Shape Memory Metal framed sail
Well, this is supposed to be an engineering SE, so let's do some calculations. 

Nitinol has a resistivity of 76\$\mu\Omega\$-cm at room temperature (higher when hot). So the resistance of that wire should be \$\rho L/A\$. 

A = \$\pi r^2\$ = 0.0081cm\$^2\$.

L = 30.5cm 

So R ~= 0.29\$\Omega\$ 

That's fairly low. For, say 20W into the wire.. we'd need about 8A, which is 2.3V.  

Let's say we use fresh alkaline AA cells which have a short circuit current of about 4-5A, so their Thevenin equivalent is a 1.5 source with 0.33 ohms in series. Three parallel strings of three AA cells (9 cells total) would yield a 4.5V supply with 0.33 ohms internal resistance (well matched for maximum power transfer). 

Current would be about 7.3A, which is pretty close to our desired 20W. If it's not enough, four strings of four (16 cells total) would give 25 or 30W.