Parachutes

The FoR is the parchutist. and I do take the acceleration of the FE into account. the acceleration of the FE is used to account for the velocity when you calculate R that is why I changed it from "v" to "at". You must remember that the FE model states that the earth accelerates up to him so the air will accelerate faster and faster by him which is well accounted for. We cannot add elements into the system to try and make it reflect reality when the model does not allow it

From the FoR of the parachutist in the RE model, the Earth is accelerating towards him.

The FoR is the parchutist. and I do take the acceleration of the FE into account. the acceleration of the FE is used to account for the velocity when you calculate R that is why I changed it from "v" to "at". You must remember that the FE model states that the earth accelerates up to him so the air will accelerate faster and faster by him which is well accounted for. We cannot add elements into the system to try and make it reflect reality when the model does not allow it

Its kind of hard to pick a FOR for this problem, as most FORs are non-inertial.

The wind will only accelerate past him at 9.8m/s² max.  Its velocity will rise past him though, until his acceleration matches the earths, which happens when the airs velocity reaches around 200kph.  

The FoR is the parchutist. and I do take the acceleration of the FE into account. the acceleration of the FE is used to account for the velocity when you calculate R that is why I changed it from "v" to "at". You must remember that the FE model states that the earth accelerates up to him so the air will accelerate faster and faster by him which is well accounted for. We cannot add elements into the system to try and make it reflect reality when the model does not allow it

No, you do not account completely for the acceleration of the FE toward the parachutist. From his FoR, the surfaces of both the FE and the RE are accelerating toward him at 1g. You put the force in the RE equation, but omit it, in error, from the FE equation.

Quote from: cbarnett97 on August 28, 2007, 08:17:26 PM

The FoR is the parchutist. and I do take the acceleration of the FE into account. the acceleration of the FE is used to account for the velocity when you calculate R that is why I changed it from "v" to "at". You must remember that the FE model states that the earth accelerates up to him so the air will accelerate faster and faster by him which is well accounted for. We cannot add elements into the system to try and make it reflect reality when the model does not allow it

No, you do not account completely for the acceleration of the FE toward the parachutist. From his FoR, the surfaces of both the FE and the RE are accelerating toward him at 1g. You put the force in the RE equation, but omit it, in error, from the FE equation.

In FE the Force of the surface of the earth is not in the system while the force of gravity is acting upon him in the RE system. how can you be pushed by an unconnected piece? if a car rushes by me the car will not affect me at all, now the wind from it may affect me but not to the extent the car would if it had hit me

What I think is going on here is that you are trying to calculate his velocity relative to the surface of the earth and I am calculating the force acting upon his body.

Quote from: Gulliver on August 28, 2007, 08:22:24 PM

Quote from: cbarnett97 on August 28, 2007, 08:17:26 PM

The FoR is the parchutist. and I do take the acceleration of the FE into account. the acceleration of the FE is used to account for the velocity when you calculate R that is why I changed it from "v" to "at". You must remember that the FE model states that the earth accelerates up to him so the air will accelerate faster and faster by him which is well accounted for. We cannot add elements into the system to try and make it reflect reality when the model does not allow it

No, you do not account completely for the acceleration of the FE toward the parachutist. From his FoR, the surfaces of both the FE and the RE are accelerating toward him at 1g. You put the force in the RE equation, but omit it, in error, from the FE equation.

In FE the Force of the surface of the earth is not in the system while the force of gravity is acting upon him in the RE system. how can you be pushed by an unconnected piece? if a car rushes by me the car will not affect me at all, now the wind from it may affect me but not to the extent the car would if it had hit me

You forget that you chose the FoR of the parachutist. You have to deal with the acceleration of the surface toward him equally in both models.

We can simplify this quickly. Imagine a vertical tube perfectly evacuated expect for a ping-pong ball at the top. As the ball falls in the RE model gravity accelerates it down. As the FE surface rises it accelerates up toward the ball. The surface meets the ball at the same time with the same speed in either model.

We don't disagree on any other point. You don't need the air resistance. You don't need the parachute. You just have to face the fact that FE mimics the force of gravity for the constant acceleration of the FE up at 1g. You must add this into your equations.

Quote from: cbarnett97 on August 28, 2007, 08:31:23 PM

Quote from: Gulliver on August 28, 2007, 08:22:24 PM

Quote from: cbarnett97 on August 28, 2007, 08:17:26 PM

The FoR is the parchutist. and I do take the acceleration of the FE into account. the acceleration of the FE is used to account for the velocity when you calculate R that is why I changed it from "v" to "at". You must remember that the FE model states that the earth accelerates up to him so the air will accelerate faster and faster by him which is well accounted for. We cannot add elements into the system to try and make it reflect reality when the model does not allow it

No, you do not account completely for the acceleration of the FE toward the parachutist. From his FoR, the surfaces of both the FE and the RE are accelerating toward him at 1g. You put the force in the RE equation, but omit it, in error, from the FE equation.

