Roundtrip by Plane: Understanding Wind & Velocity Effects

[brochesspro]

Homework Statement

Given below.

Relevant Equations

Equation for relative velocity may be required (Given below).

The question I have is that if the aero plane is traveling in the same direction as the wind, would it not increase its velocity, as in with boats and streams? So, if by chance, $w = v$, then the velocity of the aero plane would double. It feels weird as going by the same logic, if the speed of the wind happens to be in opposite direction, then the plane would be at rest. There is something wrong with my intuition/logic, is there not?

And one more thing. The speed of the plane relative to the air is $v$. Does this mean that the speed of the plane is $v$ w.r.t. a person standing on ground or is it that speed w.r.t. the wind blowing?

I can not start solving the problem before I can understand the answers to these questions.

 

[phyzguy]

So, if by chance, $w = v$, then the velocity of the aero plane would double. It feels weird as going by the same logic, if the speed of the wind happens to be in opposite direction, then the plane would be at rest. There is something wrong with my intuition/logic, is there not?

Your logic is correct. So can you now answer the question?

 

[phyzguy]

And one more thing. The speed of the plane relative to the air is $v$. Does this mean that the speed of the plane is $v$ w.r.t. a person standing on ground or is it that speed w.r.t. the wind blowing?

The speed v is with respect to the air. So the speed with respect to a person on the ground is v+w or v-w depending on the wind direction.

 

[brochesspro]

Your logic is correct. So can you now answer the question?

The speed v is with respect to the air. So the speed with respect to a person on the ground is v+w or v-w depending on the wind direction.

Thank you. I will try to answer the question now.

 

[Lnewqban]

Thank you. I will try to answer the question now.

The measured speed is relative to the mass of air, which is the closest reference that the instruments of the airplane (excluding GPS) have to estimate that value.
GPS is able to measure speed of the airplane respect to the curved surface of the Earth, regardless wind speed and direction.

 

[brochesspro]

The measured speed is relative to the mass of air, which is the closest reference that the instruments of the airplane (excluding GPS) have to estimate that value.
GPS is able to measure speed of the airplane respect to the curved surface of the Earth, regardless wind speed and direction.

I do not think we have to take the mass of air or the curvature of the Earth into account. The Earth is relatively flat. I will see you tomorrow.

 

[haruspex]

I do not think we have to take the mass of air or the curvature of the Earth into account.

@Lnewqban wrote "mass of air", i.e. the air as a body, not "mass of the air". In the same way, meteorologists speak of a moving "air mass".
Nor did he imply you need to consider the curvature of the Earth's surface.

 

[brochesspro]

Is my work till here correct?

 

[DaveC426913]

Nor did [ @Lnewqban ] imply you need to consider the curvature of the Earth's surface.

There was an implied implication:

GPS is able to measure speed of the airplane respect to the curved surface of the Earth,

 

[Lnewqban]

There was an implied implication:

Sorry about any confusion that my post may have created.
The thing is that, when cruising at high altitudes, any airplane must fly faster than its projected shadow on the ground: both have different radii respect to the center of the Earth.
Hope I have not made it even more confusing.
Please, see:
https://en.wikipedia.org/wiki/Ground_speed

https://www.grc.nasa.gov/WWW/k-12/airplane/move.html

 

[haruspex]

There was an implied implication:

On the contrary. If @Lnewqban had written "velocity [wrt] the surface of the Earth" then one would have to worry about the curvature. Specifying wrt the curved surface means that is taken care of, and we do not need to worry about it.

 

[brochesspro]

View attachment 295074
Is my work till here correct?

Why do we not discuss the topic at hand? Is the procedure I followed correct?

 

[haruspex]

View attachment 295074
Is my work till here correct?

You have correct expressions for velocities, but speed is not a vector. What is the speed in the direction B to A, assuming the plane can actually get there?
Also, next step is to consider the round trip time.

 

[brochesspro]

You have correct expressions for velocities, but speed is not a vector. What is the speed in the direction B to A, assuming the plane can actually get there?

Did I not already calculate the speed from B to A, which is $w - v$? Since this is 1D motion, the signs take care of everything, like direction.

Also, next step is to consider the round trip time.

So I need to use the formula $$Distance = Speed \times Time$$ for both the cases and add the times.

Am I correct in doing this?

 

[haruspex]

Did I not already calculate the speed from B to A, which is w−v?

Can speed be negative? What will happen if w-v is positive?

So I need to use the formula

Yes.

