This is how Gene described the conversion in his "Technical Manual" on his G8-2 engines.
"Horsepower versus Jet Thrust
Most people are used to evaluating power on the basis of an engine's horsepower rating. However, an engine that produces pure thrust (rather than torque at a rotating output shaft) is rated in pounds of thrust. Jet thrust cannot be stated in horsepower, unless a specific speed is given. By calculation, the horsepower rating of any jet engine operating in a static condition is zero, regardless of the amount of thrust. It's possible, however, to make comparisons on the basis of tests performed on conventional power plants in similar applications. This would be referred to as "Equivalent Horsepower" (HP1).
For example, if tests on a conventional reciprocating engine showed that a 10 hp engine with a fixed-pitch propeller produced a static thrust of 50 pounds at full throttle, then a value of 5 pounds of thrust per horsepower might be used. Since the G8-2-40 engine develops 40 pounds of static thrust at full power, HP1 would be 8 hp. The shortfall of this approach, however, is that the thrust of a conventional engine/propeller system declines with increased speed, whereas the thrust of the G8-2 jet increases with speed (using an air scoop). So the two power systems are not directly comparable (click to see graph). In addition, one would have to be confident that the comparison horsepower value was actually representative of a typical ICE/propeller power system...
...If a G8-2-40 jet were moving at 400 mph (586 ft/sec), the foregoing equation would result in 42.7 hp (HP = 40 x 586/550 = 42.7). The same engine moving at 600 mph (954 ft/sec) would result in 69.5 hp (HP = 40 x 954/550 = 69.5). In order to achieve meaningful results, however, the value for velocity must be appropriate for the particular application. In other words, the velocity must be based on the result of the particular thrust acting to overcome the resistance of a particular body. The speed at which the two factors become equal (thrust and resistance) will be the maximum speed attainable in the particular application. This would be the appropriate velocity to use in the foregoing equation."
Sorry but the formulas, etc would not paste properly, but hopefully this help.
I would also suggest that you run this past Luc and Viv to get their take on it as they are working full time with related concepts.