## The Power to Push a Sailboat

# The Power to Push a Sailboat

**How much power should you expect to use to push your sailboat? **

The equation used to calculate the power needed to push a boat through water is called the

resistance equation, which is given by:

P = 1/2 * rho * v^2 * A * C_D

where:

P is the power required in watts

rho (rho) is the density of water, usually taken to be 1025 kg/m^3

v is the speed of the boat in meters per second

A is the wetted surface area of the boat in square meters

A = 2 * (0.453 * LWL + 0.4425 * BEAM^0.5)^2

C_D is the drag coefficient, which is a dimensionless parameter that depends on the shape and

roughness of the boat’s hull and appendages.

This equation gives an estimate of the power required to overcome the frictional drag and wave-making

resistance of the boat. The resistance increases with the speed of the boat and the square of the speed,

making it more difficult to maintain high speeds. The hull speed of a boat is defined as the maximum

speed at which the bow wave created by the boat will not grow larger and start to break. Beyond the hull

speed, the resistance increases rapidly, making it very difficult to achieve higher speeds.

The hull speed of a boat can be estimated using the formula:

V_h = 1.34 * (LWL)^0.5

where:

V_h is the hull speed in meters per second

LWL is the waterline length of the boat in meters.

So the power required to push a boat through water will be highest at the hull speed, and will

increase as the speed of the boat approaches the hull speed.

**Using these equations to calculate values for your own boat: **

The calculator below uses the equations just discussed, to estimate your boats hull speed and then output a graph that shows how much power is required to push a given boat at a given speed (in either m/s or Kts).

Further, the graph will indicate a recommended “cruising speed” that walks down the steepest part of the curve (about 65% of the boats hull speed) to provide a reasonable tradeoff between power usage and speed.

Given this information, one can calculate the amount of battery capacity they should have onboard, for a given amount of time motoring between charges.

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Let’s go one step further and calculate the size battery we need to motor to Catalina Island from Long Beach (about 25 miles). Let’s assume NO WIND, so we’re motoring the whole way there (for this exercise, we’ll assume we’re sailing back, however we could just double the final results for a round trip).

To estimate how much energy it will take to motor our 33′ sailboat for 25 miles at a speed of 5.8 miles per hour (5 Kts), we first need to calculate how much power will be needed to push the boat at that speed.

Based on the calculator results above (using imperial, 29′ LWL and 11′ beam) we get a power requirement of 2.1 kW to push the boat at 5.8 miles per hour (5 Kts).

We use the following formula to calculate the energy required to travel 25 miles:

Energy = Power x Time

where Power is the power requirement, and Time is the time taken to travel the distance.

To calculate the time taken to travel 25 miles at a speed of 5.8 miles per hour, we can use the formula:

Time = Distance / Speed

where Distance is the distance traveled, and Speed is the speed of the boat.

Converting the distance to kilometers, we have:

Distance = 25 miles x 1.609 km/mile = 40.225 km

Converting the speed to kilometers per hour, we have:

Speed = 5.8 miles per hour x 1.609 km/mile = 9.334 km/h

Using the formula, Time = Distance / Speed, we get:

Time = 40.225 km / 9.334 km/h = 4.307 hours

Finally, using the formula, Energy = Power x Time, we get:

Energy = 2.1 kW x 4.307 hours = 9.0407 kWh

So, it will take approximately 9.04 kWh of energy to motor a 33′ sailboat for 25 miles at a speed of 5.8 miles per hour, assuming a power requirement of 2.1 kW.

To calculate the cost to charge a 10 kWh battery at a cost of electricity of $0.10/kWh, we simply need to multiply the battery capacity by the cost of electricity per kWh.

In this case, the cost to charge a 10 kWh battery is:

10 kWh x $0.10/kWh = $1.00