Perpetual motion/"free energy"

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MrKappa

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Re: Perpetual motion/"free energy"
« Reply #240 on: September 22, 2008, 08:27:36 PM »
I personally don't have a huge problem with the idea of there being some way of you telling what the local gravitational field is (the gravitational equivalent of an altimeter with respect to some 'true' zero-gravitation field), although it would mean that the strong equivalence principle was wrong.

This is interesting... it seems to be the difference between calibrating to perceived time and real-time... I have only just learned about the Terrestrial Reference Frame and I imagine this is exactly that...

You seem to be a very technical person... I did come across this thread discussion regarding a very arguable alternative mathematics for gravity... It is all I have to offer... I do not really understand it all but it might be of interest to you...

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Dr Matrix

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Re: Perpetual motion/"free energy"
« Reply #241 on: September 23, 2008, 02:10:48 AM »
You seem to be a very technical person... I did come across this thread discussion regarding a very arguable alternative mathematics for gravity... It is all I have to offer... I do not really understand it all but it might be of interest to you...

Well glad to be of service in pointing you in the direction of interesting stuff!

Thanks for the link... sadly I can debunk their claims in 5 minutes (since that what I have spent the last 5 minutes doing).  Look at this video and get hold of a stopwatch.  We will assume that the hammer and feather are being dropped from (1.5 +/- 0.3)m height (estimate, but a reasonable one with a suitably large uncertainty).

Now, I took 13 timings of the hammer drop and got a value for the time it took to reach the surface at (1.15 +/- 0.20)s, with the error there being one standard deviation.  We then plug these two facts into the well-known high school kinematic equation for motion:

d = v0t + (1/2)at2, which setting v0 = 0 and solving for a (acceleration due to gravitational attraction) gives us:

a = 2d/t2

So plug in the numbers and errors and we get a = (2.3 +/- 1.8)ms-2, which covers the accepted (Newtonian) value for the Moon of gMoon = 1.62ms-2 to within one standard deviation.  The '64% of Earth g' would be 6.28ms-2, which lies out beyond 2 standard deviations (even with my massive uncertainties).  This makes their claim unlikely to beyond 75% confidence.  Still, it's good to test these things

EDIT: calculator fail! Modified error calculation due to me fucking it up the first time round...
« Last Edit: September 23, 2008, 02:17:00 AM by Matrix »
Quote from: Arthur Schopenhauer
All truth passes through three stages. First, it is ridiculed. Second, it is violently opposed. Third, it is accepted as being self-evident.

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MrKappa

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• Math abstracts reality... it does not create it...
Re: Perpetual motion/"free energy"
« Reply #242 on: September 23, 2008, 08:37:06 PM »
So plug in the numbers and errors and we get a = (2.3 +/- 1.8)ms-2, which covers the accepted (Newtonian) value for the Moon of gMoon = 1.62ms-2 to within one standard deviation.  The '64% of Earth g' would be 6.28ms-2, which lies out beyond 2 standard deviations (even with my massive uncertainties).  This makes their claim unlikely to beyond 75% confidence.  Still, it's good to test these things

I don't doubt that Newtonian physics works. I apologize for being unable to follow your math. Unless it's written in computer code I have a hard time understanding it. ( at least you haven't used latex ) I have a simple question for you.

If two pieces of paper are suspended in a vacuum facing each other free from external influence of other mass.

One paper is 10 miles square and the other is one mile square. They face each other two miles apart.

Which object crosses the halfway mark first?

I have tried to understand a piece of your math by translating it to what I know...

function gravity(\$v,\$O,\$a,\$t){
return (\$v x \$O x \$t) + ((\$a x (\$t x \$t)) / 2) ;
}

I am actually close to being classically math notationally illiterate... It's quite embarrassing since I know a good handful of programming languages.

Is the above what d = v0t + (1/2)at2 translates to?

