The FE accelerates the air with it. It would work exactly the same...ok, now other than your inertial mass what forces counteract this upward force caused by the air?
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?
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.
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...
you are 100% correct, gravity or gravitation is not a force hence why I wrote F=mg so you see gravity is not a forceair 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?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.
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
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
air resistance is is defined as "R=1/2DpAv2
Really? What thread was that?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.
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.
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 earthOn 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
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.
in RE: yes, in FE: no
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.
no
air resistance is is defined as "R=1/2DpAv2here is where I explained about air resistance earlier in this thread
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
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
and if you could show me where the acceleration is when you calculate air resistance
F=-1/2 DpAv2
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 youand 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
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.
Did someone call?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.
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.
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 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.
you must remember you are starting at the same velocity as the earth so you will initially lose altitude you will re accelerate back up to 9.8m/s2. now in some cases you would land because you are already close enough to the ground for the earth to catch up to you, but at higher altitudes you would eventually reach your FE "terminal velocity" and you will just be floating there.Terminal velocity would be your speed toward the ground. The value would be positive. Since speed = distance/time, after a short time you'd cover the distance to the ground. You would not just be floating there.
there is a reason i put it in quotes what I meant by that is that your velocity and the velocity of the earth would be the same.Okay then... The FE would continue to accelerate. You would not accelerate as much as some of the supporting air passed through and around your parachute. The FE's ground would reach you.
initially but as the air contnues to accelerate the relative velocity of the air increases so your acceleration up will match that of the earthSorry, no. The air that gets by your chute fails to accelerate you, so your acceleration is less than the air's, and thereby less than the FE's ground. As the air gets by you faster, you lose acceleration at an even faster rate.
you must remember you are starting at the same velocity as the earth so you will initially lose altitude you will re accelerate back up to 9.8m/s2.Why? Like I said, unless the air resistance manages to accelerate you to this figure the Earth WILL catch up.
initially but as the air contnues to accelerate the relative velocity of the air increases so your acceleration up will match that of the earthThe question is CAN air resistance accelerate you to 9.8m/s2 relative to an outside observer? If it can, according to the FE model, you will remain at an equal distance away from the Earth. If your acceleration is greater than the Earth's you will be move away from Earth's surface, while if your acceleration is less than the Earth's you touchdown on the ground.
the only thing holding you back at first is your inertial mass after that your acceleration will increase until it matches the acceleration of the earth. the canopy will trap a good portion of the air so you will accelerate up to 9.8m/s2 in a relatively short period of time
the only thing holding you back at first is your inertial mass after that your acceleration will increase until it matches the acceleration of the earth. the canopy will trap a good portion of the air so you will accelerate up to 9.8m/s2 in a relatively short period of timeA "relatively short period of time"?
Um...what?the only thing holding you back at first is your inertial mass after that your acceleration will increase until it matches the acceleration of the earth. the canopy will trap a good portion of the air so you will accelerate up to 9.8m/s2 in a relatively short period of time
A good portion of the air != all the air. Therefore you won't get all of the acceleration of the air, and you will not accelerate >= 9.8m/s/s
Quoted for Excellencethe only thing holding you back at first is your inertial mass after that your acceleration will increase until it matches the acceleration of the earth. the canopy will trap a good portion of the air so you will accelerate up to 9.8m/s2 in a relatively short period of time
A good portion of the air != all the air. Therefore you won't get all of the acceleration of the air, and you will not accelerate >= 9.8m/s/s
I will partially correct myself here, you will not acclerate away from the earth but you would maintain a static distance away from the earth.
A good portion of the air does not equal all of the air. Therefore, you will not get all of the acceleration from the air, and your acceleration will not be greater than or equal to 9.8 meters per second per second.I know the notation...I don't know how you went from:
I will partially correct myself here, you will not acclerate away from the earth but you would maintain a static distance away from the earth.
Ok, I'm just thinking here, but... For you to maintain an altitude with the parachute, wouldn't the air hitting the parachute have to be 100% efficient? I mean, wouldn't air being moved around in the parachute, and sliding around it make that not 100%, so you couldn't just stay at an altitude?
first off nice to see more thatn 5 words writen here, but air, especially air accelerating at that speed can create quite a bit of force, after 5 sec. or so the relative velocity of the air will have reached hurricane force speeds. and it has the ability to accelerate a body to 9.8m/s2. If I push a car up a hill and my feet slip a little bit I am still able to push the car
Let me try an analogy.have you ever seen thos indoor skydiving places? they manage to match the acceleration of the earth using only air and they do not even have the luxury of a large parachute to increase the wind resistance
Let's say that an imaginary, special mailbox on a post. The post holds the mailbox off the FE's ground by 3 feet.
Let's envision a wicked, new insect that carries the wood of the post, one piece at a time, up over the mailbox--a termite with wings.
After the insect does its harvesting of the entire post, the mailbox will rest on the FE's ground.
Now substitute:
Mailbox for chute
air for wood
column of air for post
the nature of air for insect.
no, it will just be that not all of your energy will be put into pushing the car, but there will still be a net force acting on the car to push it up the hillfirst off nice to see more thatn 5 words writen here, but air, especially air accelerating at that speed can create quite a bit of force, after 5 sec. or so the relative velocity of the air will have reached hurricane force speeds. and it has the ability to accelerate a body to 9.8m/s2. If I push a car up a hill and my feet slip a little bit I am still able to push the car
But the car will go down the hill. Now this situation will be like you ALWAYS pushing, but also always slipping.
no, it will just be that not all of your energy will be put into pushing the car, but there will still be a net force acting on the car to push it up the hillfirst off nice to see more thatn 5 words writen here, but air, especially air accelerating at that speed can create quite a bit of force, after 5 sec. or so the relative velocity of the air will have reached hurricane force speeds. and it has the ability to accelerate a body to 9.8m/s2. If I push a car up a hill and my feet slip a little bit I am still able to push the car
But the car will go down the hill. Now this situation will be like you ALWAYS pushing, but also always slipping.
You should review the conservation laws.
I don't get it
Now in FE the force that causes us to have weight is the acceleration of the ground under us which while we are on the ground will cause the exact same result as in the RE, but as soon as we leave the ground the earth is no longer pushing up on us so now we only have mass so in F=ma the force acting upon us is zero so in FE we would have no weight.Uh, in the RE you would also have no weight.
in the Re model you could have weight theoretically, while in the FE model you must have no weight otherwise the model would not be accurateNow in FE the force that causes us to have weight is the acceleration of the ground under us which while we are on the ground will cause the exact same result as in the RE, but as soon as we leave the ground the earth is no longer pushing up on us so now we only have mass so in F=ma the force acting upon us is zero so in FE we would have no weight.Uh, in the RE you would also have no weight.
We are talking about reality. What would the scale read?No The whole post was about a flaw in FE theory so please do not try and change it, either explain to me how FE theory explains the lack of downward force when looked at from the correct FoR or give up, and while explaining the lack of a downward force go ahead and show me how it reflects reality
Now in FE the force that causes us to have weight is the acceleration of the ground under us which while we are on the ground will cause the exact same result as in the RE, but as soon as we leave the ground the earth is no longer pushing up on us so now we only have mass so in F=ma the force acting upon us is zero so in FE we would have no weight.We are talking about this statement, which seems to be the point you are lacking. In RE and the FE you would have no weight in free fall. But since you claim you would, I ask again: What would the scale read?
No The whole post was about a flaw in FE theory so please do not try and change it, either explain to me how FE theory explains the lack of downward force when looked at from the correct FoR or give up, and while explaining the lack of a downward force go ahead and show me how it reflects reality
Oh i get it. You are looking for a reason to run to your EP and I am not going to give it to you, so you need to be able to support your model with more than that.Ok. What will I weigh on the scale as I am in free fall? I promise I won't bring up the EP.
So once again how does FE Theory account for the lack of a downward force when looked at from the correct FoRWhy must there be one?
well if there isnt any explaination then FE theory does not work because the forces do not resolve correctly to match reality so we can say that FE theory does not reflect what really happensSo once again how does FE Theory account for the lack of a downward force when looked at from the correct FoRWhy must there be one?
So, you didn't read it.oh I read it and you are wrong.
it was a very well thought out explaination however the problem is that you are applying RE theory to FE theory, In RE our weight comes from the gravitational attraction between us and the earth so no matter our altitiude there is stall a foce acting upon us. Now in FE the force that causes us to have weight is the acceleration of the ground under us which while we are on the ground will cause the exact same result as in the RE, but as soon as we leave the ground the earth is no longer pushing up on us so now we only have mass so in F=ma the force acting upon us is zero so in FE we would have no weight. Now I understan %100 that this does not reflect reality but that is my whole point FE theory does not reflect reality in this case, and since you guys are such fans of calulating things locally this should be pretty easy to see, the only force acting upon the parachutist on the FE is the wind resistance and nothing more so that would mean that he would be accelerated upward, does this mean he would never land? Given enough time he would not but I really do not want to take the time to set up a numerical modeling poblem to see how long it would take him to accelerate back up to 9.8m/s2 but even if it would take 500 miles to get back to that acceleration we would still notice the effect as our relative velocity would continue to decrease as we fell back to earth, unlike what happens in reality where we reach a terminala velocity and maintain it
The air would compress, you know.
