Parachutes

Quote from: cbarnett97 on August 28, 2007, 09:52:17 PM

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

Why?

Quote from: sokarul on August 28, 2007, 09:54:17 PM

Quote from: cbarnett97 on August 28, 2007, 09:52:17 PM

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

Why?

I said it not quite right so hold on. 

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

You have been warned about this stupidity! You've added in gravity for the RE, in error. It's not a resistive force!  If you want to calculate total force than you'll have to add in FE's acceleration's effect of "mg" as well--just as you agreed to do in the case without resistive forces.

Quote from: cbarnett97 on August 28, 2007, 09:52:17 PM

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

You have been warned about this stupidity! You've added in gravity for the RE, in error. It's not a resistive force!  If you want to calculate total force than you'll have to add in FE's acceleration's effect of "mg" as well--just as you agreed to do in the case without resistive forces.

I stated that acceleration would be the same which it would be yet when you add resistive forces they are no longer equal. so here let make write it from the point of acceleration
RE: a_n=mg-a_r --> a_n=mg-(DpA/2)v^2
FE: a_n=a_e-a_r --> a_n= a_e-(DpA/2)v^2
so like I said they are not equal

Edit: the FE model is also not dimensionally sound`

Quote from: Gulliver on August 28, 2007, 10:00:25 PM

Quote from: cbarnett97 on August 28, 2007, 09:52:17 PM

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

You have been warned about this stupidity! You've added in gravity for the RE, in error. It's not a resistive force!  If you want to calculate total force than you'll have to add in FE's acceleration's effect of "mg" as well--just as you agreed to do in the case without resistive forces.

I stated that acceleration would be the same which it would be yet when you add resistive forces they are no longer equal. so here let make write it from the point of acceleration
RE: a_n=mg-a -_r --> a_n=mg-(DpA/2)v^2
FE: a_n=a_e+a_r --> a_n= a_e+(DpA/2)v^2
so like I said they are not equal

Wrong. mg in the RE formula is not resistive. It's positive. It's already in the model BEFORE you added resistive forces. Wake up.

I am dealing with accelerations

I am dealing with accelerations

Then there are two reasons the mg, a force, shouldn't be in the RE equation.

a=f/m I do not know about you but I think there is a force used when calculating an acceleration

Quote from: sokarul on August 28, 2007, 09:30:53 PM

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

I know A is not a, I just choose a number that would simplify the equation well.  It should be obvious when I say "A = 9.8 m²" that I'm not talking about acceleration, due to the units I used.

a=f/m I do not know about you but I think there is a force used when calculating an acceleration

You subtract an acceleration from a force in calculating an acceleration. That's a sophomoric mistake.
Then you're supposed to be dealing with the resistive acceleration, not the total acceleration.
Then you need to realize that a_e = g, by model definition.
Then you should be able to see that the accelerations are equal.

Quote from: Gulliver on August 28, 2007, 10:00:25 PM

Quote from: cbarnett97 on August 28, 2007, 09:52:17 PM

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

You have been warned about this stupidity! You've added in gravity for the RE, in error. It's not a resistive force!  If you want to calculate total force than you'll have to add in FE's acceleration's effect of "mg" as well--just as you agreed to do in the case without resistive forces.

I stated that acceleration would be the same which it would be yet when you add resistive forces they are no longer equal. so here let make write it from the point of acceleration
RE: a_n=mg-a_r --> a_n=mg-(DpA/2)v^2
FE: a_n=a_e-a_r --> a_n= a_e-(DpA/2)v^2
so like I said they are not equal

Edit: the FE model is also not dimensionally sound`

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.

Yes but this only calculates terminal velocity of an object moving towards a singular (ie the very closest) point on the earth's surface. Calculate the same TV over an average area and you will find discrepancies suggesting that when extrapolated suggest a round earth.

This is what happens when you try to pass off wikipedia knowledge as your own.  cbarnett97, start with the governing equation, F=ma, for each model and derive the equations for terminal velocity using a simple free body diagram.  You can do this right?
Hint:  The answer is below:

RE => mg=F_D

FE => ma=F_D

Sorry, I just got here. Have we proved the earth is flat yet?

