Re: Terminal Velocity?
« Reply #20 on: February 04, 2020, 11:51:51 AM »


Quote
Have you ever heard of indoor sky diving?
Think about how that works...

I have heard of it...and I’ve done it.  I was stationary as long as the airflow was constant and the distance between me and the floor never changed.
Me too. It’s really hard, isn’t it?
What if the airflow wasn’t constant?
What would happen if the speed of the air kept increasing?

Then the distance would continue to increase

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Re: Terminal Velocity?
« Reply #21 on: February 04, 2020, 05:56:26 PM »
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He is, relative to the air.

Is the air moving relative to the skydiver?
Yes, that is a necessary consequence of the skydiver moving relative to the air.
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Re: Terminal Velocity?
« Reply #22 on: February 04, 2020, 09:37:06 PM »
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He is, relative to the air.

Is the air moving relative to the skydiver?
Yes, that is a necessary consequence of the skydiver moving relative to the air.

So what forces are impacting both the air and skydiver to cause movement?  And how would the relative velocity be determined? 
« Last Edit: February 04, 2020, 09:52:08 PM by pricelesspearl »

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Re: Terminal Velocity?
« Reply #23 on: February 04, 2020, 10:04:20 PM »
So what forces are impacting both the air and skydiver to cause movement?
UA is pushing the Earth, which is pushing the atmolayer, which is pushing the skydiver, but only a little bit.

And how would the relative velocity be determined? 
The same way any other velocity would be determined. Pick your favourite method.
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Re: Terminal Velocity?
« Reply #24 on: February 05, 2020, 10:05:10 PM »
So what forces are impacting both the air and skydiver to cause movement?
UA is pushing the Earth, which is pushing the atmolayer, which is pushing the skydiver, but only a little bit.

And how would the relative velocity be determined? 
The same way any other velocity would be determined. Pick your favourite method.

And what would be the formula for determining terminal velocity under that model?

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Re: Terminal Velocity?
« Reply #25 on: February 05, 2020, 11:03:41 PM »
And what would be the formula for determining terminal velocity under that model?
So, when others explained to you that this would be no different than under RET, they meant it. Further questions will be pretty boring for everyone involved.
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Re: Terminal Velocity?
« Reply #26 on: February 06, 2020, 12:27:14 AM »
And what would be the formula for determining terminal velocity under that model?
So, when others explained to you that this would be no different than under RET, they meant it. Further questions will be pretty boring for everyone involved.

And "they" are wrong.  It wouldn't be the same formula for a couple of different reasons.  First, the RE formula is designed to calculate the point at which the forces are balanced.  In FE version of terminal velocity, the forces are unbalanced.  You would be solving for something entirely different.

Second, you can't include the effect of gravity in the formula because obviously, there is none.  Any effect that an accelerating earth would have is already accounted for in the drag coefficient.  In RE, the velocity of the object moving through the air is the number used in the formula.  In FE, it would need to be the relative velocity since the diver and the earth are moving towards one another.




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Re: Terminal Velocity?
« Reply #27 on: February 06, 2020, 08:22:07 AM »
Sorry, they're not wrong. As I explained before, you're confusing yourself by choosing a FoR you don't find intuitive. Solve it relative to the Earth and it becomes quite simple.
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Re: Terminal Velocity?
« Reply #28 on: February 06, 2020, 02:43:59 PM »
Sorry, they're not wrong. As I explained before, you're confusing yourself by choosing a FoR you don't find intuitive. Solve it relative to the Earth and it becomes quite simple.

FoR has nothing to do with it.  The same formula is used no matter what the FoR, and no matter the FoR, the formula indicates the velocity at which the forces are balanced.  Changing the frame of reference doesn’t change the forces from balanced to unbalanced.

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Re: Terminal Velocity?
« Reply #29 on: February 06, 2020, 04:37:41 PM »
FoR has nothing to do with it.  The same formula is used no matter what the FoR, and no matter the FoR, the formula indicates the velocity at which the forces are balanced.  Changing the frame of reference doesn’t change the forces from balanced to unbalanced.
Oh lord, here we go again. Once again you're arguing against RET (specifically, by denying the Equivalence Principle) while claiming to be arguing for it.

Look, I don't mean to be too harsh on you, but your understanding of physics is either appalling, or you're an extremely transparent troll. I don't care which it is (leaning firmly towards the latter), but I expect you to start taking the upper fora seriously from now on. I hope I made myself clear. If you want to argue against RET, do so openly and honestly, and without this backhanded circus stuff.
« Last Edit: February 06, 2020, 04:39:36 PM by Pete Svarrior »
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Re: Terminal Velocity?
« Reply #30 on: February 06, 2020, 05:50:12 PM »
Sorry, they're not wrong. As I explained before, you're confusing yourself by choosing a FoR you don't find intuitive. Solve it relative to the Earth and it becomes quite simple.
Agree?