Well, first, your FE equation is very wrong. Second, there is no downward force in the FE diagram.that is the equation for air resistance so how can it be very wrong, and yes there is no force pushing down but your inertial mass is like friction so it will always point opposite the applied force. So now that I think about it, My equations are accurate unless you want to bring in forces that are not involved in the system into play, but wait that would still be incorrect
So let me get this straight. This topic is another one that disproves the fe? It seems too. And its a kinda major one.
F=I -1/2DpAv2Check the units.
I'm not agreeing though.So let me get this straight. This topic is another one that disproves the fe? It seems too. And its a kinda major one.
No, cbarnett97 just can't understand why it isn't. Gulliver agreed earlier that FE has this one.
Please read one of my earlier posts which explains this.I'm not agreing though.So let me get this straight. This topic is another one that disproves the fe? It seems too. And its a kinda major one.
No, cbarnett97 just can't understand why it isn't. Gulliver agreed earlier that FE has this one.
A humnas terminal velocity is 120 mph. So a skydiver will hit this and then never acceelerate anymore. But he will still have a velocity and that velocity will take him into the earth. A skydiver in the fe better also have a 120 mph terminal velocity. This means he will never accelerate, and his velocity will not take him into the earth. As it will be in the same direction as the earth.
I'm not agreing though.So let me get this straight. This topic is another one that disproves the fe? It seems too. And its a kinda major one.
No, cbarnett97 just can't understand why it isn't. Gulliver agreed earlier that FE has this one.
A humnas terminal velocity is 120 mph. So a skydiver will hit this and then never acceelerate anymore. But he will still have a velocity and that velocity will take him into the earth. A skydiver in the fe better also have a 120 mph terminal velocity. This means he will never accelerate, and his velocity will not take him into the earth. As it will be in the same direction as the earth.
The only way the air could prevent you from 'falling' is if it your air were a solid, or something else completely retarded. You have weight and this weight allows you to slowly push your way 'down' through the air until you hit the ground END OF ARGUMENT.since when did air rovide no force? because those pesky hurricanes must be made of something other than air then
I'm not agreing though.So let me get this straight. This topic is another one that disproves the fe? It seems too. And its a kinda major one.
No, cbarnett97 just can't understand why it isn't. Gulliver agreed earlier that FE has this one.
A humnas terminal velocity is 120 mph. So a skydiver will hit this and then never acceelerate anymore. But he will still have a velocity and that velocity will take him into the earth. A skydiver in the fe better also have a 120 mph terminal velocity. This means he will never accelerate, and his velocity will not take him into the earth. As it will be in the same direction as the earth.
The only way the air could prevent you from 'falling' is if it your air were a solid, or something else completely retarded. You have weight and this weight allows you to slowly push your way 'down' through the air until you hit the ground END OF ARGUMENT.since when did air rovide no force? because those pesky hurricanes must be made of something other than air then
kgm/s^2, what is wrong with newtons?Your 'I' is not measured in newtons.
edit hold onYou were on the right track
No the whole thing I typed was wrong. Sadly, I get it now. At terminal "velocity" your acceleration would match the earths, 9.8m/s2. This same acceleration is what allows a person to not see air moving faster that 200kph. The earth would have a greater velocity than you and thus eventually hit you. But there is no terminal velocity in the FET, it would be called terminal acceleration.edit hold onYou were on the right track
On FE there is no counter force to eventially reach equalibrium
Dont know how to put in greek letters but Inertial Mass is "SigmaF=miaExactly. The inertial mass part is this: mi
No you were close on FE theory there is no force to negate the force of the wind resistance, so as the velocity of the wind increases your acceleration would also increase, while on RE as the velocity of the wind increases your acceleration will decreaseYeah thats what it does, but once your acceleration hits 9.8m/s2 the air no longer accelerate past you. So the airs velocity never changes. You can't accelerate quicker than the earth when the earth is whats accelerating you.
Yes I had retracted my statement that we would eventually accelerate away from the earth, I was mistaken in that regard and unlike a FE'er I will admit it if I am wrong, but given enough time we would accelerate to the same rate of acceleration as the earthNo you were close on FE theory there is no force to negate the force of the wind resistance, so as the velocity of the wind increases your acceleration would also increase, while on RE as the velocity of the wind increases your acceleration will decreaseYeah thats what it does, but once your acceleration hits 9.8m/s2 the air no longer accelerate past you. So the airs velocity never changes. You can't accelerate quicker than the earth when the earth is whats accelerating you.
Its not a total loss though. They still can't tell us why the earth, sun, moon, and stars all accelerate the same while being different sizes yet we don't get accelerated.Yes I had retracted my statement that we would eventually accelerate away from the earth, I was mistaken in that regard and unlike a FE'er I will admit it if I am wrong, but given enough time we would accelerate to the same rate of acceleration as the earthNo you were close on FE theory there is no force to negate the force of the wind resistance, so as the velocity of the wind increases your acceleration would also increase, while on RE as the velocity of the wind increases your acceleration will decreaseYeah thats what it does, but once your acceleration hits 9.8m/s2 the air no longer accelerate past you. So the airs velocity never changes. You can't accelerate quicker than the earth when the earth is whats accelerating you.
ok so there are your kg, and the "a" portion of that equation would provide the m/s2 so hence we get kgm/s2 and as we all know kgm/s2 is the same as saying a NewtonUh, yea, but inertial mass is measured in kg. Your equation explicitly stated inertial mass. Mass does not equal force.
This goes back to one of my first posts, when you look at the parachutist and draw a free body diagram which is just a vector diagram of all of the forces involved you will get 2 different diagrams when you do one for RE and one for FE. In RE you will have on vector pointing down which will be the gravitational attraction to the earth "mg" and pointing in an opposite direction would be your air resistance "R=1/2DpAv2" so you would end up with the equation F=mg-1/2DpAv2Here is how you are wrong
In FE you will have on vector pointing down which will be you inertial mass "I" and you will still have the vector pointing up as air resistance so you will get an equation of F=I -1/2DpAv2 the only thing that the acceleration of the air would be used for is to compute the velocity of the air in the air resistance formula.
"In the SI system of units, mass is measured in kilograms. Many other units of mass are also employed, such as: grams (g), tonnes, pounds, ounces, long and short tons, quintals, slugs, atomic mass units, Planck masses, solar masses, and eV/c2."A newton is a combination of SI units. The unit of a Newton(N) is kilograms x Mass/Seconds2 and what the engineer was trying to say is that the 2 sides of the equations were not equivilent because according to him when we do a dimensional analysis we would get N=kg-N unfortunately he was mistaken so the dimensional analysis would yield N=N which is fine.
I don't see Newtons.
Dont know how to put in greek letters but Inertial Mass is "SigmaF=miaWhat did you do with the acceleration. We are not just measuring mass, we are measuring the mass' resistance to being accelerated.
Thats wrong. Without a chute a person would only hit 9.8m/s2. When the shoot opens it gives a person greater acceleration than 9.8m/s2. Thats what keeps you from hitting the ground so fast.Yes I had retracted my statement that we would eventually accelerate away from the earth, I was mistaken in that regard and unlike a FE'er I will admit it if I am wrong, but given enough time we would accelerate to the same rate of acceleration as the earthNo, you won't. If all the energy of the air went into forcing your chute up, you would accelerate at 9.8m/s^2. However, some of the air does not transfer all of its energy to your chute, so you can never accelerate at the same rate as the earth.
Now, you said my explanation was wrong. I'm still waiting for you to show me where.
Inertial mass has nothing to do with acceleration. If it is sitting still, it still has mass.Dont know how to put in greek letters but Inertial Mass is "SigmaF=miaWhat did you do with the acceleration. We are not just measuring mass, we are measuring the mass' resistance to being accelerated.
Thats wrong. Without a chute a person would only hit 9.8m/s2. When the shoot opens it gives a person greater acceleration than 9.8m/s2. Thats what keeps you from hitting the ground so fast.Uh, no, not even close.
Of course you will probably say thats what you meant.
If I am understanding his contention he is saying that because the system is not %100 efficient it is impossible to reach the same acceleration at the air. but if you look at a boat in moving along with the river it will travel the same speed as the water, even though initially it was not moving with respect to the water.Thats wrong. Without a chute a person would only hit 9.8m/s2. When the shoot opens it gives a person greater acceleration than 9.8m/s2. Thats what keeps you from hitting the ground so fast.Yes I had retracted my statement that we would eventually accelerate away from the earth, I was mistaken in that regard and unlike a FE'er I will admit it if I am wrong, but given enough time we would accelerate to the same rate of acceleration as the earthNo, you won't. If all the energy of the air went into forcing your chute up, you would accelerate at 9.8m/s^2. However, some of the air does not transfer all of its energy to your chute, so you can never accelerate at the same rate as the earth.
Now, you said my explanation was wrong. I'm still waiting for you to show me where.
Of course you will probably say thats what you meant.
Exactly, now if you want to accelerate that mass in a frictionless environment how much force would it take to cause an acceleration? and Tada you have that objects inertial mass as opposed to its rest massInertial mass has nothing to do with acceleration. If it is sitting still, it still has mass.Dont know how to put in greek letters but Inertial Mass is "SigmaF=miaWhat did you do with the acceleration. We are not just measuring mass, we are measuring the mass' resistance to being accelerated.
