Parachutes

Quote from: cbarnett97 on August 27, 2007, 06:40:30 PM

if you are disconnected fron the earth you can not have any acceleration.

It seems like you are saying here that once you are in the air, you can not be accelerated, even by the air.  Did I misread this?

you can be acclerated by your resistance to the air, but in no way can you be accelerated by the FE.

Quote from: TheEngineer on August 26, 2007, 12:56:27 AM

RE =>  m*g = F_D

FE =>  m*a = F_D

Just solve for velocity from the drag equation.

and what is a again? before you answer you have already stated that air is not solid and it is impossible to be accelerated at 9.8m/s², which means that air cannot recreate gravity, so that would yield different results than what we see in reality

draw a little man, then draw a circle around him and that is everything in the system, now yes there is air accelerating by him hence the air resistance force, becuase accelerating air does have a velocity does it not, so as the air continues to accelerate past him it creates a force of a magnitude of 1/2DpAv². and unless I am mistaken i did include that force in my model.

Sorry. That's wrong. You must include the surface of the Earth and its acceleration.

and what is a again? before you answer you have already stated that air is not solid and it is impossible to be accelerated at 9.8m/s², which means that air cannot recreate gravity, so that would yield different results than what we see in reality

Er, I'm pretty sure you don't accelerate at 9.8m/s² with a parachute, even in RE.

Quote from: cbarnett97 on August 27, 2007, 06:48:48 PM

draw a little man, then draw a circle around him and that is everything in the system, now yes there is air accelerating by him hence the air resistance force, becuase accelerating air does have a velocity does it not, so as the air continues to accelerate past him it creates a force of a magnitude of 1/2DpAv². and unless I am mistaken i did include that force in my model.

Sorry. That's wrong. You must include the surface of the Earth and its acceleration.

only when you are trying to relate your height to the earth, but that is not the same as calculating the forces acting on the parachutist.

Quote from: cbarnett97 on August 27, 2007, 06:56:39 PM

and what is a again? before you answer you have already stated that air is not solid and it is impossible to be accelerated at 9.8m/s², which means that air cannot recreate gravity, so that would yield different results than what we see in reality

Er, I'm pretty sure you don't accelerate at 9.8m/s² with a parachute, even in RE.

no it does not but what the engineer is looking to do is run to his EP argument so I am letting him go there so he will be happy then I will show him how the EP will not work, then he will accuse me of trying to prove Einstein wrong

Quote from: Gulliver on August 27, 2007, 06:47:46 PM

Quote from: cbarnett97 on August 27, 2007, 06:31:32 PM

Quote from: Gulliver on August 27, 2007, 06:28:13 PM

Quote from: cbarnett97 on August 27, 2007, 06:18:38 PM

yes not once have I ever even talked about horizontal forces. but when you derive an equation you want to keep it somwhat general but not to general so it can be applied to all situations. so when deriving an air resistance problem you will need to use all components of force for that type of problem, just like if you are trying to derive a friction problem you an not just use the force over time and ignore the coefficient of static friction/Kinetic friction. you will never get reproducable results that way.

Sorry. That's wrong. Decomposition of vectors into a coordinate system is a common and practical method in problem solving in physics. The intent was to show that no matter whatever force is involved, whatever the object's mass, or whatever initial velocity FE and RE produce equal results.

Now let's remember that this result does not apply to all situations. When altitudes are high enough to change gravity or distance traveled vertically enough to change the height due to the curvature of the Earth, this derivation is overly simple.

Like I said when you derive an equation you had better be able to apply it to all situations, that is why it is called a general case. you can not say you proved something that stipulates "as long as nothing else changes I am right" No offense but that is very engineerish. you might as well just say that as long as it is looked at locally and you add in the acceleration of the earth which is not local I am right

Sorry. You're not right. Any proof has its limits, but this proof destroys all arguments that FE and RE have different local gravities.

sorry if you are trying to predict results in the real world it had better be able to reproduce every situation or that is a false model, go ahead and rederive it using the correct formulas that reflect what we are trying to predict and you will see a big difference. Remember if you are trying to show that they are the same we have a very accepted equation for terminal velocity so you would need to show that on the FE it is the same

I have shown that the Force in both models are equal. The model is accurate.

