Parachutes

how would these work on the FE, since the earth is accelerating up to us at 9.8m/s² the forces will never balance out because there is no acceleration actin on our bodies to counteract the wind rushing past us so based upon the size of our parachute we would be accelerated up and out of the atmosphere

The FE accelerates the air with it.  It would work exactly the same...


The FE accelerates the air, which is caught by your parachute.

It would work exactly the same.

The FE accelerates the air with it.  It would work exactly the same...


The FE accelerates the air, which is caught by your parachute.

ok, now other than your inertial mass what forces counteract this upward force caused by the air?

ok, now other than your inertial mass what forces counteract this upward force caused by the air?

Mm, the air resistance and parachute are not enough to counteract the accelerating FE. But I don't really understand your question; you'll have to dumb it down for me.

Its impossible for the air to be moving up at the same speed as the earth. Its fluid, its density is extraordinarily small compared to the ground and the water. Furthermore, as it is a gas, it has no surface tension. It would be flattened against the surface of FE immediately and pushed off the edge by the forces.

Its impossible for the air to be moving up at the same speed as the earth. Its fluid, its density is extraordinarily small compared to the ground and the water. Furthermore, as it is a gas, it has no surface tension. It would be flattened against the surface of FE immediately and pushed off the edge by the forces.

What we're saying assumes the FE had an answer for why the air wasn't pushed off the edge.

air resistance is is defined as "R=1/2DpAv²
now in RE when you jump out a plane you experience 2 forces: the gravitational attraction to the earth and the resistive forces from the air, so you end up with F=mg-1/2DpAv² now R is negative due to the fact that I took "down" to be positive.
In FE that equation would become F=-1/2 DpAv² which tells us that the net force on an object will continue to push it up futher away from earth

I hope that helps

Ah, i see. In that case, it'd work pretty much the sam, though i expect there would be some variations too minute for laymen like us to find.

air resistance is is defined as "R=1/2DpAv²
now in RE when you jump out a plane you experience 2 forces: the gravitational attraction to the earth...

False.  Gravity, and gravitation, as forces do not exist in either model.  Gulliver and I came to that conclusion in another thread.

Quote from: cbarnett97 on August 23, 2007, 01:24:50 PM

air resistance is is defined as "R=1/2DpAv²
now in RE when you jump out a plane you experience 2 forces: the gravitational attraction to the earth...

False.  Gravity, and gravitation, as forces do not exist in either model.  Gulliver and I came to that conclusion in another thread.

you are 100% correct, gravity or gravitation is not a force hence why I wrote F=mg so you see gravity is not a force

Oh, nevermind, I see it.

it is the same equation I just learned it using different letters
R=1/2DpAv²
R= resitive force
D=drag Coefficient
p(roh)=density of fluid
A=cross sectional area of object

So you see it is the same I learned to R instead of F to make it easier to keep track of terms while appying the equation

Quote from: cbarnett97 on August 23, 2007, 01:24:50 PM

air resistance is is defined as "R=1/2DpAv²
now in RE when you jump out a plane you experience 2 forces: the gravitational attraction to the earth...

False.  Gravity, and gravitation, as forces do not exist in either model.  Gulliver and I came to that conclusion in another thread.

Really? What thread was that?

I hold that in classical physics gravity is a force in the RE model. I don't see any need to invoke the power and complications of GR in dealing with this challenge.

I'm still failing to see why you take out m and g. They are both relevant, as m is the mass of the person and g is the acceleration of the FE. Why would you remove them?

in the system of the parachutist a in the case of FE is not in it so the first part of that equation "ma" would be zero because the earth is accelerating at you not you to it. same as if a car accelerates at a wall and it hits it, if you look at it from the walls perspective it sat there until the car got to it

in the system of the parachutist a in the case of FE is not in it so the first part of that equation "ma" would be zero because the earth is accelerating at you not you to it. same as if a car accelerates at a wall and it hits it, if you look at it from the walls perspective it sat there until the car got to it

But that doesn't change that you need to figure out the acceleration to find the force.