No I was right. But actually you were also right if you were talking about a RE. I was talking about a FE.Thats wrong. Without a chute a person would only hit 9.8m/s2. When the shoot opens it gives a person greater acceleration than 9.8m/s2. Thats what keeps you from hitting the ground so fast.Uh, no, not even close.
Of course you will probably say thats what you meant.
and Tada you have that objects inertial mass as opposed to its rest massWhich is still mass. It suddenly does not become a force.
No I was right. But actually you were also right if you were talking about a RE. I was talking about a FE.No, you were wrong in both situations.
"Inertial mass. This is mainly defined by Newton's law, the all-too-famous F = ma, which states that when a force F is applied to an object, it will accelerate proportionally, and that constant of proportion is the mass of that object. In very concrete terms, to determine the inertial mass, you apply a force of F Newtons to an object, measure the acceleration in m/s2, and F/a will give you the inertial mass m in kilograms."The nice thing about math is you can calculate any unknown as long as you know some of the items in the equation, they are saying the same thing just going about it from the opposite direction, I am using the mass to calculate toe force and they are using the force to calculate the mass
The nice thing about math is you can calculate any unknown as long as you know some of the items in the equation, they are saying the same thing just going about it from the opposite direction, I am using the mass to calculate toe force and they are using the force to calculate the massBut you are only using the mass in your equation.
Umm no.No I was right. But actually you were also right if you were talking about a RE. I was talking about a FE.No, you were wrong in both situations.
In the feYes.
Terminal "velocity" is actually terminal acceleration right?
This acceleration would be at 9.8m/s2 right?No.
But it took you longer to reach the 9.8m/s2 right?Longer than what?
So the earths velocity compared to yours is faster right?Yes.
So when you pull the chute you have to gain velocity as to not hit the earth at 200kph right?Yes.
So how does one speed up compared to the earth velocity while accelerating slower?You are not going faster than the earth.
When a person on the fe hits the earth they are nto accelerating right?With respect to the earth, yes.
So thus they have to be accelerating at 9.8m/s2 too right?No.
So in the fe a person acceleration would start at zero and then clime to 9.8m/22 to hit terminal "velocity". The they would pull their chute and be accelerated to a number much greater than 9.8m/s and than slowly deaccelerate back to 9.8m/s2.No.
Otherwise they would eventually start to gain altitude.No.
Do you understand?Your logic, no.
I will make it this much easier for you, the only opposing force to the air resistance in the FE model would be the counter force measured in newtons that is caused by the mass's resistance to being accelerated that I represented with the symbol "I" in the equation. Now please show me how my equation was flawed.You have mass equal to a force.
that is why i changed it just for you to be a force, you can go ahead an read a physics textbook and you will learn all about it, but if it will make you happy i can remove it since according to you it should not be there, but that will only help prove my point since the acceleration in the same direction of the earth would only increase faster. Even though the equation would then be wrongI will make it this much easier for you, the only opposing force to the air resistance in the FE model would be the counter force measured in newtons that is caused by the mass's resistance to being accelerated that I represented with the symbol "I" in the equation. Now please show me how my equation was flawed.You have mass equal to a force.
that is why i changed it just for you to be a force, you can go ahead an read a physics textbook and you will learn all about it, but if it will make you happy i can remove it since according to you it should not be there, but that will only help prove my point since the acceleration in the same direction of the earth would only increase faster. Even though the equation would then be wrongLet me ask you this: What acceleration did you use for your original 'I'?
Well being that I have not run any numbers I have not put any numbers into it, but you can use any number you want because it is a proportion, but If I were to run some numbers I would use a very small acceleration in this case because I am only trying to find the force needed to get the object moving.that is why i changed it just for you to be a force, you can go ahead an read a physics textbook and you will learn all about it, but if it will make you happy i can remove it since according to you it should not be there, but that will only help prove my point since the acceleration in the same direction of the earth would only increase faster. Even though the equation would then be wrongLet me ask you this: What acceleration did you use for your original 'I'?
In the feYes.
Terminal "velocity" is actually terminal acceleration right?QuoteThis acceleration would be at 9.8m/s2 right?No.QuoteBut it took you longer to reach the 9.8m/s2 right?Longer than what?QuoteSo the earths velocity compared to yours is faster right?Yes.QuoteSo when you pull the chute you have to gain velocity as to not hit the earth at 200kph right?Yes.QuoteSo how does one speed up compared to the earth velocity while accelerating slower?You are not going faster than the earth.QuoteWhen a person on the fe hits the earth they are nto accelerating right?With respect to the earth, yes.QuoteSo thus they have to be accelerating at 9.8m/s2 too right?No.QuoteSo in the fe a person acceleration would start at zero and then clime to 9.8m/22 to hit terminal "velocity". The they would pull their chute and be accelerated to a number much greater than 9.8m/s and than slowly deaccelerate back to 9.8m/s2.No.QuoteOtherwise they would eventually start to gain altitude.No.QuoteDo you understand?Your logic, no.
So if the skydiver's acceleration is 9.8m/s^2 and the earth's acceleration is 9.8m/s^2, how will they ever meet?That was my original point, and that is what is wrong with the FE theory in this case.
So if the skydiver's acceleration is 9.8m/s^2 and the earth's acceleration is 9.8m/s^2, how will they ever meet?The skydivers acceleration went to zero when he jumped out of the plane. Thus the earth gains on him. As the air around him accelerates, it accelerates him. But this is no where near 9.8m/s2 yet. Terminal velocity is actually when his acceleration finally reaches the earths acceleration. So he does not accelerate towards it. How could he stop accelerating towards the earth is he never even matches the earth's acceleration?
Did you read my original explanation?So if the skydiver's acceleration is 9.8m/s^2 and the earth's acceleration is 9.8m/s^2, how will they ever meet?That was my original point, and that is what is wrong with the FE theory in this case.
The skydivers acceleration went to zero when he jumped out of the plane. Thus the earth gains on him. As the air around him accelerates, it accelerates him. But this is no where near 9.8m/s2 yet. Terminal velocity is actually when his acceleration finally reaches the earths acceleration. So he does not accelerate towards it.Now do you see how your original explanation was wrong?
How could he stop accelerating towards the earth is he never even matches the earth's acceleration?Did you read my original explanation?
I guess you missed this post.The skydivers acceleration went to zero when he jumped out of the plane. Thus the earth gains on him. As the air around him accelerates, it accelerates him. But this is no where near 9.8m/s2 yet. Terminal velocity is actually when his acceleration finally reaches the earths acceleration. So he does not accelerate towards it.Now do you see how your original explanation was wrong?
No the whole thing I typed was wrong. Sadly, I get it now. At terminal "velocity" your acceleration would match the earths, 9.8m/s2. This same acceleration is what allows a person to not see air moving faster that 200kph. The earth would have a greater velocity than you and thus eventually hit you. But there is no terminal velocity in the FET, it would be called terminal acceleration.
I guess no one remembered when I said that any experiment or observation of this kind (remember the lollipop example) will not yield you the results you desire. The EP (I sure as hell hope my parachute is larger than a closet!) assures us that you would not be able to tell the difference between 'gravity' and acceleration.yes I read it but your mistakes are this: first you invoke EP which will only apply when asked "what did you feel" and it is not valid when using a model to predict behavior. Secondly you assume that air can provide only a very minimal force when, and lastly you assume that since the system is not %100 efficient then it will never be accelerated. I hope that helps you
Now, Gulliver is correct (gasp!) in that you are taking the air to be a rigid body that does not escape the canopy. You would have the greatest acceleration only when (unpowered) you were rigidly connected to the Earth. Such as if your chute got stuck in some trees and you were left hanging, accelerating at 9.8m/s^2. However, when you are in the airplane you are accelerating at the same rate as the Earth. Once you jump out of the plane, you begin to decelerate. There is a point at which you reach terminal velocity, at which your acceleration towards the earth is zero, but you still have a certain velocity. As the earth continues to accelerate, it pushes on your chute with a greater force, thus allowing you to maintain your relative velocity to the Earth. Now, since the air is not a rigid body, it escapes around your chute, ensuring that your current upwards velocity is just slightly less than that of the Earth. Therefore, you slowly approach the Earth, and land nice and easy.
yes I read it but your mistakes are this: first you invoke EP which will only apply when asked "what did you feel" and it is not valid when using a model to predict behavior. Secondly you assume that air can provide only a very minimal force when, and lastly you assume that since the system is not %100 efficient then it will never be accelerated. I hope that helps you"What did you feel?" What the hell are you talking about? The EP is great at finding a model to predict behavior. That's how much of the FE stuff works. When did I say the air can only provide a 'very minimal' force? I assume the system is not 100% efficient because it is not. I hope that helps you.
Ok by not answering my post I'm going to assume you figured out you were wrong.yes I read it but your mistakes are this: first you invoke EP which will only apply when asked "what did you feel" and it is not valid when using a model to predict behavior. Secondly you assume that air can provide only a very minimal force when, and lastly you assume that since the system is not %100 efficient then it will never be accelerated. I hope that helps you"What did you feel?" What the hell are you talking about? The EP is great at finding a model to predict behavior. That's how much of the FE stuff works. When did I say the air can only provide a 'very minimal' force? I assume the system is not 100% efficient because it is not. I hope that helps you.
