Airplane Range in North Wind: Prove and Max Value

[trumpsnuffler]

Homework Statement

An airplane has a speed v and a flying range (out and back) of R in calm
weather.

(a) Prove that in a north wind of speed n (n < v) its range is

R(phi) = R(v^2 - n^2) / [v(srqt(v^2 - n^2 sin^2(phi))]

in the direction whose true bearing from north is phi (Assume that the plane travels to the destination and back in the same straight line).

[so essentially, prove that the range of an airplane flying to and from a destination, with wind 'n' blowing north = the above equation]

(b) What is the maximum value of this range and in what direction may it be attained.

Homework Equations

a^2 + b^2 = c^2 ?

The Attempt at a Solution

I really have no idea how to attempt to tackle this problem. I know people don't like it when people post up here with no attempt at a solution, but i spent 7 hours on this last night and got nowhere.

I'm just looking for someone to help me understand how to go about doing it. Am i right in thinking it's a relative velocity problem? I can see from the denominator that the '(srqt(v^2 - n^2 sin^2(phi))' involves Pythagoras's theorem. But then i don't see where the other 'v' comes from in the denominator.

Please, any help would be greatly appreciated.

 

[tiny-tim]

hi trumpsnuffler! welcome to pf!

(have a phi: φ and a square-root: √ and try using the X² tag just above the Reply box )

Am i right in thinking it's a relative velocity problem?

that's right!

ok, always approach a question like this in logical steps

first: what did you get for the actual speeds, relative to the ground, for the outward and return sections (let's call them u and w)?

 

[trumpsnuffler]

Thanks for your reply Tim, but once again i find myself here at 6am in the morning after slogging it out with this question for 7 hours.

Okay. so I'm obviously missing something.

The velocity of the Plane with respect to air = Vpa = v
The velocity of Air with respect to the ground = Vag = n

Thus, the actual speed of the airplane relative to the ground = v + n. (i say 'thus', but i have a feeling it isn't as basic as I've just made it out to be...)

Because the numerator of the given equation is R(v²-n²), i can't help but feel this is related to (v + n) [journey there] and (v - n) [journey back], in that (v + n)(v - n) = v² - n². I'm not sure if there is any relation in this, and if not, i think it might be throwing me off chasing the right answer.

As it's a trip in a straight line there and back. Do we visualise this as essentially one journey, where midway, the wind changes direction? and if so, is it less a case of adding to separate journeys together, but instead multiplying them? It's just the more i look at the given equation, the more i want to separate the fraction into 2 fractions multiplied together...?(!)

My notepad's last 10-15 pages or so are just literally the same vector diagram drawn again and again, it's starting to dawn on me that I might be miles out here..

 

[tiny-tim]

hi trumpsnuffler!

(just got up :zzz: …)

Thanks for your reply Tim, but once again i find myself here at 6am in the morning after slogging it out with this question for 7 hours.
…
Thus, the actual speed of the airplane relative to the ground = v + n. (i say 'thus', but i have a feeling it isn't as basic as I've just made it out to be...)

Because the numerator of the given equation is R(v²-n²), i can't help but feel this is related to (v + n) [journey there] and (v - n) [journey back], in that (v + n)(v - n) = v² - n². I'm not sure if there is any relation in this, and if not, i think it might be throwing me off chasing the right answer.

goodness, do you work through every night?

look, this isn't a sensible use of your time …

if you're totally stuck on something for more than one hour, then stop, come here and ask for help, go to sleep, and in the morning everything will be clearer!

ok, yes you're correct … it's v + n and v - n, of course using vector addition not ordinary addition

and once you've found the speeds, you need to add the times (because that's fixed by the amount of fuel … can you see why?) …

and time = ?/?

