jim_mich wrote:Rebalancing the weights is no problem. I had that part solved a long time ago. The weights move and the wheel turns 180 ready to start again. The CG rides a circle from top around to bottom, then spikes back up to the top as the weights swing, ready for the next 180. (Or maybe it won't work?)
The Center of Gyration experiments you did a while back using the hammer men parallelograms on single pivots seems to offer the best hope (with limited knowledge of your final design). CF's are then free to influence the 'Pairs of Pairs' to flatten the parallelograms because of the non linear CF development. Trouble is I can't see at this stage how to get it to morph back to its original shape & positioning. Perhaps it uses gravity to assist & counter the CF's near the top of the wheel but it still seems symmetrical to me but then I'm only guessing here about your design. Good luck anyway.
jim_mich wrote:The excess energy seems to come from the CF of two weights moving at different speeds. Didn't Bessler say they gain energy from their swinging?
This to me was a very vague statement & probably deliberately so. If you look at a one-system then it is very hard to see how to break the symmetry's from swinging weights. If you think it might be a two part system, with a prime mover to lift/reposition weights & an OOB system to create the imbalance to turn the wheel, then it seems to be more feasible & understandable the way he wrote it.
Momentum and energy are two way streets.
Momentum can receive energy and increase.
Or it can give up energy and decrease.
When energy moves something then work is done and the energy becomes momentum.
When momentum moves something then work is done and momentum becomes energy.
Work in its simplest sense means to move something.
Energy in its simplest sense means able to move something
Momentum in its simplest sense means energy stored as inertia.
Jim
Thank you very much, that is as clear as I have seen it written and I will ponder this to understand what you are doing.
Whatever it is, I also hope as Fletcher that you gain momentum on your idea each day.
Interesting thread, here. One thing I find amazing, is the two people on the train, and their ability to walk end to end and back on a train going in circles, one foot in front of the other, without falling down. The real-time processing and signaling that the brain effortlessly calculates in order to perform this not so simple (as this thread has revealed) task is incredible. Honda's ASIMO could only dream of such ability and agility. In fact, ASIMO cannot even do that.
Energy and momentum are not the same.
I like to use metric, so I'll explain with that.
Kinetic Energy = mass * velocity^2 / 2
Potential Energy = mass * acceleration * distance
aka mass * gravity * height
Momentum = mass * velocity.
velocity = distance / time
acceleration = distance / time^2 = velocity / time
Force = mass * acceleration
Work = Energy = Force * distance
Power = Energy / time
C Accel = v^2/radius
C Force = mass * C Accel = mass * v^2/radius
These are your basic physics calculation.
Momentum is a different cookie, and can get quite complicated to calculate when you have impact or cohesion of two bodies. Add a coefficient of restitution to the mix and I get headaches.
When we look at CP force, it is just that, a force like gravity.
Let us take our train travelling at 10 m/s around a circle with a radius of 200m.
100kg person at train speed
CP Force = 50 Newton
CP Accel = .5 m/s^2.
Kinetic energy = 5000 joule.
Momentum = 1000 kg*m/s
This same person will move forward at a speed of 5 m/s. Total is 15 m/s.
CP Force= 112.5 N
CP Accel= 1.125 m/s^2
Kinetic NRG= 112500 joule (correction 11250)
Momentum = 1500 kg*m/s
Another 100 kg person moves backward at 5 m/s. Total 5 m/s.
CP Force= 12.5 N
CP Accel= .125 m/s^2
Kinetic NRG= 1250 joule
Momentum = 500 kg*m/s
In a perfectly effecient system, 107500(correction 10750) joule are required to increase velocity by 5 m/s with 3750 joule are required to decrease 5 m/s.
Yet the mometum increases and decreases proportionally with velocity,
while the energy, force, and acceleration increase and decrease exponentially.
This is according to the formulas in my Schaum's Physics book.
To double your velocity you must quadruple the energy.
To double your momentum you need to double your velocity.
