The Wind Powered Sail-less Boat

Guillermo:

Wow ... finally you answered the question that I asked you many pages ago. Excellent.

Now I'd be interested in knowing why you continue to ask for more and more complicated analysis when the cart sail functions exactly as an ice-boat (or other) sail on a broad reach.

Other than the faster rotation (the world is a larger diameter than our shaft after all), put a fly on the tip of the cart sail and one on the tip of a traditional sailing rig with a downwind VMG of greater than 1.0 and they will tell the EXACT same story. They both "exceeded the speed of the wind (VMG) by tacking downwind..."

This problem here is not one of physics Guillermo -- the physics aspects of sailing have been done to death. The problem is you won't trace the path of the cart's propsail in your mind or on paper and ask yourself ... "how does the path of that sail differ from the path of the traditional sailing rig?"

You will find that it differs only in diameter -- the ice-boat sail takes a very large, world sized spiral path, and the cart sail takes a much smaller, shaft sized spiral path.

JB

 

Guillermo, would you care to elaborate on your 'completeness' remark?

If you replied to my 'is this a joke?' question, I missed it, sorry.

 

We are advancing....
I'll duly answer your recent comments soon, gentlemen.

In the mean time let me put some other question.

I think Mark’s reasoning has the problem of not saying where the energy to make the boat go faster than the wind comes from, but he asumes the premise of the boat is already in that condition.

Now, let’s do some insight into his demonstration.

When the boat speed is the same as the wind speed, the net thrust available has to be Fet = 0, obviously.

Then:

( V / V −W ) ηt ηg ηp -1 = 0

To simplify the writing, let’s call ηt ηg ηp = α

So
( V / V −W ) α -1 = 0
( V / V −W ) α = 1
V α = V – W
W = V (1 - α)

But, as V = W, we need α to be 0 in this condition, so a 0% efficiency of the system. How do we eat this?

Of course we have a singularity at V = W and we should solve such singularity using the limits theory, not linear algebra.

I would like Mark to address this problem.

Thanks in advance.

 

Not true. He shows that there is excess energy available, and this is true from 0 mph to greater than wind speed.

It may be "obvious" but it ain't true.

How does your ice-boat have any thrust when it's downwind VMG is exactly 1.0 wind speed?

 

Why Fnet must be zero at V=W? For a conventional sailboat it needs to be negative, but for propeller turbine boat it can be very positive and nothing special happens at V=W. It is just a singularilty in formulations that bases the efficiency to apparent velocity.

Since the definition of the propeller efficiency has apparent velocity as a multiplier, the efficiency is zero at V=W. This actually means that the thrust can be infinite at any finite power without violating the formulation. However, in practise it is impossible to have an infinite thrust and the propeller needs to be very big in order to have a good thrust.

Joakim

 

Nope. He doesn't demonstrate that. He asumes that as an intrinsic needing of the boat in a DDWFTTW condition (as it is, of course). But he doesn't explain where that excess thrust comes from.

Because it is not going DDW.

 

First of all, for the benefit of the other readers, as the vehicle accelerated, it goes through the following states in order:
1) V-W<0 , the vehicle is slower than the wind
2) V-W=0 , is the DDWFTTW threshold
3) V-W>0 , the vehicle is faster than the wind

The analysis does allow V-W<0, but within limits. If the vehicle isn't moving and V-W is at its most negative value, the prop isn't even turning, so certainly the prop analysis is bogus --- it's simply acting like a parachute. But once the prop tip speed ramps up to a significant fraction of |V-W|, then the analysis should be OK.

 

Not quite. In this case of V-W=0, there is no air drag. But there's still the water drag. Or for a wheeled vehicle there's still the rolling resistance.

 

This is ABSOLUTELY AND DEAD WRONG.

When the boat speed is the same as windspeed the water turbine is already pumping power into the air propeller. The near field airflow past the blades is rearwards and the prop is producing forward thrust.

If you cannot appreciate this then you are denying that any propeller is capable of producing static thrust.

As I have stated before, stand behind an aircraft when it is testing brakes and confirm that the nearfield airflow can be greater than the farfield flow. It is simply a matter of applying power to the air to accelerate it past the blades. If the boat is doing 7kts in a 7kt breeze it can extract ample power from the water to power the prop providing the gearing is set up so the turbine overcomes the prop. A big slow moving prop will be best because it acts on a large area meaning the nearfield velocity is lower so thrust to power is high.

But at least we have got to the specific point where your understanding limits your grasp of the concept.

Rick W

 

That the available energy for the same control volume is not equal. And it has to be and produce the same effects in both experiments for them to be equivalent.

 

Here is a link for a simple formulas to calculate the approximate thrust/power for the V=W case: http://www.mh-aerotools.de/airfoils/prpstati.htm

 

Prove it. (where the energy comes from to reach V = W).

 

There is no physical singularity at V=W. The ηp/(V-W) = 0/0 singularity is there in the equations only because of the ηp definition, as Joakim points out. What I did in the derivation is in effect to "undo" the ηp definition, which resolved the 0/0 ratio into its actual limit.

I could have started without ever introducing ηp , and jumped straight to the actuator-disk theory stuff on page 2, but I decided that that was actually less physically clear. The initial stuff on page 1 is relatively simple and useful I think, even though it has the mathematical 0/0 problem.

 

Of course it is. The ice craft tacking downwind is extracting its energy from a bigger amount of wind than when going DDW, that's why the VMG is faster than the wind. But to justify the DDW thing you need to use an extra energy nobody has demonstrated where it comes from.

 

This applies to the the farfield condition. It is the same as saying a tug operates at zero efficiency when it is doing a bollard pull. I bet the builders and owners would not like to hear this. But it is correct however irrelevant to comparing performance of such vessels. And likewise irrelevant to analysing the V=W condition.

What is more important is the the thrust available for the applied power to the prop. The minimum power required to achieve a given thrust under static conditions is:
Power = (4/pi/9/rho)^0.5 x Thrust^(3/2)/D in Watts
Where rho is air density and D is prop diameter (in metric standard units). In practice the power will be higher because of blade losses. This shows the significance of diameter.

Rick W