As it's a trip in a straight line there and back. Do we visualise this as essentially one journey, where midway, the wind changes direction? and if so, is it less a case of adding to separate journeys together, but instead multiplying them? It's just the more i look at the given equation, the more i want to separate the fraction into 2 fractions multiplied together...?(!)

uh-uh, don't try to over-simplify, it tends not to work

My notepad's last 10-15 pages or so are just literally the same vector diagram drawn again and again, it's starting to dawn on me that I might be miles out here..

i don't understand why you haven't got the speeds (or maybe you have got them, and you don't know how to combine them?)

describe (or copy) your https://www.physicsforums.com/library.php?do=view_item&itemid=99"and show us what speeds you got

 

[trumpsnuffler]

Hi tim, thanks for your reply again.

Okay, so I would post my vector diagram up here, but I'm not too sure how to (forgive me, I'm new here as you know).

But this is how it looks, as best described as i can:

graph, x and y axis.

the vector v comes out of the origin at 45degrees (irrelevant as could be any angle), coming straight up from v is the vector 'n'.

i then drew a line from the origin to the top of the n vector to form a triangle.

φ is the angle from the y-axis to the resultant vector drawn. from this, i can deduce that φ is also the angle at the top of the vector triangle due to 'Z-angles' (i'm not too sure of the proper terminology here - i only remember calling them 'z-angles' from my school days...!).

The vertical component of the vector i got as n + vcosφ.
And the horizonatal component i got as vsinφ.

But neither of those look right? I can see, that to get the 'n sinφ' i have to split my vector triangle and then the resulting split (which starts inbetween v and n, and goes perpendicular to the resulting vector) is n sin(φ).

where am i going wrong??

 

[trumpsnuffler]

Okay, so the speed of the plane relative to the ground (journey there):

Vpa = v

Vag = n

therefore Vpg = Vag + Vpa

= v + n

And for the return journey.

Vpa = v
Vag = -n
So Vpg = v - n

But this suggests that this is the two speeds for the journeys, whereas i know that it's the distances?

I can see from here that it's a case of time = speed/distance, but I'm just struggling to get the two speeds, unless of course v+n and v-n ARE the two speeds.

Thanks in advance for your help tim.

 

[trumpsnuffler]

sorry, * time = distance/speed.

 

[tiny-tim]

sorry, I've been out all evening

on the bus home, i worked out the answer, and now I'm not surprised you've been having trouble with this …

in my considerable experience, this is the nastiest question of its type I've ever come across!

ok, the way to go is to call the range 2Q (for the angle φ), and to draw the vector of length Q at an angle φ to North …

now draw two vector triangles, both including that vector, to represent the two legs of the journey (you don't know the lengths of the sides, but you do know the ratio, n/v of the lengths) …

what are the angles of these two triangles? ​

(that's difficult enough … let's get that right before trying anything more )

 

[trumpsnuffler]

I'm going to be honest with you Tim, I've literally spent 10 hours a day for the past 3 days trying to solve this one question, so it is with some slight relief that you agree that it is a particularly difficult problem. I'm about ready to go mad...

I'm a bit lost. so is φ the angle made from the north to the vector v, or the angle made from north to the vector Q?

and how do i know the ratio of n/v? having drawn out the two vector diagrams, i can't see how to make a correlation between the ratio of the two sides n and v of the two triangles?

 

[tiny-tim]

… so is φ the angle made from the north to the vector v, or the angle made from north to the vector Q?

and how do i know the ratio of n/v? having drawn out the two vector diagrams, i can't see how to make a correlation between the ratio of the two sides n and v of the two triangles?

φ is the angle in the question, from North to the line of flight, ie from North to Q

and you do know exactly what n and v are … they're fixed by the question (ok, I know they're not actual numbers, but they're still fixed) … and since the engine and the wind obviously act for the same length of time, that means the two sides must have the ratio n/v

(and now I'm off to bed :zzz:)

 

[trumpsnuffler]

ahhh i see. i wasn't quite sure if the question was saying that the plane leaves in the direction φ with speed = v, or if φ was the direction of the resultant vector Q.

i still don't really understand how the ratio of the two sides is n/v though. both the sides must be the same length in order to return to the start point of the journey, so how is there a ratio between them? surely they're the same length??

sorry if I'm being a bit slow...

also, is it not possible to find out the resultant vector through trigonometry and pythagoras's theorem?

