Parachutes

where does mg come from. now look at the FE model carefully, when you jump out of a plane you are no longer being accelerated by the FE so it catches up to you. So once again do not think of reality when you look at the problem. The FE model is not an accurate model for predicting behavior because of this simple fact of the missing mg.

The mg comes from the acceleration upwards of the FE. I don't agree that the Earth accelerates upward, but I can put aside my disbelief long enough to understand their argument.

As the FE accelerates upward, the velocity at which the parachutist approaches the surface increase. You must deal with it, or admit a bias that you can't overcome.

Oh there is no confusion here. According to the FE model what would happen is that when oyu jum out of a plane you are no longer accelerating with the FE but at the same time there is a constant force put on your body and that force will continue to accelerate in the same direction as the FE until your acceleration matched that of the FE then because of that short period of time where the accelerations were not equal the velocity of the earth would be greater so it would eventually catch up to you.

Now this is close to what we see in reality, but according to their model the only thing that should effect the terminal velocity of an object would be its air resistance. Now this would be correct under the FE model because if you look at the case of zero air resistance your terminal velocity would be zero and the earth would accelerate up to you at 9.8m/s². How it is wrong is that we know that air resistance is not the only thing to affect terminal velocity. For example under the FE model a ball that weighs 1N would have the same Terminal velocity as a ball that weighs 10N. and we know that the only time this is true in reality is if you drop the balls in outer space. when in the presence of resistive forces the terminal velocities would be and are different.

Quote from: TheEngineer on August 28, 2007, 03:09:00 PM

Why do you keep ignoring my post with the terminal velocity equations?

because they are wrong

Really?  Show me.

your equations do not reflect the rules set forth by the model. in FE there is no F=ma acting upon the skydiver only a F=R

Oh there is no confusion here. According to the FE model what would happen is that when oyu jum out of a plane you are no longer accelerating with the FE but at the same time there is a constant force put on your body and that force will continue to accelerate in the same direction as the FE until your acceleration matched that of the FE then because of that short period of time where the accelerations were not equal the velocity of the earth would be greater so it would eventually catch up to you.

Now this is close to what we see in reality, but according to their model the only thing that should effect the terminal velocity of an object would be its air resistance. Now this would be correct under the FE model because if you look at the case of zero air resistance your terminal velocity would be zero and the earth would accelerate up to you at 9.8m/s². How it is wrong is that we know that air resistance is not the only thing to affect terminal velocity. For example under the FE model a ball that weighs 1N would have the same Terminal velocity as a ball that weighs 10N. and we know that the only time this is true in reality is if you drop the balls in outer space. when in the presence of resistive forces the terminal velocities would be and are different.

The bold part is wrong.  Terminal velocity is never zero.   

The last part about balls being different weights.  In the FE the will still have each have different terminal "velocities".  The lighter ball will take a slower  wind speed to reach the 9.8m/s² acceleration, just like in the RE lighter objects have slower terminal velocities.   

your equations do not reflect the rules set forth by the model. in FE there is no F=ma acting upon the skydiver only a F=R

Wouldn't that only be true if you were looking from outside the Earth?

Quote from: cbarnett97 on August 28, 2007, 03:52:49 PM

Oh there is no confusion here. According to the FE model what would happen is that when oyu jum out of a plane you are no longer accelerating with the FE but at the same time there is a constant force put on your body and that force will continue to accelerate in the same direction as the FE until your acceleration matched that of the FE then because of that short period of time where the accelerations were not equal the velocity of the earth would be greater so it would eventually catch up to you.

Now this is close to what we see in reality, but according to their model the only thing that should effect the terminal velocity of an object would be its air resistance. Now this would be correct under the FE model because if you look at the case of zero air resistance your terminal velocity would be zero and the earth would accelerate up to you at 9.8m/s². How it is wrong is that we know that air resistance is not the only thing to affect terminal velocity. For example under the FE model a ball that weighs 1N would have the same Terminal velocity as a ball that weighs 10N. and we know that the only time this is true in reality is if you drop the balls in outer space. when in the presence of resistive forces the terminal velocities would be and are different.

The bold part is wrong.  Terminal velocity is never zero.   

The last part about balls being different weights.  In the FE the will still have each have different terminal "velocities".  The lighter ball will take a slower  wind speed to reach the 9.8m/s² acceleration, just like in the RE lighter objects have slower terminal velocities.   

