Vector Plane

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A plane flys due east at a constant airspeed of 151ms^-1. The wind blows at a constant speed of 64ms^-1 from the north. Calculate the resultant speed and the direction of the aircraft.

So I was thinking. V^2=151^2+64^2 . V=164km h^-1


Tan θ= 64/151 θ=23degrees The air speed is 164km in the direction 23degrees south-east.


let's say the aircraft came somewhere from California and went to Kansas. On the return journey, the pilot wants to fly directly back from Kansas to California . The air speed of the plane is 151ms^-1 and the wind still blows at a constant speed of 64ms^-1 from the north.

In what direction must the plane head to reach California?
So I believe the pilot has to fly at an angle of 23degrees N of W to correct for the wind

What will be the resultant speed? Would this just be the same?
 
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Dr.Peterson

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Check it out for yourself! If he flies in that direction (at his usual air speed), what will be the resultant?

If you find that it's different from what you expect, think about why, perhaps by comparing the figures for the two problems.

Of course, the problem is nonsense, as at that scale, the earth is not flat and wind velocity can't be constant. Ignore that.
 
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won't his resultant speed just be the same but in the other direction?
 

Dr.Peterson

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Have you sketched the picture? It isn't clear to me whether you are actually carrying out the check or not, and what you mean by "the resultant speed".

For the first problem, you will have a right triangle whose legs are the given airspeed and wind velocity, and the hypotenuse is the resultant. Right?

For the second, you are using the direction of that resultant as the hypotenuse in the opposite direction, but its length will be the given airspeed, not the resultant speed from the first problem. When you add the wind velocity vector, the resultant is what you hope will be directly east. Is it? Are these triangles congruent?
 
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(For number 1) Ok, The plane is initially traveling due east at a constant speed of 151 ms^1 with a constant wind of 64ms^1 from the north.

I used V^2= 151^2+64^2 = 164 ms^1 and I used tan θ= 64/151 θ= 23degrees South-East. So the resultant speed and direction is 164 ms^1 south-east.?

(For number 2) The pilot wants to fly directly back from where he came. So he would be now flying the opposite way. due west? But since were flying at an angle of 23degrees to correct for the wind it has to be 23degrees North-west?

Aircraft resultant speed now. So are you saying the plane is now flying at 164ms^1 airspeed instead of 151ms^1 if this is the case, V^2=164^2+64^2 = 176.045448677ms^1 23degrees North-west?
 

Dr.Peterson

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(For number 2) The pilot wants to fly directly back from where he came. So he would be now flying the opposite way. due west? But since were flying at an angle of 23degrees to correct for the wind it has to be 23degrees North-west?

Aircraft resultant speed now. So are you saying the plane is now flying at 164ms^1 airspeed instead of 151ms^1 if this is the case, V^2=164^2+64^2 = 176.045448677ms^1 23degrees North-west?
No, I'm saying that's why your answer is wrong. By your supposition, he flies at his airspeed of 151 m/s, on a heading of 23 degrees north of west. But if you add the wind velocity vector, you won't get the due-west course you want. So that is not the right angle. And the resultant speed will be a leg of the right triangle, not the hypotenuse.

Why not just solve the second problem from scratch, and see that the angle is not the same as in the first? The plane's vector is at a yet-unknown angle WNW with length 151, and that plus the vector due south at 64 has to sum to a vector due west. What angle must it be?
 
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Drawing a right-angled triangle with the hypotenuse being 151 and leg b being 64 I get the leg of a; or W-E(E-W) ( the top) to be 136.77? so using this V^2=136.77^2+64^2 = 151.00342016ms^1 the angle is basically the same 23degrees North-West?

If this is also wrong I'll have a better look a bit later. Thanks
 

Dr.Peterson

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The speed is right, but not the angle. Did you use the inverse tangent again, without thinking? The correct angle is a little more.

By the way, if these problems are from a book, they may be asking for the directions to be given in a different form, as your "__ degrees north-west" style is not how headings are usually described, even though it's perfectly understandable.
 
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I'll fix the angle and what would be the correct phrase to say the direction?
 

Dr.Peterson

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what would be the correct phrase to say the direction?
That depends on what you've been taught; there are several ways they might do it in a given textbook. If your book says it the way you did, then there's nothing wrong.

The two main forms I am familiar with are an angle clockwise from north (e.g. 090° means east, and 225° means southwest), and an angle from the nearest north or south direction (e.g. N90°E means east, S45°W means southwest). For examples of these and some others, see here and here.
 
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So the direction the pilot need to fly on the return journey is 67.03° North-west? or is it something very similar to 23degrees I have to results I get
 

Dr.Peterson

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Please show your work, not just your answer. As it is, I can't be sure where you went wrong. The correct answer is much closer to 23 than to 67. It looks like you just took the complement of 22.97, rather than using a different trig function entirely.
 
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D I use cos(θ)=a−djacent/h−ypotenuse ?
 

Dr.Peterson

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Why?

Here's the picture I think you should be making:

FMH116690.png

For the angle "northwest" (i.e. degrees north of west), you want the green angle α; what trig function is that?

For the standard bearing (degrees clockwise from north), you want the red angle β; what trig function is that (a little harder)?
 
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Sin because I would know the SOH opposite and hypotenuse?
 

Dr.Peterson

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Yes.

Now do it! You don't need to ask about every little step.
 
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If I use the inverse of sin. sin^-1(64/151=25degrees I think this one is correct.

So the new resultant speed is 151ms^-1 and the direction is 25dgrees North-West.
 

Dr.Peterson

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Correct angle; but what is the resultant speed (that is, the actual speed as measured from land)? You stated it once before.
 
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the 151.00342016ms^1 if so I just rounded it. Or is it 164ms^-1
 
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Dr.Peterson

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No. Look again at my picture. How can the resultant (the vector going west) be the same length as the hypotenuse (air speed = 151)?

Check post #7, where you gave the wrong answer (that 151.003...) and also the right answer, which you get from 151^2 - 64^2. I failed to point out which number was correct. Your wrong answer was just a matter of getting back to the 151 as the hypotenuse, rather than sticking with the leg, which is the answer you want.

I get the impression you are not thinking clearly about which vector is which. That's actually why I didn't explicitly show the resultant in my picture, just the two vectors that are being added.

Recall the problem:
let's say the aircraft came somewhere from California and went to Kansas. On the return journey, the pilot wants to fly directly back [due west] from Kansas to California . The air speed of the plane is 151ms^-1 and the wind still blows at a constant speed of 64ms^-1 from the north.

In what direction must the plane head to reach California?
So we have the airspeed in an unknown direction (the 25 degrees we worked out), added to the known windspeed (the 64 m/s southward), giving the resultant (due west). The resultant is not the hypotenuse (as it was in the first problem), but the leg. Of course, you weren't asked to find that.
 
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