> If you drop a object from a 100 meter height,

This message came from kloskie@.... I am guessing from the style

> given there is no resistance.

>

> how long will it take to hit

> the ground?

> answer....t= 10.2040816327 seconds

>

> How fast will it be going when it hits?

> answer....v = 100.0000000005 m/sec^2

>

and the tone that this message must also be from Paul Baughn, although I was

confused for a bit. Paul, am I right? Are you using two or three different

e-mail accounts?

I haven't been asking these questions in order to embarrass you.

I have been trying to demonstrate where your theoretical knowledge is shaky.

I have repeatedly stated that your fluid flow calculations are in error; I

have been trying to go back to first principles to demonstrate to you *why*

they are in error.

On this level, the conversation is not about me or you. You called me a

"dimwit". That's an opinion, and you are welcome to it. Your opinion of my

intelligence doesn't make one bit of difference to me, nor does it make a

single shard of difference to that dropped object.

You calculations about that 100 meter drop are, in point of incontrovertable

fact, wrong.

When an object at rest is suddenly subjected to a constant acceleration a,

its velocity after time T will be

V = a T

The distance it will travel is

d = 1/2 a T^2

This is elementary stuff; with constant acceleration the velocity is the

time integral of acceleration, and the distance is the time integral of

velocity.

Paul -- for the third time: Look in the first couple of chapters of your

physics text. It will be there. Guaranteed. Failing that, ten seconds of

searching the web gave me this: http://physics.webplasma.com/physics04.html

The same equations I've given are there; if you look at them, keep in mind

that for simplicity I've intentionally been ignoring Vo and Xo (the initial

velocity and position) by setting them to zero.

Applying them here, where a falling object is subjected to a constant

acceleration of 9.8 m/s^2 gives a falling time of 4.52 seconds, and a final

velocity of 44.27 meters per second.

It appears that you attempted to apply those equations, and made the

following error:

Instead of

d = 1/2 a T^2 transforming to T = sqrt(2d/a)

You somehow have decided that T = d/a.

This is just plain wrong. d is in meters; a is m/s^2. Dividing d by a

gives you dimensions of sec^2. You can't give a time in seconds squared; it

has to come out to seconds. Any calculation which does so *has* to be

wrong.

I have been at pains to explain what the principles and equations are when I

calculate a numerical result. You have not returned the courtesy. I ask

again: Please justify, with theoretical analysis and/or references that I

can check, why you think an object dropped from 100 meters in a 9.8 m/s^2

gravitational field, will be moving at 100 m/s^2 [sic] after 10.2 seconds.

And you simply *must* be open to the possibility that you are wrong.

In fact, we need to thrash out once and for all, why do you think you can

call a velocity "100 m/sec^2" Velocity is *defined* as a change in distance

divided by a change in time;- bob

im having a little problem with your calculation below

((((

The 500 Joules of Kinetic Energy in the moving 1000kg mass will be

converted

entirely into the Potential Energy of the lifted 1 kg mass.

500 kg-m^2/s^2 = 1kg * 9.8 m/s^2 * height.

))))

could you please explain.

I get the following.

500 kg-m^2/s^2 = 1kg * 9.8 m/sec^2 * h

500/1 = 1*9.8*10

500/1 = 9.8*10

500 = 98

500 and 98 are not equal.

--- In free_energy@y..., "gearjammer123us" <paul@w...> wrote:

> you ever go fishing bob.

> sometimes when you use rotten bait your catch will be greater

> than when you use fresh bait.

> it depends on the fish.

>

> I intentionally gave you wrong answers to see if I could get you to

> admit to the correctness or validity of my proposed system.

>

> this mission was successful.

>

> this clip is in reference to bobs post # 2621

> here he gives a problem and then he gives his solution to

> the problem.

> the problem is a mere mass vs mass problem.

> the correct answer will provide the distance that

> mass 1 will pull mass 2.

>

> it is a simple problem.

> note: he is using "frictionless wheels"

> **********************************

> note also:

> water flowing through a smooth wall plastic pipe

> is met with a resistance of only .002 which can be

> expressed as a percentage of the gravity that causes

> resistance in the form of friction.

>

> ie..a

> 1000 kg mass of fluid

> vs a mass equivalent to

> .002 percent of the 1000 kg mass.

> ***************************************

> to continue with bobs exercise.

>

> and the mass has a velocity of 1 m/s

>

> Here's the question: At the exact moment that the rising weight

comes

> to a halt, how high will it be over the ground? How long will it

take

> to get there?

>

> here are bobs answers:

>

> The height of the 1kg mass at the moment the masses stop moving

will

> be 51 meters.

> The time it will take for the 1kg mass to reach that height will be

> 102 seconds.

> *** this is correct ***

> this proves that bob knows physics.