I've asked before, but I'll repeat myself: What is specifically wrong with my proof. How doesn't the Force for both forces match?

Are the initial velocities the same? Yes.
Are the cross-sectional area of the parachutes the same? Yes.
Are the air densities the same? Yes.
Are the wind velocities pushing by the parachutist the same? Yes.

Are the forces therefore the same? Yes.

Is the proof correct? I think so.

Quote from: cbarnett97 on August 27, 2007, 06:52:07 PM

Quote from: Gulliver on August 27, 2007, 06:47:46 PM

Quote from: cbarnett97 on August 27, 2007, 06:31:32 PM

Quote from: Gulliver on August 27, 2007, 06:28:13 PM

Quote from: cbarnett97 on August 27, 2007, 06:18:38 PM

yes not once have I ever even talked about horizontal forces. but when you derive an equation you want to keep it somwhat general but not to general so it can be applied to all situations. so when deriving an air resistance problem you will need to use all components of force for that type of problem, just like if you are trying to derive a friction problem you an not just use the force over time and ignore the coefficient of static friction/Kinetic friction. you will never get reproducable results that way.

Sorry. That's wrong. Decomposition of vectors into a coordinate system is a common and practical method in problem solving in physics. The intent was to show that no matter whatever force is involved, whatever the object's mass, or whatever initial velocity FE and RE produce equal results.

Now let's remember that this result does not apply to all situations. When altitudes are high enough to change gravity or distance traveled vertically enough to change the height due to the curvature of the Earth, this derivation is overly simple.

Like I said when you derive an equation you had better be able to apply it to all situations, that is why it is called a general case. you can not say you proved something that stipulates "as long as nothing else changes I am right" No offense but that is very engineerish. you might as well just say that as long as it is looked at locally and you add in the acceleration of the earth which is not local I am right

Sorry. You're not right. Any proof has its limits, but this proof destroys all arguments that FE and RE have different local gravities.

sorry if you are trying to predict results in the real world it had better be able to reproduce every situation or that is a false model, go ahead and rederive it using the correct formulas that reflect what we are trying to predict and you will see a big difference. Remember if you are trying to show that they are the same we have a very accepted equation for terminal velocity so you would need to show that on the FE it is the same

I have shown that the Force in both models are equal. The model is accurate.

I've asked before, but I'll repeat myself: What is specifically wrong with my proof. How doesn't the Force for both forces match?

Are the initial velocities the same? Yes.
Are the cross-sectional area of the parachutes the same? Yes.
Are the air densities the same? Yes.
Are the wind velocities pushing by the parachutist the same? Yes.

Are the forces therefore the same? Yes.

Is the proof correct? I think so.

forces are not the same:
RE: F=mg-R
FE: F= -R

Quote from: Username on August 27, 2007, 06:52:15 PM

Quote from: cbarnett97 on August 27, 2007, 06:40:30 PM

if you are disconnected fron the earth you can not have any acceleration.

It seems like you are saying here that once you are in the air, you can not be accelerated, even by the air.  Did I misread this?

you can be acclerated by your resistance to the air, but in no way can you be accelerated by the FE.

You must include the acceleration of the FE toward the parachutist.

Quote from: cbarnett97 on August 27, 2007, 06:54:05 PM

Quote from: Username on August 27, 2007, 06:52:15 PM

Quote from: cbarnett97 on August 27, 2007, 06:40:30 PM

if you are disconnected fron the earth you can not have any acceleration.

It seems like you are saying here that once you are in the air, you can not be accelerated, even by the air.  Did I misread this?

you can be acclerated by your resistance to the air, but in no way can you be accelerated by the FE.

You must include the acceleration of the FE toward the parachutist.

Not in the system. We can not add things to the system just to make it correct. this is fron the FAQ
"Q: "What about gravity?"