Where did you get \$v, \$O , \$a from? I only see \$t which is 1.15ms...
« Last Edit: September 23, 2008, 09:17:58 PM by MrKappa »

Parsifal

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Re: Perpetual motion/"free energy"
« Reply #243 on: September 23, 2008, 11:23:15 PM »
I have tried to understand a piece of your math by translating it to what I know...

function gravity(\$v,\$O,\$a,\$t){
return (\$v x \$O x \$t) + ((\$a x (\$t x \$t)) / 2) ;
}

I am actually close to being classically math notationally illiterate... It's quite embarrassing since I know a good handful of programming languages.

Is the above what d = v0t + (1/2)at2 translates to?

Where did you get \$v, \$O , \$a from? I only see \$t which is 1.15ms...

No, 0 is a subscript. v0 means v when t=0.
I'm going to side with the white supremacists.

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MrKappa

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• Math abstracts reality... it does not create it...
Re: Perpetual motion/"free energy"
« Reply #244 on: September 23, 2008, 11:29:52 PM »
No, 0 is a subscript. v0 means v when t=0.

Yes... Okay... But from a physics and science perspective... I have only seen the notation used in chemistry...

H20

Hydrogen plus two Oxygen atoms...

Essentially... does it hold the same in other areas of physics?

v0t

v = velocity?
t = time?

v + (0 x t)

(t x 0) + v?

Parsifal

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Re: Perpetual motion/"free energy"
« Reply #245 on: September 23, 2008, 11:33:14 PM »
Yes... Okay... But from a physics and science perspective... I have only seen the notation used in chemistry...

H20

Hydrogen plus two Oxygen atoms...

Essentially... does it hold the same in other areas of physics?

v0t

v = velocity?
t = time?

v + (0 x t)

(t x 0) + v?

No. To use computer terminology, it's more like v is another function (think of it as a black box function; you don't know how it works) that you can pass a parameter to that represents a certain value of t, and it returns the value of v for that value of t. So v0 is just saying v(0).
I'm going to side with the white supremacists.

?

MrKappa

• 448
• Math abstracts reality... it does not create it...
Re: Perpetual motion/"free energy"
« Reply #246 on: September 23, 2008, 11:49:04 PM »
Yes... Okay... But from a physics and science perspective... I have only seen the notation used in chemistry...

H20

Hydrogen plus two Oxygen atoms...

Essentially... does it hold the same in other areas of physics?

v0t

v = velocity?
t = time?

v + (0 x t)

(t x 0) + v?

No. To use computer terminology, it's more like v is another function (think of it as a black box function; you don't know how it works) that you can pass a parameter to that represents a certain value of t, and it returns the value of v for that value of t. So v0 is just saying v(0).

Okay... So...

H20

Translates to...

H = Hydrogen new function('Whatever parameters Hydrogen requires'){return \$Hydrogen_Atom';}
O = Hydrogen new function('Whatever parameters Oxygen requires'){return \$Oxygen_Atom';}

Water = H + (2 x Oxygen)

Where as...

v0t

translates to...

v = Velocity new function('Whatever value velocity is in kilometers'){return \$kilometers;} // really this should be distance...
t = Time new function('Whatever parameters time requires in seconds'){return \$seconds;}

v0t

d = v +  (0 x t)

Since the multiplication factor = zero... then velocity is the only thing that matters and time is thrown away from the equasion?

This seems to be like we do not understand how to measure velocity... I apologize for strange assumptions I just cannot understand how this as supposed to work out and I have been trying for a while now...

What is d supposed to equal? Distance?
« Last Edit: September 23, 2008, 11:51:03 PM by MrKappa »

Parsifal

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Re: Perpetual motion/"free energy"
« Reply #247 on: September 23, 2008, 11:53:05 PM »
Okay... So...

H20

Translates to...

H = Hydrogen new function('Whatever parameters Hydrogen requires'){return \$Hydrogen_Atom';}
O = Hydrogen new function('Whatever parameters Oxygen requires'){return \$Oxygen_Atom';}

Water = H + (2 x Oxygen)

Where as...

v0t

translates to...

v = Velocity new function('Whatever value velocity is in kilometers'){return \$kilometers;} // really this should be distance...
t = Time new function('Whatever parameters time requires in seconds'){return \$seconds;}

v0t

Answer = v +  (0 x t)

Since the multiplication fact = zero... then velocity is the only thing that matters and time is thrown away from the equasion?

