The Wind Powered Sail-less Boat

[philSweet]

Good grief! My equasions are reversed.

[masalai]

As I see it the wind turbine acts as a big wing/sail as in a helicopter (rotary wing) which initially pushes the vehicle downwind, and since the wheels are connected this then rotates the wind turbine but the rotation must be in the other direction to force the turbine to blow air aft (backwards) once this occurs then the craft is pushed by the velocity of the air produced by the air turbine driven by the wheels and also by the tail wind also pushing the "rotating wing/sail area" and something does not quite add up as what changes the wind turbine rotational direction from a driven force to make the wheels turn to a driving force by the movement of the wheels causing the wind turbine to become a fan which pushes the vehicle????? Now I am confused - which is easy to do.....

[Joakim]

Originally Posted by Guillermo

That makes no sense to me. You're asuming a boat is sailing faster downwind than the wind to demonstrate it can sail faster downwind than the wind. That's circular. And you're saying it's possible to extract energy from nothing. The only physics where it's possible to extract energy from nothing is on quantum mechanics, not in everydays life.

The cart thing, which video I posted (as well as several others) before, is neatly a fake. Bringing its reasoning to the extreme it is like saying that if in absence of wind we push the cart fast enough it will create its own wind and run forever. If that's not perpetual movement, let God come here and see it. I'm with Boston in this one.

Sad to read this, since I have read your many posts with good respect and really should be understandable for a NA. I hope you can open your mind and think beyond limitations of a sail.

I didn't assume the vessel was going faster than the wind, I just calculated the force balance in that situation and it clearly shows that having clearly more thrust than drag is possible, thus it would be easy to maintain such a speed. Now you can calculate some intermediate situations you think are uncrossable.

You really don't seem to understand the cart thing. It gets it's driving force from the velocity difference of the treadmill and air. It will not work without that velocity difference just as no sailing vessel will move through water in any direction without a velocity difference between air and water.

Can you have a velocity through water if the geographic wind velocity is zero but there is a 5 kn current? Is it different from sailing in 5 kn wind without current? Will you notice any difference in your instruments or sailing performance (except GPS)? Compare this to treadmill.

Joakim

[Guest625101138]

Phil and Jim
I am not awarding half points. You have to have both answers correct in terms of direction and velocity. Hint - it is possible for the vehicle to travel in the opposite direction to the direction of the sliding plate.

It is good to see someone actually putting their mind to it. I appreciate your effort and with a bit more thinking you should get it. I am certain if you understand this simple problem you have the keys to work out sailing directly into the wind and faster than the wind down wind.

Joakim - offer still open to you. At least there are some actually using their reasoning ability.

Rick W

[Joakim]

Originally Posted by Rick Willoughby

Joakim - offer still open to you. At least there are some actually using their reasoning ability.

I don't want to spoil others change for thinking. I already asked for the equation in post #64 and prior to asking did the simple maths.

Joakim

[philSweet]

The device as drawn cannot travel in the direction opposite to the plate.

by inspection-

1. for buggy velocity < 2, upper idler must turn cw
2. v > 2, upper idler must turn ccw
3. v < 0, lower idler must turn ccw
4. v > 0, lower idler must turn cw

as drawn, upper idler and lower idler must both either travel cw or both ccw. Thus only 0 < v < 2 are feasible (case 1 and 4 satisified) cases 2 and three cannot both be satisified.

[clmanges]

Case 1:
Plate moves 2 m/sec relative to cart
Cart moves 2/3 m/sec relative to ground
Plate moves 2 2/3 m/sec relative to ground, from left to right

Case 2:
Plate moves 2 m/sec relative to cart
Cart moves 6 m/sec relative to ground
Plate moves 4 m/sec relative to ground, from right to left, since cart is outrunning plate by the difference of the two speeds.

There, is that better?
You should have stated the question properly to begin with.

[philSweet]

clmanges, no, it is not better. regarding your case two, apply the logic from my preceding post.

if plate is moving from right to left at four m/s and cart is moving at six m/s (right to left implied), then upper idler is going cw and the lower one ccw. the device is not shown this way, thus, regardless of whether the 2 m/s is taken to be relativ to the cart or the ground, your case two is not correct.

[Tiny Turnip]

Rick states that the plate velocity is 2m/s relative to the ground, not the trolley.

I am finding this one very difficult to grasp, and have been giving it a lot of careful, step by step thought. In both Rick's pictures, I find the top left wheel wanting to rotate at a rotational (angular?? rusty physics again) velocity different that is from that of the plate when is in contact with it, and and so I get concerned about the nature of the interface between plate and wheel - slippery, or high friction/mechanical?

In Rick's case 2, if you follow the gearing round, the trolley wants to scoot to the right at 6m/s, but this would mean top left wheel rotating anti clockwise.If it is driven by the plate, we already know it is rotating clockwise at 2m/s angular/rotational velocity. So I don't think it goes anywhere, and probably doesn't drive, whether it is a high or low friction interface. Or, does it resolve into 2m/s to the right?


Jim, you state that in your analysis, the output velocity is the buggy velocity. Could you enlarge on what you mean/take by input velocity please?

as Vin + Vout = 2m/s, then you are not referring to the plate velocity as the input velocity?

Joakim's illustration in post#64 I understand very readily.

The velocity of Joakim's cart seems to me would vary depending on the relative positions of cart, ground and 'wind'.