A: The Earth is accelerating upwards at 1g (9.8m/s^2) along with every star, sun and moon in the universe. This produces the same effect as gravity.

Q: "Isn't this version of gravity flawed? Wouldn't planes/helicopters/paragliders crash into the Earth as the Earth rises up to them?"

A: No. By the same argument, we could ask why planes/helicopters/paragliders don't crash into the Earth as they accelerate down towards them.  The reason that planes do not crash is that their wings produce lift, which, when the rate of acceleration upwards equals that of gravity's pull downwards, causes them to remain at a constant altitude.

The same thing happens if the Earth is moving up. The plane is accelerating upwards at the same rate as the Earth, which means the distance between them does not change. Therefore, the plane stays at the same height and does not crash.
 so as you can see is that planes do not crash because they are producing an acceleration equal to the earth, now the earth can cause that same acceleration on a parachutist?

Quote from: Gulliver on August 27, 2007, 06:58:31 PM

Quote from: cbarnett97 on August 27, 2007, 06:48:48 PM

draw a little man, then draw a circle around him and that is everything in the system, now yes there is air accelerating by him hence the air resistance force, becuase accelerating air does have a velocity does it not, so as the air continues to accelerate past him it creates a force of a magnitude of 1/2DpAv². and unless I am mistaken i did include that force in my model.

Sorry. That's wrong. You must include the surface of the Earth and its acceleration.

only when you are trying to relate your height to the earth, but that is not the same as calculating the forces acting on the parachutist.

That is the point! This is a relative issue. The height is measured from the parachutist to the ground.

You're correct that the forces acting on the parachutist are not equal between the models. In RE, gravity accelerates the parachutist down. In FE, the UA magically accelerates the Earth up. But since the parachutist, or anyone else, can't determine which is being accelerated, the models provide the same results.

Quote from: cbarnett97 on August 27, 2007, 07:00:55 PM

Quote from: Gulliver on August 27, 2007, 06:58:31 PM

Quote from: cbarnett97 on August 27, 2007, 06:48:48 PM

draw a little man, then draw a circle around him and that is everything in the system, now yes there is air accelerating by him hence the air resistance force, becuase accelerating air does have a velocity does it not, so as the air continues to accelerate past him it creates a force of a magnitude of 1/2DpAv². and unless I am mistaken i did include that force in my model.

Sorry. That's wrong. You must include the surface of the Earth and its acceleration.

only when you are trying to relate your height to the earth, but that is not the same as calculating the forces acting on the parachutist.

That is the point! This is a relative issue. The height is measured from the parachutist to the ground.

You're correct that the forces acting on the parachutist are not equal between the models. In RE, gravity accelerates the parachutist down. In FE, the UA magically accelerates the Earth up. But since the parachutist, or anyone else, can't determine which is being accelerated, the models provide the same results.

no, if the forces are not equal they will not produce the same results which is something that we could see

Quote from: Gulliver on August 27, 2007, 07:12:18 PM

Quote from: cbarnett97 on August 27, 2007, 06:54:05 PM

Quote from: Username on August 27, 2007, 06:52:15 PM

Quote from: cbarnett97 on August 27, 2007, 06:40:30 PM

if you are disconnected fron the earth you can not have any acceleration.

It seems like you are saying here that once you are in the air, you can not be accelerated, even by the air.  Did I misread this?

you can be acclerated by your resistance to the air, but in no way can you be accelerated by the FE.

You must include the acceleration of the FE toward the parachutist.

Not in the system. We can not add things to the system just to make it correct. this is fron the FAQ
"Q: "What about gravity?"

A: The Earth is accelerating upwards at 1g (9.8m/s^2) along with every star, sun and moon in the universe. This produces the same effect as gravity.

Q: "Isn't this version of gravity flawed? Wouldn't planes/helicopters/paragliders crash into the Earth as the Earth rises up to them?"