In FE the Force of the surface of the earth is not in the system while the force of gravity is acting upon him in the RE system. how can you be pushed by an unconnected piece? if a car rushes by me the car will not affect me at all, now the wind from it may affect me but not to the extent the car would if it had hit me

You forget that you chose the FoR of the parachutist. You have to deal with the acceleration of the surface toward him equally in both models.

We can simplify this quickly. Imagine a vertical tube perfectly evacuated expect for a ping-pong ball at the top. As the ball falls in the RE model gravity accelerates it down. As the FE surface rises it accelerates up toward the ball. The surface meets the ball at the same time with the same speed in either model.

We don't disagree on any other point. You don't need the air resistance. You don't need the parachute. You just have to face the fact that FE mimics the force of gravity for the constant acceleration of the FE up at 1g. You must add this into your equations.

It is in the presence of these resistive forces that throws the FE theory out of whack. In the absence of resistive forces yes the acceleration in the 2 models would be the same but as soon as you add resistive forces into the mix the results are skewed. this is mainly due to where the forces come from. I suspect you are relating it to the EP and while it is correct that when I jump out of plane I would not be able to tell if I was falling or if the earth was rising up to meet me, however when we develop a model to explain what is happening and predict behavior we can not use EP because we need to state what forces are involved and how they are affecting us.

Quote from: Gulliver on August 28, 2007, 08:38:09 PM

Quote from: cbarnett97 on August 28, 2007, 08:31:23 PM

Quote from: Gulliver on August 28, 2007, 08:22:24 PM

Quote from: cbarnett97 on August 28, 2007, 08:17:26 PM

The FoR is the parchutist. and I do take the acceleration of the FE into account. the acceleration of the FE is used to account for the velocity when you calculate R that is why I changed it from "v" to "at". You must remember that the FE model states that the earth accelerates up to him so the air will accelerate faster and faster by him which is well accounted for. We cannot add elements into the system to try and make it reflect reality when the model does not allow it

No, you do not account completely for the acceleration of the FE toward the parachutist. From his FoR, the surfaces of both the FE and the RE are accelerating toward him at 1g. You put the force in the RE equation, but omit it, in error, from the FE equation.

In FE the Force of the surface of the earth is not in the system while the force of gravity is acting upon him in the RE system. how can you be pushed by an unconnected piece? if a car rushes by me the car will not affect me at all, now the wind from it may affect me but not to the extent the car would if it had hit me

You forget that you chose the FoR of the parachutist. You have to deal with the acceleration of the surface toward him equally in both models.

We can simplify this quickly. Imagine a vertical tube perfectly evacuated expect for a ping-pong ball at the top. As the ball falls in the RE model gravity accelerates it down. As the FE surface rises it accelerates up toward the ball. The surface meets the ball at the same time with the same speed in either model.

We don't disagree on any other point. You don't need the air resistance. You don't need the parachute. You just have to face the fact that FE mimics the force of gravity for the constant acceleration of the FE up at 1g. You must add this into your equations.

It is in the presence of these resistive forces that throws the FE theory out of whack. In the absence of resistive forces yes the acceleration in the 2 models would be the same but as soon as you add resistive forces into the mix the results are skewed. this is mainly due to where the forces come from. I suspect you are relating it to the EP and while it is correct that when I jump out of plane I would not be able to tell if I was falling or if the earth was rising up to meet me, however when we develop a model to explain what is happening and predict behavior we can not use EP because we need to state what forces are involved and how they are affecting us.

There is no resistive forces in the fe. 

There would be your resistance to the accelerating air. That would not go anywhere in either model

Quote from: Gulliver on August 28, 2007, 08:38:09 PM

Quote from: cbarnett97 on August 28, 2007, 08:31:23 PM

Quote from: Gulliver on August 28, 2007, 08:22:24 PM

Quote from: cbarnett97 on August 28, 2007, 08:17:26 PM

The FoR is the parchutist. and I do take the acceleration of the FE into account. the acceleration of the FE is used to account for the velocity when you calculate R that is why I changed it from "v" to "at". You must remember that the FE model states that the earth accelerates up to him so the air will accelerate faster and faster by him which is well accounted for. We cannot add elements into the system to try and make it reflect reality when the model does not allow it

No, you do not account completely for the acceleration of the FE toward the parachutist. From his FoR, the surfaces of both the FE and the RE are accelerating toward him at 1g. You put the force in the RE equation, but omit it, in error, from the FE equation.

