Parachutes

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TheEngineer

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Re: Parachutes
« Reply #150 on: August 25, 2007, 09:40:28 PM »
Since when is mass measured in Newtons?


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cbarnett97

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Re: Parachutes
« Reply #151 on: August 25, 2007, 09:42:51 PM »
Inertial mass, not just mass. And inertial mass is the "resistance of a mass to being accelerated" so yes inertial mass is measured in Newtons. You of course already knew this since it is one of the stepping stones that Einstein pondered over when he developed his theory of relativity
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TheEngineer

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Re: Parachutes
« Reply #152 on: August 25, 2007, 09:52:01 PM »
Uh, no.  Inertial mass is measured in kg.


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CommonCents

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Re: Parachutes
« Reply #153 on: August 25, 2007, 09:57:22 PM »
Do you have any evidence to support your outlandish claim?
OMG!

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cbarnett97

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Re: Parachutes
« Reply #154 on: August 25, 2007, 09:58:26 PM »
Dont know how to put in greek letters but Inertial Mass is "SigmaF=mia
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sokarul

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Re: Parachutes
« Reply #155 on: August 25, 2007, 09:58:51 PM »
edit hold on

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cbarnett97

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Re: Parachutes
« Reply #156 on: August 25, 2007, 10:01:05 PM »
edit hold on


You were on the right track
On FE there is no counter force to eventially reach equalibrium
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sokarul

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Re: Parachutes
« Reply #157 on: August 25, 2007, 10:12:43 PM »
edit hold on


You were on the right track
On FE there is no counter force to eventially reach equalibrium
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. 
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cbarnett97

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Re: Parachutes
« Reply #158 on: August 25, 2007, 10:15:52 PM »
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 decrease
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TheEngineer

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Re: Parachutes
« Reply #159 on: August 25, 2007, 10:18:20 PM »
Dont know how to put in greek letters but Inertial Mass is "SigmaF=mia
Exactly.  The inertial mass part is this: mi


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cbarnett97

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Re: Parachutes
« Reply #160 on: August 25, 2007, 10:20:18 PM »
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 Newton
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sokarul

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Re: Parachutes
« Reply #161 on: August 25, 2007, 10:21:30 PM »
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 decrease
Yeah 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.  
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cbarnett97

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Re: Parachutes
« Reply #162 on: August 25, 2007, 10:23:46 PM »
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 decrease
Yeah 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 earth
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sokarul

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Re: Parachutes
« Reply #163 on: August 25, 2007, 10:27:20 PM »
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 decrease
Yeah 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 earth
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. 
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TheEngineer

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Re: Parachutes
« Reply #164 on: August 25, 2007, 10:28:52 PM »
Quote
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 Newton
Uh, yea, but inertial mass is measured in kg.  Your equation explicitly stated inertial mass.  Mass does not equal force.
« Last Edit: August 26, 2007, 01:29:58 AM by TheEngineer »


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cbarnett97

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Re: Parachutes
« Reply #165 on: August 25, 2007, 10:29:35 PM »
No, this one is a total loss. The only thing they can say is either a: "you are wrong", and give no evidence of why or b: "the math is wrong" and give no supporting evidence
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divito the truthist

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Re: Parachutes
« Reply #166 on: August 25, 2007, 10:31:57 PM »
"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."

I don't see Newtons.
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cbarnett97

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Re: Parachutes
« Reply #167 on: August 25, 2007, 10:32:57 PM »
The definition of inertial mass is a force, just because mass is in the title does not mean anything. You should know this since you are an expert in Relativity.

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.

Here is how you are wrong
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TheEngineer

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Re: Parachutes
« Reply #168 on: August 25, 2007, 10:34:58 PM »
Inertial mass = mi.  How is that a force?


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cbarnett97

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Re: Parachutes
« Reply #169 on: August 25, 2007, 10:37:15 PM »
"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."

I don't see Newtons.
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.
« Last Edit: August 25, 2007, 10:39:29 PM by cbarnett97 »
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cbarnett97

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Re: Parachutes
« Reply #170 on: August 25, 2007, 10:39:02 PM »
Dont know how to put in greek letters but Inertial Mass is "SigmaF=mia
What did you do with the acceleration. We are not just measuring mass, we are measuring the mass' resistance to being accelerated.
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sokarul

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Re: Parachutes
« Reply #171 on: August 25, 2007, 10:39:04 PM »
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 earth
No, 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.
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.  
Of course you will probably say thats what you meant.
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TheEngineer

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Re: Parachutes
« Reply #172 on: August 25, 2007, 10:41:49 PM »
Dont know how to put in greek letters but Inertial Mass is "SigmaF=mia
What did you do with the acceleration. We are not just measuring mass, we are measuring the mass' resistance to being accelerated.
Inertial mass has nothing to do with acceleration.  If it is sitting still, it still has mass.


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TheEngineer

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Re: Parachutes
« Reply #173 on: August 25, 2007, 10:42:32 PM »
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. 
Of course you will probably say thats what you meant.
Uh, no, not even close.


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cbarnett97

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Re: Parachutes
« Reply #174 on: August 25, 2007, 10:43:09 PM »
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 earth
No, 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.
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.  
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.
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cbarnett97

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Re: Parachutes
« Reply #175 on: August 25, 2007, 10:44:40 PM »
Dont know how to put in greek letters but Inertial Mass is "SigmaF=mia
What did you do with the acceleration. We are not just measuring mass, we are measuring the mass' resistance to being accelerated.
Inertial mass has nothing to do with acceleration.  If it is sitting still, it still has mass.
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 mass
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sokarul

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Re: Parachutes
« Reply #176 on: August 25, 2007, 10:45:49 PM »
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. 
Of course you will probably say thats what you meant.
Uh, no, not even close.
No I was right.  But actually you were also right if you were talking about a RE.  I was talking about a FE.  
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TheEngineer

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Re: Parachutes
« Reply #177 on: August 25, 2007, 10:48:15 PM »
and Tada you have that objects inertial mass as opposed to its rest mass
Which is still mass.  It suddenly does not become a force.


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TheEngineer

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Re: Parachutes
« Reply #178 on: August 25, 2007, 10:49:01 PM »
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.


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Re: Parachutes
« Reply #179 on: August 25, 2007, 10:50:22 PM »
"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."
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