A: No. By the same argument, we could ask why planes/helicopters/paragliders don't crash into the Earth as they accelerate down towards them.  The reason that planes do not crash is that their wings produce lift, which, when the rate of acceleration upwards equals that of gravity's pull downwards, causes them to remain at a constant altitude.

The same thing happens if the Earth is moving up. The plane is accelerating upwards at the same rate as the Earth, which means the distance between them does not change. Therefore, the plane stays at the same height and does not crash.
 so as you can see is that planes do not crash because they are producing an acceleration equal to the earth, now the earth can cause that same acceleration on a parachutist?

You cannot omit a feature of the FE model to make your challenge right. Sorry, but you need to face the music.

I have ommited nothing. Th FE version of "gravity" is this: the reason we are stuck to the groung is not because gravity is pushing us down but rather it is because the earth is pushing us up, so that means that if we are not on the surface of the earth we are no longer being pushed, so we do not really ever fall we hang there until the earth accelerates up to us, that is the model that this site has put forth to explain gravity, now going off of that how exactly does air all of the sudden exert enough force to recreate the "force" of gravity when you are trying to predict behavior.

how exactly does air all of the sudden exert enough force to recreate the "force" of gravity when you are trying to predict behavior.

Fill up your bathtub. Place you hand on the bottom of the tub, palm up. Force your hand up quickly and watch what happens to the surface of the water. Apply same concept to accelerating FE.

I have ommited nothing. Th FE version of "gravity" is this: the reason we are stuck to the groung is not because gravity is pushing us down but rather it is because the earth is pushing us up, so that means that if we are not on the surface of the earth we are no longer being pushed, so we do not really ever fall we hang there until the earth accelerates up to us, that is the model that this site has put forth to explain gravity, now going off of that how exactly does air all of the sudden exert enough force to recreate the "force" of gravity when you are trying to predict behavior.

You omit that the surface of the Earth accelerates toward the parachutist.

Quote from: cbarnett97 on August 27, 2007, 07:34:02 PM

how exactly does air all of the sudden exert enough force to recreate the "force" of gravity when you are trying to predict behavior.

Fill up your bathtub. Place you hand on the bottom of the tub, palm up. Force your hand up quickly and watch what happens to the surface of the water. Apply same concept to accelerating FE.

oh I have no doubt that the air rushes past and it can exert a force on your body, hence air resistance, where I do have a problem is that when it is stated like this
RE: F=mg, FE: F=ma see freely falling bodies follow the same behavior in both models.

Quote from: cbarnett97 on August 27, 2007, 07:34:02 PM

how exactly does air all of the sudden exert enough force to recreate the "force" of gravity when you are trying to predict behavior.

Fill up your bathtub. Place you hand on the bottom of the tub, palm up. Force your hand up quickly and watch what happens to the surface of the water. Apply same concept to accelerating FE.

Of course, you impeach FE very well with your analogy. Did you intend that?

Since the FE moves up like your hand in the water and since the water flows off over the edge, we're faced with a renewed challenge to FE on how it retains its atmosphere.

Quote from: cbarnett97 on August 27, 2007, 07:34:02 PM

I have ommited nothing. Th FE version of "gravity" is this: the reason we are stuck to the groung is not because gravity is pushing us down but rather it is because the earth is pushing us up, so that means that if we are not on the surface of the earth we are no longer being pushed, so we do not really ever fall we hang there until the earth accelerates up to us, that is the model that this site has put forth to explain gravity, now going off of that how exactly does air all of the sudden exert enough force to recreate the "force" of gravity when you are trying to predict behavior.

You omit that the surface of the Earth accelerates toward the parachutist.

after I calculate the forces that are acting on the parachutist then if I want to calculate his height after a given time then I would include the acceleration of the earth only to calculate height. but in no instance is the acceleration of the earth taken into account with relation to the parachutist himself.
If i am pushing a car down the street and a car is accelerating towards me does that change the force acting on the car I am pushing?

Ok lets clear some things up.

In the RE a person can never accelerate more than 9.8m/s².
In the FE a person can accelerate more than 9.8m/s².