This seems to be like we do not understand how to measure velocity... I apologize for strange assumptions I just cannot understand how this s supposed to work out and I have been trying for a while now...

What is d supposed to equal? Distance?

Um, no, that isn't how it works at all. Velocity is measured in metres per second, not kilometres, and time is an independent variable, not comparable to a function.

The zero isn't a multiplier, it's a parameter. v(0) might be equal to zero metres per second, or it might be equal to sixty metres per second, or it might be equal to the speed of light. As I said, it can be considered to be a black box function for the purposes of this discussion.

Yes, d is distance - or, more correctly, displacement (though for the purposes of this discussion, they can be considered the same thing).

Also, the subscript belongs to the v, not to the t. In H20, there are two hydrogens, not two oxygens. Subscripts nearly always refer to the thing before them.
I'm going to side with the white supremacists.

?

MrKappa

• 448
• Math abstracts reality... it does not create it...
Re: Perpetual motion/"free energy"
« Reply #248 on: September 24, 2008, 12:00:49 AM »
Also, the subscript belongs to the v, not to the t. In H20, there are two hydrogens, not two oxygens. Subscripts nearly always refer to the thing before them.

Oh my god... Thank you so much for dumbing this down for me...

Let me try again...

H2O

Translates to

Water = Hydrogen(2) + Oxygen(1)

v0t

Translates to...

Displacement = Velocity(0) + Time(1)

I think I understand this notation now...
« Last Edit: September 24, 2008, 12:04:48 AM by MrKappa »

Parsifal

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Re: Perpetual motion/"free energy"
« Reply #249 on: September 24, 2008, 12:03:22 AM »
v0t

Translates to...

Distance = Velocity(0) + Time(1)

I think I understand this notation now...

Almost, but it really doesn't mean anything to pass Time a parameter, since it has a given value. It's more like \$time is a variable, and velocity is a function. Acceleration may also be considered a function, so to properly translate the equation, we would have:

function distance(\$time){
return (velocity(\$time) * \$time) + ((acceleration(\$time) * (\$time * \$time)) / 2) ;
}

Incidentally, I don't know how much you know about calculus, but acceleration would always be the derivative function of velocity. That is, acceleration(\$time) is the rate of change of velocity(\$time) as \$time increases.
« Last Edit: September 24, 2008, 12:04:54 AM by Osama bin Laden »
I'm going to side with the white supremacists.

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MrKappa

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• Math abstracts reality... it does not create it...
Re: Perpetual motion/"free energy"
« Reply #250 on: September 24, 2008, 12:11:41 AM »
v0t

Translates to...

Distance = Velocity(0) + Time(1)

I think I understand this notation now...

Almost, but it really doesn't mean anything to pass Time a parameter, since it has a given value. It's more like \$time is a variable, and velocity is a function. Acceleration may also be considered a function, so to properly translate the equation, we would have:

function distance(\$time){
return (velocity(\$time) * \$time) + ((acceleration(\$time) * (\$time * \$time)) / 2) ;
}

Incidentally, I don't know how much you know about calculus, but acceleration would always be the derivative function of velocity. That is, acceleration(\$time) is the rate of change of velocity(\$time) as \$time increases.

Quote
\$time is a variable

True... but time being a constant is only something we perceive at the moment... There is one thing I fundamentally understand... General Relativity was an attempt to free the "code" of physics from constants... any programmer will tell you that having constants in your code makes a big mess... it destroys your function...

The product of time itself is a measurement... the actual function of time could very well be influenced by other variables...

Again... Thank you for explaining the notation... Understanding math notation is not easy for me...

Parsifal

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Re: Perpetual motion/"free energy"
« Reply #251 on: September 24, 2008, 12:14:29 AM »
Oops, that was actually wrong. I got confused myself.

