A.T. said:
You can even outrun the moving airmass along the downwind direction:
I can't arrive at that. Imagine a cart on a frictionless track moving in a horizontal plane (top view) inclined at some angle ##\theta##, and a ##90^{\circ}## vane (sail).
Begin by applying the "Momentum Equation" for flow with constant cross-sectional properties:
$$ \sum F = \frac{d}{dt} \int_{cv} \rho v ~d V\llap{-} + \sum_{cs} \dot m_o v_o - \sum_{cs} \dot m_i v_i $$
##cv## is the control volume
##cs## is the control surface - dashed box
the subscripts ##o,i##on the momentum terms refer to the outflow, inflow respectively.
This becomes two equations in the direction ##x,y##:
Force
$$ \sum F_x = -N \sin \theta $$
Accumulation
$$ \frac{d}{dt}\int_{cv} \rho v ~d V\llap{-} = \frac{d}{dt} \left( v_x\int_{cv} \rho ~ d V\llap{-}\right) = M\frac{d}{dt}( v_x )$$
##M## is the total mass of the control volume ( cart + flow ) and it is constant assuming incompressible flow (continuity)
Momentum
$$ \sum_{cs} \dot m_o v_o - \sum_{cs} \dot m_i v_i = \dot m_{o_x} v_x - \dot m_{i_x} w $$
under continuity ##\dot m_{o_x} = \dot m_{i_x} = \dot m##
and w.r.t. the inertial frame the flow momentum outflow is non-zero ##\dot m v_{x}## ( the flow is following the cart in the ##x## direction).
Putting it all together for the ##x##direction:
$$ -N \sin \theta = M \frac{d v_x}{dt} + \dot m v_x - \dot m w \tag{1}$$
Repeat for the ##y## direction ( final result shown):
$$ N \cos \theta = M \frac{dv_y}{dt} - \dot m ( w - v_y ) \tag{2}$$
The objective is to solve the resulting system of linear first order ODE's by substitution using the track constraint:
$$\frac{v_y}{v_x} = \tan \theta \tag{3} $$
$$(3) \implies \frac{d v_y}{dt} = \frac{d v_x}{dt} \tan \theta \tag{4} $$
I solved (2) for ##N## and sub into (1), then sub (3) and (4) ( eleminating ##v_y## and its derivative) to get the following ODE:
$$ M \frac{d v_x}{dt} ( 1 + \tan^2 \theta ) + \dot m ( 1 + \tan \theta ) ( v_x - w) = 0 \tag{5}$$
Correction to (5)
$$ M \frac{d v_x}{dt} ( 1 + \tan^2 \theta ) + \dot m ( 1 + \tan \theta )v_x - \dot m w ( 1+ \tan \theta) = 0 \tag{5} $$
Edit: The solution is not valid as mass flow rate has dependecy on ##v_x## as well.
With the substitution:
$$ u = \dot m ( 1 + \tan \theta ) ( v_x - w) \tag{6}$$
Implies
$$ \frac{du}{dt} = \dot m ( 1 + \tan \theta ) \frac{d v_x}{dt} \tag{7} $$
Sub (6) and (7) into (5):
$$\frac{M ( 1 + \tan^2 \theta)}{\dot m ( 1+ \tan \theta) } \frac{du}{dt} + u = 0 \tag{8}$$You get a solution of the form:
$$ u_f = u_o e^{-\beta t} \tag{9} $$
The conclusion of all this is in the limit as ##t \to \infty## the RHS ##\to 0##, hence:
$$ \lim_{t \to \infty} u_f = 0 \implies \dot m ( 1 + \tan \theta ) ( v_x - w ) = 0 $$
In other words ##v_x \to w## as ##t \to \infty##.
We don't have to find a solution to (5), just observe the steady state response:$$ \cancel{ M \frac{d v_x}{dt} ( 1 + \tan^2 \theta )}^0 + \dot m ( 1 + \tan^2 \theta )v_x - \dot m w ( 1+ \tan \theta) = 0 \tag{6} $$
$$ \lim_{\frac{dv_x}{dt} \to 0} \implies v_x \to w\frac{1 + \tan \theta}{1 + \tan^2 \theta} \tag{7}$$
the limits of (7) in terms of ##\theta##:
$$\theta \to 0 , v_x \to w$$
$$ \theta \to 90^{\circ} , v_x \to 0 $$
So, as far as I can see it's never going to outrun the wind ##w## in the direction of the wind.
##v##(the magnitude of the cart velocity) is given by:
$$v = \frac{w ( 1 + \tan \theta)}{ \cos \theta (1 + \tan^2 \theta)} \tag{8}$$
Where have I blundered that allows the cart to go faster than the wind in the direction of the wind?