This is what happens when you try to pass off wikipedia knowledge as your own.  cbarnett97, start with the governing equation, F=ma, for each model and derive the equations for terminal velocity using a simple free body diagram.  You can do this right?
Hint:  The answer is below:

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

RE => mg=F_D

FE => ma=F_D

Ahhhh that is cute, you take what I accuse you of and try and turn it on me. If You draw a free body diagram you will not always start with F=ma especially in the FE model. In the FE model as soon as you leave the surface of the earth F=ma goes out the window

Let us look at it like this: according to EP we can not tell what is going on hence in effect you guys are right, the earth in either model could be described as coming up to meet us. Now if you base your mathematical model on that premise and try to use that said model to predict behavior it will not yield accurate results, and after careful testing that model would need to be looked at from a different angle ie: maybe we fall back to the earth and a new mathematical model is created and we find that this model will actually allow us to predict behavior we see in real life. And what I am saying is this the Forces that are supposed to act upon a skydiver based upon the FE model do not reflect reality, you can try all oyu want to add in the acceleration of the earth into your model but the simple fact is that according to the FE model he does not accelerate. so that means no acceleration, however the air does accelerate around him creating an increasing velocity. and when you take this into account, the Forces do not add up.
Now does this prove that the earth is round? Not really, at the very least it shows a major hole in the FE theory that needs to be reevaluated. The difficulty in this though is that RE theory reflects all conditions we have thought of so far, While FE theory only works when you are not dealing with freely falling bodies, so somehow they would need to come up with a model that reflects both conditions. Or they could take the easy way out and either a) just add what they need into the model to make it true without changing the model or b) invent some magical force that somehow recreates the RE theory, so far you guys have went with option a but I am sure one of the real FE'ers will go with option b sometime soon

Ahhhh that is cute, you take what I accuse you of and try and turn it on me. If You draw a free body diagram you will not always start with F=ma especially in the FE model. In the FE model as soon as you leave the surface of the earth F=ma goes out the window

So you don't know basic physics, then.

Let us look at it like this: according to EP we can not tell what is going on hence in effect you guys are right, the earth in either model could be described as coming up to meet us. Now if you base your mathematical model on that premise and try to use that said model to predict behavior it will not yield accurate results, and after careful testing that model would need to be looked at from a different angle ie: maybe we fall back to the earth and a new mathematical model is created and we find that this model will actually allow us to predict behavior we see in real life. And what I am saying is this the Forces that are supposed to act upon a skydiver based upon the FE model do not reflect reality, you can try all oyu want to add in the acceleration of the earth into your model but the simple fact is that according to the FE model he does not accelerate. so that means no acceleration, however the air does accelerate around him creating an increasing velocity. and when you take this into account, the Forces do not add up.
Now does this prove that the earth is round? Not really, at the very least it shows a major hole in the FE theory that needs to be reevaluated. The difficulty in this though is that RE theory reflects all conditions we have thought of so far, While FE theory only works when you are not dealing with freely falling bodies, so somehow they would need to come up with a model that reflects both conditions. Or they could take the easy way out and either a) just add what they need into the model to make it true without changing the model or b) invent some magical force that somehow recreates the RE theory, so far you guys have went with option a but I am sure one of the real FE'ers will go with option b sometime soon

I disagree, and have voiced this before, with the bold part of your post. I've provided a proof that shows that mathematical models for both FE and RE are the same within reasonable assumptions.

We've walked you through 20 pages right to the door of your error and now you avoid our points. Is there a reason you have responded to this post?:

Quote from: cbarnett97 on August 28, 2007, 10:54:19 PM

a=f/m I do not know about you but I think there is a force used when calculating an acceleration

You subtract an acceleration from a force in calculating an acceleration. That's a sophomoric mistake.
Then you're supposed to be dealing with the resistive acceleration, not the total acceleration.
Then you need to realize that a_e = g, by model definition.
Then you should be able to see that the accelerations are equal.

Quote from: cbarnett97 on August 28, 2007, 10:14:12 PM

Quote from: Gulliver on August 28, 2007, 10:00:25 PM

Quote from: cbarnett97 on August 28, 2007, 09:52:17 PM

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

You have been warned about this stupidity! You've added in gravity for the RE, in error. It's not a resistive force!  If you want to calculate total force than you'll have to add in FE's acceleration's effect of "mg" as well--just as you agreed to do in the case without resistive forces.

I stated that acceleration would be the same which it would be yet when you add resistive forces they are no longer equal. so here let make write it from the point of acceleration
RE: a_n=mg-a_r --> a_n=mg-(DpA/2)v^2
FE: a_n=a_e-a_r --> a_n= a_e-(DpA/2)v^2
so like I said they are not equal

Edit: the FE model is also not dimensionally sound`