Wind Velocity: Findings & Insight Needed

In summary: The velocity of the wind relative to the ground would actually be 13 km/h, 39 degrees east of south. Thanks for the question.
  • #1
Iniuria12
10
0
Having a little bit of trouble with this and not sure if my findings\thoughts are right and would like some insight.

If a boat is traveling south at 10 km/h and a person on the deck measures the wind velocity to be 8 km/h east, would the actual wind velocity (velocity of wind relative to the ground) still be 8 km/h east or would the answer be resolved by drawing a vector diagram and having a velocity roughly 6 km/h at heading 53 degrees east of south??

Haven't been able to find a definitive answer in correlation to this problem and not sure if I'm over-examining the question or not.

Appreciate any insight.
Thanks
 
Physics news on Phys.org
  • #2
Iniuria12 said:
Having a little bit of trouble with this and not sure if my findings\thoughts are right and would like some insight.

If a boat is traveling south at 10 km/h and a person on the deck measures the wind velocity to be 8 km/h east, would the actual wind velocity (velocity of wind relative to the ground) still be 8 km/h east or would the answer be resolved by drawing a vector diagram and having a velocity roughly 6 km/h at heading 53 degrees east of south??

Haven't been able to find a definitive answer in correlation to this problem and not sure if I'm over-examining the question or not.

Appreciate any insight.
Thanks

It depends on how the question is phrased... if the wind is 8km/h east relative to the boat, then the wind velocity relative to the ground is different. I think in your example, since they explicity said that someone on the deck measured the wind velocity, they mean to say that the 8km/h is relative to the boat and not to the ground... But I don't get 6km/h... how did you get that?

Suppose two objects are moving... x is the velocity of object 1 relative to the ground (or any reference)... y is the velocity of object 2... (note that x and y are vectors)

the velocity of y relative to x is y-x. The velocity of x relative to y is x-y.
 
Last edited:
  • #3
Ok.. So now I'm royally confused. I presume I'm just working myself up over nothing and just confusing which calculation method I should be using.

Drawing a vector diagram, I would come up with results, boat:10 km/h south, wind:8 km/h east, and resultant:13 km/h, 39 degrees, east of south..

Now with the velocity of y relative to x being y-x, that would be 8 km/h - 10 km/h = -2 km/h.

In other words, I'm just really unsure as to what I would use as my absolute answer, thus being actual wind velocity relative to the ground.
 
  • #4
Iniuria12 said:
Ok.. So now I'm royally confused. I presume I'm just working myself up over nothing and just confusing which calculation method I should be using.

Drawing a vector diagram, I would come up with results, boat:10 km/h south, wind:8 km/h east, and resultant:13 km/h, 39 degrees, east of south..

Now with the velocity of y relative to x being y-x, that would be 8 km/h - 10 km/h = -2 km/h.

In other words, I'm just really unsure as to what I would use as my absolute answer, thus being actual wind velocity relative to the ground.

Well... this is how I understand the problem... You're given the velocity of the boat relative to the ground... I'll call it [tex]\overrightarrow{v_b}[/tex] = 10km/h south. You're also given the velocity of the wind relative to the boat. I'll call this I'll call it [tex]\overrightarrow{v_{wb}}[/tex] = 8km/h east.

And what you want to calculate the velocity of the wind relative to the ground... I'll just call that [tex]\overrightarrow{v_w}[/tex].

We know that [tex]\overrightarrow{v_{wb}}[/tex]= [tex]\overrightarrow{v_{w}}[/tex] - [tex]\overrightarrow{v_b}[/tex] (I hope this makes sense, let me know if it doesn't).

So [tex]\overrightarrow{v_w}[/tex] = [tex]\overrightarrow{v_{wb}}[/tex] + [tex]\overrightarrow{v_b}[/tex]

So calculating the above resultant using vector geometry should give your answer. Actually that's exactly what you calculated... So your answer 13 km/h, 39 degrees, east of south is the velocity of the wind relative to the ground...