We are arguing how you think that by accelerating slower than something you can safely land on it.I find it very funny that someone who claims to 'know' the EP is arguing the whole parachute idea in the first place.
So, you you understand how your argument is wrong?
Ok by not answering my post I'm going to assume you figured out you were wrong."Patience is a virtue often lost."
And the FE theory is very flawed so therefore my statement stands, And I agree that the system is not %100 efficient but I disagree with your conclusionyes I read it but your mistakes are this: first you invoke EP which will only apply when asked "what did you feel" and it is not valid when using a model to predict behavior. Secondly you assume that air can provide only a very minimal force when, and lastly you assume that since the system is not %100 efficient then it will never be accelerated. I hope that helps you"What did you feel?" What the hell are you talking about? The EP is great at finding a model to predict behavior. That's how much of the FE stuff works. When did I say the air can only provide a 'very minimal' force? I assume the system is not 100% efficient because it is not. I hope that helps you.
And the FE theory is very flawed so therefore my statement stands, And I agree that the system is not %100 efficient but I disagree with your conclusionAnd yet you have failed to provide this flaw, as it sure is not in this thread.
First of all you logged off for a little bit.Ok by not answering my post I'm going to assume you figured out you were wrong."Patience is a virtue often lost."
Ok.We are arguing how you think that by accelerating slower than something you can safely land on it.No, we are arguing how you think that by accelerating at the same rate as something (from the same initial velocity, no less!) you can reach it.
So, you you understand how your argument is wrong?
Think about it like this: You are in your car, you step on the gas and start to accelerate. You and your front bumper have the same acceleration. Using your logic, you would slowly approach the bumper. That would make long trips very uncomfortable.
I find it very funny that someone who claims to 'know' the EP is arguing the whole parachute idea in the first place.
so what force counteracts the air resistance to produce a state of equalibrium so we achieve a terminal velocity? and therefore have a model that reflect realityNo force. Once you reach terminal acceleration, the air cannot push and harder.
that is not the same thing as what happens on the REso what force counteracts the air resistance to produce a state of equalibrium so we achieve a terminal velocity? and therefore have a model that reflect realityNo force. Once you reach terminal acceleration, the air cannot push and harder.
Yeah, its not. But you can't tell the difference.that is not the same thing as what happens on the REso what force counteracts the air resistance to produce a state of equalibrium so we achieve a terminal velocity? and therefore have a model that reflect realityNo force. Once you reach terminal acceleration, the air cannot push and harder.
Ok.Yep. I think I was getting everything way too convoluted. Thanks for bringing the argument back on track.
If you jump out of a plane in the FE you will loose your acceleration while the earth will not, right?
Ok, so you see how your post was wrong?Yep.
Ok, let's do this again, a little simpler. I think sokarul has finally realized the reason as to why the EP still holds, like I said, in my first post.but according to the FE model that is not what it states, there is a constant net force acting on the object so according to the FE model it is Impossible, so the only way you can be correct is you add elements of RE or introduce forces into the system that should not be there
You are in a plane, accelerating at the same rate as the earth. You exit said plane and stop accelerating. The earth continues to accelerate and accelerates the air along with it. You body experiences drag and starts to accelerate upwards. As you accelerate (again after you open your chute) your apparent acceleration with the earth slows. When your acceleration with the earth is at zero, or very close to it, you have reached your terminal velocity. However, your overall velocity is less than that of the earth. Therefore, you slowly approach the earth.
That's exactly what it states.Ok, let's do this again, a little simpler. I think sokarul has finally realized the reason as to why the EP still holds, like I said, in my first post.but according to the FE model that is not what it states
You are in a plane, accelerating at the same rate as the earth. You exit said plane and stop accelerating. The earth continues to accelerate and accelerates the air along with it. You body experiences drag and starts to accelerate upwards. As you accelerate (again after you open your chute) your apparent acceleration with the earth slows. When your acceleration with the earth is at zero, or very close to it, you have reached your terminal velocity. However, your overall velocity is less than that of the earth. Therefore, you slowly approach the earth.
In Fe what is accelerating the parachutist? Think before you answer because it is not the earth. In the FE model nothing can accelerate the guy, everything accelerates around himUh, the air is accelerating him.
ok if the air is accelerating him what is his velocity? Not his velocity relative to the earth because in the case of the skydiver there is no velocity according to the FE model.In Fe what is accelerating the parachutist? Think before you answer because it is not the earth. In the FE model nothing can accelerate the guy, everything accelerates around himUh, the air is accelerating him.
RE => Upwards force = downwards force thus, m*g=FDalso you are not looking for an equality you are looking for it to be zero
FE => Acceleration = Upwards force/mass thus, a=FD/m
FD= .5(area)(Cd)(rho)v^2
All is known except velocity. Solve for velocity and you may be surprised at the equation you get.
also you are not looking for an equality you are looking for it to be zeroYou don't understand physics at all, do you?
ok if the air is accelerating him what is his velocity? .It depends.
Not his velocity relative to the earth because in the case of the skydiver there is no velocity according to the FE modelSure there is.
well since this is not really physics, just algebra. and also since you love to nitpick every little thing so much I figured that you would remember to set it to zero.also you are not looking for an equality you are looking for it to be zeroYou don't understand physics at all, do you?
FE model states that he can hove no velocity. Remember we are talking about the different models and their ability to predict reality. so you can not add an acceleration into a system when the model tells you that you cannotok if the air is accelerating him what is his velocity? .It depends.QuoteNot his velocity relative to the earth because in the case of the skydiver there is no velocity according to the FE modelSure there is.
well since this is not really physics, just algebra. and also since you love to nitpick every little thing so much I figured that you would remember to set it to zero.So you don't understand algebra, then?
Remember we are talking about the different models and their ability to predict reality. so you can not add an acceleration into a system when the model tells you that you cannotSince when does the FE say you are not allowed to have an acceleration?
But the air is accelerating him, remember?
FE accelerates up to you, so in the case of the guy he sits there until the earth accelerate up to him hence he has no velocity. so like i said you cannot add elements to a system just to make sure you are correct.The air accelerates the parachutist, but not at the same rate as the Earth is accelerating, therefore the Earth catches up.
Yes it can.But the air is accelerating him, remember?
...but the air isn't a solid and can't accelerate him at 9.8m/s2 because air does get pushed out of the way so the earth catches the parachutist YAY WE'RE DONE!
Yes it can.But the air is accelerating him, remember?
...but the air isn't a solid and can't accelerate him at 9.8m/s2 because air does get pushed out of the way so the earth catches the parachutist YAY WE'RE DONE!
The air accelerates the guy to 9.8m/s2. Then he pulls his chute and gets accelerated to some number above 9.8m/s2. Then as he is floating he then deaccelerates back to 9.8m/s2
He never goes above the accleration of the earth he would just match it. Unless you want to be like the engineer and violate the Model and add an acceleration into the system that does not exist. and then claim in one post that the acceleration of the air can not recreate the acceleration of gravity and then later claim that it recreates it just fineThis thread has been going on for so long that I no longer understand your point. What is your point in all of this?
The guy is sitting in air moving at 200kph. Once the chute opens, the skydiver needs less that the 200kph to keep the 9.8m/s2 so he is able to accelerate to way more than 9.8m/s2.I don't know where you're getting the "The guy is sitting in air moving at 200kph" assumption from and I don't understand your logic in the rest of your post.
My point is this: The FE model of gravity especially with regards to freely falling objects will not produce realistic results. If I want to show on paper how far a golf ball will fly or find out the terminal velocity of an object and I use the FE model I will not get an accurate result.He never goes above the accleration of the earth he would just match it. Unless you want to be like the engineer and violate the Model and add an acceleration into the system that does not exist. and then claim in one post that the acceleration of the air can not recreate the acceleration of gravity and then later claim that it recreates it just fineThis thread has been going on for so long that I no longer understand your point. What is your point in all of this?
Edit: Also, the acceleration caused by air resistance will not match the acceleration of the Earth.
Terminal velocity.The guy is sitting in air moving at 200kph. Once the chute opens, the skydiver needs less that the 200kph to keep the 9.8m/s2 so he is able to accelerate to way more than 9.8m/s2.I don't know where you're getting the "The guy is sitting in air moving at 200kph" assumption from and I don't understand your logic in the rest of your post.
Don't get angry at me for not understanding, but please clarify.
Terminal velocity.The guy is sitting in air moving at 200kph. Once the chute opens, the skydiver needs less that the 200kph to keep the 9.8m/s2 so he is able to accelerate to way more than 9.8m/s2.I don't know where you're getting the "The guy is sitting in air moving at 200kph" assumption from and I don't understand your logic in the rest of your post.
Don't get angry at me for not understanding, but please clarify.
Indoor skydiving wind speeds are around 120 mph or roughly 200kph. Tell me, what would happen if a indoor skydiver pulled a parachute?
Its apparent you do not understand what happens to a skydiver in the FE. So what would you like me to explain?Terminal velocity.The guy is sitting in air moving at 200kph. Once the chute opens, the skydiver needs less that the 200kph to keep the 9.8m/s2 so he is able to accelerate to way more than 9.8m/s2.I don't know where you're getting the "The guy is sitting in air moving at 200kph" assumption from and I don't understand your logic in the rest of your post.