For example, on the journey there:

I drew the vector diagram, and linked the resultant vector Q to the top of n. Next i drew a line perpendicular from Q to where vector v meets vector n, thus splitting the triangle into two right angled triangles.

the angle at the top between n and Q is φ, and from this i found that the line splitting the triangle into two right angled triangles was nsin(φ).

Using this and the vector v, i deduced through pythagoras's theorem that the first part of the vector Q = √[v² - n²sin²(φ)]. And then the other part of the vector Q can likewise be found pythagoras's theorem to be √[n²-n²sin²(φ)].

So then the vector Q must be of length √[v² - n²sin²(φ)] + √[n²-n²sin²(φ)].

Am i even close...?

 

[tiny-tim]

good morning!

let's simplify this by drawing real triangles on the ground (then we can't go wrong!)

imagine that the wind and the engine act at different times …

first the plane flies for a time t₁ at speed v (with no wind), then it stops and let's the wind blow it South for a time T₁ at speed n, and ends up at distance S along direction φ from North …

(we can do this because everything is a vector, and we can move vectors around before combining them)

(of course, a North wind blows South )

so the https://www.physicsforums.com/library.php?do=view_item&itemid=99"is now a real triangle on the ground: its sides have lengths vt₁ nT₁ and Q

all you know is the angle φ, the length Q, and the ratio of the two other sides (= n/v)

ok, now do the same thing for the return trip, draw another real triangle on the ground, with one of the sides (Q) the same …

you now have two triangles joined together

i] what are the other angles of these triangles? (you'll need either the sine or cosine law)
ii] from where on these triangles can we tell the (fixed) amount of fuel used?

 

[trumpsnuffler]

Is there anyway i can attach a drawing to one of these posts, so i can show you why I'm getting so confused through a vector diagram?

I can try and explain it through words, but it'd be a lot easier if i could post a diagram on here.

Say, for example, the vector v goes off in a north-eastern direction, and then the vector n, brings it down south. The angle φ makes with the northern axis (or y-axis), is not an angle within the triangle, as only a percentage of the angle is within the triangle, the other part is outside?

(but yeah, as i said, is there anyway i can attach drawings to this...?)

 

[tiny-tim]

Say, for example, the vector v goes off in a north-eastern direction, and then the vector n, brings it down south. The angle φ makes with the northern axis (or y-axis), is not an angle within the triangle, as only a percentage of the angle is within the triangle, the other part is outside?

yes, that's right, except φ is in the "third" corner of the triangle (either φ or 180°-φ), isn't it?

now draw two triangles like that, one for each leg of the journey (all the arrows will be "backwards" on the return leg) …

what are the other angles?

(i've never attached a diagram, so i don't know how to do it … i know some people use image hosting sites like photobucket … there is a way of directly attaching diagrams, but i think they have to be vetted first, and that can take hours)

 

[trumpsnuffler]

hmmm, I'm not too sure I'm either a) working out the angles correctly, or b) working out the correct angles..

but okay. the angles i got for the internal angles of triangle vQn are:

the angle between n and Q is (180 - φ)

then the angle at the top, called, say, A, between v and n, through use of the cosine rule is arccos[(v² + n² - Q²)/2vn].

and then from this, the remaining angle is 180 - (180 - φ) - arccos[(v² + n² - Q²)/2vn].

however i can't help but feel that as I've said, I've gone about this wrong, as i don't see how I'm going to get to the answer from here...

 

[trumpsnuffler]

hmmm, I'm not too sure I'm either a) working out the angles correctly, or b) working out the correct angles..

but okay. the angles i got for the internal angles of triangle vQn are:

the angle between n and Q is (180 - φ)

then the angle at the top, called, say, A, between v and n, through use of the cosine rule is arccos[(v² + n² - Q²)/2vn].

and then from this, the remaining angle is 180 - (180 - φ) - arccos[(v² + n² - Q²)/2vn].

however i can't help but feel that as I've said, I've gone about this wrong, as i don't see how I'm going to get to the answer from here...