 

[brochesspro]

Can speed be negative? What will happen if w-v is positive?

Note that $w - v$ is a vector. $w$ is in the positive direction and $v$ is in the negative direction. So basically, we have to look at the magnitude of the vector difference. Thus, the speed is $| w - v |$.

Am I right?

 

[haruspex]

Note that $w - v$ is a vector. $w$ is in the positive direction and $v$ is in the negative direction. So basically, we have to look at the magnitude of the vector difference. Thus, the speed is $| w - v |$.

Am I right?

w and v are given as speeds, not vectors. If we assume the plane can get back to A, how can we simplify | w - v |?

 

[brochesspro]

w and v are given as speeds, not vectors. If we assume the plane can get back to A, how can we simplify | w - v |?

Since this is 1D motion, we can write the velocities in the form of vectors.

Am I wrong to think so?

 

[brochesspro]

This is another version of the work I did after starting fresh. How do you think it fares? I think there is something fishy about it but I just can not seem to point my finger at it. Please evaluate it, @haruspex. Thank you.

 

[jbriggs444]

This is another version of the work I did after starting fresh. How do you think it fares? I think there is something fishy about it but I just can not seem to point my finger at it. Please evaluate it,

The result, $\frac{2dw}{w^2-v^2}$ has some of the right behavior. But yes, it is fishy. Somehow the variables seem to have been swapped.

One assumes that $w$ is the wind speed and that $v$ is the airplane's velocity relative to the air.

If $w=v$, $w^2 = v^2$, the denominator becomes zero and an infinite round trip time is computed. This makes sense since the upwind leg of the journey will indeed take forever.

If $v=0$, we get a result of $\frac{2d}{w}$. That would be the correct number for a round trip at velocity $w$. But $w$ is the wind speed.

Let us back up and see if we can spot the error...

You start with some stuff that I find hard to follow because you are not documenting your variable names. That is a pet peeve of mine. Slowing down to decipher that, I get:

$\vec{V_\text{pw}}$ is the velocity of the plane with respect to the wind.
$\vec{V_\text{pm}}$ is the velocity of the plane with respect to the mass of the Earth (I guess).
$\vec{V_\text{wm}}$ is the velocity of the wind with respect to the mass of the earth.

You assert that the velocity of the plane with respect to the wind is the vector difference between the plane's velocity and the wind's velocity, both relative to the fixed ground frame: $\vec{V_\text{pw}} = \vec{V_\text{pm}} - \vec{V_\text{wm}}$. That is a valid assertion.

You then carefully look at your sign conventions and use the positive speed $v$ to replace $\vec{V_\text{pw}}$ and the positive wind speed $w$ for $\vec{V_\text{wm}}$.

The conclusion is that the plane's ground velocity ($\vec{V_\text{pm}}$) for the downwind leg is $v+w$. Indeed, that seems obviously correct.

You proceed to re-use the same variable names for the upwind leg. Again, you are careful with the sign conventions. This time the plane's velocity relative to the ground is the negation of its ground speed.

The conclusion is that the plane's ground velocity ($\vec{V_\text{pm}}$) for the upwind leg is $w-v$. This is the negation of $v-w$ and since the plane is leftward moving this time, that seems obviously correct again.

Ahhh, there it is. I think I see the error.

You assert that the total time is the sum of the downwind flight time plus the upwind flight time. But before writing that assertion down, you immediately rewrite the downwind flight time as $\frac{d}{v+w}$ and the upwind flight time as $\frac{d}{w-v}$.

But that second term is not correct. The displacement to be traversed is not $d$. It is $-d$. A negative displacement covered at what will (hopefully) be a negative velocity.

[My hat is off to teachers who have to reverse engineer this kind of stuff every time they grade papers. It boggles the mind that you do not all go bonkers]

 

[brochesspro]

The result, $\frac{2dw}{w^2-v^2}$ has some of the right behavior. But yes, it is fishy. Somehow the variables seem to have been swapped.

One assumes that $w$ is the wind speed and that $v$ is the airplane's velocity relative to the air.

If $w=v$, $w^2 = v^2$, the denominator becomes zero and an infinite round trip time is computed. This makes sense since the upwind leg of the journey will indeed take forever.

If $v=0$, we get a result of $\frac{2d}{w}$. That would be the correct number for a round trip at velocity $w$. But $w$ is the wind speed.

Let us back up and see if we can spot the error...

