I had expected when I asked if it was possible to use the 10x6 prop and a 6 hp motor to power the boat, that the answer would be based on the available torque and power of the engine to turn the prop. What I got was that it may be possible for the boat to make that speed with only 6 hp, and that the prop pitch could not have been 6". According to the nomogram in Dave Gerr's "The Nature of Boats" a 6 hp engine should be able to turn a 10" prop at 1800 rpm. The nomogram also indicates that the prop should be 8" diameter if it is turning 2250 rpm. The Fn for a 17' (4.1 m) hull moving at 14.8 mph (6.6 m/sec) is 1.01, so it is not a displacement hull and the 20-30% slip estimates may be too high.
Next question, does the following argument work at all as an initial estimate of the pitch needed to attain the observed speed?
The known values are prop diameter and boat speed. The gear ratio and pitch are unknown, but need to be determined to make the boat perform as advertised. I thought that it would be routine to estimate the thrust for a given hp and speed and a trivial matter to estimate the requisite pitch. All I found was one rough estimation of thrust at a given speed on this forum posted by Daiquiri. Slip could be accounted for in Newton's action/reaction law by multiplying the thrust of the prop by the volume it passes through in 1 second to find how fast that volume should be moving backwards in the immediate vicinity of the prop. That should account for the fact that slip decreases at higher speeds. I varied the pitch by trial and error to find where the swept volume accelerated by the prop was moving at the same velocity as the slip. The assumptions being that slip may be accounted for by the fact that the water the prop is interacting with is moving in the opposite direction, and the slip is zero relative to that imaginary cylinder.
Using these calculations:
If the Happy Clam does make 14.8 mph (6.6 m/sec) at 225o rpm, the slip should be a function of the prop thrust and mass of water accelerated the prop.
1. The The 25.4 cm diameter propeller moving at 6.6 m/sec will act on 335 liters (speed x blade area) of water.
2. If the prop is estimated to be 50% efficient, the thrust works out to 356 N (79.8 lb)
3. The water acted on by the prop in one second is accelerated to 1.1 m/sec (356/335).
4. Making the rash assumption that the water the prop is acting on is moving backwards at 1.1 m/sec, and the effective slip is 0, a pitch of 8.1 inches would produce the observed speed of 6.6 m/sec with 14% overall slip. Shifting gears, as suggested by Gonzo; at 1800 rpm with a pitch of 10.1, the water is still moving backwards at 1.1 m/sec with a 14% slip.
Next question, does the following argument work at all as an initial estimate of the pitch needed to attain the observed speed?
The known values are prop diameter and boat speed. The gear ratio and pitch are unknown, but need to be determined to make the boat perform as advertised. I thought that it would be routine to estimate the thrust for a given hp and speed and a trivial matter to estimate the requisite pitch. All I found was one rough estimation of thrust at a given speed on this forum posted by Daiquiri. Slip could be accounted for in Newton's action/reaction law by multiplying the thrust of the prop by the volume it passes through in 1 second to find how fast that volume should be moving backwards in the immediate vicinity of the prop. That should account for the fact that slip decreases at higher speeds. I varied the pitch by trial and error to find where the swept volume accelerated by the prop was moving at the same velocity as the slip. The assumptions being that slip may be accounted for by the fact that the water the prop is interacting with is moving in the opposite direction, and the slip is zero relative to that imaginary cylinder.
Using these calculations:
If the Happy Clam does make 14.8 mph (6.6 m/sec) at 225o rpm, the slip should be a function of the prop thrust and mass of water accelerated the prop.
1. The The 25.4 cm diameter propeller moving at 6.6 m/sec will act on 335 liters (speed x blade area) of water.
2. If the prop is estimated to be 50% efficient, the thrust works out to 356 N (79.8 lb)
3. The water acted on by the prop in one second is accelerated to 1.1 m/sec (356/335).
4. Making the rash assumption that the water the prop is acting on is moving backwards at 1.1 m/sec, and the effective slip is 0, a pitch of 8.1 inches would produce the observed speed of 6.6 m/sec with 14% overall slip. Shifting gears, as suggested by Gonzo; at 1800 rpm with a pitch of 10.1, the water is still moving backwards at 1.1 m/sec with a 14% slip.