That equation only works with constant acceleration, so that should be a constant, not a function, and instead of velocity(\$time) you want velocity(0), since we want the velocity when t=0, not when t=\$time.

Hope that makes sense.
I'm going to side with the white supremacists.

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MrKappa

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• Math abstracts reality... it does not create it...
Re: Perpetual motion/"free energy"
« Reply #252 on: September 24, 2008, 12:41:12 AM »
Oops, that was actually wrong. I got confused myself.

That equation only works with constant acceleration, so that should be a constant, not a function, and instead of velocity(\$time) you want velocity(0), since we want the velocity when t=0, not when t=\$time.

Hope that makes sense.

No... Gravity makes no sense to me in terms of mathematics... Velocity is the product of two measurements... Time and distance traveled... the product being meters per second...

When your trying to figure out the distance using only the time it took to travel... of course your going to need to throw in velocity as a constant... considering it's trusted to work within certain parameters... Is that what you meant?

Acceleration equals velocity for the purposes of gravity seeing as it is a constant.

I am not trying to understand gravity in full... I have learned a little bit about notation and I think that is a big victory for me today...

Parsifal

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Re: Perpetual motion/"free energy"
« Reply #253 on: September 24, 2008, 12:43:14 AM »
Acceleration equals velocity for the purposes of gravity seeing as it is a constant.

No, acceleration doesn't equal velocity. Acceleration is the rate of change of velocity. To put it another way, acceleration is to velocity as velocity is to distance.

If you start in a stationary car, and one second later you are moving at 10 metres per second, then your average acceleration across that distance was 10 metres per second per second.
I'm going to side with the white supremacists.

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MrKappa

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• Math abstracts reality... it does not create it...
Re: Perpetual motion/"free energy"
« Reply #254 on: September 24, 2008, 12:50:21 AM »
Acceleration equals velocity for the purposes of gravity seeing as it is a constant.

No, acceleration doesn't equal velocity. Acceleration is the rate of change of velocity. To put it another way, acceleration is to velocity as velocity is to distance.

If you start in a stationary car, and one second later you are moving at 10 metres per second, then your average acceleration across that distance was 10 metres per second per second.

I understand that but have a tendancy to mash things together sometimes... ( bad habit)... the acceleration is expressed in "the power of two"... the velocity doubling itself every second...

one second... 9.8m/s
two seconds... 19.8m/s
three seconds... 39.6m/s

until terminal velocity which should be the speed of light correct in a vacuum? What external force counter acts a body in space under a gravitational pull?

So essentially...

Displacement = Velocity+ time;

Time = Displacement / Velocity2;

But if the Velocity is doubling every second...

Do we not have to do something along the lines of this? ( I do not expect this to be the right answer )

Time-2 = Displacement / Velocity;

« Last Edit: September 24, 2008, 01:02:47 AM by MrKappa »

Parsifal

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Re: Perpetual motion/"free energy"
« Reply #255 on: September 24, 2008, 01:01:39 AM »
I understand that but have a tendancy to mash things together sometimes... ( bad habit)... the acceleration is expressed in "the power of two"... the velocity doubling itself every second...

one second... 9.8m/s
two seconds... 19.8m/s
three seconds... 39.6m/s

No, it doesn't double each second, it increases by the rate of acceleration multiplied by one second every second. In the case of gravitation on Earth, that means it is going 9.8, 19.6, 29.4, 39.2, etc.

until terminal velocity which should be the speed of light correct in a vacuum? What external force counter acts a body in space under a gravitational pull?

If we are discussing purely Newtonian physics (which I think we should, since trying to explain Special Relativity to you on top of this would be a nightmare), there is nothing stopping you exceeding the speed of light. That is, of course, wrong, but Newtonian physics serves well for speeds encountered in everyday life.

So essentially...

Displacement = Velocity + time;

Time = Displacement / Velocity;

Almost. Displacement = velocity * time, not + time, if and only if velocity is constant. If velocity is not constant, you would need to use calculus (or Newtonian equations of motion, if acceleration is constant) to perform such operations.
I'm going to side with the white supremacists.