You start with some stuff that I find hard to follow because you are not documenting your variable names. That is a pet peeve of mine. Slowing down to decipher that, I get:

$\vec{V_\text{pw}}$ is the velocity of the plane with respect to the wind.
$\vec{V_\text{pm}}$ is the velocity of the plane with respect to the mass of the Earth (I guess).
$\vec{V_\text{wm}}$ is the velocity of the wind with respect to the mass of the earth.

You assert that the velocity of the plane with respect to the wind is the vector difference between the plane's velocity and the wind's velocity, both relative to the fixed ground frame: $\vec{V_\text{pw}} = \vec{V_\text{pm}} - \vec{V_\text{wm}}$. That is a valid assertion.

You then carefully look at your sign conventions and use the positive speed $v$ to replace $\vec{V_\text{pw}}$ and the positive wind speed $w$ for $\vec{V_\text{wm}}$.

The conclusion is that the plane's ground velocity ($\vec{V_\text{pm}}$) for the downwind leg is $v+w$. Indeed, that seems obviously correct.

You proceed to re-use the same variable names for the upwind leg. Again, you are careful with the sign conventions. This time the plane's velocity relative to the ground is the negation of its ground speed.

The conclusion is that the plane's ground velocity ($\vec{V_\text{pm}}$) for the upwind leg is $w-v$. This is the negation of $v-w$ and since the plane is leftward moving this time, that seems obviously correct again.

Ahhh, there it is. I think I see the error.

You assert that the total time is the sum of the downwind flight time plus the upwind flight time. But before writing that assertion down, you immediately rewrite the downwind flight time as $\frac{d}{v+w}$ and the upwind flight time as $\frac{d}{w-v}$.

But that second term is not correct. The displacement to be traversed is not $d$. It is $-d$. A negative displacement covered at what will (hopefully) be a negative velocity.

[My hat is off to teachers who have to reverse engineer this kind of stuff every time they grade papers. It boggles the mind that you do not all go bonkers]

Thank you a lot for your hard work in writing your answer. I have a request. I will write my answers to the questions asked in the problem. Please evaluate them and correct my errors, if any.

a) The roundtrip will take greater time with wind, as the time taken with wind is $\frac {2dv} {v^2-w^2}$, while the time taken without wind is $\frac {2d} v$. On evaluating, the former is greater than the latter.

b) If the wind speed became so high that $w = v$, the speed when traveling from B to A would be zero and hence, the object would not move.

c) If the wind direction is reversed, then the situation is the opposite of before. Now, the speed while traveling from A to B would be zero and the plane would not move.

That is it. See you in about 8 and a half hours.

 

[haruspex]

a) The roundtrip will take greater time with wind, as the time taken with wind is $\frac {2dv} {v^2-w^2}$,

I see you at last have the signs right. @jbriggs444 corrected it by taking the fully vectorial view and noting that the B to A displacement is -d. I had been trying to get you to see that in the speed and distance view the return speed is v-w, not w-v. Same result.
One could describe your error as mixing the two views, treating v and w as vectors but d as a distance. A useful lesson.

On evaluating, the former is greater than the latter.

Does that mean you tried some specific numbers? It wouid be more persuasive if you could show it is always greater (under certain assumptions).

 

[brochesspro]

One could describe your error as mixing the two views, treating v and w as vectors but d as a distance. A useful lesson.

I will be careful from next time onwards.

Does that mean you tried some specific numbers? It wouid be more persuasive if you could show it is always greater.

Is this not enough to prove that the time taken to travel with wind is greater than the time taken to travel without wind? In the last step, $w^2$ is always positive, and hence, $-w^2$ is always negative, and thus, by definition, less than zero. Thank you for your effort.

Am I right in doing so?

 

[haruspex]

I will be careful from next time onwards.View attachment 295135
Is this not enough to prove that the time taken to travel with wind is greater than the time taken to travel without wind? In the last step, $w^2$ is always positive, and hence, $-w^2$ is always negative, and thus, by definition, less than zero. Thank you for your effort.

Am I right in doing so?

looks good. Note that your algebra (second step) assumes v>0 and v>w (or neither).

 

[brochesspro]

looks good. Note that your algebra (second step) assumes v>0 and v>w (or neither).

So all my answers to the original problem statement are correct, or are there any errors?

 

[haruspex]

So all my answers to the original problem statement are correct, or are there any errors?

Looks fine.

 

[brochesspro]

Looks fine.

Sorry, I did not see this reply. Thank you to all who contributed to this thread.