The Wind Powered Sail-less Boat

You have had no way to know this clmanges, so this isn't a criticism, but you have it backwards -- we built the darn things and recorded the videos *after* years of discussion on other forums. The videos were to *back up* and demonstrate our previous explanations. This is why in our videos you hear us using names like "Steve" and "George" ... we made most of those videos in response to specific concerns of people on those forums.

We never made the videos for general consumption and never expected then to be picked up on forums all over the world.

Interestingly enough, while the chassis of the cart sees a 180 degree switch while passing through the referred to point, the propsail sees only an infinitesimally small change in relative wind at that moment. Following is a link to an image which demonstrates the above:



JB

 

As I remind everyone when discussing these efficiencies -- the wind is very powerful -- we can't go wasting power willy-nilly, but if merely wishing to crack 1.0, it's not necessary to be anal about it either.

Remember that our small cart uses a simple right angle drive from an RC heli and an off the shelf RC plane prop. Additionally with that design, the prop is off angle from horizontal by ~15 degrees.

JB

 

I think it is totally over complicating the issue to start talking about the efficiencies. It has been demonstrated that it works with relatively crude models so the efficiency is not critical.

What troughs people off is that they so quickly dismiss it being possible that they start reading faster than they think and skip essential points. I have followed pages of discussion where skeptics still think that the propeller drives the wheels - when it has been pointed out several times that it is the opposite.

As people think that the propeller is a turbine they quickly come to conclusion that as the vessel reaches the wind speed the apparent wind on the prop becomes zero and it can't catch energy. This would happen but the apparatus doesn't work like that - again - the wheels drive the prop.

You need to think that the propeller works as a parachute (sail, umbrella whatever) and as speed is picked up it also work as a propeller.

There is no quantum mechanics or breaking of any fundamental laws here.

And boston - JB's question simply is do you agree that a iceboat can reach down wind marker faster than a traveling straight downwind same speed as the wind. This seems to be generally accepted - do you accept it?

 

I am but a flee on the sore of grammar

Mr T, seems like I got it close
just looked it up and you are right, a dam nice set up gives a number of 1.5
a crappy one can be as large as 20% energy loss
thus proving once again
when those nether parts develop a twinge
it pays to listen
well
.5 percent off
but thats not bad for dreaming up something I learned thirty years ago
I will however make an attempts to be more accurate in the future

JB
sorry but this isnt my baby
you got an idea its up to you to analyze it.
Im happy to nod and clap at the appropriate moments but its all you brother,
my beliefs are irrelevant to the validity of the physics involved.
I wish you the best and I certainly will be reading along, but the only thing thats going to convince me; or anyone else with a working knowledge of physics is a detailed mathematical analysis including but not limited to the suggested details.
Nobody is just going to go with a video and a note on this


Im all for you JB Gods know Ive had my share of crazy ideas
but its your game, your ball and your court
and I'am standing outside the fence eagerly awaiting the next play.
If that means you have nothing to present
so be it

In the end your going to need to dazzle me with details
should you present your work to a sufficiently detailed analytical degree finding itself without flaw, then it can begin the task of standing up to a few simple proofs

ps

I think you offered whatever level of detail required
sooooo
you don't need my consent or agreement to supply it
start singing
Im sure Rick and a few others are dying to see this

B

 

I think that after ten pages of this it's high time to take this discussion way beyond trying to convince rock solid skeptics. Therefore it is also time to start asking what are the limits....

In high performance sailboats the limits are pretty much described by the sum of the above and under water gliding angles, which will determine the upper limits of V/W (boat speed / wind speed) .
In the case of the turbine boat it seems the limits can be found by analysing the energy losses altogether.

Therefore i don't think it is anal to want a realistic number for these losses.

If we use 0.82 as air prop eff. 0.82 water prop eff. and 0.985 trans. eff. we get to total eff. of approx 2/3. A nice easy number which is also pretty realistic.

So what would be the maximum V/W if total efficiency is 2/3 (0.6667) ??

I will rework the wheels on ceiling and floor model from an energy transfer point of view when i get a moment and see what happens. This is important as with the zero loss model it is possible to plug in whatever gear ratio one needs to end up with whatever V/W is desired.

I would be interested to see with eff. = 2/3 what max value V/W gets - 2? 3? 2.5? Maybe someone with maths ability will beat me to the answer.

My other big question is what happens when the gear ratio is unity.

