This means I can convert mechanical work into electricity with insane efficiency. NO, not more than 100% efficiency, so you free energy nutjobs can sit back down now.
*** Achieving 100% efficiency is the limit set for those that do not fully understand physics, those that cannot think outside the limits that their school education has set.
Roger
In an open system, you’re right. In a closed system, you’re dead wrong.
He’s been at it for more than twenty years. Not surprisingly, he’s conned a lot of people out of their money and has left them with nothing to show for it.
Maxwell's equations describe both electric ('electrostatic') and magnetic fields, both constant and varying, and how they interact, under all conditions. They implied the existence of electromagnetic radiation, ie radio waves, which was an unexpected prediction of the equations, subsequently verified of course. It was later proved that light is also a form of electromagnetic radiation. So they are really fundamental equations of physics, covering all aspects of electric and magnetic fields, apart from quantum effects.
The first law of thermodynamics basically says that energy is conserved under all conditions, ie, it can neither be created or destroyed. It is modified by Einstein's equations to be applied to the mass-energy total, allowing for mass to be equivalent to energy. This allows for the energy effectively stored as mass.
The closed system thing applies to the second law, about entropy, which is a separate issue.
The energy in an electric field at close to the breakdown strength in air, about 3 megavolt per metre, is about 40 joules per cubic meter, ie enough to output 40 watts for one second. To imagine what this means, think of two metal plates one meter square, one meter apart. If one is connected to ground, and the other is connected to a high-voltage power supply, it would take an average 40 watts from the supply to bring the voltage on the charged plate up to 3 million volts in one second, ignoring resistive losses. At this voltage in air, we would be on the verge of a flash-over discharge between the plates.
You could convert that energy back to physical work by disconnecting them from the power supply and allowing the electrostatic attraction between the plates to pull them together.
An electric field is entirely different to a gravitational field. Gravitation acts on everything with mass, electric fields only act on charged particles. The mass equivalent of the energy in a field of 3MV/m is 4.4 * 10^-16 kilogram per cubic meter, ie about four ten-thousandths of a trillionth of the density of air, so I think we can ignore its gravitational effects here.
The gravitational field keeping the Moon in orbit is doing no work at all, any more than a rope holding a suspended weight at a constant height above the ground, or a compressed spring, are doing any more than storing energy, not consuming energy. If the rope breaks, or the spring is released, stored energy will be converted to the equivalent amount of kinetic energy of motion.
If you swing a weight on the end of a string in a circle, that is even closer to what is happening with the Moon, where the string is doing what the force of gravity is doing. The only work required there is to keep the weight in motion against air resistance and other frictional losses.
The Moon stays at a fairly constant average distance from the Earth, so the force keeping it in orbit is doing no 'work', ie no energy is required (work = force * distance moved (in the direction of the force).
The Moon in fact is slowly getting further from the Earth due to the pull of the tidal bulge of the oceans, which in turn is slowing down the rotational speed of the Earth, ie the day is slowly getting longer.
This message has been edited. Last edited by: bobfromoz,
Originally posted by heresjonny: Electric fields will also act on molecules that are dipoled, like water.
Yes, that's because the assymmetries in those molecules cause the distribution of the charged particles (negatively charged electrons and positively charged nucleuii of the atoms mean that the forces exerted on them by the electric field don't balance out.
Even non-polar molecules, and even individual atoms will become weakly polarised in an electric field because the electrons and protons experience opposite forces from the field.
IOW these effects are still ultimately due to the electric field acting on electrons and protons.
Maxwell's equations describe both electric ('electrostatic') and magnetic fields, both constant and varying, and how they interact, under all conditions
*** Can you point me to the Maxwell's equations that deal with electrostatic fields and how they do/perform work?
A mental experiment: I have a stationary charged object and I hold a neutral one in my hand. The magnitude of the charge on the object is such that 1 Joule of work is done by my hand in moving the object through 0.2 meters say against the pull of the charge on the neutral object that I am holding.
It will therefore mean that I will do 1 Joule of work pulling against the pull of the charge in moving the neutral object 0.2 m away from it and 1 Joule of work in restraint against its pull in moving the object 0.2 m towards it.
If I do say 5 such 'to and fro' movements I will have done 5 x 2 = 10 Joules of work.
Question:
(i) How much work will the charged object have done in this process, and
(ii)Assuming no charge leakes from the charged object, what will be its charge at the end of the process: Less than at the beginning; same as at the beginning; greater than at the beginning?
