Solving For Maximum CPA

You are underway on course 50*T and speed of 12knots. the eye of the storm is moving towards 265*T at 22knots. if you maneucer at 12knots to avoid huriccane. what could be the cpa? ans 63miles

This is a question from one who follows this blog. 

I already tackled a problem like this on this site. Here's the link Storm Avoidance: Maximum CPA

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How To Solve For Geostrophic Wind Speed

I hope u can help me with this problem I encountered on my review material, " if the distance between 2 consecutive isobars( drawn at intervals of 4mb) is found to be 100nm and the latitude is 50 degrees, what is the Geostrophic wind speed?" thankyou in advance :)

This is a question submitted to us. I found this question new to me. Can you help us how to solve this problem? Use the comment box for your solution. Thanks.

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How To Determine Storm's Position

Hello dear readers,

It's been good that many are following this blog of mine, and many of them interacted with yours truly. Having made this site (to help those who review to become marine deck officer) doesn't mean I'm all knowing.

So since there are questions submitted to me which I am incapable to answer to, therefore I posted these questions from time to time hoping others may see ,and, if they know, would give us the answer. Or who knows it is you who can help us here.

And this is one those questions. This is regarding how to determine the storm's position. If you know, use the comment box below and we would be glad to read it.

a rough guide, in the absence of an accurate meteorological data, if the corrected barometer reading is 5 hPa (0.15 in.) lower than the local normal and the wind force is about 8, the tropical cyclone's center will be about how far away? answer:100nm

 thanks in advance

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Storm Avoidance: Maximum Distance

You are underway on course 050°T and your maximum speed is 13 knots. The eye of a hurricane bears 100°T, 120 miles from your position. The hurricane is moving towards 275°T at 25 knots. If you maneuver at 13 knots to avoid the hurricane, what could be

Storm Avoidance: Maximum Distance

Given:    Own course = 50°T
               Own speed = 13 knots
     Hurricane bearing = 100°T
      Hurricane course = 275°T
       Hurricane speed = 25 knots

 What is asked: Maximum distance

Solution:

We have to solve first for the course to steer to have the maximum CPA before we can solve for the maximum distance.

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)

         13 ÷ 25 = 0.52 inv cosine = 58.67° + 275° = 333.67°

We can now proceed in solving for the maximum distance in avoiding the storm.

1. We have to add 180° to the course to steer, and then minus 360°.
2. The answer in #1 will be subtracted by the storm bearing.
3. Multiply the cosine of your answer in #2 times the distance of the hurricane from your position.

333.67° + 180° = 513.67° - 360° = 153.67° - 100° = 53.67° and then 120 miles × Cos 53.67° = 71 miles is the final answer.


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Solving For True Wind Direction

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: 

1. True wind direction
2. 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

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

Solving For Apparent Wind Speed

Your vessel is on course 135°T, speed 18 knots. From the appearance of the sea you estimate the speed of the speed of the true wind as 24.5 knots. The apparent wind is 40° on the starboard bow. Determine the speed of the apparent wind. 

Given:         Own speed = 18 knots
            True wind speed = 24.5 knots
      Apparent wind Angle = 40°

What is asked: Apparent wind speed

Solution:  Own speed² + True wind speed² + 2 x Own speed x True wind speed x Cos 40° and then square root the answer to get the apparent wind speed. This the formula for apparent wind speed: 

Apparent wind speed = √18² + 24.5² + 2 x 18 x 24.5 x Cos 40° = 39.99 knots  

Well, as you notice, there's a little discrepancy if we are going to compare our computed answer with the official answer which is 36.0 knots. So what shall we do? My instructor during my review used to advice us to choose the nearest answer in cases like this. That is to select the best answer, so to speak.

Apparent wind speed

This is very problem taken from my reviewer book. 

Feel free to comment.


See Related Problem

• How To Solve For True Wind Velocity
• Solving For True Wind Direction 

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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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
• Panofsky and Brier Formula for Wind Speed

         

        

 

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
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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
• How To Determine The Storm Central Pressure

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.

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• How To Solve For Beaufort Force