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The missile and aircraft the saga goes on

  • Thread starter ozzy
  • Start date
  • #1
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Well, I am sorry to bother you again, but it seems that after solving my last problem nobody enters that topic anymore :rolleyes: I hope I won't get on moderators nerves here...
Well, anyway. As one may remember, I asked about this question:

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.


The solution was made by balakrishnan_v, and did I mentioned that he is THE KING? :wink:
But since then I was thinking about the solution and here what I discovered:


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] - Where Am I Wrong????? Please help!!! :frown:
 

Answers and Replies

  • #2
since you are using an integral surely you are failling to answers the question.

isnt the answer is a classical curve that be done without calculus?
 
  • #3
12
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The answer for the original question is found here:

balakrishnan_v said:
Consider any instant at which the missile is making an angle A(t) 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 [tex]\int_{0}^{T} (V - Ucos{A(t)})dt = d[/tex]
or [tex]VT-U\int_{0}^{T} (cos{A(t)})dt = d[/tex]..(1)
and the other equation is that the horizontal distance moved by the plane and missile are same
so we have [tex]\int_{0}^{T} (Vcos{A(t)})dt = UT[/tex]
or [tex]\int_{0}^{T} (cos{A(t)})dt = \frac{UT}{V}[/tex]
substituting this in (1) we have
[tex]VT-U(UT/V)=d[/tex]
or [tex]T=\frac{HV}{(V^2-U^2)}[/tex]
But please help me with the current, new question, which is MY question, and not from the book....
 
  • #4
70
1
ozzy said:
[tex]\int_{0}^{T} (Usin{f(t)})dt = 0[/tex]
therefore [tex]\int_{0}^{T} (sin{f(t)})dt = 0[/tex] - Where Am I Wrong????? Please help!!! :frown:
If you consider the positive x-axis for the reference system of the missile as pointing in the direction of V and the y-axis as being perpendicular to it, you can see that the plane always lies on the positive x-axis as seen from the missile. Although at each instant it has a velocity component in the y-direction of -U*sin(theta), where theta is the angle the missile's velocity makes with the horizontal, the missile simply turns to keep the plane on its positive x-axis, thereby keeping the plane's velocity component in the y-direction of the missile's coordinate system effectively zero.

In short, the rotation rate of the missile's coordinate system cancels out the -U*sin(theta) component of the plane's velocity in that coordinate system.
 
  • #5
12
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Brinx said:
If you consider the positive x-axis for the reference system of the missile as pointing in the direction of V and the y-axis as being perpendicular to it, you can see that the plane always lies on the positive x-axis as seen from the missile. Although at each instant it has a velocity component in the y-direction of -U*sin(theta), where theta is the angle the missile's velocity makes with the horizontal, the missile simply turns to keep the plane on its positive x-axis, thereby keeping the plane's velocity component in the y-direction of the missile's coordinate system effectively zero.

In short, the rotation rate of the missile's coordinate system cancels out the -U*sin(theta) component of the plane's velocity in that coordinate system.
Yes, I also have this intuition, and regarding what you say, the last integral, with zero outcome is the right one, since the plane in missile system doesn't travel over Y axis. But since the second integral is the right one, and not fourth, it seems that the plane travels over distance [tex]\frac{HU}{V}[/tex] in Y missile direction...
how can you explain it?

Or actually what you try to say is that there is no velocity component in Y direction of the plane in missile system??? But how it can be? Then the initial speed (not velocity) of the plane in missile system is zero???
I don't get...Maybe this is the reason why I am not a mechanist, but an electronicker:)
 
Last edited:
  • #7
12
0
gnpatterson said:
yes and no...
yes in terms of expanding the problem to something universal...
no in terms that I still can't understand how it is possible that the second integral is not zero, but the fourth suppose to be a zero...
I understand that there is some problem with the assumptions...I just don't get what is the problem to assume, that in missile system the plane travels overall zero distance in vertical axis....
 

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