Don't get angry at me for not understanding, but please clarify.
Indoor skydiving wind speeds are around 120 mph or roughly 200kph. Tell me, what would happen if a indoor skydiver pulled a parachute?
We're not talking about indoor skydivers. In the real world, the earth isn't a fan.
try using the correct equations when you derive it, you should end up withRemember, I did that for you already?
vt=sqrt(2mg/DpA)
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
try using the correct equations when you derive it, you should end up withOkay. Tell me what incorrect equations I used (and the reason they're incorrect), and I'll fix it. Otherwise, I'll just have to assume that you're not following along.
vt=sqrt(2mg/DpA) and then you would solve it with numerical modeling in the case of the FE.
Its apparent you do not understand what happens to a skydiver in the FE. So what would you like me to explain?Terminal velocity.The guy is sitting in air moving at 200kph. Once the chute opens, the skydiver needs less that the 200kph to keep the 9.8m/s2 so he is able to accelerate to way more than 9.8m/s2.I don't know where you're getting the "The guy is sitting in air moving at 200kph" assumption from and I don't understand your logic in the rest of your post.
Don't get angry at me for not understanding, but please clarify.
Indoor skydiving wind speeds are around 120 mph or roughly 200kph. Tell me, what would happen if a indoor skydiver pulled a parachute?
We're not talking about indoor skydivers. In the real world, the earth isn't a fan.
1) V=1/2at2 not just "at"Thank you for the reply.
2) you decided to over simplify the force of velocity by omitting the components of R which is not time specific just velocity specific and since in a real world experiment ther would be more than one integral involved it would become a numerical modeling problem specifically we would use the euler method which if I remember right is a(v,x,t)=sigmaF(v,x,t)/m (you guys may want to check me on that), and that is where the time would come into play.
3) this one is very minor but it does come into play when deriving the problem, we are looking for zero acceleration not zero force (yes I know thet when there is no acceleration there is no net force so please do not waste time telling me that)
Ah shit you are right on the first one, my bad. however you can not integrate one on a formula that has more than one variable not to mention that a integral of F=ma will yeild a differnet result than F=1/2DpAv2. and the only reason finding a zero acceleration instead of a zero force matters is the steps you would take while deriving the equation. It is close but it is too much of a over simplification, think about what you are saying and then asl yourself if can be applied to every situation horizontal, vertical and a combination of the 2.1) V=1/2at2 not just "at"Thank you for the reply.
2) you decided to over simplify the force of velocity by omitting the components of R which is not time specific just velocity specific and since in a real world experiment ther would be more than one integral involved it would become a numerical modeling problem specifically we would use the euler method which if I remember right is a(v,x,t)=sigmaF(v,x,t)/m (you guys may want to check me on that), and that is where the time would come into play.
3) this one is very minor but it does come into play when deriving the problem, we are looking for zero acceleration not zero force (yes I know thet when there is no acceleration there is no net force so please do not waste time telling me that)
1) Sorry. I disagree. I believe that you've confused the distance formula with the velocity formula.
2) Sorry. I disagree. I believe that the F(t) is quite generic. If the force depends on any other variable, the definite integral on dt, would include its effect.
3) Since there is no division by F(t), it's being zero does not impact the derivation. As a result, I don't understand your third concern enough to deal with it.
The rules of integration are quite clear. As long as done properly parametric variables, such as v, can be under the integral.Ah shit you are right on the first one, my bad. however you can not integrate one on a formula that has more than one variable not to mention that a integral of F=ma will yeild a differnet result than F=1/2DpAv2. and the only reason finding a zero acceleration instead of a zero force matters is the steps you would take while deriving the equation. It is close but it is too much of a over simplification, think about what you are saying and then asl yourself if can be applied to every situation horizontal, vertical and a combination of the 2.1) V=1/2at2 not just "at"Thank you for the reply.
2) you decided to over simplify the force of velocity by omitting the components of R which is not time specific just velocity specific and since in a real world experiment ther would be more than one integral involved it would become a numerical modeling problem specifically we would use the euler method which if I remember right is a(v,x,t)=sigmaF(v,x,t)/m (you guys may want to check me on that), and that is where the time would come into play.
3) this one is very minor but it does come into play when deriving the problem, we are looking for zero acceleration not zero force (yes I know thet when there is no acceleration there is no net force so please do not waste time telling me that)
1) Sorry. I disagree. I believe that you've confused the distance formula with the velocity formula.
2) Sorry. I disagree. I believe that the F(t) is quite generic. If the force depends on any other variable, the definite integral on dt, would include its effect.
3) Since there is no division by F(t), it's being zero does not impact the derivation. As a result, I don't understand your third concern enough to deal with it.
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.
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 rightyes 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.
you must also remember that you are applying the acceleration of the FE to this problem and by definition of the FE model there can be no acceleration on the parachutist if he is not standing on the surface being pushed by the earth. otherwise there would have to be some sort of gravity acting on the person. Like I said earlier if you draw a free body diagram of the forces acting upon the skydiver it will moke more sense.Sorry, I disagree. I've drawn the force diagrams. I stand by my proof.
This goes back to one of my first posts, when you look at the parachutist and draw a free body diagram which is just a vector diagram of all of the forces involved you will get 2 different diagrams when you do one for RE and one for FE. In RE you will have on vector pointing down which will be the gravitational attraction to the earth "mg" and pointing in an opposite direction would be your air resistance "R=1/2DpAv2" so you would end up with the equation F=mg-1/2DpAv2
In FE you will have on vector pointing down which will be you inertial mass "I" and you will still have the vector pointing up as air resistance so you will get an equation of F=I -1/2DpAv2 the only thing that the acceleration of the air would be used for is to compute the velocity of the air in the air resistance formula.
yes but they forget their model and revert right back to the RE model and if you are disconnected fron the earth you can not have any acceleration. It is in the definition of the FE model, If a car is accelerating at me I somehow also get that acceleration?you must also remember that you are applying the acceleration of the FE to this problem and by definition of the FE model there can be no acceleration on the parachutist if he is not standing on the surface being pushed by the earth. otherwise there would have to be some sort of gravity acting on the person. Like I said earlier if you draw a free body diagram of the forces acting upon the skydiver it will moke more sense.Sorry, I disagree. I've drawn the force diagrams. I stand by my proof.
I suspect that you're forgetting the "trick" that FE regularly uses to make n00bs looks bad. The distance, velocity, and force are relative to the surface of the Earth. So the acceleration of the FE up produces exactly the same effect as the object's acceleration down by RE gravity.
Sorry. You're not right. Any proof has its limits, but this proof destroys all arguments that FE and RE have different local gravities.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 rightyes 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.
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 sameSorry. You're not right. Any proof has its limits, but this proof destroys all arguments that FE and RE have different local gravities.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 rightyes 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.
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?
RE => m*g = FD
FE => m*a = FD
Just solve for velocity from the drag equation.
you can be acclerated by your resistance to the air, but in no way can you be accelerated by the FE.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?
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 realityRE => m*g = FD
FE => m*a = FD
Just solve for velocity from the drag equation.
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
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.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.
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 wrongand 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 realityEr, I'm pretty sure you don't accelerate at 9.8m/s2 with a parachute, even in RE.
I have shown that the Force in both models are equal. The model is accurate.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 sameSorry. You're not right. Any proof has its limits, but this proof destroys all arguments that FE and RE have different local gravities.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 rightyes 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.
forces are not the same:I have shown that the Force in both models are equal. The model is accurate.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 sameSorry. You're not right. Any proof has its limits, but this proof destroys all arguments that FE and RE have different local gravities.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 rightyes 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.
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.
You must include the acceleration of the FE toward the parachutist.you can be acclerated by your resistance to the air, but in no way can you be accelerated by the FE.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?
Not in the system. We can not add things to the system just to make it correct. this is fron the FAQYou must include the acceleration of the FE toward the parachutist.you can be acclerated by your resistance to the air, but in no way can you be accelerated by the FE.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?
That is the point! This is a relative issue. The height is measured from the parachutist to the ground.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.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.
no, if the forces are not equal they will not produce the same results which is something that we could seeThat is the point! This is a relative issue. The height is measured from the parachutist to the ground.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.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.
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.
You cannot omit a feature of the FE model to make your challenge right. Sorry, but you need to face the music.Not in the system. We can not add things to the system just to make it correct. this is fron the FAQYou must include the acceleration of the FE toward the parachutist.you can be acclerated by your resistance to the air, but in no way can you be accelerated by the FE.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?
"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?
how exactly does air all of the sudden exert enough force to recreate the "force" of gravity when you are trying to predict behavior.
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.
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 thishow 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?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.
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.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.
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.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 thishow 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.
RE: F=mg, FE: F=ma see freely falling bodies follow the same behavior in both models.
Ok lets clear some things up.Sorry. I'm confused.
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.
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.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.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 thishow 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.
RE: F=mg, FE: F=ma see freely falling bodies follow the same behavior in both models.
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.Ok lets clear some things up.Sorry. I'm confused.
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.
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.
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!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.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.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 thishow 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.
RE: F=mg, FE: F=ma see freely falling bodies follow the same behavior in both models.
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.
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.
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.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.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 thishow 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.
RE: F=mg, FE: F=ma see freely falling bodies follow the same behavior in both models.
