How much horsepower to weight is required for a 1:1 thrust ratio? I did a math experiment today, but wanted to verify the results.

I wanted to figure out how much horsepower is needed for an aircraft to hover in midair with no wings/lifting body (1:1 thrust/weight).

So to start I found that a unit of horsepower is the amount of work required to move 550 pounds 1 foot every second. 

Then I wanted to find out how many foot/pounds of force gravity places on 1 pound of material. With a static acceleration of 9.8m/s (32f/s), it seems that 1 pound has 32 foot pounds of gravity acting on it.

Now for my equation: 

Hp = 550 (f/p)
Gravity = 32 (f/p)
Hp / Gravity = 17.1875


So it seems 1 hp can hold 17 pounds in the air, but to incorporate propeller inefficiency, I just used 70% of that. This leaves me with 1 hp being able to vertically lift 12 pounds when accounting for the inefficiency of a propeller. 

Is this a relatively correct assumption? I know thrust changes as the aircraft accelerates, but in a hovering state the acceleration is 0 f/s as a whole. So will these numbers apply?

Edit: I initially used a specific ultralight to prove how high the numbers were, but I used wrong data from the spec sheet. So I removed that reference from my question.
Horsepower cannot be converted into thrust without knowing the speed at which this is done. In your case this is zero, so the special equation for static thrust applies. With $T$ for thrust, $P$ for power, $\rho$ for air density and $d_P$ and $\eta_P$ the propeller diameter and efficiency, respectively, this is $$T_0 = \sqrt[\LARGE{3\:}]{P^2\cdot\eta_{P}^2\cdot\pi\cdot d_P^2\cdot\frac{\rho}{2}}$$
With this equation you cannot simply say how many HP are required to lift that many pounds; instead, you need to add the propeller geometry. If we assume a very efficient propeller with a large diameter, we can combine blade size and mass into disk loading – here 8.5 kg/m² or 41.5 lbs/ft² is a good value. Now we can write, using standard atmospheric density at sea level:
$$\frac{T_0^2}{P^2\cdot\eta_{P}^2} = \frac{2\cdot1.225\,\frac{\text{kg}}{\text{m}^3}}{8.5\,\frac{\text{kg}}{\text{m}^2}\cdot g} = 0.0294\,\frac{\text{s}^2}{\text{m}^2}$$
assuming for simplicity a 1 m propeller diameter. If we now use 70% for propeller efficiency, as you did in the question, the power to mass ratio becomes 8.33 Watt power per N of weight force or 81.7 Watt per kg. Converted into funny units this is 0.05 HP per pound of mass. 1 HP can hold 20 lbs in the air, but that will not be enough to even spin up the propeller unless it can adjust blade pitch. For practical usability I recommend to double that number. If the propeller diameter shrinks and disk loading increases, efficiency will drop and the power required will rise.