Gravity Mill - any comments to this idea?

[prajna]

tbird,

ratio is 4 to 1, not 3 to 1, right?

Yes, absolutely. Must've been tired and hung up over that original 1 atmosphere or something,  I'll fix it. It knocks on to the following calculations too.

the only thing you would have to do is divide by 2 at the end.

Surely just subtract 1 at the end (or am I tired and confused this morning too?)  Whatever the absolute pressure is in the container the gauge pressure will read 1 atmosphere less because the gauge is zeroed to 1 atmosphere, yes?  Scuba diving?  I've done it twice: once with a friend who taught me a handsignal underwater that nearly had me coughing up my regulator because it looked so funny and once when I was in the army where they seemed more concerned at beasting you through physical exercises to prove just how 'hard' Royal Engineers divers are! Neither seemed, to my poor recall, to have spent much energy on familiarising me on the niceties of the diffference between absolute pressure and gauge pressure. As to tire pressures, most people read the pressure chart in their handbook or the label on the door post and fill the tire until it reads that pressure.  Oh dear.  I think I will have to print the pressure as gauge pressure and indicate, for poor, confused characters like myself that the absolute pressure is 1 bar more than the gauge pressure.

It sounds as if you are now confused between neutral pressure at 10m and neutral pressure at sea level.  The pressure required to double the displacement at any particular depth will be twice the neutral pressure at that depth and the gauge pressure will be 1 bar less.  Are we there yet?

why not give the user the max height, using as a standard 10% of shuttle pipe size, or something reasonable, automaticly.

Because I want to allow the user to experiment with the effect of changing the height of the head.  All of the inputs are there so that the user can experiment with the effect that changing their values has on the various results.  It seemed logical to me that one of the first things you might want to decide is 'how high can I pump the water?' and the head height is the main determinant for that.

The user doesn't have to provide a value for this, since it is a default.  I understand what you are saying though: it would be nice to have the most efficient system specified as the defaults but calculating the most efficient system is not a trivial task.  Perhaps later I will change some of the default values.

it would only have to use both strokes.  water delivery just doubled.

Yes, the original ELSA.  I spent some time looking at John's chaotic drawings last night and that is certainly the way to go.  Designing this system has gone a long way towards helping me understand all of the factors involved so that I could prove to myself (and others, of course) that the physics really do work.  When this version is complete and correct I will do another calculator based on the 'true' ELSA - hopefully with rather less effort than this has taken since I am now pretty familiar with what is involved.  I would very much like to come up with some elegant way to recompress the shuttle for ELSA though.  All of the rest of John's design is nice and simple but the recompression methods are all a bit ugly and Heath-Robinson.  Perhaps that is why he has had difficulty introducing the concept to the world.

raising the delivery height increases the energy stored.  wouldn't time have to be considered somewhere?

Certainly.  At a later date I would like to introduce fluid dynamics to the mix and friction so that we can calculate the stroke speed.  That will allow proper output values to be calculated and we will know if you really can power your house with one of these in your yard.

[prajna]

Ok guys, I think that is as far as I will go with this design of ELSACALC (unless tbird wants me to add metric to imperial conversion in this version).  Next I will work on John's double acting ELSA so that we can calculate the output from that.

[ooandioo]

Hi all.
New message from John Herring:

Yes you may post if you wish.
Plesae includ my e-mail address, gravity_machine@yahoo.com in your posting`s
I am going in the hospital tomorrow and hope to be home by Monday.
sorry no time today !
Sincerely
John

Just real Productions wrote:
Hello Mr. Herring.

Thanks for your reply. We are looking forward, discussing and working with you. Maybe you would like to join our foru m board at
http: //www.overunity.com/index.php/topic,570.0.html.

Thank you for the image ? we already know it from the internet.

I?m looking forward, hearing from you. Kind regards, Andreas Lutz.

PS: May I post your replies in th e overunity.com forum?

[2tiger]

Hi all
Please take a look at this graphics. It is a simulation from the software Interactive Physics.
The marbles (1cm diameter) are representing the water. The orange one is representing the shuttle (50% more weight).

As you can see the 50 % more weight on the right side is only able to lift the blue marble on the left side half his height - 5 mm. If the marbles were water, the water will rise 5 mm too. But if you want the water to overcome the 5 mm so that it can flow out, you have to choose a tube that is 4 mm high OR you have to make the shuttle 50%+10% = 60% heavier (by same volume of water)!

So the relation between the weight of the shuttle and the max. lift is 10 % more weight (by same volume of water) for 1mm lift up.  1cm (marble diameter) = 100%, 2cm =200% and so on. And remember you have to lift this weight with your "balloon" only by doubling the volume of the shuttle.

Now take a look at the lower graphic the ratio between both tubes is 1:1.
What is the ratio in the ELSA design? Let us say 1:5 (lift tube : shuttle piston tube).
So if we don?t want to violate the hydrostatic paradoxon that stefan pointed out in his earlier posts, there have to be on the shuttle side, 5 times more downforce (for example press the water 1 mm down) in order to lift the water in the lift-tube 1 mm up.
 
Now we know the relation 10% more weight (shuttle) for one 1mm lift. But when the ration between the tubes is
1:5 we will need 5 times 10% more weight (shuttle) for lift the water 1 mm up.

Stefan - did you try what i write in my last post?
From the beginning you have been right with your assumptions, that this won?t work.
If you now are still believing that ELSA works, I will send you my "barbyfamilytube" for free (without shiping costs).
I live near by Bad Segeberg (Karl May) in Germany.

Next i will try to analyse the upstroke of ELSA. But the downstroke definitive won?t work in that manner.

CU
2Tiger

[prajna]

2Tiger,

Water at sealevel has a force of 14.5psi or 1.02kg acting on it.  If I have a 'U' tube such as the one in your drawing that has an internal diameter of 3.57cm it will hold 1 litre/m of tube.  Let's assume that there is enough water in the tube so that the bend at the bottom is under water.  If I then pour 1 litre of water into the tube then the water level in the tube will rise 50cm.  Equally, if I put a 1kg weight in one leg the water level will raise 50cm (since 1 litre of water weighs 1kg).  Is that what you were trying to say?

What will happen if there is just 40cm of height of tube over the level of the water and I put a weight of 1kg in one leg?  The water will flow out of the other leg (because we know that 1kg will push the water up 50cm) and the weight will continue to sink, pushing even more water out of the tube until it gets to the bend at the bottom.  It will have pushed nearly half of the water out of the tube (considerably more than half a kg of water) even though the weight I put in is only 1kg.  This is how ELSA works when the weight is going down.