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.
Since your accelerations aren't relative, I don't see how your point is germane to this discussion.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.Ok lets clear some things up.Sorry. I'm confused.
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.
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.
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.
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.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.
Since your accelerations aren't relative, I don't see how your point is germane to this discussion.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.Ok lets clear some things up.Sorry. I'm confused.
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.
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.
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.
See?????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.
I don't know how to explain it any more clearly. The effect of RE's gravity is mimicked by the effect of the FE's acceleration of the Earth's surface toward the parachutist. You must include this effect in the FE model in order to have a complete solution.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.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.
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.
How? If i am water skiing do I need to hang on to the tow rope for the boat to accelerate me, I cant just stand in front of it and let the wind push me I need to be connected to the system for it to accelerate meI don't know how to explain it any more clearly. The effect of RE's gravity is mimicked by the effect of the FE's acceleration of the Earth's surface toward the parachutist. You must include this effect in the FE model in order to have a complete solution.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.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.
RE => m*g = FD
FE => m*a = FD
Just solve for velocity from the drag equation.
How? If i am water skiing do I need to hang on to the tow rope for the boat to accelerate me, I cant just stand in front of it and let the wind push me I need to be connected to the system for it to accelerate me
I am making an analogy not a relation.How? If i am water skiing do I need to hang on to the tow rope for the boat to accelerate me, I cant just stand in front of it and let the wind push me I need to be connected to the system for it to accelerate me
You're trying to equate the force of an accelerating Earth to a boat? One is a big, flat expanse of matter (FE obviously..). The other is an aerodynamic craft.
So now air can recreate the force of gravity? so do you stand by your statement that it can never accelerate you to 9.8m2
RE => m*g = FD
FE => m*a = FD
Just solve for velocity from the drag equation.
Let's make sure that we the analogy straight.How? If i am water skiing do I need to hang on to the tow rope for the boat to accelerate me, I cant just stand in front of it and let the wind push me I need to be connected to the system for it to accelerate meI don't know how to explain it any more clearly. The effect of RE's gravity is mimicked by the effect of the FE's acceleration of the Earth's surface toward the parachutist. You must include this effect in the FE model in order to have a complete solution.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.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.
The problem is that the earth already was gaining on him. In the FE when the skydiver hits the ground he is accelerating at 9.8m/s2, but in the same direction of the earth. If he was accelerating less than that, the skydiver would accelerate as he falls. Thats a bad thing, in both models.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.
If that statement is wrong, you equate the air and parachute to a pole connected to the Earth. If anything, you'll reach 9.8m/s2 after the initial deployment of the parachute, and I can't imagine it lasting for more than a few seconds. To which the Earth will significantly gain on the skydiver's position.
So now air can recreate the force of gravity? so do you stand by your statement that it can never accelerate you to 9.8m2If a FE skydiver at some point doesn't accelerate faster than 9.8m/s2 he will die. (unless he jumps out of the plane with his chute open. )
Fixed it for youLet's make sure that we the analogy straight.How? If i am water skiing do I need to hang on to the tow rope for the boat to accelerate me, I cant just stand in front of it and let the wind push me I need to be connected to the system for it to accelerate meI don't know how to explain it any more clearly. The effect of RE's gravity is mimicked by the effect of the FE's acceleration of the Earth's surface toward the parachutist. You must include this effect in the FE model in order to have a complete solution.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.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.
In RE:
A motorboat accelerates a skier toward the shore.
The motorboat is gravity.
The skier is the parachutist.
The shore is the surface of the Earth.
In FE:
The shore accelerates toward the skierThe acceleration of the shore is the UA.
The boat accelerates away from the shore toward the skier pulling the shore with it is the UA
The shore is the surface of the Earth.
The skier is the parachutist.In both models the relative velocities, the distance between the shore and the skier, and the time the skier reaches the shoreare all the same.
The forces acting upon the skier are not equal so the skiers reach the shore at different times and at different relative velocitiesMy proof stands.Needs Improvement
Let's make sure that we the analogy straight.
In RE:
A motorboat accelerates a skier toward the shore.
The motorboat is gravity.
The skier is the parachutist.
The shore is the surface of the Earth.
In FE:
The shore accelerates toward the skierThe acceleration of the shore is the UA.
The boat accelerates away from the shore toward the skier pulling the shore with it is the UA
The shore is the surface of the Earth.
The skier is the parachutist.In both models the relative velocities, the distance between the shore and the skier, and the time the skier reaches the shoreare all the same.
The forces acting upon the skier are not equal so the skiers reach the shore at different times and at different relative velocitiesMy proof stands.Needs Improvement
Fixed it for you
FixedFixed it so its wrong for you.Let's make sure that we the analogy straight.How? If i am water skiing do I need to hang on to the tow rope for the boat to accelerate me, I cant just stand in front of it and let the wind push me I need to be connected to the system for it to accelerate meI don't know how to explain it any more clearly. The effect of RE's gravity is mimicked by the effect of the FE's acceleration of the Earth's surface toward the parachutist. You must include this effect in the FE model in order to have a complete solution.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.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.
In RE:
A motorboat accelerates a skier toward the shore.
The motorboat is gravity.
The skier is the parachutist.
The shore is the surface of the Earth.
In FE:
The shore accelerates toward the skierThe acceleration of the shore is the UA.
The boat accelerates away from the shore toward the skier pulling the shore with it is the UA
The shore is the surface of the Earth.
The skier is the parachutist.In both models the relative velocities, the distance between the shore and the skier, and the time the skier reaches the shoreare all the same.
The forces acting upon the skier are not equal so the skiers reach the shore at different times and at different relative velocitiesMy proof stands.Needs Improvement
I am not saying it, you are ae equating the acceleration of the earth to be tha same as gravity when the earth itself is unable to accelerate the person, the only thing that can accelerate him would be his air resistance, and that acceleration would only speed him up to match the acceleration of the earth to the point where his relative position to the surface of the earth would not change. Imagine this: go to a river near you with a buddy, have him get on a raft and let him accelerate to the same speed as the river. Then get a distance ahead of him and then you get into the river on your own raft and just the the current accelerate you. now ask yourself "will he ever catch you?" and the answer will be yes if you are close enough to where he could catch you while the river was accelerating you but if oyu were far enough away initially he would never catch you because you are both traveling at the same rate.Sorry. I can't understand your rambling. I'll try to respond to your comments one at a time.
I am not saying it.Yes, you did.
You are equating the acceleration of the earth to be the same as gravity when the earth itself is unable to accelerate the person.I was quite clear: the acceleration of the shore toward the skier is what causes the skier to accelerate toward the shore.
The only thing that can accelerate him would be his air resistance.No. That's wrong. There would also be water resistance.
That acceleration would only speed him up to match the acceleration of the earth to the point where his relative position to the surface of the earth would not change.Wrong. Even when acceleration is zero, velocity can carry the skier to shore.
I am not saying it, you are ae equating the acceleration of the earth to be tha same as gravity when the earth itself is unable to accelerate the person, the only thing that can accelerate him would be his air resistance, and that acceleration would only speed him up to match the acceleration of the earth to the point where his relative position to the surface of the earth would not change. Imagine this: go to a river near you with a buddy, have him get on a raft and let him accelerate to the same speed as the river. Then get a distance ahead of him and then you get into the river on your own raft and just the the current accelerate you. now ask yourself "will he ever catch you?" and the answer will be yes if you are close enough to where he could catch you while the river was accelerating you but if oyu were far enough away initially he would never catch you because you are both traveling at the same rate.
I am not saying it, you are ae equating the acceleration of the earth to be tha same as gravity when the earth itself is unable to accelerate the person, the only thing that can accelerate him would be his air resistance, and that acceleration would only speed him up to match the acceleration of the earth to the point where his relative position to the surface of the earth would not change. Imagine this: go to a river near you with a buddy, have him get on a raft and let him accelerate to the same speed as the river. Then get a distance ahead of him and then you get into the river on your own raft and just the the current accelerate you. now ask yourself "will he ever catch you?" and the answer will be yes if you are close enough to where he could catch you while the river was accelerating you but if oyu were far enough away initially he would never catch you because you are both traveling at the same rate.Wrong.
ah hell you are right I was doing the same thing you guys are doing, I was mixing the two models. you are correct he would still land but that will still not match reality which is my whole point to begin with the accelerations will not match up to realityYou need to tell us the reason that the accelerations will not match up to reality.
the forces do not equate. in RE we are sped up and that is causes the increasing air resistance so the force of us speeding up cna me matched in an equal and opposite direction, While in FE the increasing air resistance comes from the air speeding up around us so instead of getting a balance of forces we get a net force that acts on us so our acceleration will never be zero, the best that can be had is that our acceleration will match that of the air.ah hell you are right I was doing the same thing you guys are doing, I was mixing the two models. you are correct he would still land but that will still not match reality which is my whole point to begin with the accelerations will not match up to realityYou need to tell us the reason that the accelerations will not match up to reality.
