Problems with FE's "gravity"

Quote from: sokarul on September 13, 2007, 06:21:38 PM

Ok you see, the earth accelerating upwards accelerates the air which accelerates the plane.  How else could the plane reach terminal "velocity"?  In the FE for a plane to reach terminal velocity it has to match the earth acceleration. 
So who doesn't understand fe physics?  That's right, YOU!

So... the air that pushes the plane UP increase the FALLING speed of the plane, down towards Earth... Ok, thank you.

Quote from: sokarul on September 12, 2007, 08:28:35 PM

Plus if you think that (in the FE) a falling plane has no acceleration, then as stated before, get your money back.  A falling plane has no acceleration at T₀ but then starts to gain acceleration. 

Yes, I still think that a falling plane (on FE) has no acceleration. Sure for a few seconds it will be some acceleration because of friction with the air, but that doesn't change the ultimate result.

Quote from: SoNic on September 13, 2007, 06:17:28 PM

A, and speed can be a noun for many different things, including the one you call velocity. Or a controlled substance.

While this is ultimately true, in physics, they are in essence two different things.

Quote from: divito on September 13, 2007, 06:22:12 PM

Quote from: SoNic on September 13, 2007, 06:17:28 PM

A, and speed can be a noun for many different things, including the one you call velocity. Or a controlled substance.

While this is ultimately true, in physics, they are in essence two different things.

Actually if you look on the picture it is "speed vectors"... Doesn't say "speed IS a vector". It was ment to be as in "speed direction"...
I like to mess with them thou...

Quote from: Gulliver on September 02, 2007, 10:19:18 AM

The one 'g' acceleration is the felt acceleration on the FE, not the actual acceleration from a valid SR FoR. FET uses an infinite set of accelerated FoRs to examine Nature as seen on the FE. However, since the FE has accelerated out of that FoR, it's immediately invalid to do so. Now from an outside observer, the FE is not accelerating at 1g and does not increase in speed so much as to exceed the speed of light. To this observer, the FE's acceleration is not constant, but rather is constantly decreasing.

So, how is that not relative, again?

What about that guy not on the Earth?

Quote from: Gulliver on September 02, 2007, 10:19:18 AM

The one 'g' acceleration is the felt acceleration on the FE, not the actual acceleration from a valid SR FoR. FET uses an infinite set of accelerated FoRs to examine Nature as seen on the FE. However, since the FE has accelerated out of that FoR, it's immediately invalid to do so. Now from an outside observer, the FE is not accelerating at 1g and does not increase in speed so much as to exceed the speed of light. To this observer, the FE's acceleration is not constant, but rather is constantly decreasing.

So, how is that not relative, again?

Quote from: TheEngineer on September 13, 2007, 06:16:30 PM

Quote from: Gulliver on September 02, 2007, 10:19:18 AM

The one 'g' acceleration is the felt acceleration on the FE, not the actual acceleration from a valid SR FoR. FET uses an infinite set of accelerated FoRs to examine Nature as seen on the FE. However, since the FE has accelerated out of that FoR, it's immediately invalid to do so. Now from an outside observer, the FE is not accelerating at 1g and does not increase in speed so much as to exceed the speed of light. To this observer, the FE's acceleration is not constant, but rather is constantly decreasing.

So, how is that not relative, again?

Again all observers measure the same value for the felt acceleration.

So you can't answer it then?

Ok, for people that want to complicate thinghs more with the air resistance (noted here r):
For plane: S_p=S₀+r(a(t-t₀))² - because the air resistance varies with square of the speed (I am simplyfing here). See, the air is accelerating the plane upwards, and the friction force will add to the original speed.
For Earth S=S₀+a(t-t₀)
Now the relative speed of plane  to Earth will be S_p-S
That means Delta S= r(a(t-t_o)²-a(t-t₀)
When this relative speed will be constant? When it will not be variable in time. So derivative of this function with the time is zero.
That means r(a(t-t_o)/2-a=0
So the time when speed will be terminal (constant) t=2/r+t₀ - so depends only of the resistance of the plane in air (given by it's shape and position).
So, again, how is this not working in FE?

Still doesn't make the picture right. 

It was for him:

Quote from: sokarul on September 13, 2007, 06:44:56 PM

Still doesn't make the picture right. 

Planes fall from sky on FE too... accelerated air (pushed by Earth) won't save the plane...

I never said it would.  I just wanted to make it clear that the plane will start to accelerate up to 9.8m/s²

Quote from: TheEngineer on September 14, 2007, 12:21:54 PM

So you can't answer it then?

So you can't read the answer then?

Quote from: sokarul on September 14, 2007, 06:21:36 PM

I never said it would.  I just wanted to make it clear that the plane will start to accelerate up to 9.8m/s²

And I proved to you that it won't... several posts ago. It will fall with a terminal velocity that depends of the shape of the falling object. Of course you need to understand basic math and physics for that :p
PS: [Easy mode on] The air is not a solid under the plane. It will not accelerate the plane up with 9.81m/^s.[Easy mode off]

You didn't prove anything. 
Here, to find the right answers, answer these questions.
1. How is terminal velocity defined in the RE?
2. How is terminal velocity defined in the FE?
3. How is terminal velocity different between the two models?
4. What has to happen for an object to reach terminal velocity in the FE?
Answer these right and you will see why you are wrong. 

Quote from: sokarul on September 15, 2007, 10:56:51 AM

You didn't prove anything. 
Here, to find the right answers, answer these questions.
1. How is terminal velocity defined in the RE?
2. How is terminal velocity defined in the FE?
3. How is terminal velocity different between the two models?
4. What has to happen for an object to reach terminal velocity in the FE?
Answer these right and you will see why you are wrong. 

I don't have to answer to questions regarding RE... that's for you to do.
Questions 2 and 4... read my mathematical demonstration above. Or is to much math for you to comprehend?

Wrong.
Simply answer question 2. 

Ok, for people that want to complicate thinghs more with the air resistance (noted here r):
For plane: Sp=S0+r(a(t-t0))2 - because the air resistance varies with square of the speed (I am simplyfing here). See, the air is accelerating the plane upwards, and the friction force will add to the original speed.
For Earth S=S0+a(t-t0)
Now the relative speed of plane  to Earth will be Sp-S
That means Delta S= r(a(t-to)2-a(t-t0)
When this relative speed will be constant (terminal velocity)? When it will not be variable in time. So derivative of this function with the variation of time is zero.
That means r(a(t-to)/2-a=0
So the time when speed will be terminal (constant) t=2/r+t0 - so depends only of the resistance of the plane in air (given by it's shape and position).

spamthink

Quote from: sokarul on September 15, 2007, 05:15:29 PM

Wrong.
Simply answer question 2. 

In the post above it is stated what means terminal velocity - in any model (FE or RE) is the moment when the delta speed between Earth and plane is constant.
You want the whole quote? Here, with red marks:

Quote

Ok, for people that want to complicate thinghs more with the air resistance (noted here r):
For plane: Sp=S0+r(a(t-t0))2 - because the air resistance varies with square of the speed (I am simplyfing here). See, the air is accelerating the plane upwards, and the friction force will add to the original speed.
For Earth S=S0+a(t-t0)
Now the relative speed of plane  to Earth will be Sp-S
That means Delta S= r(a(t-to)2-a(t-t0)
When this relative speed will be constant (terminal velocity)? When it will not be variable in time. So derivative of this function with the variation of time is zero.
That means r(a(t-to)/2-a=0
So the time when speed will be terminal (constant) t=2/r+t0 - so depends only of the resistance of the plane in air (given by it's shape and position).