Parachutes

no

Did you forget about air resistance, dumbshoe?

Because air is fluid and such, cbarnett is right in saying it'd not be the same. However, i'm not convinced myself that it'd be different enough to merit particular notice.

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

here is where I explained about air resistance earlier in this thread

There's one problem with your explanation.  That problem is gravity.  Gravity isn't a force.

There's one problem with your explanation.  That problem is gravity.  Gravity isn't a force.

gravity is not a force and no where in my equation is gravity a force, hence why the equation states mg to show the force due to gravity, not the force of gravity. so no it is fine

There is no force due to gravity.


Fail.

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

so how do parachutes work again if gravity does not exist?

closed container + fluid...


The fluid is accelerating up at 9.8 m/s/s with the Earth (close container).  g (the 'acceleration' towards the Earth in relation to the Earth) is still 9.8m/s/s.  Air resistance.

Air resistance is defined as F=-1/2 DpAv2 so as you can see the net force would push you up higher in the absence of gravity

You forget that you have to apply the resistance due to air to your acceleration compared to the Earth.  Your acceleration compared to the Earth is the replacement for 'gravity'.

Let me try to describe it another way.  Normally, you jump and the Earth catches up to you.  Your relative acceleration is 9.8m/s/s.

With a parachute, you jump and the Earth starts catching up to you.  As it is accelerating, it is accelerating the air.  The accelerated air pushes on the parachute, but not at 9.8m/s/s or more, because fluids don't work that way.  Your relative acceleration is less than 9.8m/s/s.

Sorry in FE, there is no acceleration when it come to the parachute, the wind rushing past will be the velocity when you look at the system correctly. If you are still not sure just draw a free body diagram and look at the forces involved.

and if you could show me where the acceleration is when you calculate air resistance

F=-1/2 DpAv2

You are completely fucking hopeless.

like I said show me

and if you could show me where the acceleration is when you calculate air resistance

F=-1/2 DpAv2

v = 9.8

Quote from: cbarnett97 on August 23, 2007, 05:13:43 PM

and if you could show me where the acceleration is when you calculate air resistance

F=-1/2 DpAv2

v = 9.8m/s²

very close but to properly solve this equation you need to use numerical modeling and calculate R for every second due to the fact that it is 9.8m/s² and not just 9.8m/s. and the reason you would never hit the ground is that once R overcame your inertial mass it would begin to lift you

very close but to properly solve this equation you need to use numerical modeling and calculate R for every second due to the fact that it is 9.8m/s² and not just 9.8m/s. and the reason you would never hit the ground is that once R overcame your inertial mass it would begin to lift you

You are lifted in a parachute, but you still lower to the Earth eventually.

initialy the earth would catch up but after a period of time your elevation in relation to the earth you would climb, the same as if you tried to push a car up a small incline, initially the car would push you back but as you added more force you would end up moving the car up the incline and as you continued to add more and more force the car would speed up

I still don't see what force is magically lifting the skydiver.

Replace air with water, and think of a plate (FE) coming up underneath a falling rock with a parachute (with holes) in the water. Other than water being a lot more dense than air, it's close to the same thing (if you add the holes). The rock will not magically shoot out of the water, and the rock will still meet with the plate.

nothing magically lifts the parachute. the parachute just becomes a large sail. imagine a sail boat trying to sail into a strong headwind, it will not reach its port because the wind would push it back.

nothing magically lifts the parachute. the parachute just becomes a large sail. imagine a sail boat trying to sail into a strong headwind, it will not reach its port because the wind would push it back.

Like: http://en.wikipedia.org/wiki/Kitesurfing

I don't know, I'll wait for Gulliver to look at this and try to explain it and see if he agrees with you.

now imagine those kitesurfers had very large kites, they would gain altitude until the wind slowed up

Quote from: cbarnett97 on August 23, 2007, 05:39:41 PM

nothing magically lifts the parachute. the parachute just becomes a large sail. imagine a sail boat trying to sail into a strong headwind, it will not reach its port because the wind would push it back.

Like: http://en.wikipedia.org/wiki/Kitesurfing

I don't know, I'll wait for Gulliver to look at this and try to explain it and see if he agrees with you.

Did someone call?

I remind everyone of the EP: The primary effects of acceleration and gravity are locally indistinguishable.

Yes, a parachute provides an upward force, just like a plane's wing provides lift, just like a post provides support for a mailbox. You can think as the air underneath the parachute as a (collapsing) support.

No, the force does not, as a rule, overwhelm gravity or the FE alleged acceleration.

divito, would you be a gatekeeper to prevent these errant threads about the primary effects of gravity and acceleration from going too far, please?

cb, you've presented the best analysis yet on the parachute problem. You just need to see that the FE is accelerating towards the parachute so the "height" is dropping.

in the presence of no resistive forces yes the distance between the 2 will decrease. however the earth(in the case of the FE) has this wind that gets to you beforehand that in the presence of resistive forces this wind will push you away from it. imagine going outside on a windy day, it is possible to just stand there even in a moderate wind, but when you increase your wind resistance it becomes very difficult if not impossible to just stand there because the wind will pull you along with it.

in the presence of no resistive forces yes the distance between the 2 will decrease. however the earth(in the case of the FE) has this wind that gets to you beforehand that in the presence of resistive forces this wind will push you away from it. imagine going outside on a windy day, it is possible to just stand there even in a moderate wind, but when you increase your wind resistance it becomes very difficult if not impossible to just stand there because the wind will pull you along with it.

What you say is well thought out and accurate. If all of the air stayed between you and the accelerating FE, your parachute would keep you from falling farther. If more air came under your parachute, you'd actually climb.

However, as a rule, the air gets by your parachute a little at a time and you descend.

In your wind analogy:
most of the time you can make headway-->you descend
sometimes you can't move forward and just stand against the wind-->you stay at the same height
rarely you actually get blown backward-->you rise

Does that answer you? I'd be happy to answer any challenge or question. You've been helpful to me in the past.

in my analogy though there will never be any noticeable time where you could make "headway" against the wind, after 1 sec the wind is going past you at 9.8m/s after 2 sec it is moving past you 19.6m/s and so on. now just imagine standing on your front lawn with a deployed parachute and a gust of wind comes along at almost 20m/s would you be able to make headway?

in my analogy though there will never be any noticeable time where you could make "headway" against the wind, after 1 sec the wind is going past you at 9.8m/s after 2 sec it is moving past you 19.6m/s and so on. now just imagine standing on your front lawn with a deployed parachute and a gust of wind comes along at almost 20m/s would you be able to make headway?

You're not the one making the headway, it would be another object comming at 20m/s.  As long as some air escapes that parachute, it will reach you.

in the presence of no resistive forces yes the distance between the 2 will decrease. however the earth(in the case of the FE) has this wind that gets to you beforehand that in the presence of resistive forces this wind will push you away from it. imagine going outside on a windy day, it is possible to just stand there even in a moderate wind, but when you increase your wind resistance it becomes very difficult if not impossible to just stand there because the wind will pull you along with it.

To be fair (I am not pro-FE), if what you say is true, the air resistance would to have to produce enough force as to accelerate you at a rate of < -9.8 m/s^2 (opposite of the Earth's motion) such that you begin to displace further and further away from the [flat] Earth's surface. Otherwise, if your acceleration is > -9.8 m/s^2, the Earth will eventually catch up.

EDIT: signs.
EDIT #2: All of the accelerations I've posted are relative to a stationary observer in space.

My apologies, double post.