According to the FE model terminal velocity would be zero, now we know that is not the case in reality which is my whole point, and according to the FE model mass does not matter only the cross section of the area exposed to the resistance, so as long as the 2 balls were of equal size they should have the same terminal velocity, which once again does not reflect reality

Oh there is no confusion here. According to the FE model what would happen is that when oyu jum out of a plane you are no longer accelerating with the FE but at the same time there is a constant force put on your body and that force will continue to accelerate in the same direction as the FE until your acceleration matched that of the FE then because of that short period of time where the accelerations were not equal the velocity of the earth would be greater so it would eventually catch up to you.

Now this is close to what we see in reality, but according to their model the only thing that should effect the terminal velocity of an object would be its air resistance. Now this would be correct under the FE model because if you look at the case of zero air resistance your terminal velocity would be zero and the earth would accelerate up to you at 9.8m/s². How it is wrong is that we know that air resistance is not the only thing to affect terminal velocity. For example under the FE model a ball that weighs 1N would have the same Terminal velocity as a ball that weighs 10N. and we know that the only time this is true in reality is if you drop the balls in outer space. when in the presence of resistive forces the terminal velocities would be and are different.

You continue to forget that terminal velocity is dependent not on the "absolute" velocity of the parachutist but on relative velocity with the surface. You have NEVER dealt with the acceleration of the FE, yet you must.

Quote from: cbarnett97 on August 28, 2007, 03:59:34 PM

your equations do not reflect the rules set forth by the model. in FE there is no F=ma acting upon the skydiver only a F=R

Wouldn't that only be true if you were looking from outside the Earth?

no it is only true if you add the surface of the earth to the system of the parachutist

Quote from: sokarul on August 28, 2007, 04:05:31 PM

Quote from: cbarnett97 on August 28, 2007, 03:52:49 PM

Oh there is no confusion here. According to the FE model what would happen is that when oyu jum out of a plane you are no longer accelerating with the FE but at the same time there is a constant force put on your body and that force will continue to accelerate in the same direction as the FE until your acceleration matched that of the FE then because of that short period of time where the accelerations were not equal the velocity of the earth would be greater so it would eventually catch up to you.

Now this is close to what we see in reality, but according to their model the only thing that should effect the terminal velocity of an object would be its air resistance. Now this would be correct under the FE model because if you look at the case of zero air resistance your terminal velocity would be zero and the earth would accelerate up to you at 9.8m/s². How it is wrong is that we know that air resistance is not the only thing to affect terminal velocity. For example under the FE model a ball that weighs 1N would have the same Terminal velocity as a ball that weighs 10N. and we know that the only time this is true in reality is if you drop the balls in outer space. when in the presence of resistive forces the terminal velocities would be and are different.

The bold part is wrong.  Terminal velocity is never zero.   

The last part about balls being different weights.  In the FE the will still have each have different terminal "velocities".  The lighter ball will take a slower  wind speed to reach the 9.8m/s² acceleration, just like in the RE lighter objects have slower terminal velocities.   

According to the FE model terminal velocity would be zero, now we know that is not the case in reality which is my whole point, and according to the FE model mass does not matter only the cross section of the area exposed to the resistance, so as long as the 2 balls were of equal size they should have the same terminal velocity, which once again does not reflect reality

You are constantly making the sophomoric mistake of failing to state your frame of reference when dealing with height, velocity, and acceleration. If you'd take the time to state the frame each time, you'd see your mistake.

Quote from: cbarnett97 on August 28, 2007, 04:09:17 PM

Quote from: sokarul on August 28, 2007, 04:05:31 PM

Quote from: cbarnett97 on August 28, 2007, 03:52:49 PM

Oh there is no confusion here. According to the FE model what would happen is that when oyu jum out of a plane you are no longer accelerating with the FE but at the same time there is a constant force put on your body and that force will continue to accelerate in the same direction as the FE until your acceleration matched that of the FE then because of that short period of time where the accelerations were not equal the velocity of the earth would be greater so it would eventually catch up to you.

Now this is close to what we see in reality, but according to their model the only thing that should effect the terminal velocity of an object would be its air resistance. Now this would be correct under the FE model because if you look at the case of zero air resistance your terminal velocity would be zero and the earth would accelerate up to you at 9.8m/s². How it is wrong is that we know that air resistance is not the only thing to affect terminal velocity. For example under the FE model a ball that weighs 1N would have the same Terminal velocity as a ball that weighs 10N. and we know that the only time this is true in reality is if you drop the balls in outer space. when in the presence of resistive forces the terminal velocities would be and are different.