 

[tiny-tim]

try using the sine rule instead

 

[trumpsnuffler]

sorry for the delay in a reply, had to drop my housemate off to pick up his car.

okay, so from the sine rule, i get:

sin (180-φ)/v = sin(C)/n

and from this, so C = arcsin(nsin(180-φ)) <-------- can this be canceled to give n(180-φ)?

and then the final angle is just 180 - (180 - φ) - the above answer from C.

and then from there do i use sine/cosine rule to find out Q?

 

[tiny-tim]

okay, so from the sine rule, i get:

sin (180-φ)/v = sin(C)/n

yes, except you can use sin(180°-φ) = sinφ,

so sinC₁ = (n/v)sinφ …

leave it like that!

ok, there are two triangles, so, in the other triangle, what is sinC₂ ?

and from where on these triangles can we tell the (fixed) amount of fuel used?

 

[trumpsnuffler]

isn't it just the same?

the angle made from the north axis to the resultant vector Q (in the return journey) must be φ in order for the plane to return to its original place.

so, from this, and the 'z rule' of angles, the top angle, between n and Q must therefore be equal to φ.

so sin(φ)/v = sin(c_{2})/n

and so sin(c_{2}) = (n/v)sinφ

and I'm not too sure about the fuel...

is it the ratio of the times spent to do the respective parts of the journey?

 

[trumpsnuffler]

hmmm. latex doesn't seem to work for me..

 

[tiny-tim]

(don't bother with the latex, it's easier to use the X₂ tag just above the Reply box )

isn't it just the same?
…
and so sin(c_{2}) = (n/v)sinφ

that's right!

now, since the fuel is fixed, and since the fuel is used at a constant rate (ie litres per second), that means the total time taken is also fixed …

and the total time shows up where on the triangles? ​

 

[trumpsnuffler]

the vectors leading to Q? i.e. v and n?

 

[tiny-tim]

the total time for both journeys?

 

[trumpsnuffler]

so would this be the (v + n) from the journey there, and then (v - n) for the journey back?

or (v - n) for the way there, and (v + n) for the way back (now i know that a north wind blows north and not south...)

 

[trumpsnuffler]

i think at the rate this question is taking me, you might break 15,000 posts by the end tim...

 

[tiny-tim]

since the range with no wind is R, that means the engine can run for a time R/v

if t₁ and t₂ are the times on the two legs, then t₁ + t₂ = R/v, so Q/R = Q/(vt₁ + vt₂) …

where can we read that fraction off the triangles?

 

[trumpsnuffler]

hmm. what is this new variable S...?

(sorry to be so slow, please be patient with me!)

 

[tiny-tim]

oops!

oops, i forgot which letter i was using!

(i used S originally, then decided it looked too much like 5, so went the other way and used Q instead )

i meant Q … I've gone back and edited it now

sorry!

 

[trumpsnuffler]

hmmm, okay, would the time be calculated via pythagoras's theorem for the individual x and y components of v and n, for each journey repectively?

(n/v)sin(φ)..?

i'm a bit lost, i can see why the numerator is R(v+n)(v-n) and i can see that the v on the denominator comes from R/v, but i don't really get the last part, I'm still not sure where the pythagorian part of the denominator (√(v² - n²sin²φ)) and i guess I'm just clutching at straws a bit, trying to piece things together to make this last part.

God help me if a question like this comes up in my exam in a couple of days..

 

[tiny-tim]

stop trying to work backwards from the answer …

work forwards!

and no, you don't need Pythagoras, just the sine rule and a bit of trig

(to make it easier to write, since C₁ = C₂, call them both θ, get an expression in φ θ and φ ± θ, and find cosθ in terms of n and v)

oh, and go to bed early tonight! :zzz:​

 

[AC130Nav]

hmmm, okay, would the time be calculated via pythagoras's theorem for the individual x and y components of v and n, for each journey repectively?