It is modified by Einstein's equations to be applied to the mass-energy total, allowing for mass to be equivalent to energy. This allows for the energy effectively stored as mass.
I will not say much on this because there is too much in your post that I think needs sorting out to dwell on this issue. I will only say that in my view this entire energy mass equivalence concept is very flawed in that mass is actually being equated to kinetic energy in the analysis.
The energy capacity of a single particle: electron, proton, nucleus etc is infinity. But we can discuss this at another time after we have trashed out the other more urgent matters of your post like the question above and comments in my posts below.
To imagine what this means, think of two metal plates one meter square, one meter apart. If one is connected to ground, and the other is connected to a high-voltage power supply, it would take an average 40 watts from the supply to bring the voltage on the charged plate up to 3 million volts in one second, ignoring resistive losses.
*** This doesn't sound like it makes sense to me. I = P/V and Q = It
=> It takes only (40/ 1,000,000) x 1
ie 4 x 10**-5 Ampere seconds [Coulombs if that is the right unit] to charge a 1m x 1m metal plate of unspecified thickness?
You need to clarify this or was it just an illustrative example you were giving?
You could convert that energy back to physical work by disconnecting them from the power supply and allowing the electrostatic attraction between the plates to pull them together.
*** Assuming the plates are 10 mm thick and coefficient of friction between plate and support is 0.2, we get
Work Done in moving plate thru 1 meter = 0.2 x Weight x 1 [ie Friction Force x distance]
=> W.D = 0.2 x [1 x 1 x 0.01 x 9000 x 9.81] x 1
(the density of the metal is taken as 9000 kg/m**3) = 180 Joules
The electrostatic energy required to charge the plate was 40 Joules according to your information [ie 40 W x 1 sec]
If the plates are pulled together just once the so called efficiency will therefore be (180/ 40) x 100
ie Efficiency = 450 %
And the charge on the plate will not be consumed by doing this work so that unless so called corona or other discharge takes place the plate may do this work over and over many times which is effectively an infinite energy device.
An electric field is entirely different to a gravitational field. Gravitation acts on everything with mass, electric fields only act on charged particles.
*** Name one thing that gravity can act on that electrostatic fields cannot act on.
quote:
The mass equivalent of the energy in a field of 3MV/m is 4.4 * 10^-16 kilogram per cubic meter, ie about four ten-thousandths of a trillionth of the density of air, so I think we can ignore its gravitational effects here.
*** OK and what is the mass equivalent of 1 cubic meter of the earth's gravity field?
I hope you understand that what you have just attempted to give me is the mass or mass equivalence of a meter cubed of an electrostatic field and although I cannot verify the figure you gave me I will expect that that of the earth's gravity field to be similar.
The gravitational field keeping the Moon in orbit is doing no work at all...If you swing a weight on the end of a string in a circle, that is even closer to what is happening with the Moon, where the string is doing what the force of gravity is doing. The only work required there is to keep the weight in motion against air resistance and other frictional losses.
*** You sound like a fairly intelligent person.
But may I draw on some of your critical thinking ability and knowledge of math, physics and dynamics of moving objects.
Can you think it through to understand that for an object moving in a straight line with constant velocity to become an object orbiting the earth say that acceleration, accompanied by motion towards the earth equal probably to four times the radius of orbit for every cycle, is required?
quote:
The Moon stays at a fairly constant average distance from the Earth, so the force keeping it in orbit is doing no 'work', ie no energy is required (work = force * distance moved (in the direction of the force).
*** The moon has mass, centripetal acceleeration towards the earth, and travels a distance of roughly 4 x radius of orbit towards the earth for every orbit it makes around the earth. [The accompanying movement of the moon is perpendicular to the earth. These movements in two orthogonal directions allows it to describe a circle around the earth]
The work done per orbit is therefore M x a x (4 x R) where M = mass of moon, a = acceleration and R = radius of orbit.
But if there was no gravitational force involved, and there was a giant chain between the Earth and Moon keeping the Moon from flying off in a straight line, would you say that chain was somehow channelling all that energy you describe?
Does the engine in your car have to work a lot harder when you steer it around a curve? Apart from the extra energy loss in the tyres when cornering, of course.
I can send a steel ball rolling around the wall of a metal bowl in a circle and it will keep rolling for quite a while before friction gradually slows it down. Where is all the energy your argument would require to accelerate it toward the center of the bowl coming from? The smoother the bowl the longer it keeps going - sure looks like the only energy drain is the small amount lost to friction.
