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The missile and the aircraft

  • Thread starter ozzy
  • Start date
12
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For two weeks I can't solve this question with my friends. Please help us!!!

an aircraft flights at height H with a constant velocity U.
When it passes right above the missile, the missile is fired with a constant velocity V (V>U), but the vector of the velocity of the missile points directly to the aircraft all the time of the flight of the missile.
The question is - what is the time of the total flight of the missile, till it gets the aircraft?
The solution should be done without any integrals.
Sorry for my bad English:)
 
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Answers and Replies

HallsofIvy
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Taking t= 0 at the moment the airplane passes over the missile, the airplanes position is given by (Ut, H). (x-axis horizontal, y-axis vertical)

If we call the missile's position at time t (x(t),y(t)) then dy/dx= (y-H)/(x-Ut).

The missile's speed (not velocity) is constant:[tex]\sqrt{(1+ (\frac{dy}{dx})^2}= V[/tex] or [tex]\frac{dy}{dx}= \sqrt{V^1- 1}[/tex]. As y approaches H, in order for that to be constant, we must have x= Ut but since we don't know x, I don't see any immediate way to find t.
 
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HallsofIvy said:
A small correction:

The missile's speed (not velocity) is constant:[tex]\sqrt{(1+ (\frac{dy}{dx})^2}= V/(\frac{dx}{dt})[/tex]...
I solved these equations numerically in Matlab :
Vx = dX/dt = v*(u*t - X) / sqrt((u*t - X)^2 + (H - Y)^2)
Vy = dY/dt = v*(H - Y) / sqrt((u*t - X)^2 + (H - Y)^2)

=> dX/dY = (u*t - X) / (H - Y)
v^2 = (dX/dt )^2 + (dY/dt)^2

=>

H=10;
u=5;
v=6;

X(1) = 0;
Y(1) = 0;

dt = 0.001;

t = 0:dt:H / ( v * ( 1 - u^2/v^2))+10*dt;

for i = 1:length(t)

Vx(i) = v*((u*t(i) - X(i)) / sqrt((u*t(i) - X(i))^2 + (H - Y(i))^2));
Vy(i) = v*((H - Y(i)) / sqrt((u*t(i) - X(i))^2 + (H - Y(i))^2));

X(i+1) = X(i) + Vx(i)*dt;
Y(i+1) = Y(i) + Vy(i)*dt;

end

and these are a corresponding graphs:
http://foto.mail.ru/mail/illuha77/1/i-4.jpg [Broken]
http://foto.mail.ru/mail/illuha77/1/i-2.jpg [Broken]
 
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70
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HallsofIvy said:
[tex]\frac{dy}{dx}= \sqrt{V^1-1}[/tex]
HallsofIvy, you probably made a typo there. Following from your calculation I'd expect that to be [tex]\frac{dy}{dx}= \sqrt{V^2-1}[/tex]. In any case, that leads to dy/dx being constant since V is constant, i.e. the rocket flies off in a straight line to the upper right (if you take the aircraft to be flying towards the right, of course). The missile is supposed to be flying towards the momentary position of the plane at all times, though.

I've arrived at some none-too-helpful integrals myself. I'm not sure whether those are of any use, but I'll post them here anyway:

[tex] \int_{0}^{T} \frac{Ut - x(t)}{\sqrt{(Ut - x(t))^{2} + (H - y(t))^{2}}} \cdot V dt = UT [/tex]

and

[tex] \int_{0}^{T} \frac{H - y(t)}{\sqrt{(Ut - x(t))^{2} + (H - y(t))^{2}}} \cdot V dt = H [/tex]

T here means the time of interception, and all other symbols are adopted from HallsofIvy's post. This exercise reminds me of calculating the arc length of a parabola, although it's not entirely the same.
 
Dont integrate at all.
You will get
[tex]\frac{Hv}{v^2-u^2}[/tex]
 
Astronuc
Staff Emeritus
Science Advisor
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One is trying to solve for a family of curves, with the constraints t = x/U, the altitude if fixed, H, velocity of the missile along (tangent to) the trajectory is V, and the fact that simultaneity requires the missile and plane get to the same point at the same time.

As balakrishnan_v indicates, the time is then expressed as a function of altitude and velocities, U and V.
 
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balakrishnan_v, how did you solve it? I'm curious to see your derivation.
 