In FE the Force of the surface of the earth is not in the system while the force of gravity is acting upon him in the RE system. how can you be pushed by an unconnected piece? if a car rushes by me the car will not affect me at all, now the wind from it may affect me but not to the extent the car would if it had hit me

You forget that you chose the FoR of the parachutist. You have to deal with the acceleration of the surface toward him equally in both models.

We can simplify this quickly. Imagine a vertical tube perfectly evacuated expect for a ping-pong ball at the top. As the ball falls in the RE model gravity accelerates it down. As the FE surface rises it accelerates up toward the ball. The surface meets the ball at the same time with the same speed in either model.

We don't disagree on any other point. You don't need the air resistance. You don't need the parachute. You just have to face the fact that FE mimics the force of gravity for the constant acceleration of the FE up at 1g. You must add this into your equations.

It is in the presence of these resistive forces that throws the FE theory out of whack. In the absence of resistive forces yes the acceleration in the 2 models would be the same but as soon as you add resistive forces into the mix the results are skewed. this is mainly due to where the forces come from. I suspect you are relating it to the EP and while it is correct that when I jump out of plane I would not be able to tell if I was falling or if the earth was rising up to meet me, however when we develop a model to explain what is happening and predict behavior we can not use EP because we need to state what forces are involved and how they are affecting us.

How can resistive forces throw FE out of whack?

Try this:
Take the parachutist model. Remove air resistance and the parachute. The models agree since there are no resistive forces.

Now add a parachute that magically transforms to provide a constant force of 1N up during the entire descent. The force would be the same in both models. Since equals added to equals are equal, the models are still equal.

Now add air resistance and the parachute. The forces would again be equal. The models are still equal.

There would be your resistance to the accelerating air. That would not go anywhere in either model

Ok, so you are falling and have -F_air in FE and in RE you have F_gravity-F_air.

Can you state an experiment that would show us where the prediction model would vary?

The thing is though you must remove resistive forces for them to be equal. when you add resistive forces they are no longer equal forces. If I have 2=2 and then I add a 3 so it becomes 2=2+3 they are no longer equal. and that is what is happening here, we are trying to use a model to predict behavior so we will need to add in resistive forces. we can not assume they don't exist

-F_air

Those are my resistive forces

Quote from: cbarnett97 on August 28, 2007, 08:49:58 PM

There would be your resistance to the accelerating air. That would not go anywhere in either model

Ok, so you are falling and have -F_air in FE and in RE you have F_gravity-F_air.

Can you state an experiment that would show us where the prediction model would vary?

Calculate the terminal velovity of an object

The thing is though you must remove resistive forces for them to be equal. when you add resistive forces they are no longer equal forces. If I have 2=2 and then I add a 3 so it becomes 2=2+3 they are no longer equal. and that is what is happening here, we are trying to use a model to predict behavior so we will need to add in resistive forces. we can not assume they don't exist

Tell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.

There would be your resistance to the accelerating air. That would not go anywhere in either model

Thats A force not forces.  Second it doesn't throw anything out of whack. 

Quote from: Username on August 28, 2007, 09:00:58 PM

Quote from: cbarnett97 on August 28, 2007, 08:49:58 PM

There would be your resistance to the accelerating air. That would not go anywhere in either model

Ok, so you are falling and have -F_air in FE and in RE you have F_gravity-F_air.

Can you state an experiment that would show us where the prediction model would vary?

Calculate the terminal velocity of an object

RE:
V_t = sqrt( 2 m g / p A C _d )

Lets say m = 1 kg
g = 9.8 m/s
p = 1 kg/m³
A = 9.8 m²
C = .5 m ²

sqrt ( 2 * 1 * 9.8 / 1 * 9.8 * .5 ) = 1m/s

FE:
V_t = sqrt( 2 m a / p A C _d )

Again,
m = 1 kg
g = 9.8 m/s
p = 1 kg/m³
A = 9.8 m²
C = .5 m ²

sqrt ( 2 * 1 * 9.8 / 1 * 9.8 * .5 ) = 1m/s

Edit: ignor that "ahh" it was pretty retarded.

Quote from: cbarnett97 on August 28, 2007, 09:03:41 PM

Quote from: Username on August 28, 2007, 09:00:58 PM

Quote from: cbarnett97 on August 28, 2007, 08:49:58 PM

There would be your resistance to the accelerating air. That would not go anywhere in either model

Ok, so you are falling and have -F_air in FE and in RE you have F_gravity-F_air.

