dennis@home wrote:

"Rick Cavallaro" wrote in message
...

Energy is a particularly nasty little item. Very few people seem to
stop and realize that it's not an intrinsic property that something
has. The amount of kinetic energy something has depends entirely on
the frame of reference from which you measure it. Any frame will do
to give you the same results, but you sort of have to pick a frame and
stick with it.

For once I can agree with rick (shock + horror), use the same frame of
reference and do the sums.


try this example..

assume a wind of -10 m/s at the cart
assume the ground is doing -20 m/s


What does this mean you ask..

Do we? It seems obvious enough.

well its a cart going at twice the wind speed down wind.
Its the same as the wind being 10 m/s and the cart travelling down wind at
20 m/s which I will use for the example as it makes the maths easier. You
can use any other figures you like, the conclusions are the same.

OK

This gives a kinetic energy to the relative wind passing the cart of 100
units (-10 x -10)

To what wind? There is -10 m/s wind all around the cart, but we're only
interested in that part of it which is going through the prop. We need to
know how much that is. Suppose you put 6kg of air through it every second.
Then the kinetic energy of those 6kg of air before they go through the prop
will be (1/2)x(6kg)x(-10m/s)^2 which is 300 kg m^2/s^2, or 300 J.

but do notice it is going backwards and is being
generated by ricks prop.

No, it's there already, it's the headwind resulting from the cart's
ground speed being 10 m/s more than the wind's ground speed. What the
prop does is make the 10 m/s backwards relative wind go even faster
backwards.

Ah look at the energy needed, we are expending energy to make the wind go
backwards.

Yes, of course we are expending energy to make the relative wind go faster
backwards, but how much energy? To what speed are you wanting the prop
to accelerate the wind? Suppose you want to accelerate that 6kg of air
from -10 m/s to -20 m/s.

This would increase its kinetic energy to 1200 J, which means adding 900 J
to the 300 J it already had.

However we are not slowing the real wind at all (it is still -10 m/s) so
it is not losing any energy to anywhere

We are slowing both the relative wind ("slowing" in the sense of making
its speed more negative, i.e. from -10 m/s to -20 m/s), and also slowing
the real wind (from +10 m/s to 0 m/s). But beware, you are switching
frames when changing from relative to real wind. Both winds experience
a reduction in speed by 10 m/s, but in the two frames, the kinetic energy
changes which accompany the speed changes are not the same. In the moving
frame it increases by 900 J (from 300 J to 1200 J), in the stopped ground
frame it decreases by 300 J (from 300 J to 0 J).

Momentum change rules tell us that if you accelerate 6kg by 10 m/s each
second, this will give 60 N of thrust. If you didn't have the propeller
and instead had a mule pulling the cart using a string with 60 N tension,
without accelerating the cart, and if the cart is moving at 20 m/s, then
the mule would be doing 1200 J of work each second (i.e. delivering
1200 W of power). Obviously if you're applying a force to the cart and it
isn't accelerating, then the force must be doing something else, and in
this case it is operating a generator driven by the wheels. The generator's
1200 W power output could be used to boil an on-board kettle. Or we could
divert 900 W of this power to run a motor to drive the prop to provide the
thrust to make the mule redundant. The spare 300 W is available for losses
and air drag.

Is it a coincidence that the spare 300 J is the same energy as the real
wind is giving up?