Airplane Range in North Wind: Prove and Max Value

[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.