The point is that at every point in its orbit, the force of gravity acting on the Moon is perpendicular to its velocity, therefore it does no 'work'. There is in principle no energy required to change the direction of motion of an object, only to increase the magnitude of its velocity. The kinetic energy of motion of the moon does not change, therefore no energy source is required
The outer parts of every rotating object are also being restrained from flying off in a straight line by the inter-atomic forces holding it together - is there a continuous drain of energy from somewhere to hold it together?
The outer parts of every rotating object are also being restrained from flying off in a straight line by the inter-atomic forces holding it together - is there a continuous drain of energy from somewhere to hold it together?
There is one that keeps atoms from collapsing in on themselves, or just falling apart period.
But if there was no gravitational force involved, and there was a giant chain between the Earth and Moon keeping the Moon from flying off in a straight line, would you say that chain was somehow channelling all that energy you describe?
*** I am not sure what you mean by "channelling all that energy" but if you are asking whether the chain is doing work as defined by mechanics then I guess maybe not. Mechanical work is said to be done when the point of application of force moves thru a distance but this does not happen for the chain.
That case you highlighted is similar to towing a vehicle the tow chain does no mechanical work but the towing vehicle does work.
In my concept of things the chain does work just as your vehicle does work if you run the engine while it in gear on a steep slope to keep it from rolling down.
But we need not consider these cases since the orbit of the moon is better simulated by a number of chains spaced at angles of 10 degrees say. If the moon starts off on the y-axis at 1000 Km [arbitrary figure] from the earth and moving in the x-direction then by the time it reaches the second chain it will be [(1000/Cos.10) - 1000] Km from the earth. Therefore in order to keep it at a radius of 1000 Km from the earth the second chain must pull the moon inwards a distance of [(1000/Cos.10) - 1000] Km and so the chain does mechanical workwork continuously.
quote:
Does the engine in your car have to work a lot harder when you steer it around a curve? Apart from the extra energy loss in the tyres when cornering, of course.
*** Yes it works harder: the 'steeper' the curve the harder it must work. The energy loss in the tyres on a road surface that is not 'banked' represents that work and so your tyres wear faster and there is 'scrubbing' of the road surface that causes the road surface also to wear faster.
You cannot corner on a flat surface without scrubbing and if the surface is banked to prevent scrubbing then the work is done against gravity in a way similar to that described below for your steel ball in a metal bowl.
quote:
I can send a steel ball rolling around the wall of a metal bowl in a circle and it will keep rolling for quite a while before friction gradually slows it down. Where is all the energy your argument would require to accelerate it toward the center of the bowl coming from? The smoother the bowl the longer it keeps going - sure looks like the only energy drain is the small amount lost to friction.
*** All tenergy is coming from the elastic strength of the bowl material: each time the ball passes a point the metal deflects ie stretches and returns to its unstrained shape after the ball passes.
Mechanical work is therefore done continuously by the walls of the bowl on the steel ball even though this work does not cause the ball to stop orbiting: it is friction that slows it down.
In this system, like the earth/ moon system, the forces doing the work brings the object to rest along one axis, accelerates it back up to its original speed but in the opposite direction, brings it to rest in that direction, then reaccelerates it to its original velocity.
Simultaneously work is being done in a similar manner along the orthogonal axis.
quote:
The point is that at every point in its orbit, the force of gravity acting on the Moon is perpendicular to its velocity, therefore it does no 'work'.
*** Perpendicular to its instantaneous velocity but not to its original velocity. As explained above the moon is brought to rest along its original direction of motion while it is accelerated to a velocity of the same magnitude in a direction perpendicular to its original direction of motion.
The instantaneous velocity is a combination of these two velocities.
quote:
There is in principle no energy required to change the direction of motion of an object, only to increase the magnitude of its velocity. The kinetic energy of motion of the moon does not change, therefore no energy source is required
*** Kinetic energy is a vector therefore changing direction means changing Ke and hence doing work.
Roger
The outer parts of every rotating object are also being restrained from flying off in a straight line by the inter-atomic forces holding it together - is there a continuous drain of energy from somewhere to hold it together?[/quote]
The outer parts of every rotating object are also being restrained from flying off in a straight line by the inter-atomic forces holding it together - is there a continuous drain of energy from somewhere to hold it together?
[/quote]
*** There is a continuous DEMAND on energy from the inter atomic forces but like I said this energy is infinite and therefore is not depleted.
