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Author Topic: Energy conservation?  (Read 9689 times)
Group: Guest
I have a problem solving this simple set up.  It keeps giving me the wrong answer and I cannot find where did I go wrong.  

We have a pulley and two masses on each side.  m1 is twice m2  and both at the same distance h above ground.  They both were initially at rest.  The energy in this state is:

m(1)gh + m(2)gh = 2m(2)gh + m(2)gh = 3m(2)gh

Then we let it go.  The heavy mass go down and the less heavy mass go up.  When m1 reach bottom it has 0 potential energy and possesses kinetic energy.  m2 possesses kinetic energy and now has twice potential energy.  

1/2 m(1)v^2 + 1/2 m(2)v^2 + m(2)g2h = m(2)v^2 + 1/2 m(2)v^2 + 2m(2)gh

Now compare the total initial energy and final energy.  They should be equal since energy is conserved:

3m(2)gh = m(2)v^2 + 1/2 m(2)v^2 + 2m(2)gh

We can go ahead and eliminate m2 to make it less confusing:

3gh = v^2 + 1/2 v^2 + 2gh = 3/2 v^2 + 2gh

gh = 3/2 v^2


How can this be?  mgh = 1/2 mv^2 ===> gh = 1/2v^2 !!
   
Group: Guest

mgh=1/2mv^2 applies only if v is the speed of the center of gravity of your system.

Consider the fall of the center of gravity of your system. You have a closed system of mass 3m. The kinetic energy is 1/2 3m v'^2 but v' is the speed of the center of gravity, which is slower than the speed v of the masses because m goes upward at v and 2m goes downward at v.
I think the mistake is around this point. I will see it again tomorrow (it's now late, here).

   
Group: Guest


Yes, you are correct.  The speed of this system is slower than the actual mgh = 1/2mv^2.  Energy is conserved.  Thanks for taking a look at it.



   
Hero Member
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Posts: 805
I found a very good website that has a lot of historical information about Bessler and his wheel and all the troubles he went through.  It's fascinating reading!   This link is just one page, once you click on it, on the left are multiple links to different "chapters"     I'm so absorbed into this right now, my belief gage is pegged to the max ...

http://orffyre.tripod.com/bessler/id22.html


   
Group: Guest
Bessler represents pure experiment with little math.
Newton represents pure math with little experiment.
One cannot go without the other.  


I've come up with a picture below.  One look at it in any static position and see equilibrium (ignore the rod weight).  However, when it is rotating at certain speed the center of gravity hop up periodically.  



EDIT:  Never mind, the length factor would not make it hop.

   
Sr. Member
****

Posts: 462
When mass is given from outside energy is not conserved ? We all know from school the http://en.wikipedia.org/wiki/Inclined_plane, right ?

What about the case Tesla described in his autobiography ? SNOW BALL EFFECT !

What is bothering me a lot is the question : is that how every radio receiver operate ? what if I can build WHOLE range energy sucker antenna receiver !!!!????

What if I can make self-regulating resonant circuit which only need to acomodate SMALL losses from external source to accumulate energy in very fast rate like Mhz or Ghz and then using that energy in Hz rate with GAIN !!!?????


 :o
   
Hero Member
*****

Posts: 805
I've drawn a wheel, that I'm sure others have tested already, and probably found that it does not work, but that's ok, it only stimulates thought.  


Here's some aspects of the Beesler wheel that were observed:

1)  eight weights made a banging noise on the descending side
2)  springs are used inside, not to store energy, but to enable a flexable structure, or arms, levers, etc..
3)  the machine achives a set RPM when unloaded


number 3 is very interesting.  If the machine is realy unbalanced, it should rotate faster and faster and accelerate to high speeds, but something prevents it from accelerating to infinity, and that tells me it depends on falling rate of objects inside,   if they fall too slow, it has torque one way, if they fall too fast they have torque the other way, so somewhere is the optimum velocity for it.



Since this topic is on energy, here's another idea I have:

 I believe energy concervation needs to be applied ONLY PER WEIGHT, NOT PER ASSEMBLY.  

I've given this thought a lot of attention lately, and I believe if there is a relative angular "slip" between the weights and the wheel, than we can get more energy out of the assembly than in!  Each weight will loose a portion of energy and give it to the wheel, and as a result will "lag" behind the wheel and slip constantly as the wheel revolves.

For example, as the wheel turns,  each weight moves and trades positions around the wheel, and eventually completes one revolution relative to the wheel as the wheel perhaps has rotated 8 times relative to ground.  This slip is regardless of the  structure inside the wheel, or the path of the weights take inside.   The problem is to envision the right mechanism.


EM01 wheel

if those levers with the weights are spring loaded, they will swing out on the left and overbalance the wheel on that side, but the spring will eventually pull them in at the bottom.   At the bottom, if the weight starts to lift up under spring tension, its effective pivot point moves back, since the weight is decellerated as it moves to the red inner circle.

[edit:]  if the wheel has too high a velocity, the weight at the bottom will be held down longer by the centrifugal forces, so the strength of the springs vs the velocity, seem to dictate what steady state RPM it should have.

EM
   
Group: Guest
...
what if I can build WHOLE range energy sucker antenna receiver !!!!????
...

