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How many feet to reach terminal velocity

how many feet to reach terminal velocity

The power absorbed by a hull is its resistance (wave drag, skin friction, etc.) multiplied by the speed at which it is travelling.

The actual power produced by the engine is substantially larger, as much of the engine power is lost to inefficiencies at various points in the system. For example, a gearbox might take 3% off the engine's brake horsepower, and a typical propeller would convert 50-60% of the horsepower that reaches it into useable thrust. (Really good, well-matched props can sometimes be 70%+ efficient.)

When you look at a trolling motor rated for something like "55 lb thrust", the figure given is often the maximum thrust that motor will give. Let's say for the sake of argument that you'll be going at 2 knots, or 1 m/s with the trolling motor at full power. If you're getting the full 55 lbf of thrust at this speed, then your hull has 55 lbf total resistance at this speed. 55 lbf = 0.245 kN, so (0.245 kN)*(1 m/s) = 0.245 kNm/s = 0.245 kW. That's about 1/3 hp.

Now, once again for the sake of argument (these figures might not be representative of your actual setup) let's say your propeller efficiency on that trolling motor is 50%. Then it's actually producing 0.49 kW or 2/3 hp. Electric motors aren't perfect, let's say our example motor

is 80% efficient at full power, so we're up to 0.62 kW of electric power that's actually being sent into the thing. At 24 volts that's about 26 amps; at 12 volts, more like 52 amps.

To compare against a gasoline engine is a little harder. An outboard's SHP rating is usually at full power (very near WOT) and usually measured at the propshaft. To get the same 55 lbf of thrust as above, again with a 50% propeller efficiency, we get the exact same 0.49 kW or 2/3 hp. Shaft power has the exact same meaning regardless of what is turning the shaft.

The question becomes, then, what gas engine will produce 2/3 hp at the prop shaft. You'd probably find that a 2 hp outboard at about one-half to two-thirds throttle would be roughly equivalent.

Now, all of what I've just said rests on a big assumption- that we actually know what thrust the trolling motor is making at a given speed. Short of putting strain gauges on its mounting bracket, we generally don't- and can't- know this value accurately.

But the key message in all that math I just posted is this: You cannot make a direct comparison between thrust (force) and power, unless you also know the speed of the vessel when it is absorbing that thrust or that power.

Category: Forex

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