# Free Energy | searching for free energy and discussing free energy

## Gravity powered devices => Gravity powered devices => Topic started by: johnny874 on June 16, 2012, 03:38:23 PM

Title: W = M*D/T
Post by: johnny874 on June 16, 2012, 03:38:23 PM
In considering the current explanation of work and it's relationship to energy, it seems that time is not a consideration.
While they give examples of how to calculate how much work is performed under specific conditions, it seems that if
a mass (1kg) is moved 9.8M in one second, then it would be the equivalent of 9.8N^2 or 96.04 Newtons of work. And if
the same 1kg of mass were moved in .5 (1/2) seconds, then it would be 1kg*9.8M/.5 = 192.08 Newtons of force.
It seems considering something like this helps to understand the energy requirements of a system or how much energy
could be expected to be derived from a system. This is something I have thought of while working on Bessler's Wheel.
The basic idea is if a weight drops a distance we will call x and I want to move water a distance y and in a certain amount
of time, t, then how much energy does x require to operate the system ?
In a sense, this is bringing science into engineering because mass and velocity are taken into consideration.

John
http://en.wikipedia.org/wiki/Work_(physics (http://en.wikipedia.org/wiki/Work_(physics))

edited to correct spelling and add that a 1kg weight has 9.8N(Newtons) of force, gravity's effect on it (why it has mass).
So if you consider that xy/t = work or it's equivalent energy in Newtons. Of course, md/t would be what everyone would understand.

Title: Re: W = M*D/T
Post by: johnny874 on June 16, 2012, 05:40:32 PM
@All,
In a sense, I have combined Newton's work with gravity and force and found a way that they
might work better with engineering. This would be by considering force in Newtons and relating
work to 1 second of gravity. It would be a basic standard that could possibly be more easily
applied to engineering than needing to convert Joules to understand how much energy a system
has, needs or can have extracted from it.

Jim
Title: Re: W = M*D/T
Post by: johnny874 on June 16, 2012, 06:52:12 PM
Last post for the day on this, okay ?
What I now realize is that 1N/w or Newton of work is 1kg*9.8m/1s = 96.04 Newtons of Force.
This is a basic standard which could be applied to many things. for example, 1N of air is about 3.6 psi.
This is because 1kg/9.8 = .102. When converted to SAE it is the equivalent of 3.6 psi. Neat, huh ?
102 g = 3.59794 oz
http://www.metric-conversions.org/weight/ounces-conversion.htm (http://www.metric-conversions.org/weight/ounces-conversion.htm)

A simple way to discuss any idea using common values. What this allows for is being able to understand when
the amount of work that can be expected from a system. And if it is not realized, then it could be asked if it is
entropy, 2nd Law of Thermodynamics, resistance or proper flow has not been established. It does allow for
establishing a guideline based on accepted principles in science by using a standard valuation for force or energy
in an engineered system.

Bon Appetite
Jim

Something to consider, if 36 psi (liquid or gas) is acting on a piston with a surface area of 10^2", then it has a total value of 360.
If this value is divided by 3.6, then the answer is 100 newtons. This force would be equal to a 1kg weight moving
at 9.8m/s. Okay, there would be 3.96newtons left over  ;)

Title: Re: W = M*D/T
Post by: MileHigh on June 16, 2012, 07:18:53 PM

http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html#mechcon
Title: Re: W = M*D/T
Post by: johnny874 on June 16, 2012, 07:38:32 PM

http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html#mechcon (http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html#mechcon)

Hi Milehigh,
I think for this section of the forum, that might be a bit more than what people want.
Most of the values we work with are for bodies (mass) moving slower than 9.8m/s.
The link seems to be for those people who can relate to 300,000 kph (speed of light/magnetic behavior)
as a basis for their observations.

Jim
Title: Re: W = M*D/T
Post by: johnny874 on June 16, 2012, 07:55:37 PM
@All,
I made a mistake converting grams.
3.6 psi is actually ounces.
There about 4.44 newtons per 1psi.

Jim
Title: Re: W = M*D/T
Post by: johnny874 on June 17, 2012, 01:06:49 AM
@Milehigh,
something to think about, electrons and photons behave? little mass but r quite fast.
it seems slower moving bodies have more mass, it may have something to do with plasma physics,

jim
Title: Re: W = M*D/T
Post by: johnny874 on June 17, 2012, 02:10:25 AM
aw c,mon people,
you haven,t studied physics ?
Title: Re: W = M*D/T
Post by: MileHigh on June 17, 2012, 03:38:55 AM
Jim:

I gave you a link to the Hyperphysics web site because I could see that your units were incorrect.  It's not about relativistic speed effects and all that stuff.

For example, the unit of work is a Newton of force times one meter of displacement, a Newton-meter.  (They want you to say "Joule" because Newton-meter is used for torque.)

So a Newton is equal to a kg*m/s^2.  So one Joule is a kg*m^2/s^2.

MileHigh
Title: Re: W = M*D/T
Post by: johnny874 on June 17, 2012, 04:36:17 AM
Hi Milehigh,
I am glad you posted. I wish more people were willing to discuss things.
With what I posted, it was more for mechanical engineering.
As for Joules, it is like you said, mass is squared. I think this added step increases the level of difficulty.
Stefan has asked that things be put in terms anyone can understand.
In a way, he is asking someone like you to become a teacher.

Jim
Title: Re: W = M*D/T
Post by: johnny874 on June 17, 2012, 05:49:06 AM
@All,
Milehigh is someome you can probably relate to. He is smart. The only thing I did was to consider that it takes time to perform work.
Title: Re: W = M*D/T
Post by: johnny874 on June 19, 2012, 03:24:54 AM
Milehigh,
one reason why I started this thread is because gravity governs what we do.
With joules, I think it is better for calculating the mass of an element going through a mass spectrometer or the charge of an electron dependent on Heisenbergs Uncertainty Principle.
I have opened a few books over the years.
In here, I think Stefan was right when he asked that things be put in laymans terms.
With the math, I was trying to simplify it as much as porrible.
I hop ynu understand.

Jim