"The Drag equation calculates the force experienced by an object moving through a fluid at relatively large velocity. The equation is attributed to Lord Rayleigh, who originally used   in place of (L being some length). The force on a moving object due to a fluid is:

          see derivation

where

    Fd is the force of drag,
    ρ is the density of the fluid (Note that for the Earth's atmosphere, the density can be found using the barometric formula. It is 1.293 kg/m3 at 0°C and 1 atmosphere.),
    v is the speed of the object relative to the fluid,
    A is the reference area,
    Cd is the drag coefficient (a dimensionless constant, e.g. 0.25 to 0.45 for a car), and
    is the unit vector indicating the direction of the velocity (the negative sign indicating the drag is opposite to that of velocity).

The reference area A is related to, but not exactly equal to, the area of the projection of the object on a plane perpendicular to the direction of motion (i.e., cross sectional area)."

you do not need an acceleration, only a velocity and in the case of the FE you would use the velocity of the air moving past you at 9.8m/s² to calculate R

Also look at the equation there is no acceleration in it

Well that's what I mean, v would be equal to 9.8m/s². I'm still missing why that changes because of the FE. The Earth eventually meets you in both models. Air is the same density in both models and air resistance is the same. You accelerate the same speed in both models.

On a side note if you wanted to calculate the acceleration you can transform the original equation to
a=g-(DpA/2m)v² and once again you can see the acceleration would be up since g=0 in the case of FE

On a side note if you wanted to calculate the acceleration you can transform the original equation to
a=g-(DpA/2m)v² and once again you can see the acceleration would be up since g=0 in the case of FE

That doesn't make sense to me. g can't be 0.

air resistance is is defined as "R=1/2DpAv²

sorry I forgot the mg portion of it, so add mg to that equation

Quote from: CommonCents on August 23, 2007, 01:26:28 PM

Quote from: cbarnett97 on August 23, 2007, 01:24:50 PM

air resistance is is defined as "R=1/2DpAv²
now in RE when you jump out a plane you experience 2 forces: the gravitational attraction to the earth...

False.  Gravity, and gravitation, as forces do not exist in either model.  Gulliver and I came to that conclusion in another thread.

Really? What thread was that?

I hold that in classical physics gravity is a force in the RE model. I don't see any need to invoke the power and complications of GR in dealing with this challenge.

http://theflatearthsociety.org/forum/index.php?topic=16234.msg269821#msg269821

You agree with me, but don't want to call it an 'apparent' acceleration.  That's a matter of opinion, I believe, but since it does change direction in our observation, I will call it 'apparent' acceleration.

Quote from: cbarnett97 on August 23, 2007, 01:57:23 PM

On a side note if you wanted to calculate the acceleration you can transform the original equation to
a=g-(DpA/2m)v² and once again you can see the acceleration would be up since g=0 in the case of FE

That doesn't make sense to me. g can't be 0.

Of course it makes no sense, that is my point. but when looked by FE theory g must be zero otherwise that would tell us that we accelerate to the earth. since there would be a force directed at the surface of the earth

air resistance is is defined as "R=1/2DpAv²

Wouldn't it be R=-1/2DpAv²?

Of course it makes no sense, that is my point. but when looked by FE theory g must be zero otherwise that would tell us that we accelerate to the earth.

We do accelerate to the Earth.

in RE: yes, in FE: no

in RE: yes, in FE: no

The point is, you cannot tell the difference. It's the same in both models. You'd only know if you were outside the FE's FoR.

well the FE model is that the earth accelerates up to us, so based on that parachutes should not work. In the RE model we accelerate to the earth and parachutes do work. along with related items such as being able to calculate R in the first place as well as calculating termonal velocity.


And the negative sign in the equation is only needed to indicate a direction so it could be written either way, I have a buddy that always makes "down" negative no matter what so if he was showing these equations R would be positive

well the FE model is that the earth accelerates up to us, so based on that parachutes should not work. In the RE model we accelerate to the earth and parachutes do work.

Parachutes will work EXACTLY the same.

no