The actual propeller load tells you little about the percentage load on the engine, as these things are normally defined. What you have calculated in your table is the power the engine is producing, compared to what it is capable of at maximum rpm. Usually percentage load is defined as the actual torque, compared to the maximum torque available at a given rpm. Torque is a measure of the forces on the pistons, bearings, etc. The Cummins data like the one I've posted does not show fuel consumption at maximum power output, which could be used to construct your table.
If I installed a prop on the QSB in the table that only allowed it to reach 1500 rpm at full "throttle", it would burn only about 10 gal/hr compared to 20 gal/hr the same engine will do at maximum power output. So even though the engine is producing all the torque and power of which it is capable (at that rpm), would you call that 50% load? Note that it will also be near its peak torque output. I would call that 100% load. So would the engine instruments on an electronic engine.
For example, on my QSB, I cruise at 1300 rpm burning 1.9 gal/hr. The Cummins ECU reports this as 33% load. That is correctly illustrated in the Cummins graph, which shows about 50 hp absorbed by the prop and 150 available from the engine. But using the formula that Eric suggests, it would be 1.9/20.1 g/h or 9%. I believe the Cummins reported figure is correct.
If I installed a prop on the QSB in the table that only allowed it to reach 1500 rpm at full "throttle", it would burn only about 10 gal/hr compared to 20 gal/hr the same engine will do at maximum power output. So even though the engine is producing all the torque and power of which it is capable (at that rpm), would you call that 50% load? Note that it will also be near its peak torque output. I would call that 100% load. So would the engine instruments on an electronic engine.
For example, on my QSB, I cruise at 1300 rpm burning 1.9 gal/hr. The Cummins ECU reports this as 33% load. That is correctly illustrated in the Cummins graph, which shows about 50 hp absorbed by the prop and 150 available from the engine. But using the formula that Eric suggests, it would be 1.9/20.1 g/h or 9%. I believe the Cummins reported figure is correct.