Also it takes around 200kph air speed to counter act the RE's 9.8m/s² acceleration due to gravitation.  In the FE the 200kph air speed is the speed at which sets a person to 9.8m/s².


Ok glad we cleared that up.  

Quote from: divito on August 27, 2007, 07:38:25 PM

Quote from: cbarnett97 on August 27, 2007, 07:34:02 PM

how exactly does air all of the sudden exert enough force to recreate the "force" of gravity when you are trying to predict behavior.

Fill up your bathtub. Place you hand on the bottom of the tub, palm up. Force your hand up quickly and watch what happens to the surface of the water. Apply same concept to accelerating FE.

oh I have no doubt that the air rushes past and it can exert a force on your body, hence air resistance, where I do have a problem is that when it is stated like this
RE: F=mg, FE: F=ma see freely falling bodies follow the same behavior in both models.

I've explained carefully the formulas for the velocity, relative to the surface of the Earth, for both models. You've not successfully impugned any part of the proof. Yes, the relative velocities are the same. Yes, the time of touchdown is the same.

Ok lets clear some things up.

In the RE a person can never accelerate more than 9.8m/s².
In the FE a person can accelerate more than 9.8m/s².

Also it takes around 200kph air speed to counter act the RE's 9.8m/s² acceleration due to gravitation.  In the FE the 200kph air speed is the speed at which sets a person to 9.8m/s².


Ok glad we cleared that up. 

Sorry. I'm confused.
Are these numbers all relative to the surface of the Earth?
If so, then I don't see how you reached theses conclusions. You must be making unstated assumptions.
If not, then I don't see how your point is relevant.

Quote from: cbarnett97 on August 27, 2007, 07:42:04 PM

Quote from: divito on August 27, 2007, 07:38:25 PM

Quote from: cbarnett97 on August 27, 2007, 07:34:02 PM

how exactly does air all of the sudden exert enough force to recreate the "force" of gravity when you are trying to predict behavior.

Fill up your bathtub. Place you hand on the bottom of the tub, palm up. Force your hand up quickly and watch what happens to the surface of the water. Apply same concept to accelerating FE.

oh I have no doubt that the air rushes past and it can exert a force on your body, hence air resistance, where I do have a problem is that when it is stated like this
RE: F=mg, FE: F=ma see freely falling bodies follow the same behavior in both models.

I've explained carefully the formulas for the velocity, relative to the surface of the Earth, for both models. You've not successfully impugned any part of the proof. Yes, the relative velocities are the same. Yes, the time of touchdown is the same.

you are making a mistake in assuming that the forces acting upon the parachutist is the same in both models and since they are different nothing else will be the same. You cannot say that a ferrari and a pinto accelerate at the same rate because they both start at zero and then end up at 60. and not once did I ever try to relate him to the surface of the earth, I have only talked about the forces acting upon the parachutist.

Quote from: sokarul on August 27, 2007, 07:45:40 PM

Ok lets clear some things up.

In the RE a person can never accelerate more than 9.8m/s².
In the FE a person can accelerate more than 9.8m/s².

Also it takes around 200kph air speed to counter act the RE's 9.8m/s² acceleration due to gravitation.  In the FE the 200kph air speed is the speed at which sets a person to 9.8m/s².


Ok glad we cleared that up. 

Sorry. I'm confused.
Are these numbers all relative to the surface of the Earth?
If so, then I don't see how you reached theses conclusions. You must be making unstated assumptions.
If not, then I don't see how your point is relevant.

No, the accelerations are just general.  Although the RE acceleration would be compared to the surface.  The FE has alot more changes in acceleration than the RE.   


There has been a few people who have stated in the FE the max acceleration was less that 9.8m/s².  One of them did realized that was wrong. 

Quote from: Gulliver on August 27, 2007, 07:47:59 PM

Quote from: cbarnett97 on August 27, 2007, 07:42:04 PM

Quote from: divito on August 27, 2007, 07:38:25 PM

Quote from: cbarnett97 on August 27, 2007, 07:34:02 PM

how exactly does air all of the sudden exert enough force to recreate the "force" of gravity when you are trying to predict behavior.