Remember that these are vectors so when adding or subtracting you can't just deal with the magnitudes... So to calculate [tex]\overrightarrow{x}[/tex] - [tex]\overrightarrow{y}[/tex], you would be calculate this resultant:
[tex]\overrightarrow{x}[/tex] + (-[tex]\overrightarrow{y}[/tex]) using vector geometry.

Did you post the question exactly as it is written? The question asks to find the velocity of the wind relative to the ground?
 
  • #5
Question: Boat traveling south at 10 km/h. A seaman on deck measure wind velocity to be 8 km/h east. What is the actual wind velocity? (the velocity of the wind relative to the ground)

Basically your statement [tex]\overrightarrow{v_{wb}}[/tex] = [tex]\overrightarrow{v_{w}}[/tex] - [tex]\overrightarrow{v_b}[/tex] would be the same as saying True Wind Speed = Apparant Wind Speed + Boat Wind Speed. correct??

I essentially just plotted the above points onto a graph, and using the formula squareroot((10^2)+(8^2)) came up with the answer 13 km/h. And then using the tangent function = 8/10 = 39 degrees.

Should I be doing something else, or is the method I am using correct?

Thanks
 
  • #6
Iniuria12 said:
Question: Boat traveling south at 10 km/h. A seaman on deck measure wind velocity to be 8 km/h east. What is the actual wind velocity? (the velocity of the wind relative to the ground)

Basically your statement [tex]\overrightarrow{v_{wb}}[/tex] = [tex]\overrightarrow{v_{w}}[/tex] - [tex]\overrightarrow{v_b}[/tex] would be the same as saying True Wind Speed = Apparant Wind Speed + Boat Wind Speed. correct??

You meant boat speed, not boat wind speed right? We should use velocity (since velocity is the vector and speed is the scalar)...

True Wind Velocity = Apparant Wind Velocity + Boat Velocity. Yes, that's right.

I essentially just plotted the above points onto a graph, and using the formula squareroot((10^2)+(8^2)) came up with the answer 13 km/h. And then using the tangent function = 8/10 = 39 degrees.

Should I be doing something else, or is the method I am using correct?

Yes, your method is right. You got the right magnitude and direction, so I think everything is correct. Hope I didn't confuse you too much. :smile:
 
Last edited:
  • #7
Thanks a bunch for your help\insight. That is definatly a weight off of my shoulders. For the past few hours I had the correct answer the entire time, and I've been teaching myself physics and haven't had anyone to verify what I've done. Appreciate your time and help once again.

Cheers.
 
  • #8
Iniuria12 said:
Thanks a bunch for your help\insight. That is definatly a weight off of my shoulders. For the past few hours I had the correct answer the entire time, and I've been teaching myself physics and haven't had anyone to verify what I've done. Appreciate your time and help once again.

Cheers.

You're welcome. :smile:
 

1. What is wind velocity?

Wind velocity is a measure of the speed and direction of air movement in a specific location. It is typically measured in miles per hour (mph), kilometers per hour (km/h), or meters per second (m/s).

2. How is wind velocity measured?

Wind velocity can be measured using various instruments such as anemometers, wind vanes, and doppler radar. These instruments measure the speed and direction of the wind at different heights above the ground.

3. What factors affect wind velocity?

Several factors can affect wind velocity, including temperature, air pressure, and topography. Wind also tends to be stronger in open areas with fewer obstacles, and can be influenced by the rotation of the Earth and the Coriolis effect.

4. Why is understanding wind velocity important?

Understanding wind velocity is important for a variety of reasons. It can help predict weather patterns, aid in air and water transportation, and inform decisions for renewable energy sources such as wind turbines. It can also impact agriculture, construction, and aviation industries.

5. How can wind velocity be used to improve our daily lives?

By understanding wind velocity, we can plan and prepare for potential weather events, make more informed decisions about energy usage, and improve the efficiency and safety of transportation. Additionally, wind velocity data can be used for research and development of new technologies and sustainable solutions.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
965
  • Introductory Physics Homework Help
Replies
6
Views
829
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
4K
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
26
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
985
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
651
  • Introductory Physics Homework Help
Replies
5
Views
3K
Back
Top