Terminal velocity is the maximum velocity attainable by an object as it falls through a fluid (air is the most common example). It occurs when the sum of the drag force (Fd) and the buoyancy is equal to the downward force of gravity (FG) acting on the object. Since the net force on the object is zero, the object has zero acceleration.

Re: Terminal Velocity?
« Reply #31 on: February 06, 2020, 06:06:01 PM »
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Solve it relative to the Earth and it becomes quite simple.

Fine..I can do that.

I’ll use a 5 lb. sphere 10 in2 in area.

In both RE and FE, we can use all the same variables, except for “acceleration due to gravity”.  In FE, we have to use “acceleration due to UA” and from the FoR of the earth, the sphere wouldn’t be accelerating toward the earth at 9.8m/s2.  It would be accelerating at the relative acceleration between the sphere and the earth.  I don't know exactly what that is because I have no way of knowing how much the atmowhateveryouwanttocallit is accelerating the sphere, beyond your description of "a little bit". However, I do know that it would be something less than 9.8 m/s2 (if it wasn’t the distance would never decrease).  So, let’s say the relative acceleration of the sphere is half of 9.8 m/s2. in FE…4.9 m/s2.

A 5lb. sphere with a drag coefficient of .5, (https://www.engineeringtoolbox.com/drag-coefficient-d_627.html) average air density of 1.275 kg/m3 (https://www.theweatherprediction.com/habyhints2/444/), area of 10 in2 and accelerating at 9.8m/s2, has a TV of 103.96 m/s.

A 5lb. sphere with a drag coefficient of .5, average air density of 1.275 kg/m3, area of 10 in2 and accelerating at 4.9 m/s2, has a TV of 73.511 m/s.

Clearly, the same formula from the same FoR, gives different speeds for FE and RE.  Of course, the FE number is meaningless anyway because it represents the speed at which the forces are balanced, and we know that in FE TV, the forces are not balanced.

The EP has nothing to do with it.


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Re: Terminal Velocity?
« Reply #32 on: February 06, 2020, 06:09:43 PM »
However, I do know that it would be something less than 9.8 m/s2 (if it wasn’t the distance would never decrease).  So, let’s say the relative acceleration of the sphere is half of 9.8 m/s2. in FE…4.9 m/s2.
Drag force is not a constant, buddy. The whole point of terminal velocity is that eventually it will be equal 9.8m/s2

You've been given plenty of chances. This is your last. Behave or begone.
« Last Edit: February 06, 2020, 06:13:56 PM by Pete Svarrior »
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Re: Terminal Velocity?
« Reply #33 on: February 06, 2020, 06:19:26 PM »
Drag force is not a constant, buddy. The whole point of terminal velocity is that eventually it will be equal 9.8m/s2

You've been given plenty of chances. This is your last. Behave or begone.
[/quote]

Why would you know that, when your own wiki says otherwise?

Quote
In the Round Earth model, terminal velocity happens when the acceleration due to gravity is equal to the acceleration due to drag. In the Flat Earth model, however, there are no balanced forces: terminal velocity happens when the upward acceleration of the person is equal to the upward acceleration of the Earth
.

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Re: Terminal Velocity?
« Reply #34 on: February 06, 2020, 06:53:44 PM »
Why would you know that, when your own wiki says otherwise?

Quote
In the Round Earth model, terminal velocity happens when the acceleration due to gravity is equal to the acceleration due to drag. In the Flat Earth model, however, there are no balanced forces: terminal velocity happens when the upward acceleration of the person is equal to the upward acceleration of the Earth
That... does not state otherwise. Indeed, it states that the two would eventually be equal. You even quoted that part yourself.

Your fixation with balanced forces has nothing to do with this.

Let's try once more. Pretty please stop trolling the upper fora.

Agree?
Well, no, but your position is easily fixable.

Terminal velocity is the maximum velocity attainable by an object as it falls moves through a fluid (air is the most common example). It occurs when the sum of the drag force (Fd) and the buoyancy (where applicable) is equal to the downward force of gravity (FG) acting on the object the force causing the motion in question. Since the net force on the object is zero, the object has zero acceleration.

There is nothing about air resistance that makes it specific to falling.
« Last Edit: February 06, 2020, 07:05:14 PM by Pete Svarrior »
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Re: Terminal Velocity?
« Reply #35 on: February 06, 2020, 07:11:04 PM »
However, I do know that it would be something less than 9.8 m/s2 (if it wasn’t the distance would never decrease).  So, let’s say the relative acceleration of the sphere is half of 9.8 m/s2. in FE…4.9 m/s2.
Drag force is not a constant, buddy. The whole point of terminal velocity is that eventually it will be equal 9.8m/s2

You've been given plenty of chances. This is your last. Behave or begone.
Depends on the local value of g.

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Re: Terminal Velocity?
« Reply #36 on: February 06, 2020, 10:36:38 PM »
Depends on the local value of g.
Both sides of the conversation used 9.8m/s2 with the obvious intention of it meaning g. Do you think your comment is particularly helpful here?
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