(n/v)sin(φ)..?

i'm a bit lost, i can see why the numerator is R(v+n)(v-n) and i can see that the v on the denominator comes from R/v, but i don't really get the last part, I'm still not sure where the pythagorian part of the denominator (√(v² - n²sin²φ)) and i guess I'm just clutching at straws a bit, trying to piece things together to make this last part.

God help me if a question like this comes up in my exam in a couple of days..

Never a diagram to be seen here.

The wind triangles are actually simple (as long as you remember you add vectors head-to-tail). I guess I should've made two separate ones also.

 

[trumpsnuffler]

morning all, opted to get an early night in favour of an early rise!

okay, the above vector diagram's a bit misleading. on the return journey, the plane doesn't necessarily come fly off westwards to form a right angled triangle with Q v and n.

right the x and y components for the individual journeys are as follows

way there:

x component: v sinφ

y component: vcosφ - n

way back:

x component: v sinφ

y component vcosφ + n


although I'm guessing i must have done something wrong here as i have nothing involving θ

 

[tiny-tim]

hi trumpsnuffler! good morning!

okay, the above vector diagram's a bit misleading. on the return journey, the plane doesn't necessarily come fly off westwards to form a right angled triangle with Q v and n.

yes, i don't like that diagram at all, but mostly because it puts the two wind vectors on different lines

flip that bottom triangle over, so that you have a nice double-triangle, with both parts to the left of the wind vectors

although I'm guessing i must have done something wrong here as i have nothing involving θ

what you've done is ok, just not very helpful …

you can always use components, but it'll be a lot easier (in this case) to use θ and the sine rule …

as i said, what is cosθ (n terms of n and v)? ​

 

[trumpsnuffler]

and with regards to the second part of the question 'find the maximum range and for which direction', would i have to go through the normal means of finding a maximum (i.e differentiating), and if so, differentiating with regards to what?

or do i just have to make n²sin²φ = 0, which would give the answer to be either φ = 0 or φ=180 (which would make sense in that to get the maximum range, the pilot has to fly either directly into the wind, or with the wind directly behind his back to eliminate any y component)?

 

[trumpsnuffler]

okay. as (n/v)sin(φ) = sin(θ)

(n/v)cos(φ) = cos(θ)

wasn't too sure if i could make this jump, but bearing in mind that the graphs of sin and cos are so similar, surely the above expression for cos(θ) holds to be true?

 

[tiny-tim]

okay. as (n/v)sin(φ) = sin(θ)

(n/v)cos(φ) = cos(θ)

what?? not even close … sin is increasing and cos is decreasing!

use cos² + sin² = 1 ​

 

[trumpsnuffler]

oh wow. no idea why i didn't think of that..

okay. so cos² = 1 - sin²

cos² = 1 - sin²
= 1 - [(n/v)sin(φ)]²
= 1 - (n²/v²)sin²(φ)

therefore cosθ = √[1 - (n²/v²)sin²(φ)]

is the above expression the value of [t][/1] then? and [t][/2] = the same, but a positive (n²/v²)sin²(φ) instead?

 

[trumpsnuffler]

sorry, still getting to grips with using this forum. t1 and t2 were meant to be written in subscript..

 

[tiny-tim]

therefore cosθ = √[1 - (n²/v²)sin²(φ)]

yup!

though to match the given answer, you might prefer to write it as (1/v)√[v² - n²sin²(φ)]

is the above expression the value of [t][/1] then?

no of course not, it's cosθ … why would that be t₁?

(for subscripts, use the X₂ icon just above the Reply box )

if t₁ and t₂ are the times on the two legs, then t₁ + t₂ = R/v, so Q/R = Q/(vt₁ + vt₂) …

where can we read that fraction off the triangles?

 

[AC130Nav]

In the accompanying diagram air miles are proportional to time, of course. You might check out when b = 0. a = 0. and both a and b are sqr root of 2 times d.