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klaudio
New Member
1 Post |
 Posted - 03/02/2007 : 19:18:56
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Hello, my name is Claudio Bianco, I am an inventor and I have just finished my last work which started as a simple electrical generator and finally turned into a flying saucer. I invite you to see its design in www.cuerdacontinua.com
[Moved to the General Skepticism folder - Dave W.]
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GeeMack
SFN Regular
USA
1093 Posts |
Posted - 03/02/2007 : 19:46:09 [Permalink]
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Welcome to SFN, klaudio. Funny stuff. Apparently you missed our Humor category! 
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HalfMooner
Dingaling
Philippines
15831 Posts |
Posted - 03/02/2007 : 20:59:40 [Permalink]
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Welcome to SFN, klaudio! Well, I must say that the Web site certainly was professional looking. The artwork is neat, attractive, and orderly. It doesn't have the usual appearance of the work of a screwball.
I cannot say I fully understand what you are designing, but I get the impression you have designed, on paper, a wave-powered generator, and a river-flow-powered generator. These two at least look feasible, though they would have to compete with some rather sophisticated existing designs. (A set of submerged tidal and/or river-flow turbines are now operating off Manhattan, in the East River, I believe.)
But you completely lose me with your flying saucer and your other perpetual motion machines. Those are impossible, and only madness and failure lie that way. My grandfather worked on a perpetual motion machine for many years, and failed. May I ask: How did I, Claudius (one of my favorite novels, along with Claudius the God) inspire you? Because the main character is your namesake?
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“Biology is just physics that has begun to smell bad.” —HalfMooner
Here's a link to Moonscape News, and one to its Archive. |
| Edited by - HalfMooner on 03/02/2007 21:14:24 |
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McQ
Skeptic Friend
USA
258 Posts |
Posted - 03/02/2007 : 21:22:18 [Permalink]
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| I really liked the design of the website. It has a very pleasing look to it. The contents are beyond me, but the site looks and navigates wonderfully! |
Elvis didn't do no drugs!
--Penn Gillette |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/03/2007 : 00:08:30 [Permalink]
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From the site:Before you think that I am crazy or that I just got around the law of energy conservation, it is good that you know that these calculations are far from being the most adequate. The equation to calculate the work for the inflation is calculated as if the work would inflate a balloon that would be attached to the bell. Moreover, this formula would be the ideal compression of gases that is not real, since it would vary according to the efficiency of each compressor.
Nor do they take into account the mechanical friction particular to the transmission system in the generator, and the dynamic friction of the water.
Another assumption is that the temperature is equal in all sides, water and exterior air. Also, as the processes and movements are very slow for there to not be losses due to heat due to the compression and expansion of gases.
Better calculations will have to be made, that include all the aspects. The above, I think, shows the proper self criticism and doubt. But then just a few sentences later:Why do the calculations come out right?
Either I am wrong or I am not taking something into account in the formulas... And then all doubt is gone. Blanco is sure that this rig will "generate a surplus," despite all the things he admits to not having taken into account because they "surpass" his knowledge of physics.
But the assumptions made are unreasonable, and the factors being ignored are important. Too important to ignore.
But it's not just his expectations, it's also the work he already shows. Take this equation, for example (with 's' replacing the Spanish abbreviation for seconds, 'seg', and the proper abbreviation for kilowatt-hours):| We = D×G×H×Vf = 7848000 watt×s = 2.18 kWh |
Let's just work through the units on these values:D is in units of kg/m3
G is in units of m/s
Multiply them together and we have kg/m2/s
H is in m
Multiply it in, and so far we have kg/m/s
And then Vf is in m3
Which we multiply in to get kg×m2/s Now, a joule is a kg×m2/s2
And a watt is one joule per second, or kg×m2/s3
One watt-second (one watt times one second) is one joule, but Blanco doesn't have the units in joules, he's got them in joule-seconds (watt-seconds-seconds, a unit I've never seen used). Please note this isn't a joule per second (which is a watt), but a joule times a second.
Now that was the easier equation. In the equation above that on his page, we see the same D×G×H term, which we've already determined to be in units of kg/m/s, and to that he's adding a value in units of kg/m2. Since the units don't match, this addition would appear to be a problem (what is one second plus one kilogram?). This conglomeration he multiplies by Vf, which is again in m3, so the result should be in units of kg×m2/s+kg×m, and once again (after a multiplication by a unitless logarithm), he expects to get kilowatt-hours (which would be in units of kg×m2/s2) out of it. The math is just a mess. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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HalfMooner
Dingaling
Philippines
15831 Posts |
Posted - 03/03/2007 : 09:28:51 [Permalink]
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Dave W. asked: quote:
...what is one second plus one kilogram?
Some serious Thanksgiving dinner eating, would be my answer.
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“Biology is just physics that has begun to smell bad.” —HalfMooner
Here's a link to Moonscape News, and one to its Archive. |
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JohnOAS
SFN Regular
Australia
800 Posts |
Posted - 03/04/2007 : 21:20:06 [Permalink]
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I would think that for those familiar with the appropriate slang , this description was more or less self-evident.