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Dr Matrix

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Re: Perpetual motion/"free energy"
« Reply #256 on: September 24, 2008, 01:08:56 AM »
I have a simple question for you.

If two pieces of paper are suspended in a vacuum facing each other free from external influence of other mass.

One paper is 10 miles square and the other is one mile square. They face each other two miles apart.

Which object crosses the halfway mark first?

Osama seems to be doing pretty well at answering your other questions so I'll just quickly answer this one - assuming no other interactions than gravitation between the two pieces of paper, that the paper is perfectly rigid and that effects such as the Casimir effect can be ignored, and additionally that the larger piece of paper is more massive (different thickness or density could change that), then the smaller piece will appear to cross the half way point before the bigger piece.  This is assuming we have some kind of rigid measuring stick reaching between the two with the initial half way point marked on it, since otherwise neither observer would be able to tell who had 'moved more' due to neither piece of paper experiencing any acceleration.
Quote from: Arthur Schopenhauer
All truth passes through three stages. First, it is ridiculed. Second, it is violently opposed. Third, it is accepted as being self-evident.

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MrKappa

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• Math abstracts reality... it does not create it...
Re: Perpetual motion/"free energy"
« Reply #257 on: September 24, 2008, 01:10:25 AM »
Almost. Displacement = velocity * time, not + time, if and only if velocity is constant. If velocity is not constant, you would need to use calculus (or Newtonian equations of motion, if acceleration is constant) to perform such operations.

OKay...

H2O

Water = Hydrogen(2) + Oxygen(1);

(H x 2) + (O x 1)

Where as...

d = v0t

Displacement = Velocity(0) x Time(1);

Is this the exception in chemistry? Are they using different notation or is

H2O

really meant to be... ( for some unknown reason to me )

(H x 2) x (O x 1)

This is where I think my issues lay... just deciphering the notations...

If anybody has a good cheat sheet which is equivalent to the periodic table of the elements but only for math notation and physics constants I would be interested...
« Last Edit: September 24, 2008, 01:34:48 AM by MrKappa »

Parsifal

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Re: Perpetual motion/"free energy"
« Reply #258 on: September 24, 2008, 01:56:44 AM »
If anybody has a good cheat sheet which is equivalent to the periodic table of the elements but only for math notation and physics constants I would be interested...

I'll try to put one together for how various physical terms relate to each other. For now, just remember that while in Chemistry the lack of a subscript means the same thing as a subscript of 1, in mathematics subscripts are not always necessary, so the lack of one doesn't mean a 1, it means there isn't one.

Also, subscripts can be arbitrary. v0 is used to notate velocity when t=0, which is why I said that if velocity is a function of time it means the same thing as velocity(0), but there are other ways of writing it. When I was in high school, we would always use "u" to mean initial velocity (when t=0), and "v" to mean final velocity. In that case, u would mean the same thing that v0 does here.
« Last Edit: September 24, 2008, 01:58:30 AM by Osama bin Laden »
I'm going to side with the white supremacists.

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Dr Matrix

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Re: Perpetual motion/"free energy"
« Reply #259 on: September 24, 2008, 03:08:09 AM »
If anybody has a good cheat sheet which is equivalent to the periodic table of the elements but only for math notation and physics constants I would be interested...

This is the only cheat sheet for physics you will ever need

Quote from: Arthur Schopenhauer
All truth passes through three stages. First, it is ridiculed. Second, it is violently opposed. Third, it is accepted as being self-evident.

?

MrKappa

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• Math abstracts reality... it does not create it...
Re: Perpetual motion/"free energy"
« Reply #260 on: September 24, 2008, 10:14:07 PM »
If anybody has a good cheat sheet which is equivalent to the periodic table of the elements but only for math notation and physics constants I would be interested...

I'll try to put one together for how various physical terms relate to each other. For now, just remember that while in Chemistry the lack of a subscript means the same thing as a subscript of 1, in mathematics subscripts are not always necessary, so the lack of one doesn't mean a 1, it means there isn't one.