12
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Brinx said:
HallsofIvy, you probably made a typo there. Following from your calculation I'd expect that to be [tex]\frac{dy}{dx}= \sqrt{V^2-1}[/tex] . In any case, that leads to dy/dx being constant since V is constant, i.e. the rocket flies off in a straight line to the upper right (if you take the aircraft to be flying towards the right, of course). The missile is supposed to be flying towards the momentary position of the plane at all times, though.

I've arrived at some none-too-helpful integrals myself. I'm not sure whether those are of any use, but I'll post them here anyway:

[tex] \int_{0}^{T} \frac{Ut - x(t)}{\sqrt{(Ut - x(t))^{2} + (H - y(t))^{2}}} \cdot V dt = UT [/tex]

and

[tex] \int_{0}^{T} \frac{H - y(t)}{\sqrt{(Ut - x(t))^{2} + (H - y(t))^{2}}} \cdot V dt = H [/tex]

T here means the time of interception, and all other symbols are adopted from HallsofIvy's post. This exercise reminds me of calculating the arc length of a parabola, although it's not entirely the same.
If you check my second post, you will see that i mentioned that HallsofIvy made a mistake!
it should be: [tex]\sqrt{ (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 } = V[/tex]
and so [tex]\sqrt{ 1+ (\frac{dy}{dx})^2 } = \frac{V}{(\frac{dx}{dt}) }[/tex]
or
[tex]\frac{dy}{dx} = \sqrt{(\frac{V}{\frac{dx}{dt}})^2 - 1}[/tex]
 
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balakrishnan_v said:
Dont integrate at all.
You will get
[tex]\frac{Hv}{v^2-u^2}[/tex]
this is the right answer:)
but I could find it in the answers of the book...
can you please give us some guiding lines to the solution?
 
Consider any instant at which the missile is making an angle A with the horizontal
Consider along the direction in which the plane is moving
We have the relative velocity=
V-Ucos(A)
this is the rel vel along the line joining
missile and plane
so we have I(V-Ucos(A))dt,t:0->T)=d..(I is integral)
or VT-UI(cos(A)dt:t:0->T)=d..(1)
and the other equation is that the horizontal distance moved by the plane and missile are same
so we have I(Vcos(A)dt):t:0->T=UT
or I(Cos(A)dt)=UT/V
substituting this in (1) we have
VT-U(UT/V)=d
or T=VH/(V^2-U^2)
 
12
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balakrishnan_v said:
Consider any instant at which the missile is making an angle A with the horizontal
Consider along the direction in which the plane is moving
We have the relative velocity=
V-Ucos(A)
this is the rel vel along the line joining
missile and plane
so we have I(V-Ucos(A))dt,t:0->T)=d..(I is integral)
or VT-UI(cos(A)dt:t:0->T)=d..(1)
and the other equation is that the horizontal distance moved by the plane and missile are same
so we have I(Vcos(A)dt):t:0->T=UT
or I(Cos(A)dt)=UT/V
substituting this in (1) we have
VT-U(UT/V)=d
or T=VH/(V^2-U^2)
THANX!!!!
It all seems very clear and right, except the last line - how can you see that d=H?
 
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ozzy said:
THANX!!!!
It all seems very clear and right, except the last line - how can you see that d=H?
Oh, I think I see it - the initial distance between the plane and the missile is H, therefore the relative distance that a missile has to cover is D=H...

YOU ARE THE KING!!!!
 
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70
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balakrishnan_v, that's quite a neat calculation. Thanks for making it clear!
 
12
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Well, thinking about the problem a little bit further, I have another problem:

the relative velocity between the missile and the plane in Cartesian axis is:
Vx(t) = V*cos{f(t)} - U
Vy(t) = V*sin{f(t)} - 0
and initial position,that should preserve is R = (0 , H)
therefore:
[tex]\int_{0}^{T} (Vcos{f(t)} - U)dt = 0[/tex]
and [tex]\int_{0}^{T} (Vsin{f(t)})dt = H[/tex]
therefore [tex]\int_{0}^{T} (sin{f(t)})dt = H/V[/tex] - which is turns to be right, as checked it in MATLAB.

Though, in the axis system of the missile we get:
the relative velocity between the missile and the plane in missile axis is:
Vs(t) = V - U*cos{f(t)}
Vp(t) = 0 - (-U*sin{f(t)})
and initial position,that should preserve is R = (H , 0)
therefore:
[tex]\int_{0}^{T} (V - Ucos{f(t)})dt = H[/tex]
and [tex]\int_{0}^{T} (Usin{f(t)})dt = 0[/tex]
therefore [tex]\int_{0}^{T} (sin{f(t)})dt = 0[/tex] - A CONTRADICTION?????
 
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