Fill up your bathtub. Place you hand on the bottom of the tub, palm up. Force your hand up quickly and watch what happens to the surface of the water. Apply same concept to accelerating FE.

oh I have no doubt that the air rushes past and it can exert a force on your body, hence air resistance, where I do have a problem is that when it is stated like this
RE: F=mg, FE: F=ma see freely falling bodies follow the same behavior in both models.

I've explained carefully the formulas for the velocity, relative to the surface of the Earth, for both models. You've not successfully impugned any part of the proof. Yes, the relative velocities are the same. Yes, the time of touchdown is the same.

you are making a mistake in assuming that the forces acting upon the parachutist is the same in both models and since they are different nothing else will be the same. You cannot say that a ferrari and a pinto accelerate at the same rate because they both start at zero and then end up at 60. and not once did I ever try to relate him to the surface of the earth, I have only talked about the forces acting upon the parachutist.

And that's your problem. You need to be looking at acceleration on the surface of the Earth as well. That acceleration is key to understanding your error. You assume that since gravity is not working on the parachutist that the parachutist's height is not decreasing at an accelerated rate. That's the key!

Of course, you impeach FE very well with your analogy. Did you intend that?

Since the FE moves up like your hand in the water and since the water flows off over the edge, we're faced with a renewed challenge to FE on how it retains its atmosphere.

Figures. I'll have to think of something. And I have to find time to work out those gravitational numbers too.

You cannot say that a ferrari and a pinto accelerate at the same rate because they both start at zero and then end up at 60. and not once did I ever try to relate him to the surface of the earth, I have only talked about the forces acting upon the parachutist.

In RE, parachuter accelerates down to stationary Earth. Air provides resistance.

In FE, Earth accelerates up to stationary parachuter. Air provides resistance.

Quote from: Gulliver on August 27, 2007, 07:47:59 PM

Quote from: cbarnett97 on August 27, 2007, 07:42:04 PM

Quote from: divito on August 27, 2007, 07:38:25 PM

Quote from: cbarnett97 on August 27, 2007, 07:34:02 PM

how exactly does air all of the sudden exert enough force to recreate the "force" of gravity when you are trying to predict behavior.

Fill up your bathtub. Place you hand on the bottom of the tub, palm up. Force your hand up quickly and watch what happens to the surface of the water. Apply same concept to accelerating FE.

oh I have no doubt that the air rushes past and it can exert a force on your body, hence air resistance, where I do have a problem is that when it is stated like this
RE: F=mg, FE: F=ma see freely falling bodies follow the same behavior in both models.

I've explained carefully the formulas for the velocity, relative to the surface of the Earth, for both models. You've not successfully impugned any part of the proof. Yes, the relative velocities are the same. Yes, the time of touchdown is the same.

you are making a mistake in assuming that the forces acting upon the parachutist is the same in both models and since they are different nothing else will be the same. You cannot say that a ferrari and a pinto accelerate at the same rate because they both start at zero and then end up at 60. and not once did I ever try to relate him to the surface of the earth, I have only talked about the forces acting upon the parachutist.

At terminal velocity the forces are the same.  They are just in different directions between the models.

Quote from: Gulliver on August 27, 2007, 07:43:50 PM

Of course, you impeach FE very well with your analogy. Did you intend that?

Since the FE moves up like your hand in the water and since the water flows off over the edge, we're faced with a renewed challenge to FE on how it retains its atmosphere.

Figures. I'll have to think of something. And I have to find time to work out those gravitational numbers too.

Quote from: cbarnett97 on August 27, 2007, 07:53:36 PM

You cannot say that a ferrari and a pinto accelerate at the same rate because they both start at zero and then end up at 60. and not once did I ever try to relate him to the surface of the earth, I have only talked about the forces acting upon the parachutist.