>

> ******************

> so bob is saying here that a 1000kg mass that has a velocity of 1

> meter per second can lift a 1kg mass to a height of 51meters.

> even though the 1000kg mass is only traveling at a velocity

> of 1 meter per second when it starts to pull the 1 kg mass.

> and it will take 102 seconds to lift it 51meters.

>

> ******************

>

> although this exercise can in no way show a comparison to the system

> in which we are discussing.

> mainly because you cannot tie ropes to water.

> and in the system we are discussing the fluid is in constant

> circulation.

> and circulating through different sized pipes.

> and it has two different constant velocities.

>

> here he wants to use an example that describes a

> stationary mass and a mass with velocity.

> and the two masses can only travel at the same velocity.

> once they are both in motion.

> because of the rope that binds there movement to one another.

>

> in my re-structured example that would have provided a better

> description of the fluids in motion I had both masses moving at a

> velocity of 1m/sec.

> were anyone to look at the illustration

> ( that I provide at the bottom of this post )

> it clearly shows that neither of the masses in question are

> stationary.

> and it clearly shows that the two masses in question will

> have two separate velocities.

> and it clearly shows that the two masses will be traveling through

> two different size pipes.

>

> in bobs calculations he first uses Kinetic energy.

> ie..

>

> The Kinetic Energy of the 1000kg mass moving at 1 m/s is

> 0.5 M V^2 = 500 Joules.

> 500 * 1 = 500 Joules.

>

> I would like to add here that.

> if observed while the system is in operation.

> the 1 kg mass moving at a velocity of 10m/sec is as follows.

>

> The Kinetic Energy of the 1kg mass moving at 10 m/s is

> 0.5 M V^2 = 500 Joules.

> .5 * 100 = 50 Joules.

>

> here it is clear that the 1000kg mass has (10) times the energy

> of the 1kg mass

> ie..

>

> 500J vs 50J

>

> we could have stopped at this point.

> and assured ourselves that the 1000kg mass had more energy.

> than the 1kg mass.

>

> however in true skeptic form he then moves on to convert the

Kinetic

> energy to a distance..

> ie..

>

> The 500 Joules of Kinetic Energy in the moving 1000kg mass will be

> converted

> entirely into the Potential Energy of the lifted 1 kg mass.

>

> 500 kg-m^2/s^2 = 1kg * 9.8 m/s^2 * height.

>

> Do the division, and height = 500/9.8 meters = 51.02 meters.

> ...this is correct....

>

> so he has used KE to provide distance.

>

> he then moves on to the second mass

> and here he uses the F=m*a equation.

> his answer here states clearly that the 1kg mass

> will cause a deceleration of 0.0098 m/s^2

> the velocity of the 1000kg mass is 1m/sec

> in a situation where there is no friction.

> as in "frictionless wheels"

> the first second the 1000kg mass will decelerate to 0.9902m/sec^2

> the second second the 1000kg mass will decelerate to 0.9804/sec^2

> so on and so on.

> ....this is correct...

>

> The pulley system will exert a constant 9.8 Newtons

> (1kg * 9.8 m/s^2) of horizontal force on the 1000kg mass.

>

> F = m * a, which means a = F / m = 9.8 N / 1000 kg = 0.0098 m/s^2

>

> .... this is correct.....

>

>

>

> ****************

> Here's the exercise: Attach the 1kg mass to a length of massless

rope.

> (Every physics lab has an infinite supply of massless rope.) Run

the

> rope

> though a frictionless pulley way up high. (The frictionless pulleys

> are in

> the cabinet next to the frictionless rope locker.) Run the rope

> through

> another pulley at ground level, so that you pull horizontally on

the

> rope to

> raise the weight off the floor.

>

> The door opens, and in comes a 1000kg shopping cart, rolling on the

> perfectly level floor on frictionless wheels at 1 m/s. Lasso the

cart

> with

> the free end of the rope.

>

> As the rope tightens, the moving 1000kg mass starts to lift the 1kg

> mass.

> The work that it is doing -- or, if you insist, the resistance that

> it is

> feeling, although I cringe to say it that way -- causes the

shopping

> cart to

> slow down. Eventually the shopping cart will come to a halt.

>

> At that moment, of course, the 1kg weight will start to return

> earthward,

> dragging the shopping cart backwards. But let's ignore that for a

> moment.

>

> Here's the question: At the exact moment that the rising weight

comes

> to a

> halt, how high will it be over the ground? How long will it take to

> get

> there?

>

> I sincerely hope that you don't regard this calculation as

> irrelevant. It

> is absolutely central to understanding your pipe system.

>

> The answers to my two questions are:

>

> The height of the 1kg mass at the moment the masses stop moving

will

> be 51

> meters.

>

> The time it will take for the 1kg mass to reach that height will be

> 102

> seconds.

>

> I am either right or wrong. Which is it? And if I am wrong, what

are

> the

> right answers?

