Solving Aircraft Guidance in a Crosswind

[jonathanM111]

I want to piggyback from this thread:

https://www.physicsforums.com/threads/aircraft-guidance-in-a-crosswind.198836/

this is the question I am attempting to solve, however this thread has been locked

I have found that you can express the two velocities the following way at any point in the planes path:

V(ynet)= (velocity of aircraft)(sin(theta))+V(wind) where theta is the angle between the direction of the aircraft and the x axis, this angle changes as the aircraft reaches its destination

V(xnet)=(velocity of aircraft)(sin(theta))

I suppose now I can use the hint and divide these two equations? how would I go about doing this?
thank you

 

[.Scott]

I can't really set up your problem statement.
Let's use the terms air speed and ground speed - and be careful in differentiating velocities (vectors) from speeds (scalars).

So:
Va: Aircraft velocity relative to the air mass (vector).
Vg: Aircraft velocity relative to ground (vector).
Vw: Airmass velocity (vector, wind).
Then: Vg = Va+Vw

Each of these velocities can be separated into X and Y (lat/long) coordinates and even Z. The addition works in each case.
Vgx = Vax+Vwx
Vgy = Vay+Vwy

If you do have speed and bearing versus velocity, then convert:
Sa: Aircraft speed through the air mass.
Ha: Heading (clockwise from North) of the aircraft.
Vax = Sa*cos(Ha)
Vay = Sa*sin(Ha)

 

[jonathanM111]

I made a mistake, to find the V in the x-axis I put sin(theta). It should be cos(theta), other than that our equations are pretty much the same, only difference is that we already know that Vwx=0 because the vector of the wind is perpendicular to the initial direction of the aircraft. Take a look at the original problem format and my drawing.

 

[.Scott]

I will try to give you enough clues without spilling everything:

I would take the speed of the aircraft to be 1 and the wind speed to be γ.
So at any point x, y, the velocity will be:
dx/dt = [??]/sqrt(x²+y²)
dy/dt = [??] [+/-] [??]/sqrt(x²+y²)
dy/dx = ([??] [+/-] [??]∙sqrt(x²+y²))/[??]

Then, according to www.woframalpha.com:
y(x) = [??] sinh(γ [??] + c)

Plot for γ=0.5 is:
https://www.wolframalpha.com/input/?i=Plot[x+Sinh[0.3466+-+0.5+Ln[x]],+{x,+0,+2}]

 

[jonathanM111]

great hints, okay so I let aircraft speed be 10 and wind speed be 1 so as to satisfy the ratio of 0.1 for γ.
this is what I get
dy/dt = 10 sin(θ)+1
dx/dt = 10 cos(θ)
sin and cos can be expressed as ratios so:
dy/dt = 10 y/(sqrt(x² + y²)) +1
dx/dt = 10 x/(sqrt(x² + y²))
and then using a hint from the book i get
dy/dx =(y/x+sqrt(x² + y²)/10x)
this is a homogeneous equation, with a little bit of algebra i make it into the following form
dy/dx = y/x + (sqrt(1+(y/x)²)/10)
let
v=y/x
v + (sqrt(1+v²)/10)
and
dy/dx=v+x dv/dx
so
v+x dv/dx = v + (sqrt(1+v²)/10)
v cancels
x dv/dx = (sqrt(1+v²)/10)
seperate variables we end up with the two integrals which come out to be
10sinh^-1(y/x)= ln(x)+c
are there any mistakes? how do I take y out of the hypebolic sin?
P.s. I'll get back to you in maybe 2 days, I have other things to take care of, I appreciate it

 

[jonathanM111]

I'm back, I then I take the sinh of both sides and get

y/x =sinh(10ln(x)+c)

plug in initial conditions of y=0 and x=2,
0=sinh(10ln(2)+c)
therefore c must be -10ln(2)

the final equation is then y=xsinh(10ln(x)-10ln(2))

this doesn't look like what you graphed earlier, am I missing something?

 

[jonathanM111]

I realized I have the wrong number, instead of 10 it should be 1/10. when I fix that I get the same thing you got but with the signs the other way around. How did you get a negative infront of the .5ln(x), considering that we need the inside of the trig function to be zero then the c has to equal -ln(2) when y=0 and x=2. when I graph this one, my graph looks as if the wind is blowing downward but with the same shape as yours.

 

[jonathanM111]

I realized I have the wrong number, instead of 10 it should be 1/10. when I fix that I get the same thing you got but with the signs the other way around. How did you get a negative infront of the .5ln(x), considering that we need the inside of the trig function to be zero then the c has to equal -ln(2) when y=0 and x=2. when I graph this one, my graph looks as if the wind is blowing downward but with the same shape as yours.

https://www.wolframalpha.com/input/?i=Plot[x+Sinh[(0.1)(ln(x/2))],+{x,+0,+2}]

 

[jonathanM111]

I know the culprit, at the very beginning there should be a minus sign in front of the V(velocity of wind)