The man moving to a lesser velocity loses 3750 joule of kinetic energy.
In order for the man moving forward to gain 3750 joule of energy, resulting in an energy of 8750 joule, he needs to increase his velocity by 3.23 m/s resulting in a final velocity of 13.23 m/s (1323 kg*m/s of momentum).
Remember, these people are walking in the train, they transfer energy through their feet into the train and act against it in both directions.
I hope this helps to better understand what is going on. Cheers
Last edited by spingoogl on Fri Sep 12, 2008 9:16 pm, edited 3 times in total.
I came here to chew some bubble gum, and kick some ass.....and I'm all out of bubble gum.
Quite an entrance you make, picking up on a two year old thread, I like it!
If energy is acquired from acceleration(gravity), then would it stand to reason that an acceleration engine would weigh less during operation?
A reasonable question, one that I am attempting to answer with a physical build. At present I would say that the weight will vary with the cyclic design of the machine. But whatever weight it looses it will gain back in the form of the accelerated driving mass making contact with that of lesser velocity.
You are a machinist, I will be looking forward to see what kind of "swarf" you have to offer this forum :-)
First a quick note that there is an error in your calculations. The forward moving person would have Kinetic NRG = 11250 joule instead of the 112500 Joule that you list. It seems you typed an extra zero.
Later this error affects Joules needed in the perfectly efficient system, which would be 6500 Joule instead of the 107500 that you list.
Other than these two small math errors your presentation is very good.
Spingoogl wrote:The man moving to a lesser velocity loses 3750 joule of kinetic energy.
In order for the man moving forward to gain 3750 joule of energy, resulting in an energy of 8750 joule, he needs to increase his velocity by 3.23 m/s resulting in a final velocity of 13.23 m/s (1323 kg*m/s of momentum).
This is correct.
The whole train on a curve concept concerns relativity. The energy needed to accelerate an object depends on the surface (or mass) against which the pushing is done. If the person is riding on a moving platform (such as the train) and then accelerates relative to the train the force needed is less than if the person must accelerate relative to the ground. If you measure everything using the moving platform as the reference system then the forward moving person and the rearward moving person both gain equally.
But the CP force will not cooperate. It will always be relative to the absolute space outside the wheel. Thus even though the same force is applied to accelerate each of the two people, one forward and one rearward, the resulting CP forces and kinetic energies become unbalanced. Thus the ectropy of the system is increased. Potential usable energy of the system increases. If unbalanced CP forces cause the people (or weights) to self move then just maybe the wheel will still put out the same increased potential energy (increased ectropy) without adding energy to move the people or weights.
.....as a result of the dynamic charge unbalance of a moving dipole , at : http://blackholeformulas.com/files/elec ... avity.html
...or simplified , as an oscillating basic mechanism of the atom ( tiny "perpetuum mobile"...leverage ?! ) .
The lever is the starting concept of mechanics , also...a " dipole "...
So , if the continuous motion ( " perpetuum mobile " ) is ubiquitous , have we a general idea or understanding of the same " variable / oscillatory " leverage ?!
Al_ex
It seems the best of the forum was when it was fresh .
jim_mich had some gray matter working ,
If I had a balance scale on the train , having it in the harizontal plain , facing outward , and the same mass on both sides of the fulcrum , moving the mass equal distances outward will still have the scale in a balanced state , even if one mass is going in the same direction as the train and the other in a reverse direction . This suggest to me that the cf force on both masses must stay the same inside the train . If the masses was tied to strings at the same radius as the train , but not to the train itself , but gaining or loosing velocity relative to the train , the cf would not be the same for the 2 masses .
jim_mich , you are a wonderfull old stoep kakker .
I don't profess to know what the Jew/Bigot is about but it would be hard for anyone visiting here not to notice that you lead the field by a country mile when it comes to hatred on this board CloudCamper.
Oh, yeah.....it would surely lead to a better understanding of Gravity.
"Everything you know will always equal the sum of your ignorance"
