The Wind Powered Sail-less Boat

my God he's an Aussie
I should have asked for more beer
oh well
you will still enjoy all the conversions
cmon
show me what you can do
I want to see you work this out

 

mystery question
were are these two pictures taken
and what are they of





and now for the real question

sorry I almost forgot
the part highlighted in red needs to be in ft/lb as well so you can more easily see my next point which is going to be that the energy of the plate that weighs two pounds moving at twenty miles an hour is far less than the energy of the system as a hole at the sixty miles an hour it will end up at

soo were in the equation did you get any additional energy
you didnt
so it doesnt work
does not compute
system slows to a gradual halt over delta T having never attained it max potential
system cannot end up having more energy than it started with unless it contained stored energy in the first place or unless energy was added

if you are in australia you might want the Appaloosa's e mail address so you maybe could just send em your credit card info
that way you wont have to keep making long distance calls to buy all these beers you are going to owe me



and the real answer to the question originally posted about how fast the system ends up at if the plate is moving at 2m/s isnt 6m/s
its entirely dependent on the mass of the roller device and the mass of the plate such that the system at speed in a perfectly frictionless enviroment will be nothing more than a relationship between its mass and the ttl foot pounds of the original energy imparted on the additional mass

Body slam
next
B

 

bingo

on the 3;1 ratio gear it would be 6m/s except
but there is a major loss because of mass considerations
if the weight of the plate is 2k and its going 2m/s
and the weight of the roller is 2k and its going om/s
then the top speed of this system in a frictionless environment is
given its energy
drum role kids
1m/s
or fifty percent of the original
doesn't mater what the gear ratio's are
although you will lose more to torque effort in the 3:1 system than you will in the 1:3 system

 

Lost my first post, and I can't be bothered to make this one as nice as the first.

1:3 gearing gives 0.5 m/s
3:1 gearing gives 1.5 m/s

Used the cart as a frame of reference, meant that the total velocity of the top/bottom of the wheels was 2m/s.

so V(lower)+V(upper)=2m/s

1:3 => V(lower)=1/3*V(upper)

substituting => V(lower) + 3*V(lower)=2m/s
V(lower)=0.5m/s
This means the cart (over glorified ball bearing) moves at 0.5 m/s relative to the ground.

3:1 => Just substitute V(lower)=3*V(upper) instead.

Video looks genuine to me, but who knows. It actually accelerates at a number of points and I've never seen a treadmill that slopes down towards the front. Its possible that the treadmill was adjusted by the camera man to achieve equilibrium.

 

Can Ricks question really be this difficult? TheUnLogicalOne is quite close already, but even his solving has sign problems. Remember that gearing can have a sign (lower and upper wheels spin in opposite directions) as well as velocity. Try also gearings 1.1:1 and 0.9:1 we will see how Boston reacts to those. What happens at 1:1 gearing?

Boston

How can you think something that is frequently done in ice sailing perpetual motion? They reach the leeward mark much faster than air molecules they "met" at the windward mark. They don't sail dead downwind, but they could tow a sledge that would go dead downwind faster than wind. The rotary motion of the propeller/turbine solves the problems that makes jibing necessary for conventional sail vessels.

Why would someone bother to make scientific tests and publications on something that is obvious to the science community AND has no commercial possibilities. No glory or money to be expected.

Joakim

 

I looked at several of the videos, and in one of them, they used a carpenter's level to show that the treadmill platform was level. Then, they raised the front of the treadmill, showing the device running uphill.
All the treadmill videos are pointless, but you don't get that until #7, where they show that there is no wind blowing on the treadmill from any direction. Therefor, all the driving energy for the cart is being provided by the treadmill; if the treadmill were shut off, the cart would be deprived of input energy and come to a stop as soon as its inertia was depleted.
Therefor, the treadmill vids have nothing whatsoever to do with DDWFTTW.

More presently; I'm still working on Rick's problem.

 

The wind power vehicle doesn't see any difference wether the floor (treadmill) or the wind is moving. Just fix the coordinates to the treadmill and you have downwind and the vehichle is going faster than that downwind despite upphill.

Stopping the treadmill is equal to stopping the wind. Both will obviously stop the vehicle.

Joakim

 

First problem: 1:3 drive

To make matters easier for me, I arbitrarily assigned a dimension to the wheels: a circumference of 2 m.

This means the upper wheel rotates at one rev/sec.
Due to the drive reduction, the lower wheel now rotates at 1/3 rev/sec

1/3 rev/sec times 2 m = 2/3 m/sec, left to right.

****************

Second problem: 3:1 drive

Again, the upper wheel turns one rev/sec.
The lower wheel turns 3 rev/sec

3 rev/sec times 2 m = 6 m/sec, again, left to right.

