity is 7.9 meters/sec. At that speed the drag on both objects will be 2.4 newtons. At that point in the fall, will they both be falling and accelerating at the same speed? Your claim is that they will. You are wrong. We only have to calculate the acceleration at that point to see why you're wrong. At that point, drag on both is 2.4 newtons. That means the the unopposed force on the lighter object will be 9.8 newtons of gravity – 2.4 newtons of drag, or 7.4 newtons. This means that at that point the lighter object will be accelerating at 7.4 meter/sec/sec (7.4 newtons divided by 1 kg) But the unopposed force on the heavier object will be 98 newtons of gravity – 2.4 newtons of drag, or 95.6 newtons. If you apply 95.6 newtons of force to a 10 kg object, the object will accelerate at 9.56 meters/sec/sec.
So, halfway to the lighter object's terminal velocity (in time, not distance), the lighter object is accelerating at 7.5 m/sec/sec, and the heavier one will be accelerating at 9.56 meters/sec/sec. Obviously, if one object is accelerating at a greater rate than the other, they can't possibly maintain the same velocity. And that's just a snapshot of what's happening at one instant in time. The only time the acceleration of the objects has been equal is at the precise moment of release, when velocity as zero, and thus drag was zero. At any instant after the objects have started falling, the acceleration of the heavier object will be greater than the lighter object, because of drag.
Now, I don't expect you to agree. It is clear that your brain is so tied up in the effort of not admitting that you're wrong, that you're unable to follow the explanation.
Here's what you should do. Print this entire discussion out, and take it to your physics grad student friend. Seriously, print it out, don't “explain” it too her, because you clearly don't understand it yourself, and thus are unable to describe it accurately, plus you can't seem to resist making up fictional details like helium balloons, which were not a part of the discussion. Like I said, print it all out without editing it, and hand it to her, and ask her to take it home and read it and think about it.
One of two things is going to happen:
1) she's going to immediately say, “no you misunderstood what I was saying, they do not fall with exact same identical speed until one reaches it's terminal velocity, that's not what I said.
or
2) she will, after having read and considered the explanation , say, yep sorry, I told you incorrectly, the objects will not fall at the same speed until one reaches terminal velocity.
Of course, you could save everyone a lot of effort by just admitting you were wrong. This is painfully obvious to everyone at this point, but I don't think there's any chance you're going to do that. For you it has obviously become a game of: "don't ever admit I was wrong" instead of "I want to understand this better"