Efficient cargo ships

Projects like these smell way too much like greenwashing.

Mehh...I got nothing to lose. I can always e-mail Wilson, ask for the actual numbers and the worst they can do is tell me to eff off.

 

Pilots use the data all the time:

The kite's controlling computer would use pilot data (probably RT -real time - satellite data)

At different flight levels, you have different winds.

Just use the size of the kite, like you would fill a sail at sea level for the energy you would get for the size of sail.

You would need to do a little trigonometry (Physics) to get the actual "pull" from the kite. I would reduce the amount a few percent for the 'rig.'

You should be able to get your numbers from that point.

If not let me know, or you can call the kite company that is doing this.

http://www.google.com/#hl=en&tbo=d&...a986a3276db36d&bpcl=39967673&biw=1416&bih=596

 

Not greenwashing.

Flight level data is well known. Aviation & military is what really pushed our modern weather forecasting.

THERE IS A LOT of wind up there .... just waiting for you to grab some.

 

Believe me, I'm not trying to be difficult, but that is nothing more then a sales blurb.

I was hoping to find a lot more detail. From which, I would create a Excel sheet, plug in numbers and do an analysis. Like I said in my first post, I have exact wind and direction numbers from 80 meter wind turbines. If I were given one hour energy versus wind numbers from sails or kites, I can tell you exactly how much energy it will produce in my location. In fact, I can do it from any given location if I had the yearly wind and direction numbers.

Something tells me that companies may not want to give those numbers out. Particularly if they are doing this for greenwashing. That is why I was hoping to find someone here who is really good with calculating wind in sails power. Or had existing numbers.

Anywho.....I'm not trying to argue or annoy anyone.

 

Doing simple wind x times x hours can be done but wont work with any kind of accuracy. Winds and direction vary too much over an hour. On top of that, given that it's a kite with all kinds movements in height and direction, therefore a large variation in "pull", I need something already based on empirical data. Or at the very least, an empirical based percentage of wind direction.

Given some basic numbers, I can calculate "pull" for any given moment in time, but I have no idea where the kite will be next. Will it be 60 degrees overhead and pull up more then pull forward? And how often or how long it will be in any given spot for any given time. I thought sails would be easier in that respect. And I still don't know what "pull" I have for any given size of kite. Or the relationship between wind speed, size and pull.

A wee bit complicated...but my good buddy Excel can help.

In comparison, commercial size wind turbine power is far, far easier to calculate. It rotates to face it in whatever direction the wind takes it and the power is predictable based on blade size and design. Solar even easier.

I got your e-mail and I'm game if you are. We can do it in public for peer review or we can do it by mail.

 

OK.

The kites they are using are using GPS systems (the real GPS, not sure if they can access the weapons grade). The computers are controlling where it needs to be to get good winds.

Jets airliners do the same thing, the problem is they are flying through soup at about 550 kts (usually True Air Speed TAS is MUCH greater) A kite pulling forward at 20 to 30 kts is much easier to place in good winds, and since it is not carrying passengers, it can be placed in really strong winds on purpose.

 

Hi Spartan,
By my calculations, a 2000kw diesel engine would use 10.56 tonnes of fuel in 24 hours
.
Horsepower = drag x speed.
At 5m/s (9.7 kts) boat speed the drag must be 2,000,000/5 = 400kN

Using Benellis lift theory, I very roughly calculate that 1 square metre of wing sail could create 15N of force at 5m/s, and 61.25N of force at 10m/s relative wind speed.

Thus a 320 m2 sail could produce 4.9kN of thrust at 5m/s relative wind speed.
The engine could now only have to provide 395.1kN of thrust.
395,100 x 5= 1975kw and could use 10.43 tonnes of fuel per day.

At 10m/s relative wind speed the 320 m2 sail could produce 19.6kN of thrust. The engine could then only be required to produce 1900kw and use about 10 tonnes per day.

Once you have calculated the average wind vector,and then the average relative wind vector for a voyage, you can calculate the average fuel savings for a certain wing sail.

However, you must understand that my knowledge of aerodynamics comes from watching the film "Mary Poppins", so it may all be complete nonsense.
Best Wishes,
Adam