Can you state an experiment that would show us where the prediction model would vary?

Calculate the terminal velocity of an object

RE:
V_t = sqrt( 2 m g / p A C _d )

Lets say m = 1 kg
g = 9.8 m/s
p = 1 kg/m³
A = 9.8 m²
C = .5 m ²

sqrt ( 2 * 1 * 9.8 / 1 * 9.8 * .5 ) = 1m/s

FE:
V_t = sqrt( 2 m a / p A C _d )

Again,
m = 1 kg
g = 9.8 m/s
p = 1 kg/m³
A = 9.8 m²
C = .5 m ²

sqrt ( 2 * 1 * 9.8 / 1 * 9.8 * .5 ) = 1m/s

Ahh, are you saying can't have acceleration instead of gravity because, for example, if two objects were falling of different masses the earth would have to be accelerating at different speeds for them both to exist?

basically that. also the guy does not accelerate to the earth, it is the other way around so when calculating terminal velocity you can not use "a" because he is not accelerating so his velocity will be different and then if you relate that to the surface of the earth you get differnet numbers than we see in reality. and based upon this knowledge terminal velocity is not based upon anything other that a persons resistance to the air accelerating past him.

Quote from: cbarnett97 on August 28, 2007, 09:03:41 PM

Quote from: Username on August 28, 2007, 09:00:58 PM

Quote from: cbarnett97 on August 28, 2007, 08:49:58 PM

There would be your resistance to the accelerating air. That would not go anywhere in either model

Ok, so you are falling and have -F_air in FE and in RE you have F_gravity-F_air.

Can you state an experiment that would show us where the prediction model would vary?

Calculate the terminal velocity of an object

RE:
V_t = sqrt( 2 m g / p A C _d )

Lets say m = 1 kg
g = 9.8 m/s
p = 1 kg/m³
A = 9.8 m²
C = .5 m ²

sqrt ( 2 * 1 * 9.8 / 1 * 9.8 * .5 ) = 1m/s

FE:
V_t = sqrt( 2 m a / p A C _d )

Again,
m = 1 kg
g = 9.8 m/s
p = 1 kg/m³
A = 9.8 m²
C = .5 m ²

sqrt ( 2 * 1 * 9.8 / 1 * 9.8 * .5 ) = 1m/s

Edit: ignor that "ahh" it was pretty retarded.

I think one kilogram falls faster than 1 m/s.

Quote from: Username on August 28, 2007, 09:20:09 PM

Quote from: cbarnett97 on August 28, 2007, 09:03:41 PM

Quote from: Username on August 28, 2007, 09:00:58 PM

Quote from: cbarnett97 on August 28, 2007, 08:49:58 PM

There would be your resistance to the accelerating air. That would not go anywhere in either model

Ok, so you are falling and have -F_air in FE and in RE you have F_gravity-F_air.

Can you state an experiment that would show us where the prediction model would vary?

Calculate the terminal velocity of an object

RE:
V_t = sqrt( 2 m g / p A C _d )

Lets say m = 1 kg
g = 9.8 m/s
p = 1 kg/m³
A = 9.8 m²
C = .5 m ²

sqrt ( 2 * 1 * 9.8 / 1 * 9.8 * .5 ) = 1m/s

FE:
V_t = sqrt( 2 m a / p A C _d )

Again,
m = 1 kg
g = 9.8 m/s
p = 1 kg/m³
A = 9.8 m²
C = .5 m ²

sqrt ( 2 * 1 * 9.8 / 1 * 9.8 * .5 ) = 1m/s

Edit: ignor that "ahh" it was pretty retarded.

I think one kilogram falls faster than 1 m/s.

the mistake was that A is not a. A is the cross sectional area of the object that is facing the wind

Quote from: Username on August 28, 2007, 09:20:09 PM

Quote from: cbarnett97 on August 28, 2007, 09:03:41 PM

Quote from: Username on August 28, 2007, 09:00:58 PM

Quote from: cbarnett97 on August 28, 2007, 08:49:58 PM

There would be your resistance to the accelerating air. That would not go anywhere in either model

Ok, so you are falling and have -F_air in FE and in RE you have F_gravity-F_air.

Can you state an experiment that would show us where the prediction model would vary?