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John's just this guy, you know. |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 03/04/2007 : 21:52:11 [Permalink]
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Dave, on the first equation he had the acceleration of gravity at 9.8 m/s, it think it was just a typo it should have been 9.8 m/s^2 in which case the units do work for joules.
In general the I agree that the math is a mess. It is too late for me to focus my eyes though maybe tomorrow....
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If I knew then what I know now then I would know more now than I know. |
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Dave W.
Info Junkie
USA
26020 Posts |
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BigPapaSmurf
SFN Die Hard
3192 Posts |
Posted - 03/05/2007 : 07:38:48 [Permalink]
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| The answer is 1.206 giga-zillions. Erm, thats Back to the Future, "jigga"-zillions not your standard "gig-a"-zillions |
"...things I have neither seen nor experienced nor heard tell of from anybody else; things, what is more, that do not in fact exist and could not ever exist at all. So my readers must not believe a word I say." -Lucian on his book True History
"...They accept such things on faith alone, without any evidence. So if a fraudulent and cunning person who knows how to take advantage of a situation comes among them, he can make himself rich in a short time." -Lucian critical of early Christians c.166 AD From his book, De Morte Peregrini |
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Dr. Mabuse
Septic Fiend
Sweden
9687 Posts |
Posted - 03/05/2007 : 07:47:12 [Permalink]
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kilogram-meter is a measure of potential energy.
E=mgh
Where m=mass, g=acceleration, h=height
Edited to add: using kg, m/sē, and m, you get Joule as a result. |
Dr. Mabuse - "When the going gets tough, the tough get Duct-tape..."
Dr. Mabuse whisper.mp3
"Equivocation is not just a job, for a creationist it's a way of life..." Dr. Mabuse
Support American Troops in Iraq:
Send them unarmed civilians for target practice..
Collateralmurder. |
| Edited by - Dr. Mabuse on 03/05/2007 07:48:35 |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 03/05/2007 : 07:51:04 [Permalink]
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klaudio on the work expended to inflate the bell you forgot to include the work of the compressor, which is considerable. At a depth of 100 meters you are looking at a compressor discharge pressure in excess of 9 bar or 132 psi. You also neglected to include the effects of friction in your calculations. I would expect to see viscosity and temperature terms in your calculations. The bell plowing through the water is going to have a tremendous amount of friction losses.
Perhaps you addressed these issues and I just missed them?
edited to add - yes I missed them. |
If I knew then what I know now then I would know more now than I know. |
| Edited by - furshur on 03/05/2007 13:54:36 |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 03/05/2007 : 07:52:37 [Permalink]
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quote:
Oh, dangit. Still one can't add kilogram-meters to joules and get a meaningful result, can one?
Not in my universe...
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If I knew then what I know now then I would know more now than I know. |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 03/05/2007 : 08:31:38 [Permalink]
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Mab, the problem is actually in the first term in the first equation, where he adds DGH and Po together. That'd be kg/m3×m/s2×m, or kg/m/s2 plus kg/m2. The units don't match. He then multiples the result of that addition by a number in m3, so doing the multiplication in units gives kg×m2/s2 plus kg×m. The first term of that is, indeed, joules, but the second, kg×m (kilogram-meter) isn't anything that I'm aware of, because there is no g term in there. It'd only be potential energy if it included an acceleration.
And furshur, if I read the description right, he's talking about the bell's pressure fluctuating between 10 atmospheres and 20 atmospheres, so the compressor would have to match that higher number, wouldn't it? |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 03/05/2007 : 10:40:51 [Permalink]
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Absolutely, 20 bar is about 300 psi and the storage cyliders would need to hold enough air to fill 8 cubic meters at a final system pressure of 10 bar. I have to work that out but it is not a trivial amount of energy.
I assume 2 - 3 horsepower on a 2 stage compressor.
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If I knew then what I know now then I would know more now than I know. |
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furshur
SFN Regular
USA
1536 Posts |
Posted - 03/05/2007 : 12:57:52 [Permalink]
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klaudio, as Dave has said the equation you have for the total work from the system cannot be correct. You cannot add pressure to hieght - it has no physical meaning.
The simplist way to look at this system is realize that you are converting potential energy to kenetic energy. You add potential energy to the system (pressurized air) and it is converted to kenetic energy due to bouancy. The bell then reaches the surface and you empty the air out and down it goes again converting potential engery (the height) to kenetic energy.
The losses in this system will be quite large. The friction and turbelance that will be caused by the velocity of the bell will make the conversion of PE to KE much less than 100% let alone having it MORE than 100%.
The calculations for this system are not trivial. As the bell descends it will increase in speed and for a given temperature the resultant flow will change from laminar to turbulent flow. The corresponding change in drag will not be linear and, well, the math will get just plane ugly. The same thing will happen on the ascent.
Not only does your equation violate basic math principals it does not even have the right terms in it. For "fun" I might try to give a shot at a simplistic equation, modeling what you have described, but I suspect it may be beyond my ability.
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If I knew then what I know now then I would know more now than I know. |
| Edited by - furshur on 03/05/2007 12:59:24 |
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A flying saucer design