[Tcubed]

Solving for no slip both top and bottom-

In the first case;

V=2-3v

Thus v=0.5

In the second case;

V=2-v/3

so v=1.5

Where v is M/s to the right relative to stationary plane

Now let me see what happens if the belt is twisted so the wheels turn in opposite directions....

[Tcubed]

If the machine is geared such that the wheels turn in opposite directions, then with the gear ratio of the first example

we have v=2+3v

so v=-1 (this is the case of the boat sailing straight upwind, as in going at one meter per second in the direction opposite the top plate)

and with the second gear ratio

we have v=2+v/3

so v=3 (and this would represent the boat sailing downwind faster than the wind, it is going one meter per second faster than the plate)

With all due respect Rick, didn't you mean to demonstrate how this models the turbine boat's behaviour? So did you not originally mean for the wheels to turn opposite each other?

[Tiny Turnip]

much happier now. Thankyou

Joakim, Unlogical, Phil, Jim and Tcubed. and of course Rick.

just frustrated that I am quite comfortable intuitively with the DWFTTW turbine, and I'm now happily following the maths that you gentlemen have put together (the fact that a twisted chain can be expressed with as a -ve ratio is a big help) but unless the chain is twisted I'm stumped.

[clmanges]

!@#$%^&*()_+

You should have stated the question properly to begin with.

Yeah, and I should have read it properly . . .

[Joakim]

And the easy formula for the velocity respective to ground is:

V = W / (1 + G),

Where V is the velocity, W is the wind and G is the gearing ratio. Depending how G is defined it could also be -G, 1/G or -1/G.

Now you can see, that with postive G you can not go faster than the wind downwind and also not at all to headwind. This is the normal "sailing region" where your speed comes from drag ratios when going dead downwind.

With negative G these limitations are removed and you can reach any speed by choosing G close to -1. Higher speeds require very good efficiencies, which sets limits to the speed of a propeller turbine system.

For propeller/turbine negative G means, that the propeller and the turbine have same handness and rotate to same direction. In dead downwind situation this means that both would "try to go forward".

I think Ricks rep points should go to TheUnlogicalOne, since in his post #94 he actually has it right. I didn't check the belt/chain causing the wrong sign for gearing and was only expecting the -1 and +3 m/s answers, which are more relevant for this discussion.

Joakim

[Boston]

so I left this alone for a few days to see what would shake loose
hoping to hear how folks would answer the questions I proposed
and not one true believer stepped up to the plate

I must admit I was looking forward to reading how the problem of reduced energy in, results in increased energy out
no mater what the system or gearing involved

I may have missed a post so Ill look again but I didnt notice anyone addressing the previously suggested problems with the hole perpetual motion theory

address the ratio of reduced energy available as a function of apparent wind vs vehicle acceleration on a dead run since that is the key issue limiting vehicle speed on a dead run

actually I think Teddy sorta did
but it was the same kind of smart ass answer I would have given
so it only sorta counts

if he sends me an address Ill buy him half a pint for being the only one to even make a pretense at the real issue behind why this doesnt work

Joakim's questions though had a lot of validity so I think it deserves a mention

You agree that an ice racer can reach the downwind mark faster than the wind. Thus it is not perpetual motion going faster than the wind dead downwind. This physical fact does not change wether one goes directly from A to B or by jibing.

actually there is a huge difference between tacking, running on a reach
and running directly down wind

tacking you have the alterations in the apparent wind
and the addition of centripetal forces

on a reach you have the difference in relative wind and actual wind

on a dead run
you got nothin
but the actual wind speed and a hole lot of friction

yes an ice racer can go faster than the wind powering it
but not directly down wind
do they hit the down wind mark faster than the wind itself ?
Ive never really looked into it
given that they are on such a high angle of attack
Ild have to have a diagram of there course relative to wind, wind speed and there course time and distance
Im betting it's possible, given how insanely fast those things are
but to be fair you would have to extrapolate the course as the hypotenuse of a right triangle and the wind as side "a"
if you consider the vehicle is on a 45deg angle to the wind just for the sake of argument
the vehicle would have to travel at 1.41x the speed of the wind, to hit the mark directly down wind from the start, on its 45deg course, at the same time as the wind

but the reason for this is the apparent wind is not o when the vehicle hits actual wind speed
therefor there is energy still available to offset the system friction
high school physics folks
( lets see who is first to mention my high school grammar )

course we can hoist a few anyway
but your going to have to do a lot better than the tread mill trick before Im buying

I saw some great stuff in here
but not one person could post a flicker of the mythical monster
not a shred of evidence, mathematical or actual presented
just an insistence that it works

oh and the tread mill thingy was not level
if you look at the bubble of the level they placed on the frame

nor is there any accredited independent support of the rubber band cart
no wind speed data
no cart speed data
pretty much proves that the wind can blow beach ball into the surf
and why didnt they show the road ahead of the thing at any time during the flick
hmmmmmmm

its really basic
energy out, cannot exceed energy in, minus friction

friction being the key word
if you can show that friction can be reduced to a negative on both sides of the equation at some tipping point of dynamic instability over delta t
then you got me
I was wrong
but the problems presented so far dont even come close to doing so


oh and the ice racers are definitely, not going "directly down wind"
and this hole thing is about "directly down wind"


also there is a lot to be said about powering the wheels to turn the turbine and the gear ratio's involved verses powering the turbine and turning the wheels
after all
if you take a mechanically advantaged system and run it in reverse
dont you get mechanical disadvantage ?



as for flat earthers
can anyone tell me were these pictures are taken
and what they are of
as if that isnt enough of a hint