An antenna is really a kind of "energy sucker". In a receiving antenna, there is a HF current, which feeds the receiver input. This current generates also an EM field around the antenna that adds to the ambient field from the emitter. So there are constructive and destructive interferences in the space around the antenna.
If the antenna is well designed, the constructive interferences build along the antenna conductors if it is made of linear elements, or inside the loop if it is a magnetic loop antenna (for instance a ferrite antenna). So the energy in the space around the antenna is redistributed in order to be maximised very near the antenna. Nevertheless this redistribution doesn't provide extra-energy. Interferences are only a change of the topology of energy in space.
This is well explained in this patent where the effect that I described is enhanced: the EM field generated by the receiving antenna is amplified by a negative resistance to force the EM energy to pass through the ferrite by opposing and lowering the energy outside.
http://www.google.com/patents?id=BlogAAAAEBAJ&printsec=abstract&zoom=4

   
Group: Guest
An antenna is really a kind of "energy sucker". In a receiving antenna, there is a HF current, which feeds the receiver input. This current generates also an EM field around the antenna that adds to the ambient field from the emitter. So there are constructive and destructive interferences in the space around the antenna.
If the antenna is well designed, the constructive interferences build along the antenna conductors if it is made of linear elements, or inside the loop if it is a magnetic loop antenna (for instance a ferrite antenna). So the energy in the space around the antenna is redistributed in order to be maximised very near the antenna. Nevertheless this redistribution doesn't provide extra-energy. Interferences are only a change of the topology of energy in space.
This is well explained in this patent where the effect that I described is enhanced: the EM field generated by the receiving antenna is amplified by a negative resistance to force the EM energy to pass through the ferrite by opposing and lowering the energy outside.
http://www.google.com/patents?id=BlogAAAAEBAJ&printsec=abstract&zoom=4



So what is the input to the transmitter and how do we calculate it. 


   
Group: Guest
So what is the input to the transmitter and how do we calculate it.  

I don't understand the question in the context. The "input" to the transmitter is the energy from the power supply, so it is the HF output/efficiency.
The output is P=U*I, U and I refering to the high frequency signal, supposed to feed a tuned antenna (therefore having an impedance that is only resistive).
Practical and current values of efficiency are around 60-70%.

It is not possible to calculate the power of a transmitter from a single signal of a receiving antenna without knowing the radiation pattern, the receiving antenna lobes, the absorbers that are possibly present everywhere in the beam...

« Last Edit: 2012-07-19, 08:29:34 by exnihiloest »
   
Group: Guest
I don't understand the question in the context. The "input" to the transmitter is the energy from the power supply, so it is the HF output*efficiency.
The output is P=U*I, U and I refering to the high frequency signal, supposed to feed a tuned antenna (therefore having an impedance that is only resistive).
Practical and current values of efficiency are around 60-70%.

It is not possible to calculate the power of a transmitter from a single signal of a receiving antenna without knowing the radiation pattern, the receiving antenna lobes, the absorbers that are possibly present everywhere in the beam...



Hi Exn,

The input to the transmitter is the energy from the power supply, this I agree.   P=U(instantaneous) x I(instantaneous) is the input to the transmitter if we measure it from the supply and it should match with the power from supply.  This is what we been doing with the Joule thief for a long time now.  Do you mean HF output * efficiency = input to the receiver?

I agree that it's hard to calculate power with receiving antenna so I hope we leave this as a grey area. 
   
Group: Guest
...
 Do you mean HF output / efficiency = input to the receiver?
...

I don't mean HF output / efficiency = input to the receiver (I have replaced "*" by "/" in my previous reply: typographical error due to the fact that generally, it is the useful power that we calculate from the power supply).

HF output / efficiency = power supply provided to the transmitter.

This HF output would be also the input to the receiver only if the receiver was 100% coupled to the transmitter, what is generally not the case.


   
Group: Guest
I don't mean HF output / efficiency = input to the receiver (I have replaced "*" by "/" in my previous reply: typographical error due to the fact that generally, it is the useful power that we calculate from the power supply).

HF output / efficiency = power supply provided to the transmitter.

This HF output would be also the input to the receiver only if the receiver was 100% coupled to the transmitter, what is generally not the case.




I think I see it now.  The HF output is constant regardless if we have a receiver or not.  The efficiency is ohmic loss.  Ohmic loss + HF output = apparent power*power factor.  If we do not have a receiver, HF output is loss to the universe.  Is this correct? 

   
Group: Guest
I think I see it now.  The HF output is constant regardless if we have a receiver or not.  The efficiency is ohmic loss.  Ohmic loss + HF output = apparent power*power factor.  If we do not have a receiver, HF output is loss to the universe.  Is this correct? 

Yes it is correct, in the general case when the receiver is far from the transmitter. The power is provided to the impedance of free space (http://en.wikipedia.org/wiki/Impedance_of_free_space) and the receiver extracts only a very small part of this energy.

But if the receiver antenna is very near the tansmitter antena, the receiver can receive almost all the transmitted power which is not even radiated in free space (it is a coupling as between  two coils).
   
Group: Guest
Yes it is correct, in the general case when the receiver is far from the transmitter. The power is provided to the impedance of free space (http://en.wikipedia.org/wiki/Impedance_of_free_space) and the receiver extracts only a very small part of this energy.

But if the receiver antenna is very near the tansmitter antena, the receiver can receive almost all the transmitted power which is not even radiated in free space (it is a coupling as between  two coils).

I appreciate your answer, Exn.

I see that this portion of radiated energy are like lights.  Lights radiate from a source and you can capture it at a distance (receiver).  Of course if you're near to the source, you'll be capturing more than being further away.  Having said this, you can also reflect lights/EM waves.  I think transmitter do have a portion that reflected back and seen as reactive power. 

Now in my mind, if we were to absorb EM waves to a resistance, it would be destroyed and becomes heat.  What if we were to reflected back.  Would it be reflected and in turn, electrified the reflector.  Absorption of EM waves gives momentum.  Reflection of EM waves gives twice the momentum.  So speaks my mind. 

   
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