Also, subscripts can be arbitrary. v0 is used to notate velocity when t=0, which is why I said that if velocity is a function of time it means the same thing as velocity(0), but there are other ways of writing it. When I was in high school, we would always use "u" to mean initial velocity (when t=0), and "v" to mean final velocity. In that case, u would mean the same thing that v0 does here.

This is good information... I have found the cheat sheet book and taken this from it...

After translating it I have...

Code: [Select]
` function newtons_law_of_gravitation(\$G,\$m1,\$m2,\$r12){ # ^ unit vector # r12 vector from m1 to m2 # m1 mass 1 (kg) # m2 mass 2 (kg) # F1 force on m1 ( Newtons ) \$F1 = ( (\$G * \$m1 * \$m2) / pow(\$r12,2) ) * \$r12; return \$F1; }`

Is this essentially correct? All of the items above the line are multiplied and then r12 below the line is multiplied to the powe of two and then used to divide what is above the line and is then multiplied what is next to the line?

This I'm having trouble understanding... The Big G...

I have translated it to this...

Code: [Select]
`function gravitational_constant(){    return 6.67 * pow(10,-11);}`
However I cannot understand the significance of the units...

m3 // is this mass cubed?

kg-1 // this is kg which is also mass to my knowledge... why is it negative...

s-2 // This is seconds... but it is negative as well...

How can there be three units ( two of which represent the same thing ) be lined up next to each other and some negative?

It makes no sense...

Raist

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Re: Perpetual motion/"free energy"
« Reply #261 on: September 24, 2008, 10:28:54 PM »
m3 is meters cubed, so it represents the volume.

the seconds are seconds squared, the -2 power implies inverse squared or 1/s2

Parsifal

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Re: Perpetual motion/"free energy"
« Reply #262 on: September 25, 2008, 12:59:28 AM »
Raist is correct, what that means is metres cubed per kilogram per second squared.

That function isn't entirely correct, by the way. Force is a vector quantity, meaning it has both magnitude and direction. Multipliying it by the unit vector in the direction of r12 is just giving it that direction - note that a unit vector does not have the magnitude of the variable it corresponds to, it is called a unit vector because it has a magnitude of 1 (and whatever units the measurement is in) in the direction of that vector. So, multiplying the entire thing by \$r12 is going to throw your answer out by a factor of whatever r12 is.

If you are only interested in the magnitude of the force, take out the last * \$r12 in that expression and it will work fine. If you do want the direction, there's no easy way to code that. That mathematical expression tells a human reader how the force is defined, but not how to calculate it. By contrast, a programming language needs to tell the computer exactly how to calculate that direction, so unless you're using a specialised language that can natively handle vectors, the best thing to do is just drop the * \$r12, get the computer to tell you the magnitude and then remember that it must be in the direction of the vector pointing from 1 to 2.
I'm going to side with the white supremacists.

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pieces

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Re: Perpetual motion/"free energy"
« Reply #263 on: June 22, 2021, 08:14:05 AM »
Einstein was a freemason. So was Charles Darwin.
It is all about keeping free energy from society.
The earth is flat and stationary and is basically one big magnet. There is no such thing as gravity. Only different mediums of density through which objects move.
Nor is there climate change. Just an excuse to take control over all resources on the planet by the cabal.

WHAT ON EARTH HAPPENED PARTS 1-6
https://www.bitchute.com/video/JBoVxbV6616N/?fbclid=IwAR26N_lhiBYFkmg3dG95SbU0QCp_ReILzhSmShGMjKPo72diwW27oJ7cIU4

Perpetual motion violates the law of conservation of energy, because it doesn't conserve energy; why the hell would perpetual motion be the definition of it?

Perpetual (continuing forever) - Nothing destroyed...

Laws of Conservation (continuing forever) - Nothing destroyed...

If nothing is destroyed then the energy is conserved and it continues forever.

