The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Theory => Topic started by: pricelesspearl on January 06, 2020, 05:01:40 PM
-
Terminal Velocity
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.
Q: If gravity does not exist, how does terminal velocity work?
A: When the acceleration of the person is equal to the acceleration of the Earth, the person has reached terminal velocity.
Why is terminal velocity even a thing in the UA model of FE? It is a concept applicable to free falling bodies, which according to UA don’t exist. Things don’t fall…the earth rises up to meet them.
But it is a demonstrable fact that if a person jumps out of a plane the distance between the person and ground decreases at an increasing rate until at some point (depending or air density, humidity, weight…etc.), the distance begins decreasing at a constant rate.
What causes that if the earth is rising at a constant rate and there is no air resistance (because the person is not actually “falling” but is suspended motionless) to affect the rate at which the distance between the earth and person decreases? Not to mention the fact that when a skydiver opens a chute, the rate decreases…what would cause the decrease? Again, it can’t be drag…because there is no drag on something that is not falling.
Also in the wiki...
One of the primary proofs for the Universal Accelerator is the fact that bodies fall without inertial resistance.
But in a non-vacuum environment bodies do fall with air resistance which wouldn’t be the case unless the body is actually moving through the air and not just suspended motionless until the ground reaches it.
If you release a feather and a bowling ball at the same height and both are suspended motionless until the earth meets it…why does the earth reach the bowling ball first?
-
Terminal Velocity
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.
Q: If gravity does not exist, how does terminal velocity work?
A: When the acceleration of the person is equal to the acceleration of the Earth, the person has reached terminal velocity.
Why is terminal velocity even a thing in the UA model of FE? It is a concept applicable to free falling bodies, which according to UA don’t exist. Things don’t fall…the earth rises up to meet them.
But it is a demonstrable fact that if a person jumps out of a plane the distance between the person and ground decreases at an increasing rate until at some point (depending or air density, humidity, weight…etc.), the distance begins decreasing at a constant rate.
What causes that if the earth is rising at a constant rate and there is no air resistance (because the person is not actually “falling” but is suspended motionless) to affect the rate at which the distance between the earth and person decreases? Not to mention the fact that when a skydiver opens a chute, the rate decreases…what would cause the decrease? Again, it can’t be drag…because there is no drag on something that is not falling.
Also in the wiki...
One of the primary proofs for the Universal Accelerator is the fact that bodies fall without inertial resistance.
But in a non-vacuum environment bodies do fall with air resistance which wouldn’t be the case unless the body is actually moving through the air and not just suspended motionless until the ground reaches it.
If you release a feather and a bowling ball at the same height and both are suspended motionless until the earth meets it…why does the earth reach the bowling ball first?
Not being a flat earther and having over 900 skydives I think I can take a stab at this.
If UA was a thing (and it's not) the air could not be affected by it other than the earth pushing up on it. (If UA pushed the air the same as it pushed the ground there would be consistent air pressure at all altitudes.) This would create more air pressure the lower you got due to compression. This would create the drag you are looking for. Relative to the universe outside of the UA influence you would appear to rise, not float in one spot.
A good a guess as any?
-
If UA was a thing (and it's not) the air could not be affected by it other than the earth pushing up on it. (If UA pushed the air the same as it pushed the ground there would be consistent air pressure at all altitudes.) This would create more air pressure the lower you got due to compression. This would create the drag you are looking for. Relative to the universe outside of the UA influence you would appear to rise, not float in one spot.
A good a guess as any?
Well terminal velocity is defined as the maximum falling speed you reach because the force of air pushing against you as wind resistance is the same as the force of gravity. Since UA does not have the force of gravity the phrase terminal velocity can't really be applied.
If you are asking why things reach a maximum speed when the earth is accelerating up to meet them i guess the UA could say the same thing and the RE. The force of the wind resistance.
-
Well terminal velocity is defined as the maximum falling speed you reach because the force of air pushing against you as wind resistance is the same as the force of gravity. Since UA does not have the force of gravity the phrase terminal velocity can't really be applied.
That's the point I was making when I said
Why is terminal velocity even a thing in the UA model of FE? It is a concept applicable to free falling bodies, which according to UA don’t exist. Things don’t fall…the earth rises up to meet them.
If you are asking why things reach a maximum speed when the earth is accelerating up to meet them i guess the UA could say the same thing and the RE. The force of the wind resistance.
In order to reach a "maximum speed", something has to be moving. In the FE model, things aren't moving as the earth is accelerating up to meet an object. There is no wind resistance.
-
How does a paper helicopter work if there is no wind resistance?
-
How does a paper helicopter work if there is no wind resistance?
In the absence of gravity,it doesn't. There must be something creating wind strong enough to accelerate it before there can be resistance to it. I suppose a breeze would work with a paper helicopter. Not so much with a bowling ball.
-
If there was such a thing as gravity, it would never reach terminal velocity. The closer you get to Earth the more it pulls you in. Shouldn’t gravity make an object accelerate until it comes in contact with the object.
