Parachutes

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. 

Quote from: cbarnett97 on August 26, 2007, 06:34:30 PM

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 fine

This 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.

Quote from: sokarul on August 26, 2007, 07:06:53 PM

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.

Quote from: Captain Flamingo on August 26, 2007, 07:11:32 PM

Quote from: sokarul on August 26, 2007, 07:06:53 PM

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. 

Indoor skydiving wind speeds are around 120 mph or roughly 200kph.  Tell me, what would happen if a indoor skydiver pulled a parachute? 

Quote from: sokarul on August 26, 2007, 07:34:51 PM

Quote from: Captain Flamingo on August 26, 2007, 07:11:32 PM

Quote from: sokarul on August 26, 2007, 07:06:53 PM

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. 

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 with

v_t=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 with

v_t=sqrt(2mg/DpA) and then you would solve it with numerical modeling in the case of the FE.

Quote from: Mr. Ireland on August 26, 2007, 08:23:46 PM

Quote from: sokarul on August 26, 2007, 07:34:51 PM

Quote from: Captain Flamingo on August 26, 2007, 07:11:32 PM

Quote from: sokarul on August 26, 2007, 07:06:53 PM

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. 

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.

Its apparent you do not understand what happens to a skydiver in the FE.  So what would you like me to explain? 

1) V=1/2at² not just "at"
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)

Quote from: cbarnett97 on August 27, 2007, 05:33:41 PM

1) V=1/2at² not just "at"
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)

Thank you for the reply.
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.

Quote from: Gulliver on August 27, 2007, 05:46:50 PM

Quote from: cbarnett97 on August 27, 2007, 05:33:41 PM

1) V=1/2at² not just "at"
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)

Thank you for the reply.
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.

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.

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.

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

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

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

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

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.

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.

Quote from: cbarnett97 on August 27, 2007, 06:22:05 PM

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

RE =>  m*g = F_D

FE =>  m*a = F_D

Just solve for velocity from the drag equation.