The bold part is wrong.  Terminal velocity is never zero.   

The last part about balls being different weights.  In the FE the will still have each have different terminal "velocities".  The lighter ball will take a slower  wind speed to reach the 9.8m/s² acceleration, just like in the RE lighter objects have slower terminal velocities.   

According to the FE model terminal velocity would be zero, now we know that is not the case in reality which is my whole point, and according to the FE model mass does not matter only the cross section of the area exposed to the resistance, so as long as the 2 balls were of equal size they should have the same terminal velocity, which once again does not reflect reality

You are constantly making the sophomoric mistake of failing to state your frame of reference when dealing with height, velocity, and acceleration. If you'd take the time to state the frame each time, you'd see your mistake.

my frame of reference is the parachutist, if i am running down the street at 12mph does it really matter if a car passes me at 60mph?

...
my frame of reference is the parachutist, if i am running down the street at 12mph does it really matter if a car passes me at 60mph?

You're confused. The velocities that you're quoting are not relative to the parachutist. You're changing frames regularly. If you're running down the street at 12mph and a car passes you at 60mph, you're in the frame of the street, not you. Yes, it does matter.

no no confusion, the only velocity that I am dealing with it the velocity of the air as it passes over the parachutist. Draw a free body diagram around the runner and tell me how the acceleration of the car would affect him.

no no confusion, the only velocity that I am dealing with it the velocity of the air as it passes over the parachutist. Draw a free body diagram around the runner and tell me how the acceleration of the car would affect him.

Okay, I'll do that, but I'm running out of patience. I've presented careful mathematical evidence, taking the time to carefully form my arguments.

Let's consider the following. Along a street, a runner moves at 12mph east toward a car moving at 60mph west with an initial separation of a mile

1) For the frame of reference of the street with east being positive, with 0 being the midpoint of the street:
The street moves at zero.
The runner moves at 12mph.
The car moves at -60mph
The runner is at 0.5 miles.
The car is at -0.5 miles.

2) For the frame of reference of the runner with east being positive, and with 0 being the location of the runner:
The street moves at -12mph.
The car moves at -48mph.
The runner moves at 0.
The runner is at 0.
The midpoint of the street is at 0.5 miles.
The car is at 1.0 miles.

3) For the frame of reference of the car with east being positive, and with 0 being the location of the car:
The street moves at 60mph.
The runner moves at 72mph.
The car moves at 0.
The midpoint of the street is at -0.5 miles.
The runner is at -1.0 miles.

Now, I've done what you've asked. How about answering my challenges?

your equations do not reflect the rules set forth by the model. in FE there is no F=ma acting upon the skydiver only a F=R

I think the main problem here is a basic lack of the understanding of physics principles.


Newton's second, F=ma, is the governing equation for the system.  It is not dependent on RE/FE situations.  It is what it is.  Now, in the FE, there is just one force acting on the skydiver, the drag force, while in the RE there are two, his 'weight' and the drag force.  So, like I said:

RE => mg=F_D

FE => ma=F_D

Quote from: sokarul on August 28, 2007, 04:05:31 PM

Quote from: cbarnett97 on August 28, 2007, 03:52:49 PM

Oh there is no confusion here. According to the FE model what would happen is that when oyu jum out of a plane you are no longer accelerating with the FE but at the same time there is a constant force put on your body and that force will continue to accelerate in the same direction as the FE until your acceleration matched that of the FE then because of that short period of time where the accelerations were not equal the velocity of the earth would be greater so it would eventually catch up to you.

Now this is close to what we see in reality, but according to their model the only thing that should effect the terminal velocity of an object would be its air resistance. Now this would be correct under the FE model because if you look at the case of zero air resistance your terminal velocity would be zero and the earth would accelerate up to you at 9.8m/s². How it is wrong is that we know that air resistance is not the only thing to affect terminal velocity. For example under the FE model a ball that weighs 1N would have the same Terminal velocity as a ball that weighs 10N. and we know that the only time this is true in reality is if you drop the balls in outer space. when in the presence of resistive forces the terminal velocities would be and are different.

The bold part is wrong.  Terminal velocity is never zero.   