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pieces

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Re: Perpetual motion/"free energy"
« Reply #264 on: June 22, 2021, 08:14:48 AM »
Physically impossible. Well unless you count fundamental forces and gravitation. Lolinfinite energy!!

There is no such thing gravity.

WHAT ON EARTH HAPPENED PARTS 1-6
https://www.bitchute.com/video/JBoVxbV6616N/?fbclid=IwAR26N_lhiBYFkmg3dG95SbU0QCp_ReILzhSmShGMjKPo72diwW27oJ7cIU4

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pieces

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Re: Perpetual motion/"free energy"
« Reply #265 on: June 22, 2021, 08:22:24 AM »
They can't work, they all lose energy due to friction, and have horribly inefficient ways to reuse gravitational potential energy or magnetic energy.

You are thinking inside the system. Think outside of the system. Just like people are made to believe that chemo is the only way to cure cancer etc.
If you believe the earth is flat, surely you can grasp the fact that we have lied to about so many more things

NotSoSkeptical

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Re: Perpetual motion/"free energy"
« Reply #266 on: June 22, 2021, 10:15:25 AM »
They can't work, they all lose energy due to friction, and have horribly inefficient ways to reuse gravitational potential energy or magnetic energy.

You are thinking inside the system. Think outside of the system. Just like people are made to believe that chemo is the only way to cure cancer etc.
If you believe the earth is flat, surely you can grasp the fact that we have lied to about so many more things

First, way to revive a dead topic.

Second, Energy Loss is a reality.

For a perpetual motion/"free energy" machine to work, it would have to generate energy that is greater than or equal to the energy it needs to operate.  That's not possible because energy loss is a reality.

Nothing has 100% energy efficiency.

Rabinoz RIP

boydster

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Re: Perpetual motion/"free energy"
« Reply #267 on: June 22, 2021, 11:37:52 AM »
They can't work, they all lose energy due to friction, and have horribly inefficient ways to reuse gravitational potential energy or magnetic energy.

You are thinking inside the system. Think outside of the system. Just like people are made to believe that chemo is the only way to cure cancer etc.
If you believe the earth is flat, surely you can grasp the fact that we have lied to about so many more things

First, way to revive a dead topic.

Second, Energy Loss is a reality.

For a perpetual motion/"free energy" machine to work, it would have to generate energy that is greater than or equal to the energy it needs to operate.  That's not possible because energy loss is a reality.

Nothing has 100% energy efficiency.
I'd add to this to say not only does it need to generate energy that is greater than or equal to the energy it needs to operate, but it actually needs to generate greater than 100% efficiency to be useful. Your point still stands, because a machine that generates energy equal to what it needs to operate could create a perpetual motion machine - I don't want to take away from that. But to provide any additional free energy (or in other words, to do anything actually useful), it needs to generate more than it needs to continue operating because it's the extra energy that could be used to do something useful while the machine keeps self-sustaining.

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Re: Perpetual motion/"free energy"
« Reply #268 on: June 22, 2021, 12:57:41 PM »
Agreed. It would be like creating a car engine that can 100% run without any loss, but no more. (A perpetual motion engine) Then it would be just an idle engine running. In order to make it useful to propel the car forward or backward, it would need to have greater than 100% efficiency.

We haven't found anything to be 100% efficient, let alone anything above that that could produce useful energy.
We've never really been a single entity.  We're more like a collection of rabid honey badgers stuffed into a 3 piece suit.  It occasionally bears the semblance of a man

boydster

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Re: Perpetual motion/"free energy"
« Reply #269 on: June 22, 2021, 01:07:26 PM »
It would be like creating a car engine that can 100% run without any loss, but no more. (A perpetual motion engine) Then it would be just an idle engine running.
Exactly! It's still a really cool discovery/invention that pretty much spits in the face of the laws of thermodynamics, but it is more like an art piece than a machine until it hits >100% efficiency. At 100% efficiency, I'm interested in reading the article. At 100.0000000000001%, I'm investing in the future of the company that demonstrated this is actually possible and not a scam.