Calculate the terminal velocity of an object

RE:
V_t = sqrt( 2 m g / p A C _d )

Lets say m = 1 kg
g = 9.8 m/s
p = 1 kg/m³
A = 9.8 m²
C = .5 m ²

sqrt ( 2 * 1 * 9.8 / 1 * 9.8 * .5 ) = 1m/s

FE:
V_t = sqrt( 2 m a / p A C _d )

Again,
m = 1 kg
g = 9.8 m/s
p = 1 kg/m³
A = 9.8 m²
C = .5 m ²

sqrt ( 2 * 1 * 9.8 / 1 * 9.8 * .5 ) = 1m/s

Ahh, are you saying can't have acceleration instead of gravity because, for example, if two objects were falling of different masses the earth would have to be accelerating at different speeds for them both to exist?

basically that. also the guy does not accelerate to the earth, it is the other way around so when calculating terminal velocity you can not use "a" because he is not accelerating so his velocity will be different and then if you relate that to the surface of the earth you get differnet numbers than we see in reality. and based upon this knowledge terminal velocity is not based upon anything other that a persons resistance to the air accelerating past him.

No. His velocity relative to the air is increased equally in both models, in RE by gravity, in FE by the FE's acceleration. You already agreed to this! You get the same numbers in both models.

tell me how the earth accelerates the guy when he is 5,000ft above the earth?

tell me how the earth accelerates the guy when he is 5,000ft above the earth?

Te earth pushes the air which pushes him, any questions? 

Quote from: cbarnett97 on August 28, 2007, 09:39:22 PM

tell me how the earth accelerates the guy when he is 5,000ft above the earth?

Te earth pushes the air which pushes him, any questions? 

so that accelerates him to 9.8m/s² in the same time gravity does?
or would it accelerate him to that acceleration  by  a=(DpA/2)v^2

tell me how the earth accelerates the guy when he is 5,000ft above the earth?

We've been over this when we consider the case without resistive forces. The FE accelerates towards him pushing the air by him faster. Did you forget that we've agreed that in the absence of resistive forces that the models produce the same results?

Quote from: sokarul on August 28, 2007, 09:41:05 PM

Quote from: cbarnett97 on August 28, 2007, 09:39:22 PM

tell me how the earth accelerates the guy when he is 5,000ft above the earth?

Te earth pushes the air which pushes him, any questions? 

so that accelerates him to 9.8m/s² in the same time gravity does?

It should accelerate him to 9.8m/s² in the time it take the RE air to slow a skydiver from 9.8m/s² to zero.  

Quote from: cbarnett97 on August 28, 2007, 09:43:03 PM

Quote from: sokarul on August 28, 2007, 09:41:05 PM

Quote from: cbarnett97 on August 28, 2007, 09:39:22 PM

tell me how the earth accelerates the guy when he is 5,000ft above the earth?

Te earth pushes the air which pushes him, any questions? 

so that accelerates him to 9.8m/s² in the same time gravity does?

It should accelerate him to 9.8m/s² in the time it take the RE air to slow a skydiver from 9.8m/s² to zero. 

He wants to use the parachutist's For, not the outside observer's FoR.

Quote from: cbarnett97 on August 28, 2007, 09:02:47 PM

The thing is though you must remove resistive forces for them to be equal. when you add resistive forces they are no longer equal forces. If I have 2=2 and then I add a 3 so it becomes 2=2+3 they are no longer equal. and that is what is happening here, we are trying to use a model to predict behavior so we will need to add in resistive forces. we can not assume they don't exist

Tell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.

Please answer this challenge.

Quote from: Gulliver on August 28, 2007, 09:05:18 PM

Quote from: cbarnett97 on August 28, 2007, 09:02:47 PM

The thing is though you must remove resistive forces for them to be equal. when you add resistive forces they are no longer equal forces. If I have 2=2 and then I add a 3 so it becomes 2=2+3 they are no longer equal. and that is what is happening here, we are trying to use a model to predict behavior so we will need to add in resistive forces. we can not assume they don't exist

Tell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.

Please answer this challenge.

I will be happy to show it for I think the third or fourth time.
RE: F=mg-R
FE: F=-R

and the acceleration of the air past the skydiver would account for the velocity needed to calculate the resistance

Quote from: Gulliver on August 28, 2007, 09:49:32 PM

Quote from: Gulliver on August 28, 2007, 09:05:18 PM

Quote from: cbarnett97 on August 28, 2007, 09:02:47 PM

The thing is though you must remove resistive forces for them to be equal. when you add resistive forces they are no longer equal forces. If I have 2=2 and then I add a 3 so it becomes 2=2+3 they are no longer equal. and that is what is happening here, we are trying to use a model to predict behavior so we will need to add in resistive forces. we can not assume they don't exist

Tell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.

Please answer this challenge.

I will be happy to show it for I think the third or fourth time.
RE: F=mg-R
FE: F=-R

and the acceleration of the air past the skydiver would account for the velocity needed to calculate the resistance

-R cannot = F