The last part about balls being different weights.  In the FE the will still have each have different terminal "velocities".  The lighter ball will take a slower  wind speed to reach the 9.8m/s² acceleration, just like in the RE lighter objects have slower terminal velocities.   

According to the FE model terminal velocity would be zero, now we know that is not the case in reality which is my whole point, and according to the FE model mass does not matter only the cross section of the area exposed to the resistance, so as long as the 2 balls were of equal size they should have the same terminal velocity, which once again does not reflect reality

Nope.  In the FE relative to the earth the terminal velocity would be around 200 kph and to an outside observer there is no terminal velocity.  Also in the RE mass does matter.  
Lets get another thing straight.  In the FE the air is providing the force to accelerate the skydiver.  His air resistance is not a force.  The only thing it does is say how much of the force applied by the wind actually gets applied to him.  

Let me see if I understand what you mean:
Are you are saying that, for example, if you are in a car, and a brick wall is being pushed towards you it would approach at the same speed regardless of whether or not your windows are down (which would cause drag if the car was moving)?

If thats basically what you are saying, it seems that in that model, the air is not accelerating with the wall, unlike the atmosphere which is accelerating with the fearth.

Or did I completely miss what you were saying and I'll slump back into silence for a while? heh

Let me see if I understand what you mean:
Are you are saying that, for example, if you are in a car, and a brick wall is being pushed towards you it would approach at the same speed regardless of whether or not your windows are down (which would cause drag if the car was moving)?

If thats basically what you are saying, it seems that in that model, the air is not accelerating with the wall, unlike the atmosphere which is accelerating with the fearth.

Or did I completely miss what you were saying and I'll slump back into silence for a while? heh

Yes. Your windows do not effect the speed of the air. 

Quote from: Username on August 28, 2007, 05:46:26 PM

Let me see if I understand what you mean:
Are you are saying that, for example, if you are in a car, and a brick wall is being pushed towards you it would approach at the same speed regardless of whether or not your windows are down (which would cause drag if the car was moving)?

If thats basically what you are saying, it seems that in that model, the air is not accelerating with the wall, unlike the atmosphere which is accelerating with the fearth.

Or did I completely miss what you were saying and I'll slump back into silence for a while? heh

Yes. Your windows do not effect the speed of the air. 

Well, no, if you are driving having your front windows open  and back closed would create a parachute effect, if I am not mistaken.  Thats not what I was asking.  I was wondering if cbarret was saying that the air accelerating upwards with the earth was not accelerating upwards.

Quote from: sokarul on August 28, 2007, 05:59:51 PM

Quote from: Username on August 28, 2007, 05:46:26 PM

Let me see if I understand what you mean:
Are you are saying that, for example, if you are in a car, and a brick wall is being pushed towards you it would approach at the same speed regardless of whether or not your windows are down (which would cause drag if the car was moving)?

If thats basically what you are saying, it seems that in that model, the air is not accelerating with the wall, unlike the atmosphere which is accelerating with the fearth.

Or did I completely miss what you were saying and I'll slump back into silence for a while? heh

Yes. Your windows do not effect the speed of the air. 

Well, no, if you are driving having your front windows open  and back closed would create a parachute effect, if I am not mistaken.  Thats not what I was asking.  I was wondering if cbarret was saying that the air accelerating upwards with the earth was not accelerating upwards.

Yeah it would be a parachute effect, but it doesn't effect the wind speed.  The earth is applying a force to the air and the air applies a force to you.  You drag coefficient sets how much of that force you feel.

Are you are saying that, for example, if you are in a car, and a brick wall is being pushed towards you it would approach at the same speed regardless of whether or not your windows are down (which would cause drag if the car was moving)?

If thats basically what you are saying, it seems that in that model, the air is not accelerating with the wall, unlike the atmosphere which is accelerating with the fearth.

I don't think I was clear sorry - the air is accelerating and the earth is in the common FE as I see it.   Is cbarret saying the air is not accelerating in FE?

Quote from: Username on August 28, 2007, 06:01:58 PM

Are you are saying that, for example, if you are in a car, and a brick wall is being pushed towards you it would approach at the same speed regardless of whether or not your windows are down (which would cause drag if the car was moving)?

If thats basically what you are saying, it seems that in that model, the air is not accelerating with the wall, unlike the atmosphere which is accelerating with the fearth.

