My calculations did not come back even (before the cycle was complete). My calculations showed that you had to expend energy in order to get to outer space and back again. Then when the cycle was complete you regained that energy and you are now evens.
Not in the pdf but I did account for it in my later post, it is accounted for by the change in potential energy.OK I will add to my pdf to include the change in potential energy caused by the ball rising, and it is accounted for as I have said three times now.
Smudge
I think we need to go over this again.
Post 61
Ball displacement = 449L
Weight of displaced water=44.9KGs force =440.47N
Height=9 meters
Energy required= 3964.2 joules.
Stored energy when DUT is back on earths surface
Ball displacement=449L
Weight of displaced water=449KGs force = 4404.7N
Height=9 meters
Stored energy=39642 joules.
Energy that came from somewhere=35678 joules.
Still thinking about where the energy comes from.Post 62
I think it comes down to centre of mass. Initially the mass centre is below the geometric centre then when it returns to earth the mass centre is above the geometric centre.Post 63
It's that change of centre of mass that means you don't bring the system back to its starting point. Thus Brad's statement "As our mass never changed,then all energy used to raise the mass,was returned during the descent-->energy was conserved." is untrue. Less energy is returned during the descent because the centre of mass is not returned to the starting point, it is slightly higher than it was at the start. And that difference accounts for the apparent gain.Now,here is where i disagree.
When the ball is allowed to rise to the top of the water,and the buoyant PE is transformed into dissipated energy,example-turning a generator,and driving a lightbulb,once the ball has reached the top,then that volume of water that occupied that area at the top,must now fill the void left by the ball at the bottom.So,no mater how you look at it,449 liters of water fell(equivalent to) a distance of 9 meters.
So,as i stated at the start,we had an energy gain in the buoyant PE,as when one full cycle is completed,the center of mass is the same as it was at the start,and not different like you based your math around.
So it seems to me,that this 39642.21 Joules of energy has not been accounted for in your math yet.
Let's go back to square one where you offered the thought experiment. You sent the thing into outer space and got it back again at no energy cost (or so you thought). While in outer space at 1/10 gravity you pulled the ball down at an energy cots of 1/10 that above figure (we won't argue about the exact figure since you originally used 9.8 and I used 9.81 for gravity). You then got the above energy out when you let the ball float up. So there appeared to be an energy gain of 9/10 the above figure. I then showed that there was an energy cost in going to outer space, you spent more energy getting there than you could recoup on the way back.Now,the highlighted in red.
Yes,and that cost was based around the fact that the center of mass was higher when it returned to earth.
At this very point,all energies added up to a net gain of 0.
But this is wrong.
It is wrong because the cycle is not complete.
There is the PE stored in the ball-->this you took into account,and at that very point,all energies in and out were equal.
What you did not do,is account for the PE stored in the raised water it self.
You stated your self-->the center of mass is higher when it returns back to earth.
So,when the ball rises to the top,the mass(water) that is above the ball,must fall back to the original starting position.
Only then is 1 full cycle complete
Only then can you do your math,where the center of mass will be the very same at the start and end of the cycle.
Once you have done this,you will then see that my original calculations where correct,where there is an excess of 39642 joules of energy delivered by the system.
So,i hope we are on the same page now,as the above is the very events that took place.
I can see no where that you have accounted for any energy after your PDF post,that was not already accounted for in your PDF,as your PDF accounted for all energies(except the water falling back to it's original starting position),and came back as all energies in and out added up to being equal.
So i am lost as to how you can now account for that stored energy in the water,that is above that of the starting level,when you have already given a result of E/in and E/out being equal before i mentioned this stored energy that was not taken into account
So,can we try the math again,but where 1 full cycle is complete.
This means that you can eliminate that difference in the center of mass between the start and end of the cycle,as the water level will be exactly the same at the end of the cycle,as it is at the start.
Brad
Never let your schooling get in the way of your education.
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Topic: GRAVITY: Is it a source of energy? (Read 10212 times)
