How To Solve True Wind Speed

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

• How To Solve For Apparent Wind Speed
• How To Solve True Wind Direction 

Storm Avoidance: Course To Steer To Have Maximum CPA

You are underway on course 120°T and can make 12 knots. The eye of the hurricane bears 150°T at 120 miles. The hurricane is on course 295°T at 20 knots. What course should you steer at 12 knots to have the maximum CPA?

Given:  Own course = 120°T
              Own speed = 12 knots
     Hurricane bearing = 150°T
      Hurricane course = 295°T
       Hurricane speed = 20 knots

What is asked: Course to steer

Solution:

1. We are going to divide own speed by the storm speed. 
2. And then press inverse cosine. 
3. Answer in #2 shall be added or subtracted the storm course. (Rule: + if storm bears quarter, - if storm bears bow)

So this is how to solve:  12 ÷ 20 = 0.6 inv cosine = 53° + 295° = 348°

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Converting Celsius to Fahrenheit

The thermometer reads 20ºC. Find the equivalent to Fahrenheit scale.

Given:              20ºC

What is asked: Fahrenheit equivalent

Solution:
This the formula to convert Celsius to Fahrenheit scale:  

 F =(^oC x ⁹/₅) + 32

F = (20^oC x ⁹/₅ ) + 32

F = 68º 

The step-by-step solution: 20 x 9 = 180 ÷ 5 = 36 + 32 = 68 degree fahrenheit.

Celsius to Fahrenheit formula

  

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Using Panofsky And Brier Formula To Determine The Storm Central Pressure

Using the formula developed by Panofsky and Brier, determine the storm central pressure if the approximate velocity of winds at low pressure area is 40 knots.

Given:              Wind velocity = 40 knots
What is asked: Storm central pressure 

Solution:

This is the formula: Pressure = 1016 - (^{speed - 3.9}/_3.5)²

The unit of speed in this formula is meter/sec. Since the unit of speed in the given is knots, we have to convert first 40 knots into meter/sec. To convert, divide 40 knots by 1.944. 

40÷1.944 = 21 meter/sec 

 

So we can now use the formula: Pressure = 1016- (^21-3.9/_3.5)² = 992 mb

This is the step-by-step solution:

21 - 3.9 = ans ÷ 3.5 = ans² = 24 and then 1016 - 24 = 992 mb

Panofsky and Brier formula

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See Related Problems:

• How to Solve Wind Speed Using Panofksy and Brier
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Panofsky & Brier Formula For Solving Wind Velocity

Calculate the approximate velocity of winds at low pressure area using the formula under the Classification Scheme developed by Panofsky and Brier, if the storm central pressure is 996 hpa.

Answer: 38.0 knots

Given:              Storm central pressure = 996
What is asked: Wind velocity 

Solution:

This is the formula: Wind Velocity = 3.9 + 3.5√1016 - storm central pressure
                                                     = 3.9 + 3.5√1016-996
                                                     = 3.9 + 3.5√20 
                                                     = 19.55 meter/sec 

Notice that the unit of the choices are in knots. So how can we convert 19.55 meter/sec into knots? What we shall do is to multiply it by 1.944.
                                   
                                                   So, 19.55 meter/sec x 1.944 = 38 knots

3.9, 3.5 and 1016 are constant. 996 is from the given. And this is the symbol of square root √.

See this picture:

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 See Related Problem:

• How to Determine the Storm Central Pressure
• How to Solve Wind Speed Using Panofsky and Brier

Solving Wind Speed Using Panofsky and Brier Formula

What is the approximate velocity of winds at low pressure area under Classification Scheme develop by Panofsky and Brier, if the storm central pressure is 1000 hpa?


Given:              Storm central pressure = 1000
What is asked: Wind velocity

Solution:

This is the formula we are going to use: 3.9+3.5√1016 - storm central pressure. In this formula, the unit of velocity is meter/sec.If your calculator is equation type like SVPAM, just key in this directly 3.9 + 3.5√1016-1000 = 17.9 m/s. This √ is the symbol for square root. Here's how.

Step 1. Simplify first the operation inside the square root.
            √1016-1000 = √16

Step 2. Key in or type this in your calculator in the correct order:
            3.9 + 3.5√16 = 17.9 m/s is the wind velocity.

Note: 3.9, 3.5 , and 1016 are constant. 1000 is from the given.

See below.

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See Related Problem:

• Panofsky And Brier Formula For Wind Velocity
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Calculating Beaufort Force If Wind Speed Is Given

What Beaufort Force in the Beaufort Scale indicates a mean wind speed of 41-47 knots?

Given:              Wind speed 41-47 knots
What is asked: Beaufort force

Solution:

Force = 1.5√41 = 9.6      

1.5 is constant. The speed we are going to use is of the lower range. The given wind speed range is 41-47 knots, so then we have square root 41 and then multiply it by 1.5. In our calculation we have 9.6. Just disregard the decimal point, so Force 9 is the answer.

See picture below:

Feel free to comment.  Thanks.


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• How to Solve Wind Speed in Beaufort Scale 

Finding Wind Speed In A Beaufort Scale

What average wind speed, in Beaufort Scale of wind fore, does Force 10 indicate?

Given:                  Force 10
What is asked: Wind speed

Solution:
One of the ways so we can answer this question is by memorizing the Beaufort Scale. However it is long and laborious to memorize. So is there any way to answer this question other than memorizing the Beaufort Scale? Yes, there is. This is the formula:

Wind Speed = 1.63 x Force^1.5   

Note: 1.63 is constant. Also the 1.5 power is constant.

So 1.63 x Force 10 to the power of 1.5 = 51.54 knots. Therefore, the answer is 48-55 knots because 51.54 knots is inside that range.

See picture below.

See Related Problem:

• How To Solve For Beaufort Force