I don't think I was clear sorry - the air is accelerating and the earth is in the common FE as I see it.   Is cbarret saying the air is not accelerating in FE?

I'm not even sure what he is saying anymore. 

Quote from: cbarnett97 on August 28, 2007, 03:59:34 PM

your equations do not reflect the rules set forth by the model. in FE there is no F=ma acting upon the skydiver only a F=R

I think the main problem here is a basic lack of the understanding of physics principles.


Newton's second, F=ma, is the governing equation for the system.  It is not dependent on RE/FE situations.  It is what it is.  Now, in the FE, there is just one force acting on the skydiver, the drag force, while in the RE there are two, his 'weight' and the drag force.  So, like I said:

RE => mg=F_D

FE => ma=F_D

You must ask yourself is the skydiver moving, or is everything moving around the skydiver? If you say the skydiver is moving then you are not following the FE model and gravity must exist as in the RE model, If you say that everything moves around the skydiver then the skydiver himself is not moving, so therefore there is no acceleration on the skydiver.

You must ask yourself is the skydiver moving, or is everything moving around the skydiver? If you say the skydiver is moving then you are not following the FE model and gravity must exist as in the RE model, If you say that everything moves around the skydiver then the skydiver himself is not moving, so therefore there is no acceleration on the skydiver.

This requires changing the FoR though.

Quote from: TheEngineer on August 28, 2007, 05:05:09 PM

Quote from: cbarnett97 on August 28, 2007, 03:59:34 PM

your equations do not reflect the rules set forth by the model. in FE there is no F=ma acting upon the skydiver only a F=R

I think the main problem here is a basic lack of the understanding of physics principles.


Newton's second, F=ma, is the governing equation for the system.  It is not dependent on RE/FE situations.  It is what it is.  Now, in the FE, there is just one force acting on the skydiver, the drag force, while in the RE there are two, his 'weight' and the drag force.  So, like I said:

RE => mg=F_D

FE => ma=F_D

You must ask yourself is the skydiver moving, or is everything moving around the skydiver? If you say the skydiver is moving then you are not following the FE model and gravity must exist as in the RE model, If you say that everything moves around the skydiver then the skydiver himself is not moving, so therefore there is no acceleration on the skydiver.

So there is another problem.  The skydiver is moving.  There is no reason he wouldn't be.  As soon as he jumps out of the plane his acceleration goes to zero but he is still moving with a velocity.  As he accelerates his velocity rises. 

Quote from: TheEngineer on August 28, 2007, 05:05:09 PM

Quote from: cbarnett97 on August 28, 2007, 03:59:34 PM

your equations do not reflect the rules set forth by the model. in FE there is no F=ma acting upon the skydiver only a F=R

I think the main problem here is a basic lack of the understanding of physics principles.


Newton's second, F=ma, is the governing equation for the system.  It is not dependent on RE/FE situations.  It is what it is.  Now, in the FE, there is just one force acting on the skydiver, the drag force, while in the RE there are two, his 'weight' and the drag force.  So, like I said:

RE => mg=F_D

FE => ma=F_D

You must ask yourself is the skydiver moving, or is everything moving around the skydiver? If you say the skydiver is moving then you are not following the FE model and gravity must exist as in the RE model, If you say that everything moves around the skydiver then the skydiver himself is not moving, so therefore there is no acceleration on the skydiver.

You must decide on which FoR you're going to use. You told us that you always used the FoR of the skydiver. Are you standing by that? Are you going to use the accelerated frame set (since the skydiver is accelerated in the both models)?

We shouldn't have to be guessing what FoR you're talking about!

Quote from: cbarnett97 on August 28, 2007, 04:36:21 PM

no no confusion, the only velocity that I am dealing with it the velocity of the air as it passes over the parachutist. Draw a free body diagram around the runner and tell me how the acceleration of the car would affect him.

Okay, I'll do that, but I'm running out of patience. I've presented careful mathematical evidence, taking the time to carefully form my arguments.

Let's consider the following. Along a street, a runner moves at 12mph east toward a car moving at 60mph west with an initial separation of a mile

1) For the frame of reference of the street with east being positive, with 0 being the midpoint of the street:
The street moves at zero.
The runner moves at 12mph.
The car moves at -60mph
The runner is at 0.5 miles.
The car is at -0.5 miles.