***************

Now, I'd like to add that there might be some confusion (or a sneaky trick) involved in the presentation of these problems. The illustration shows the gearing as if the ratio were stated as the number of teeth on each gear, which is the way I interpreted the problem in my solution. However, (especially if the detail of the chain drive had not been illustrated) the ratios could be interpreted as the number of input rotations to the number of output rotations. If that were the case, then the results would be just the opposite. I'm not familiar with the convention as to how gear ratios are commonly described, but I would tend to think of input as the first number in the ratio.

 

No rep points yet.

Joakim
I am happy to award you the rep points if you care to provide the answer as long as you show the workings so Curtis can understand.

Curtis - the ratios are taken as the chain lead indicates but if you gave the correct answers in reverse I will still give you the points.

The point of this exercise is that if you cannot understand this very simple problem you simply do not have the reasoning ability to understand DDWFTTW. If you do not understand you will only believe if you can experience it. Even if you see it in videos you will be looking for the piece of string or other trick. The solution requires no greater reasoning ability than that expected of a fifth grader.

Rick W

 

if the plate above is moving at 2m/s relative to the geared device
and the gearing is 3:1 the the geared device would be moving at 6m:s relative to ground and the plate would be moving at 8m/s relative to the ground

you are simply saying that you wanted the answer in terms of the plates speed relative to the ground
and we gave an answer that didnt meet your unspecified parameters

you wanted differentiation
we wanted beer

because they are tacking and were talking about a dead run
and some one said something about a turbine spinning a propeller and the system somehow changing energy flow directions which was laughable
funny thing is
you've admited they dont do it at a dead run several times now in your previous posts
and thats the point Im trying to make

if you assign a weight to the system components
and state which are at rest at time 0 then you should be able to understand that the available energy to the system is x and therefore the system as a hole cannot exceed any speed relative to x

No beer for you

 

I noticed every one ignored my conclusion that if the mass of the system components is considered and therefore there kinetic potential then it takes no great reasoning to conclude that regardless of the gearing the ttl systems kinetic energy cannot exceed the sum of its components unless acted on by some additional force

works that same way in the water folks

show me the beef
B

 

thats
directly down wind faster than the wind

its what we have been talking about the hole time
and directly down wind ( a dead run )
that turbine
mounted to that boat
when they go

the relative wind speed reduces with the systems speed increase
there is progresively less and less energy available
and the system will eventually reach equilibrium
and that equilibrium will only be some fraction of the wind speed

thank you
( beer thirty )
did you catch the address earlier
cause Im headed down there now

 

Well Rick, was I right?

If this is true, then you get no rep points -- or beer.
If you're trying to teach us something, this is a really wrong way to go about it. The lesson we're learning isn't on topic yet.

 

now I think its fare since we answered your question that you answer one of mine

why doesnt someone actually show the system they claim will go
directly down wind faster than the wind
along with an analysis of how it does it
and prove the rest of us wrong in the process

hmmmm
shouldn't be that difficult should it

if that ones to difficult maybe someone could answer this one

why is no one is willing to 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

really people if you want to prove a point then simply show us all the magic machine that can accomplish the impossible
and quit with the living room tread mills and the physics 101 questions

show me the beef

were is the film of your craft going faster than the wind on a dead run
why is there no impericle data on the process being presented
why are we being shown silly irrelevant models instead of an analysis of the vehicle that will achieve this mythical accomplishment

there is ample data concerning the efficiency of down wind tacking angles
were is yours concerning dead running

once again I am not the ogre under the bridge
or the man poking needles at the kids balloons
what I am is practical
and I am not being presented practical evidence to support this claim

I will buy the first person to do so
all the beer they can drink
in a one hour period of time
at the bar of there choice
Im not offering kudo's or flowers or big sloppy kiss to whoever can show me the space ship
just a good old fashioned celebration drunk

oh
and there was no sneaky trick
or attempt at teaching
we just took the obvious solution given the lack of parameters
and frankly made the hole thing look pretty silly
( oh, I just got back from the pub )
( I asked, and no one from Sweden called in a tab for me )
( very depressing since I did get the answer given that the plates movement relative to what, was not specified )

someone is just trying to deviate from the original statement that it was possible to go directly down wind faster than the wind
which it isnt
with some irrelevant gear ratio distraction

all the irrelevant models, tread mills and we kicked its *** gearing ratios later isnt going to change anything

the ice racers go at about a 45deg angle to the wind for a reason
course no one here wants to notice that part of the process
but thats how they do it

btw
the ice yachting speed record if I remember right is about 110 or 120 or something in there, mph or maybe knots its been a while since I looked
dam fast
but not directly down wind kids
they dont even pretend to go directly down wind

they are always on a reach of some kind and always will be

love B

peace people
all in good fun
at some point we are all wrong from time to time
grace
is when you can be wrong and can admit it
its how we learn
if Im wrong
show me the system you think can accomplish direct down wind faster than the wind
if Im not
you will not be able to
I strongly suspect the next post will not include a diagram or a film of a vehicle doing so

 

Sorry bro, but you think anyone is going to invest big bucks and time away from family/work/sailing just to prove you the case and earn some free beer