How much lift is created by a half-sized rotor?
For the hovering rotor, the stationary case, it can be safely assumed that the lift $L$ is a function of the input power $P$, the diameter $D$ of the rotor and the air density $\rho$.

Thus, $L = f(P,D,\rho)$

where $f$ is a function to be determined.

From dimensional analysis, the lift $L$ can be easily derived:

The variables are Lift $L$, dimensions $MLT^{–2}$; Power $P$, dimensions  $ML^2T^{–3}$; Rotor diameter $D$, dimensions $L$ and air density $\rho$, dimensions $ML^{–3}$

The variables form a non-dimensional product $k$

$k = L^a\cdot P^b\cdot D^c\cdot \rho^d$  where $a,b,c,d$ are numbers to be determined.

Let’s form now a parallel product $k^*$ with the dimensions:

$k^* = (MLT^{–2})^a  (ML^2T^{–3})^b  (L)^c (ML^{–3})^d$

Clearly, $k^* = M^0 L^0 T^0$... We now take the exponents for each dimension:

$a  + b + d = 0 \\
a + 2b + c – 3d = 0 \\
–2a – 3b = 0$

We make $a = 1$, since $L$ is the variable we’re going to solve for. 

$b = –2/3 \\ 
d = –1/3 \\
c = –2/3$

Then, 

$k = L^a\cdot P^b\cdot D^c\cdot \rho^d \rightarrow  k = L\cdot P^{–2/3}\cdot D^{–2/3}\cdot \rho^{–1/3}$ 

Solving for $L$

$L = k\cdot P^{2/3}\cdot D^{2/3}\cdot \rho^{1/3}$

where $k$ is a constant

Hence, for rotor diameters $D_1$ and $D_2$, and for the same power and air density, the corresponding lifts $L_1$ and $L_2$ are:

$L_1/L_2 = (D_1/D_2)^{2/3}$

For the case of $D_1 = 13 ft$ and $D_2 = 26 ft$, $L_1/L_2 = (13/26)^{2/3} = 0,63$

In other words, the smaller (13 ft) rotor gives you, for the same power and air density, just 63% of the lift attained with the larger (26 ft) rotor.

That's for the hover. For the climb, you'll need extra power. In order to move 635 kg vertically upwards at 1200 ft/min (6,09 m/s) you would need $635 \cdot 9,8 \cdot 6,09 m/s = 37,9 kW...$