You're confused. You need to look at the force causing an acceleration on the relative distance between the surface of the Earth and the parachutist. In FE, the surface accelerates towards the parachutist. The "mg" occurs then in both equations, as I have shown in the proof.the forces do not equate. in RE we are sped up and that is causes the increasing air resistance so the force of us speeding up cna me matched in an equal and opposite direction, While in FE the increasing air resistance comes from the air speeding up around us so instead of getting a balance of forces we get a net force that acts on us so our acceleration will never be zero, the best that can be had is that our acceleration will match that of the air.ah hell you are right I was doing the same thing you guys are doing, I was mixing the two models. you are correct he would still land but that will still not match reality which is my whole point to begin with the accelerations will not match up to realityYou need to tell us the reason that the accelerations will not match up to reality.
Or simply put,
RE: F=mg-R
FE: F=-R
Now who can tell us the difference between these 2
Why do you keep ignoring my post with the terminal velocity equations?because they are wrong
ewe are not relating the distance, we are comparing forces acting upon a body. and when you look at the forces acting upon the body there is no mg type force in FE. you must try to not think of what happens in real life and look at what the model predicts will happen.We are talking about comparing forces that are affecting the distance between the parachutist and the surface. You're choosing, incorrectly, to ignore the effect of the Force accelerating the FE.
no we are not we are just talking about the forces that are acting on his body, once we know what these forces are we then can relate it to the earth.ewe are not relating the distance, we are comparing forces acting upon a body. and when you look at the forces acting upon the body there is no mg type force in FE. you must try to not think of what happens in real life and look at what the model predicts will happen.We are talking about comparing forces that are affecting the distance between the parachutist and the surface. You're choosing, incorrectly, to ignore the effect of the Force accelerating the FE.
You're welcome to do so. Take it in whatever steps you need.no we are not we are just talking about the forces that are acting on his body, once we know what these forces are we then can relate it to the earth.ewe are not relating the distance, we are comparing forces acting upon a body. and when you look at the forces acting upon the body there is no mg type force in FE. you must try to not think of what happens in real life and look at what the model predicts will happen.We are talking about comparing forces that are affecting the distance between the parachutist and the surface. You're choosing, incorrectly, to ignore the effect of the Force accelerating the FE.
there is no mg in the FE model. FE'ers need to add it to make sure they are correct even though it violates their model. Then they will just claim that it is because of the EP so if you want to follow then that is fine by me but it does not make them any more rightYou're welcome to do so. Take it in whatever steps you need.no we are not we are just talking about the forces that are acting on his body, once we know what these forces are we then can relate it to the earth.ewe are not relating the distance, we are comparing forces acting upon a body. and when you look at the forces acting upon the body there is no mg type force in FE. you must try to not think of what happens in real life and look at what the model predicts will happen.We are talking about comparing forces that are affecting the distance between the parachutist and the surface. You're choosing, incorrectly, to ignore the effect of the Force accelerating the FE.
I found it quite simple to go with the relative model, but it does take an open mind.
Let me know when you put mg in the FE equation.
Yes, there is mg in the FE model, whether you agree with it or not. If you wish to impugn the FE with your example then you must add mg to your equations--or be irrelevant.there is no mg in the FE model. FE'ers need to add it to make sure they are correct even though it violates their model. Then they will just claim that it is because of the EP so if you want to follow then that is fine by me but it does not make them any more rightYou're welcome to do so. Take it in whatever steps you need.no we are not we are just talking about the forces that are acting on his body, once we know what these forces are we then can relate it to the earth.ewe are not relating the distance, we are comparing forces acting upon a body. and when you look at the forces acting upon the body there is no mg type force in FE. you must try to not think of what happens in real life and look at what the model predicts will happen.We are talking about comparing forces that are affecting the distance between the parachutist and the surface. You're choosing, incorrectly, to ignore the effect of the Force accelerating the FE.
I found it quite simple to go with the relative model, but it does take an open mind.
Let me know when you put mg in the FE equation.
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.I have already said when you jump out of a plane in the fe acceleration goes to zero. But it doe not stay there.
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.
Really? Show me.Why do you keep ignoring my post with the terminal velocity equations?because they are wrong
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.The bold part is wrong. Terminal velocity is never zero.
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.
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
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 realityOh 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.The bold part is wrong. Terminal velocity is never zero.
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 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.
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.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.
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.
no it is only true if you add the surface of the earth to the system of the parachutistyour 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?
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.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 realityOh 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.The bold part is wrong. Terminal velocity is never zero.
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 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.
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 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.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 realityOh 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.The bold part is wrong. Terminal velocity is never zero.
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 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.
...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.
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?
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.
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=RI think the main problem here is a basic lack of the understanding of physics principles.
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.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 realityOh 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.The bold part is wrong. Terminal velocity is never zero.
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 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.
Let me see if I understand what you mean:Yes. Your windows do not effect the speed of the air.
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
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.Let me see if I understand what you mean:Yes. Your windows do not effect the speed of the air.
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
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.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.Let me see if I understand what you mean:Yes. Your windows do not effect the speed of the air.
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
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)?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?
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.
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)?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?
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.
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.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=RI 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=FD
FE => ma=FD
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.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.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=RI 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=FD
FE => ma=FD
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)?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.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=RI 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=FD
FE => ma=FD
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 timeno 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?
You continue to refuse to state your choice of FoR. It's getting annoying.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 timeno 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?
Now the terminal velocity for a baseball is 43m/s but we would need to calculate "D" first so,
D=2mg/vtpA 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
So like I said Not the same thingIt 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
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 itIts kind of hard to pick a FOR for this problem, as most FORs are non-inertial.
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 itNo, you do not account completely for the acceleration of the FE toward the parachutist. From his FoR, the surfaces of both the FE and the RE are accelerating toward him at 1g. You put the force in the RE equation, but omit it, in error, from the FE equation.
In FE the Force of the surface of the earth is not in the system while the force of gravity is acting upon him in the RE system. how can you be pushed by an unconnected piece? if a car rushes by me the car will not affect me at all, now the wind from it may affect me but not to the extent the car would if it had hit meThe 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 itNo, you do not account completely for the acceleration of the FE toward the parachutist. From his FoR, the surfaces of both the FE and the RE are accelerating toward him at 1g. You put the force in the RE equation, but omit it, in error, from the FE equation.
You forget that you chose the FoR of the parachutist. You have to deal with the acceleration of the surface toward him equally in both models.In FE the Force of the surface of the earth is not in the system while the force of gravity is acting upon him in the RE system. how can you be pushed by an unconnected piece? if a car rushes by me the car will not affect me at all, now the wind from it may affect me but not to the extent the car would if it had hit meThe 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 itNo, you do not account completely for the acceleration of the FE toward the parachutist. From his FoR, the surfaces of both the FE and the RE are accelerating toward him at 1g. You put the force in the RE equation, but omit it, in error, from the FE equation.
It is in the presence of these resistive forces that throws the FE theory out of whack. In the absence of resistive forces yes the acceleration in the 2 models would be the same but as soon as you add resistive forces into the mix the results are skewed. this is mainly due to where the forces come from. I suspect you are relating it to the EP and while it is correct that when I jump out of plane I would not be able to tell if I was falling or if the earth was rising up to meet me, however when we develop a model to explain what is happening and predict behavior we can not use EP because we need to state what forces are involved and how they are affecting us.You forget that you chose the FoR of the parachutist. You have to deal with the acceleration of the surface toward him equally in both models.In FE the Force of the surface of the earth is not in the system while the force of gravity is acting upon him in the RE system. how can you be pushed by an unconnected piece? if a car rushes by me the car will not affect me at all, now the wind from it may affect me but not to the extent the car would if it had hit meThe 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 itNo, you do not account completely for the acceleration of the FE toward the parachutist. From his FoR, the surfaces of both the FE and the RE are accelerating toward him at 1g. You put the force in the RE equation, but omit it, in error, from the FE equation.
We can simplify this quickly. Imagine a vertical tube perfectly evacuated expect for a ping-pong ball at the top. As the ball falls in the RE model gravity accelerates it down. As the FE surface rises it accelerates up toward the ball. The surface meets the ball at the same time with the same speed in either model.
We don't disagree on any other point. You don't need the air resistance. You don't need the parachute. You just have to face the fact that FE mimics the force of gravity for the constant acceleration of the FE up at 1g. You must add this into your equations.
There is no resistive forces in the fe.It is in the presence of these resistive forces that throws the FE theory out of whack. In the absence of resistive forces yes the acceleration in the 2 models would be the same but as soon as you add resistive forces into the mix the results are skewed. this is mainly due to where the forces come from. I suspect you are relating it to the EP and while it is correct that when I jump out of plane I would not be able to tell if I was falling or if the earth was rising up to meet me, however when we develop a model to explain what is happening and predict behavior we can not use EP because we need to state what forces are involved and how they are affecting us.You forget that you chose the FoR of the parachutist. You have to deal with the acceleration of the surface toward him equally in both models.In FE the Force of the surface of the earth is not in the system while the force of gravity is acting upon him in the RE system. how can you be pushed by an unconnected piece? if a car rushes by me the car will not affect me at all, now the wind from it may affect me but not to the extent the car would if it had hit meThe 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 itNo, you do not account completely for the acceleration of the FE toward the parachutist. From his FoR, the surfaces of both the FE and the RE are accelerating toward him at 1g. You put the force in the RE equation, but omit it, in error, from the FE equation.
We can simplify this quickly. Imagine a vertical tube perfectly evacuated expect for a ping-pong ball at the top. As the ball falls in the RE model gravity accelerates it down. As the FE surface rises it accelerates up toward the ball. The surface meets the ball at the same time with the same speed in either model.