-
If there was such a thing as gravity, it would never reach terminal velocity. The closer you get to Earth the more it pulls you in. Shouldn’t gravity make an object accelerate until it comes in contact with the object.
Conventional physics teaches that terminal velocity is where the force of gravity is matched by an opposing force of air resistance.
-
In order to reach a "maximum speed", something has to be moving. In the FE model, things aren't moving as the earth is accelerating up to meet an object. There is no wind resistance.
Well, yes, if you want to look at it from that frame of reference, the skydiver isn't moving until the air pushes him up ever so gently. As that air becomes denser, the force of the air pushing the skydiver becomes greater. The difference between the upward acceleration of the Earth and the upward acceleration of the skydiver will therefore diminish, and eventually reach zero.
However, you are being willingly obtuse by not picking a FoR you would find more intuitive. Simply taking a non-inertial FoR in which the Earth is stationary and the skydiver is falling will eliminate your confusion.
-
In order to reach a "maximum speed", something has to be moving. In the FE model, things aren't moving as the earth is accelerating up to meet an object. There is no wind resistance.
Well, yes, if you want to look at it from that frame of reference, the skydiver isn't moving until the air pushes him up ever so gently. As that air becomes denser, the force of the air pushing the skydiver becomes greater. The difference between the upward acceleration of the Earth and the upward acceleration of the skydiver will therefore diminish, and eventually reach zero.
However, you are being willingly obtuse by not picking a FoR you would find more intuitive. Simply taking a non-inertial FoR in which the Earth is stationary and the skydiver is falling will eliminate your confusion.
Do you have measurements of air density as it varies for a skydiver. I suggest it does not significantly vary.
-
Sorry, you're right, that's obviously not a major factor. The important change is that in drag - as the skydiver's velocity relative to the air changes, so does the drag coefficient.
This error aside, my comment on pp's confusion stands. The discussion on what is "actually" moving is only likely to distract (and is arguably meaningless). It's the relative motion between the two that matters. I should have kept my response to just that, given TomIA already gave a perfectly good answer to the core question.
-
Sorry, you're right, that's obviously not a major factor. The important change is that in drag - as the skydiver's velocity relative to the air changes so does the drag coefficient.
This error aside, my comment on pp's confusion stands. The discussion on what is "actually" moving is only likely to distract (and is arguably meaningless). It's the relative motion between the two that matters. I should have kept my response to just that, given TomIA already gave a perfectly good answer to the core question.
According to FET, a skydiver's velocity is zero. That never changes. Also, according to FET, the velocity of the earth under UA never changes...therefore the relative velocity between the two should never vary.
-
Do you have measurements of air density as it varies for a skydiver. I suggest it does not significantly vary.
If I recall correctly its about 1 atmosphere of pressure change between Sea Level and 8000 feet.
-
According to FET, a skydiver's velocity is zero.
Relative to what?
Also, according to FET, the velocity of the earth under UA never changes.
That sounds like the opposite of a constant*, universal acceleration. Just, y'know, saying that an acceleration (i.e. change in velocity) leads to the velocity never changing... something's fishy about that claim. As someone who actually has an understanding of FET, I can assure you that your assertion is false. But, just to humour you: Relative to what?
* - select FoR's only, T&C's apply, see website for details
While we wait for your answers, let's explore some of the hilarious consequences of your claim:
- What we were discussing is the change of relative velocity between the skydiver and the air, not Earth. The two are not identical, unless you're prepared to claim that wind doesn't exist. I wouldn't be entirely surprised if you did.
- If, as you claim, the relative velocity between the skydiver and the Earth never changes, the skydiver will never hit the Earth. I assure you that FET does not claim things never hit the ground once they've been dropped.
- The skydiver's velocity never changing would imply that the force imposed by drag does not lead to a change in velocity. Are we to assume that you, the skydiver, are so fat and/or dense that your mass cannot be meaningfully described by the set of real numbers? That would have significant consequences on, well, everything.
As with your past comedic failures, I encourage you to rescind your claim. Secretly, however, I hope you'll double down as per usual.
-
The skydiver's velocity never changing would imply that the force imposed by drag does not lead to a change in velocity.
A skydiver's velocity changing as result of a force imposed by drag would imply that the skydiver is moving. Drag force is proportional to speed, if there is no speed there is no drag force to effect velocity. It's not that difficult to understand.
-
would imply that the skydiver is moving.
He is, relative to the air.
It's not that difficult to understand.
I agree, but yet here you are.
-
He is, relative to the air.
Is the air moving relative to the skydiver?
-
The skydiver's velocity never changing would imply that the force imposed by drag does not lead to a change in velocity.
A skydiver's velocity changing as result of a force imposed by drag would imply that the skydiver is moving. Drag force is proportional to speed, if there is no speed there is no drag force to effect velocity. It's not that difficult to understand.
Have you ever heard of indoor sky diving?
Think about how that works...
-
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.
-
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?
-
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
-
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.
-
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?
-
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.
-
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?
-
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 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.
-
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.
-
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.
-
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.
-
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.
-
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.
-
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.
-
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?
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
.
-
Why would you know that, when your own wiki says otherwise?
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.
-
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.
-
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?