"Andy Champ" wrote in message
...
toad wrote:
Care to explain why a windmill which is capable of powering itself
forward against it's own drag can only do it with a true wind? How
does it know if the wind it is 'feeling' is true or not, it has no
concept of true wind which is merely the wind speed and direction at
an arbitary stationary point.

I thought we'd done this to death.

Don't think force, think energy. Imagine the boat is still, the wind is
blowing over it, and the mill is connected to a winch to a fixed point.
There will be a level of gearing low enough somewhere, so that the boat
can wind itself forward against the winch.

As you increase the gearing, you will increase the amount of power needed
to drive the winch. (not the torque, AKA force, but the POWER).

As you increase the speed of the winch with more gears you will need more
and more power.

Be careful when you crunch the numbers on this. Drag from the mill is
proportional to the square of the apparent wind, power proportional to its
cube, and power to propel the boat proportional to boat speed times drag.
If you forget hull and aerodyamic drag and transmission losses you can
fool yourself into thinking the boat *will* keep accelerating forever.
Even so, if the true wind is zero you get no excess of power whatever you
do.

This is from an Excel spreadsheet.

Real Wind-- 10 Boat Apparent req'd Avail. Excess
speed Wind Drag power power Power (%)
1 11 121 121 1331 1000%
2 12 144 288 1728 500%
3 13 169 507 2197 333%
4 14 196 784 2744 250%
5 15 225 1125 3375 200%
6 16 256 1536 4096 167%
7 17 289 2023 4913 143%
8 18 324 2592 5832 125%
9 19 361 3249 6859 111%
10 20 400 4000 8000 100%

100 110 12100 1210000 1331000 10%


If I reset the real wind to zero, all the excess power figures go to
zero - which implies zero losses in the system.

To make it easy for anyone else, the formulae on the "9" line of that
read:
=A11+1
=A12+B$1
=B12*B12
=C12*A12
=C12*B12
=(E12-D12)/D12

I hope for your sake you are trolling, joking or both.