2) For the frame of reference of the runner with east being positive, and with 0 being the location of the runner:
The street moves at -12mph.
The car moves at -48mph.
The runner moves at 0.
The runner is at 0.
The midpoint of the street is at 0.5 miles.
The car is at 1.0 miles.

3) For the frame of reference of the car with east being positive, and with 0 being the location of the car:
The street moves at 60mph.
The runner moves at 72mph.
The car moves at 0.
The midpoint of the street is at -0.5 miles.
The runner is at -1.0 miles.

Now, I've done what you've asked. How about answering my challenges?

Lets do a very simple example of the diffence between what each model yields. To keep it simple we will use a baseball. The average baseball has a mass of .145kg and the cross section of the area is 4.2x10^-3m² and we will use T=1sec for the time
Now the terminal velocity for a baseball is 43m/s but we would need to calculate "D" first so,
D=2mg/v_tpA and when you plug in the numbers you get D=(2*.145*9.80)/(43*1.2*4.2x10^-3)=.305
Now we can calculate the force that this will yield, (note I replace V with the equivilant equation "at"
RE: F=mg-1/2DpA(at)² --> F=(.145)(9.8 )-1/2(.305)(1.2)(4.2x10^-3)(9.8 )² so we get a force of 1.35N
In FE we have  F=-1/2DpAv² --> F=-1/2(.305)(1.2)(4.2x10^-3)(9.8 )² so we end up with a force of -.074N

So like I said Not the same thing

Edit: Changed the Signs for FE to make sure the coordinate system was maintained

Quote from: Gulliver on August 28, 2007, 04:49:35 PM

Quote from: cbarnett97 on August 28, 2007, 04:36:21 PM

no no confusion, the only velocity that I am dealing with it the velocity of the air as it passes over the parachutist. Draw a free body diagram around the runner and tell me how the acceleration of the car would affect him.

Okay, I'll do that, but I'm running out of patience. I've presented careful mathematical evidence, taking the time to carefully form my arguments.

Let's consider the following. Along a street, a runner moves at 12mph east toward a car moving at 60mph west with an initial separation of a mile

1) For the frame of reference of the street with east being positive, with 0 being the midpoint of the street:
The street moves at zero.
The runner moves at 12mph.
The car moves at -60mph
The runner is at 0.5 miles.
The car is at -0.5 miles.

2) For the frame of reference of the runner with east being positive, and with 0 being the location of the runner:
The street moves at -12mph.
The car moves at -48mph.
The runner moves at 0.
The runner is at 0.
The midpoint of the street is at 0.5 miles.
The car is at 1.0 miles.

3) For the frame of reference of the car with east being positive, and with 0 being the location of the car:
The street moves at 60mph.
The runner moves at 72mph.
The car moves at 0.
The midpoint of the street is at -0.5 miles.
The runner is at -1.0 miles.

Now, I've done what you've asked. How about answering my challenges?

Lets do a very simple example of the diffence between what each model yields. To keep it simple we will use a baseball. The average baseball has a mass of .145kg and the cross section of the area is 4.2x10^-3m² and we will use T=1sec for the time
Now the terminal velocity for a baseball is 43m/s but we would need to calculate "D" first so,
D=2mg/v_tpA and when you plug in the numbers you get D=(2*.145*9.80)/(43*1.2*4.2x10^-3)=.305
Now we can calculate the force that this will yield, (note I replace V with the equivilant equation "at"
RE: F=mg-1/2DpA(at)² --> F=(.145)(9.8 )-1/2(.305)(1.2)(4.2x10^-3)(9.8 )² so we get a force of 1.35N
In FE we have  F=-1/2DpAv² --> F=-1/2(.305)(1.2)(4.2x10^-3)(9.8 )² so we end up with a force of -.074N

So like I said Not the same thing

Edit: Changed the Signs for FE to make sure the coordinate system was maintained

You continue to refuse to state your choice of FoR. It's getting annoying.

The difference is that you're applying gravity in the RE model, but refuse to apply the FE's acceleration. I don't understand your refusal. Clearly that acceleration effects the outcome. The relative velocity, the time of touchdown, and all the rest are exactly the same once you correct your model.

The force of gravity down in RE is replaced with the acceleration up of the FE.

In the "reality" of FE, the FE accelerates upward at 1g. Why would you include that fact in your FE models and equations?

So like I said Not the same thing

It might help you if you started your equations from the same point, namely Newton's second.  If you do this, you will get the exact two equations I gave you before!

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