>

> ==============

>

> My calculations:

> VERY IMPORTANT.

> NOTE ( BOBS ) CALCULATIONS.

> NOT MINE.

>

> The Kinetic Energy of the 1000kg mass moving at 1 m/s is 0.5 M V^2

=

> 500

> Joules.

> ..... CORRECT .....

>

> The 500 Joules of Kinetic Energy in the moving 1000kg mass will be

> converted

> entirely into the Potential Energy of the lifted 1 kg mass.

>

> 500 kg-m^2/s^2 = 1kg * 9.8 m/s^2 * height.

>

> Do the division, and height = 500/9.8 meters = 51.02 meters.

>

> .... ALSO CORRECT ....

>

> The pulley system will exert a constant 9.8 Newtons (1kg * 9.8

m/s^2)

> of

> horizontal force on the 1000kg mass.

>

> F = m * a, which means a = F / m = 9.8 N / 1000 kg = 0.0098 m/s^2

>

> ..... ALSO CORRECT ....

>

>

> We want to know the time necessary for that constant acceleration

to

> change the velocity by 1 m/s, so we use the equation

>

> V = a T which means that T = V / A = 1 / 0.0098 = 102.04 seconds.

>

> ..... ALSO CORRECT ......

> at this time I would like for all to note that bob knows his math.

> and he is capable of performing math in many ways.

> he has demonstrated that he can think on many levels when he is

> trying to prove a point.

>

> to continue..

>

> fluid traveling through a pipe is subject only to friction

> and viscosity if the pipe is level in orientation to the earths

> center.

> I had hoped that you would have made a issue out of this.

>

> in the below situation gravity of 9.8m/sec^2 is normally used.

> in order to use the below equation in a pipe you must use a

> "degree" of gravity.

> this degree of gravity is applied to the mass.

> this degree is .002 % of 9.8m/sec^2

> in a smooth wall plastic pipe.

> find it yourself.

>

> or 0.0196m/sec^2

> thus the weight of the mass when used in the equation

> will be the same and the amount of gravity used will be

> a percentage of the gravity constant.

> in a work equation Force is equal to mass * gravity.

>

> F = m*g

> F = 1000kg * 0.0196

> F = 19.6 kg

> thus

> E = 19.6 J

> to use your brain bob is to realize that if the amount

> of friction is reduced then the force required to move a

> object is also reduced.

> thus a 1000kg mass will have a resistance to motion

> in the form of friction of 19.6 N

> in other words the force that pushes back will be 19.6 N

> opposite the direction of the force to move the object.

>

> thus a force of 19.6 N can put the mass in motion.

> thus to move the 1000kg mass 1 meter in 1 second will

> require a amount of energy of 19.6 N * 1m = 19.6 J

>

> heres a energy comparison.

>

> energy required to "slide" the 1000kg mass 1 meter in 1 second

>

> E = F = mg = 1000kg * .0196m/sec^2 = 19.6 N * 1m = 19.6 J

>

> energy required to "lift" the 1kg to a height of 10 meters in 1

second

>

> E = F = mg = 1kg * 9.8m/sec^2 = 9.8 N * 10m = 98 J

>

> there is the perfect energy comparison.

> try it to compare mass movements that have friction.

> which my system has.

> you will see that the answer is correct.

>

> to use the exact same equation to observe the two situations.

> and to use the exact amounts of elements that will effect

> the movement of the two fluids.

> will render a correct comparison.

> and it will render a comparison that can be clearly understood.

> unless you dont want others to be capable of understanding the

> comparison which I believe is your motive.

> as others have mentioned.

>

>

> according to the above comparison the energy required to

> slide the 1000kg mass is less than the energy required to

> lift the 1kg mass

> its there in plain physics equations.

> and it cannot be denied by anyone who can use physics to determine

> the amounts energy of mass in motion.

>

> given that you know how to use physics.

> and that you know how to use physics in many ways.

>

> I stand on proof that supports the following....

>

> rather than do things a simple way you choose to do things in

> some strange way in a effort to try to confuse me.

> the above is your tools.

>

> you do not want to reveal a true energy comparison.

>

> that is your motive.

>

> it is clear that you are not trying to deliver a honest evaluation.

>

> to listen to your applied math and comparisons will deliver a

> comparison that has no visible result.

>

>

> ie....

>

> I have been trying to demonstrate where your theoretical knowledge

is

> shaky.

> I have been trying to go back to first principles to demonstrate to

> you *why* they are in error.

>

> **** even at the expense of using confusion techniques ****

>

>

> And you simply *must* be open to the possibility that you are wrong.

> **** I am always open to that ****

> are you.

>

> also a final point...

> rather than agree to use the equation you and those of your type

> will state that it is incorrect.

> or you will simply QUIT because you know that if you continue

> my system will be proven to be valid.