In RE, parachuter accelerates down to stationary Earth. Air provides resistance.

In FE, Earth accelerates up to stationary parachuter. Air provides resistance.

Nope, in the FE air provides the force to accelerate the skydiver. 

Quote from: Gulliver on August 27, 2007, 07:52:03 PM

Quote from: sokarul on August 27, 2007, 07:45:40 PM

Ok lets clear some things up.

In the RE a person can never accelerate more than 9.8m/s².
In the FE a person can accelerate more than 9.8m/s².

Also it takes around 200kph air speed to counter act the RE's 9.8m/s² acceleration due to gravitation.  In the FE the 200kph air speed is the speed at which sets a person to 9.8m/s².


Ok glad we cleared that up. 

Sorry. I'm confused.
Are these numbers all relative to the surface of the Earth?
If so, then I don't see how you reached theses conclusions. You must be making unstated assumptions.
If not, then I don't see how your point is relevant.

No, the accelerations are just general.  Although the RE acceleration would be compared to the surface.  The FE has alot more changes in acceleration than the RE.   


There has been a few people who have stated in the FE the max acceleration was less that 9.8m/s².  One of them did realized that was wrong. 

Since your accelerations aren't relative, I don't see how your point is germane to this discussion.

Quote from: Gulliver on August 27, 2007, 07:43:50 PM

Of course, you impeach FE very well with your analogy. Did you intend that?

Since the FE moves up like your hand in the water and since the water flows off over the edge, we're faced with a renewed challenge to FE on how it retains its atmosphere.

Figures. I'll have to think of something. And I have to find time to work out those gravitational numbers too.

Quote from: cbarnett97 on August 27, 2007, 07:53:36 PM

You cannot say that a ferrari and a pinto accelerate at the same rate because they both start at zero and then end up at 60. and not once did I ever try to relate him to the surface of the earth, I have only talked about the forces acting upon the parachutist.

In RE, parachuter accelerates down to stationary Earth. Air provides resistance.

In FE, Earth accelerates up to stationary parachuter. Air provides resistance.

yes air provides resistance in both cases but in RE you get a gravitaional force acting upon the body to counteract the force caused by air resistance so we can get a state of equalibrium (terminal velocity) in FE you only get the force caused by air resistance so you can never reach a state of equalibrium so oyu end up with a net force in the same direction as the accelerating earth.

Quote from: sokarul on August 27, 2007, 07:57:53 PM

Quote from: Gulliver on August 27, 2007, 07:52:03 PM

Quote from: sokarul on August 27, 2007, 07:45:40 PM

Ok lets clear some things up.

In the RE a person can never accelerate more than 9.8m/s².
In the FE a person can accelerate more than 9.8m/s².

Also it takes around 200kph air speed to counter act the RE's 9.8m/s² acceleration due to gravitation.  In the FE the 200kph air speed is the speed at which sets a person to 9.8m/s².


Ok glad we cleared that up. 

Sorry. I'm confused.
Are these numbers all relative to the surface of the Earth?
If so, then I don't see how you reached theses conclusions. You must be making unstated assumptions.
If not, then I don't see how your point is relevant.

No, the accelerations are just general.  Although the RE acceleration would be compared to the surface.  The FE has alot more changes in acceleration than the RE.   


There has been a few people who have stated in the FE the max acceleration was less that 9.8m/s².  One of them did realized that was wrong. 

Since your accelerations aren't relative, I don't see how your point is germane to this discussion.

Quote from: cbarnett97 on August 27, 2007, 06:56:39 PM

and what is a again? before you answer you have already stated that air is not solid and it is impossible to be accelerated at 9.8m/s², which means that air cannot recreate gravity, so that would yield different results than what we see in reality

Er, I'm pretty sure you don't accelerate at 9.8m/s² with a parachute, even in RE.

See??
Its relevant.  Divitio's thought process was wrong as is Mr. Ireland.  TheEngineer's was until he realized it.  I'm sure some other peoples are wrong as well.      
Them being relative acceleration does not matter.