OIC-NW Reviewer Blog: True Wind Speed

Your vessel is on course 180°T, speed 22 knots. The apparent wind is from 70° off the port bow, speed 20 knots. The true wind direction and the speed of the wind are________.

Given:                Own course = 180°T
Own speed = 22 knots
          Apparent angle = 70°
   Apparent wind speed = 20 knots

What is asked:

True wind direction
True wind speed

Solution:

Note: We have to solve first the true wind speed before we can proceed into solving true wind direction. Regarding how to solve "true wind speed", I'll refer you my other post so that I can proceed directly to solve for the true wind direction. Here's the link: solving for true wind speed.

This is the formula for true wind direction:

True Wind Directon = Apparent wind speed² - True Wind Speed² - Own Speed²
÷ 2 x True Wind Speed x Own Speed. Then inv cosine.

And then the result will be added or subtracted to Own Course.

True Wind Direction=20²-24²-22²÷2x24x22
-660÷(2x24x22) = -0.625 inv COS = 129°

Now, Own Course 180°-129°= 51° is the True Wind Direction. I subtract because the apparent wind is in the bow. You might ask, what if the apparent wind is located at the quarter? You have to add it.

Related Posts

Your vessel is on course 15°T, speed 17 knots. The apparent wind is from 40° off the starboard bow, speed 15 knots. What is the speed of true wind?

Given:     Own course = 15°T
     Own speed = 17 knots
Apparent wind bearing = 40°
Apparent wind speed = 15 knots

What is asked: True wind speed

Solution:   Own speed² + Apparent wind speed² - 2 x Own speed x Apparent wind speed x Cos 40° and then square root the answer to get the true wind speed. This the formula to solving true wind speed:

True wind speed = √17² + 15² - 2 x 17 x 15 x Cos 40° = 11 knots

For a clearer picture of the formula, see image below.

See Related Problem

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Solving For True Wind Direction

How To Solve True Wind Speed