 

Analysing the second question i return to the basic eq. for the wheel model

W/V = 1-G or G = (V-W)/V

G = 1 implies the limit to infinity of V/W which is not conceivable in the real world.

 

Ok i'm kind of in a hurry but this is what some quick number manipulations give. I would love it if someone could check it for fundamental flaws.

e is efficiency, G is gear ratio, d is distance traveled by machine in a given interval of time (assume steady state of course) Fb is force at bottom wheel , Ft force at top wheel . E is energy.

Work done top and bottom;

Eb = d*Fb

Et = e*d*Fb = (W/V)*d*Ft

Also we have Fb = G*Ft

So e*G = W/V

Going back to previous post we get W/V approach e as G nears unity. G = 1 is the limit .

So for e = 2/3 we get V/W = 3/2 = 1.5

Not all that impressive really.

Like i said this was done spur of the moment so i would not be at all surprised if i made a mistake. Lets see if someone else can take up the slack..

 

ok i already see a problem .

As e tends towards unity we should see V/W tend to infinity , which is not happening so the above eq is wrong .

I'll give it another shot later.

 

Boston:
>my beliefs are irrelevant to the validity
>of the physics involved.

If asked to explain things to *you*, your beliefs are of utmost importance -- it's called a starting point. If you don't believe that 2+2 = 4, there's not much point in moving on to bigger numbers.

You refuse to stipulate that the the laws of physics allow a device to arrive at a finish point faster than the wind that is powering it -- something that has been factually established and repeated daily for something like a century.

Essentially, you wish me to explain to you the subtleties of around the globe navigation while while not stipulating that the world it round. Thanks Boston, but no thanks.

If you don't believe the ice-boat can beat the balloon, say so and I'll direct you to links where you can bring yourself up to speed. If you do believe so and just won't say, it doesn't speak well of your conversational intent here and leaves lie to your "where does the energy come from" questions.

Best wishes.

JB

 

Tcubed, I don't think you'll find any theoretical limitations until lightspeed comes into play -- practical limitations will kick in a something quite a bit less than that.

Since we clearly have more drag than a ice-boat and yet extract energy the exact same way, I'm quite confident stating that DDWFTTW will be unable to meet the downwind VMG of a good ice-boat.

My money says 2x windspeed will fall with just a bit of work and that 3x will take more money than anyone will likely care to spend.

JB

 

Theory is just as good as practice if every factor is accounted for.

Water boats can and do sail faster downwind than the wind, so i' m interested in seeing what sorts of ratios we can expect with a given efficiency for the turbine boat.

I agree with the ratios of V/W you put forward , but i want to crack the maths behind it and would appreciate some discussion in that area.

The equations for an ice boat are the same as for a sea boat , just the coefficients change.

I am also interested in figuring out how a machine like this transitions from G > 1 through to G < 1 . Has anyone really looked into that?

 

This is true Rick but one must be careful how that information is presented -- many people object to our device *precisely* based on the above definition.

To remove energy from the air we must reduce it's speed. If a propeller *accelerates* the air, we can't be slowing it down -- right? Herein lies one of the most unintuitive aspects of DWFTTW ... we accelerate the wind relative to the vehicle, but slow it relate to the ground.

Of course that brings up a question that will stump 99% of sailors (and even most boat designers) -- by the above established definition, is a sail a "prop" or a "turbine"?

JB

 

Not sure which machine you are referring to when you say "machine like this", but I can say this -- the transition for the cart sail is indentical to the transition for any other sail and no different than any other speed increase. It's just another small change in the direction of the apparent wind.

http://www.mediafire.com/imageview.php?quickkey=ndqeit2m3gz&thumb=5

JB

 

The machine i' m talking about is the wheel on ceiling and floor model.

G is the gear ratio. I'm not talking about the V > W 'barrier' that seems to fixate a lot of members.

G = 1 would correspond to both props linked one to one and with the same pitch, which i can't imagine working. Therefore G must always not be equal to one. But G must adopt different values depending on whether the machine is going upwind (real wind) or downwind or at any other angle, where i think is where it gets interesting and so far no one has gone into that.

 

Here is a drawing I did some time ago to demonstrate the ratio differences you are describing:

http://www.mediafire.com/imageview.php?quickkey=m7yytmonird&thumb=5

Same force direction ... different speeds and directions depending on gearing.

Yes, it must not be 1:1 and the ratios on either side of 1:1 lead to upwind or downwind operation depending on which way gear. All one has to do to our cart is put smaller wheels on it and it backs into the wind (upwind) rather than going downwind.

JB