Looking for an intelligent argument. (Terminal Velocity)

OK.

I SHALL BE BACK!

I am not sure you are the one trying to apply it

Since an object moving through air will not truly be in "free fall" because of air resistance we cannot apply the EP

/\    acceleration of the earth.  /\     acceleration of the person
|                                          |
|                                          |
When these are equal, the person has reached terminal velocity.

|
\/  Acceleration of the person.  /\  Acceleration due to air resistance.
                                           |
When these are equal, the person has reached terminal velocity.

V=at regardless of FE or RE.

A_net=A_Earth-A_Object

There is a terminal velocity. Once A_net=0 the relative velocity will be constant which is the same as terminal velocity.

To barnet,



From Engy and Robbyj:

Quote from: TheEngineer on August 17, 2008, 12:22:28 AM

/\    acceleration of the earth.  /\     acceleration of the person
|                                          |
|                                          |
When these are equal, the person has reached terminal velocity.

|
\/  Acceleration of the person.  /\  Acceleration due to air resistance.
                                           |
When these are equal, the person has reached terminal velocity.

Quote from: Robbyj on August 23, 2008, 07:44:07 PM

V=at regardless of FE or RE.

A_net=A_Earth-A_Object

There is a terminal velocity. Once A_net=0 the relative velocity will be constant which is the same as terminal velocity.

so show the math that shows the accelerations are of equal magnitude at t=2, since it is such a simple concept it should be very easy

An accelerometer does not measure the '1g' of the Earth until you are at terminal velocity in either the RE or FE model. By definition free fall (without air resistance) is a relative state of rest (objects following geodesics), so if it could measure 'absolute acceleration' then you would have a means of figuring out the 'preferred rest frame' of the Universe, which is forbidden in relativity.

To clarify, an accelerometer would read (in the vertical axis, with '+' meaning 'upwards'):

1) stationary on the runway
 RE : +1g (9.81ms^-2) contact acceleration
 FE : +1a_UA (9.81ms^-2) contact acceleration

2) climbing to drop altitude
 RE : +1g + (vertical acceleration of aircraft)
 FE : +1a_UA + (vertical acceleration of aircraft)

3) level at drop alitutude
 RE : +1g - (small correction for altitude (g drops as 1/r² from surface of Earth))
 FE : +1a_UA - (small correction for altitude (source depends on your 'flavour' of FE))

4) the instant that Tom Bishop jumps out horizontally from the door
 RE : zero
 FE : zero

5) before reaching terminal velocity
 RE : zero + (air resistance as fall towards Earth at 1g acceleration [acceleration as measured from the ground])
 FE : zero + (air resistance as fall towards Earth at 1a_UA acceleration [acceleration as measured from the ground])

6) terminal velocity
 RE : +1g
 FE : +1a_UA

As you can see, the two are identical at all points. Neither accelerates faster than the other and neither 'feels' any different at any point (so long as you have a reasonable physical explanation for g dropping off with altitude, which FE provides a few alternative theories for). If I'm wrong at any point here, feel free to explain why.

Quote from: cbarnett97 on September 04, 2008, 01:15:30 PM

so show the math that shows the accelerations are of equal magnitude at t=2, since it is such a simple concept it should be very easy

Quote from: Matrix on September 01, 2008, 03:13:40 AM

An accelerometer does not measure the '1g' of the Earth until you are at terminal velocity in either the RE or FE model. By definition free fall (without air resistance) is a relative state of rest (objects following geodesics), so if it could measure 'absolute acceleration' then you would have a means of figuring out the 'preferred rest frame' of the Universe, which is forbidden in relativity.

To clarify, an accelerometer would read (in the vertical axis, with '+' meaning 'upwards'):

1) stationary on the runway
 RE : +1g (9.81ms^-2) contact acceleration
 FE : +1a_UA (9.81ms^-2) contact acceleration

2) climbing to drop altitude
 RE : +1g + (vertical acceleration of aircraft)
 FE : +1a_UA + (vertical acceleration of aircraft)

3) level at drop alitutude
 RE : +1g - (small correction for altitude (g drops as 1/r² from surface of Earth))
 FE : +1a_UA - (small correction for altitude (source depends on your 'flavour' of FE))

4) the instant that Tom Bishop jumps out horizontally from the door
 RE : zero
 FE : zero

5) before reaching terminal velocity
 RE : zero + (air resistance as fall towards Earth at 1g acceleration [acceleration as measured from the ground])
 FE : zero + (air resistance as fall towards Earth at 1a_UA acceleration [acceleration as measured from the ground])

6) terminal velocity
 RE : +1g
 FE : +1a_UA

As you can see, the two are identical at all points. Neither accelerates faster than the other and neither 'feels' any different at any point (so long as you have a reasonable physical explanation for g dropping off with altitude, which FE provides a few alternative theories for). If I'm wrong at any point here, feel free to explain why.

I seem to remember posting a set of equations that proved the models were identical.  Plus, the whole EP thing.

show that the EP is valid for all times then since it is so simple

In the physics of relativity, the equivalence principle refers to several related concepts dealing with the equivalence of gravitational and inertial mass, and to Albert Einstein's assertion that the gravitational "force" as experienced locally while standing on a massive body (such as the Earth) is actually the same as the pseudo-force experienced by an observer in a non-inertial (accelerated) frame of reference.

Experiments performed in a uniformly accelerating reference frame with acceleration a are indistinguishable from the same experiments performed in a non-accelerating reference frame which is situated in a gravitational field where the acceleration of gravity = g = -a = intensity of gravity field.

"Then occurred to me the happiest thought of my life, in the following form. The gravitational field has only a relative existence... Because for an observer falling freely from the roof of a house there exists - at least in his immediate surroundings - no gravitational field....(in this consideration air resistance is, of course, ignored)."

Go to the beach, you can see the slow curvature of the earth when you look out to sea.

Quote from: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/grel.html

Experiments performed in a uniformly accelerating reference frame with acceleration a are indistinguishable from the same experiments performed in a non-accelerating reference frame which is situated in a gravitational field where the acceleration of gravity = g = -a = intensity of gravity field.

and air resistance is not a uniformly accelerating field so therefore the EP can not be applied until it is uniform, ie. net acceleration with the air is zero

Uh, no. http://en.wikipedia.org/wiki/No-communication_theorem

This has just gone too far.

The EP is all about reference frames.  In FE, the reference frame we choose goes with the Earth.  It accelerates at 9.8 m/s^2 upwards.  This is the reference frame in which we do all of our experiments.  By the EP, any experiment we do in this reference frame is equivalent to any we do in a gravitational field of -9.8m/s^2.  It's that simple.  Really.

Quote

and air resistance is not a uniformly accelerating field so therefore the EP can not be applied until it is uniform, ie. net acceleration with the air is zero

We are not working in the frame of object which falls through the air.  We are working in the frame of the Earth.  The EP applies.

If you disagree with this, you disagree with General Relativity, because it also states that we are accelerating upwards at 9.8m/s^2.