We don't disagree on any other point. You don't need the air resistance. You don't need the parachute. You just have to face the fact that FE mimics the force of gravity for the constant acceleration of the FE up at 1g. You must add this into your equations.
How can resistive forces throw FE out of whack?It is in the presence of these resistive forces that throws the FE theory out of whack. In the absence of resistive forces yes the acceleration in the 2 models would be the same but as soon as you add resistive forces into the mix the results are skewed. this is mainly due to where the forces come from. I suspect you are relating it to the EP and while it is correct that when I jump out of plane I would not be able to tell if I was falling or if the earth was rising up to meet me, however when we develop a model to explain what is happening and predict behavior we can not use EP because we need to state what forces are involved and how they are affecting us.You forget that you chose the FoR of the parachutist. You have to deal with the acceleration of the surface toward him equally in both models.In FE the Force of the surface of the earth is not in the system while the force of gravity is acting upon him in the RE system. how can you be pushed by an unconnected piece? if a car rushes by me the car will not affect me at all, now the wind from it may affect me but not to the extent the car would if it had hit meThe 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 itNo, you do not account completely for the acceleration of the FE toward the parachutist. From his FoR, the surfaces of both the FE and the RE are accelerating toward him at 1g. You put the force in the RE equation, but omit it, in error, from the FE equation.
We can simplify this quickly. Imagine a vertical tube perfectly evacuated expect for a ping-pong ball at the top. As the ball falls in the RE model gravity accelerates it down. As the FE surface rises it accelerates up toward the ball. The surface meets the ball at the same time with the same speed in either model.
We don't disagree on any other point. You don't need the air resistance. You don't need the parachute. You just have to face the fact that FE mimics the force of gravity for the constant acceleration of the FE up at 1g. You must add this into your equations.
There would be your resistance to the accelerating air. That would not go anywhere in either modelOk, so you are falling and have -Fair in FE and in RE you have Fgravity-Fair.
-Fair
Calculate the terminal velovity of an objectThere would be your resistance to the accelerating air. That would not go anywhere in either modelOk, so you are falling and have -Fair in FE and in RE you have Fgravity-Fair.
Can you state an experiment that would show us where the prediction model would vary?
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 existTell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.
There would be your resistance to the accelerating air. That would not go anywhere in either model
RE:Calculate the terminal velocity of an objectThere would be your resistance to the accelerating air. That would not go anywhere in either modelOk, so you are falling and have -Fair in FE and in RE you have Fgravity-Fair.
Can you state an experiment that would show us where the prediction model would vary?
basically that. also the guy does not accelerate to the earth, it is the other way around so when calculating terminal velocity you can not use "a" because he is not accelerating so his velocity will be different and then if you relate that to the surface of the earth you get differnet numbers than we see in reality. and based upon this knowledge terminal velocity is not based upon anything other that a persons resistance to the air accelerating past him.RE:Calculate the terminal velocity of an objectThere would be your resistance to the accelerating air. That would not go anywhere in either modelOk, so you are falling and have -Fair in FE and in RE you have Fgravity-Fair.
Can you state an experiment that would show us where the prediction model would vary?
Vt = 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:
Vt = 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
Ahh, are you saying can't have acceleration instead of gravity because, for example, if two objects were falling of different masses the earth would have to be accelerating at different speeds for them both to exist?
I think one kilogram falls faster than 1 m/s.RE:Calculate the terminal velocity of an objectThere would be your resistance to the accelerating air. That would not go anywhere in either modelOk, so you are falling and have -Fair in FE and in RE you have Fgravity-Fair.
Can you state an experiment that would show us where the prediction model would vary?
Vt = 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:
Vt = 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.
the mistake was that A is not a. A is the cross sectional area of the object that is facing the windI think one kilogram falls faster than 1 m/s.RE:Calculate the terminal velocity of an objectThere would be your resistance to the accelerating air. That would not go anywhere in either modelOk, so you are falling and have -Fair in FE and in RE you have Fgravity-Fair.
Can you state an experiment that would show us where the prediction model would vary?
Vt = 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:
Vt = 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.
No. His velocity relative to the air is increased equally in both models, in RE by gravity, in FE by the FE's acceleration. You already agreed to this! You get the same numbers in both models.basically that. also the guy does not accelerate to the earth, it is the other way around so when calculating terminal velocity you can not use "a" because he is not accelerating so his velocity will be different and then if you relate that to the surface of the earth you get differnet numbers than we see in reality. and based upon this knowledge terminal velocity is not based upon anything other that a persons resistance to the air accelerating past him.RE:Calculate the terminal velocity of an objectThere would be your resistance to the accelerating air. That would not go anywhere in either modelOk, so you are falling and have -Fair in FE and in RE you have Fgravity-Fair.
Can you state an experiment that would show us where the prediction model would vary?
Vt = 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:
Vt = 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
Ahh, are you saying can't have acceleration instead of gravity because, for example, if two objects were falling of different masses the earth would have to be accelerating at different speeds for them both to exist?
tell me how the earth accelerates the guy when he is 5,000ft above the earth?Te earth pushes the air which pushes him, any questions?
so that accelerates him to 9.8m/s2 in the same time gravity does?tell me how the earth accelerates the guy when he is 5,000ft above the earth?Te earth pushes the air which pushes him, any questions?
tell me how the earth accelerates the guy when he is 5,000ft above the earth?We've been over this when we consider the case without resistive forces. The FE accelerates towards him pushing the air by him faster. Did you forget that we've agreed that in the absence of resistive forces that the models produce the same results?
It should accelerate him to 9.8m/s2 in the time it take the RE air to slow a skydiver from 9.8m/s2 to zero.so that accelerates him to 9.8m/s2 in the same time gravity does?tell me how the earth accelerates the guy when he is 5,000ft above the earth?Te earth pushes the air which pushes him, any questions?
He wants to use the parachutist's For, not the outside observer's FoR.It should accelerate him to 9.8m/s2 in the time it take the RE air to slow a skydiver from 9.8m/s2 to zero.so that accelerates him to 9.8m/s2 in the same time gravity does?tell me how the earth accelerates the guy when he is 5,000ft above the earth?Te earth pushes the air which pushes him, any questions?
Please answer this challenge.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 existTell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.
I will be happy to show it for I think the third or fourth time.Please answer this challenge.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 existTell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.
-R cannot = FI will be happy to show it for I think the third or fourth time.Please answer this challenge.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 existTell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.
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
Why?-R cannot = FI will be happy to show it for I think the third or fourth time.Please answer this challenge.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 existTell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.
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
I said it not quite right so hold on.Why?-R cannot = FI will be happy to show it for I think the third or fourth time.Please answer this challenge.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 existTell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.
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 will be happy to show it for I think the third or fourth time.Please answer this challenge.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 existTell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.
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
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 accelerationYou 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 will be happy to show it for I think the third or fourth time.Please answer this challenge.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 existTell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.
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
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 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 accelerationYou 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 will be happy to show it for I think the third or fourth time.Please answer this challenge.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 existTell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.
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
RE: an=mg-a -r --> an=mg-(DpA/2)v^2
FE: an=ae+ar --> an= ae+(DpA/2)v^2
so like I said they are not equal
I am dealing with accelerationsThen there are two reasons the mg, a force, shouldn't be in the RE equation.
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.the mistake was that A is not a. A is the cross sectional area of the object that is facing the windI think one kilogram falls faster than 1 m/s.RE:Calculate the terminal velocity of an objectThere would be your resistance to the accelerating air. That would not go anywhere in either modelOk, so you are falling and have -Fair in FE and in RE you have Fgravity-Fair.
Can you state an experiment that would show us where the prediction model would vary?
Vt = 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:
Vt = 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.
a=f/m I do not know about you but I think there is a force used when calculating an accelerationYou subtract an acceleration from a force in calculating an acceleration. That's a sophomoric mistake.
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 accelerationYou 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 will be happy to show it for I think the third or fourth time.Please answer this challenge.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 existTell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.
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
RE: an=mg-ar --> an=mg-(DpA/2)v^2
FE: an=ae-ar --> an= ae-(DpA/2)v^2
so like I said they are not equal
Edit: the FE model is also not dimensionally sound`
RE:Calculate the terminal velocity of an objectThere would be your resistance to the accelerating air. That would not go anywhere in either modelOk, so you are falling and have -Fair in FE and in RE you have Fgravity-Fair.
Can you state an experiment that would show us where the prediction model would vary?
Vt = 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:
Vt = 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.
RE => mg=FD
FE => ma=FD
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?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
Hint: The answer is below:RE => mg=FD
FE => ma=FD
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 windowSo 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.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.
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
a=f/m I do not know about you but I think there is a force used when calculating an accelerationYou 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 ae = g, by model definition.
Then you should be able to see that the accelerations are equal.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 accelerationYou 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 will be happy to show it for I think the third or fourth time.Please answer this challenge.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 existTell us how the resistive forces, not gravity or the FE's acceleration, aren't equal.
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
RE: an=mg-ar --> an=mg-(DpA/2)v^2
FE: an=ae-ar --> an= ae-(DpA/2)v^2
so like I said they are not equal
Edit: the FE model is also not dimensionally sound`