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Math equation simplification question.


Greg H.

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Scenario:

 

Let us suppose we have a stationary Target T being orbited in a perfect circle by an Attacker A. The circle has the radius of R in meters. T has a weapon that can track A at a set rate, T1 measured in radians per second. Given those conditions, I want to find the minimum velocity (V) in meters per second at which A can move around the circle of the fixed radius (R) and still exceed the value of T1.

Let us also suppose there are no outside influences to consider.

 

So here's how I worked this. First, we know the circumference of a circle is

[math]

2\pi R

[/math]

 

 

We can then write a formula based on the velocity that determines how long it takes A to complete the circle.

[math]

T_2 = \frac{2\pi rad}{\frac{2\pi R m}{V m/s}}

[/math]

which can be rewritten as

[math]

T_2 =\frac{ 2\pi V rad \cdot m}{2\pi R m \cdot s}

[/math]

where the top of the fraction is in terms of radian meters, and the bottom is in meter seconds.

 

Here's my question: Should the [math]2\pi [/math] and the meters cancel each other out, leaving us with a value in radians per second that is only depended on the velocity and the radius of the circle?

[math]

T_2 = \frac{V}{R} rad/sec

[/math]

 

And since we need a value of V which yields a T2 greater than T1, the final equation would be

[math]

V > T_1 \times R

[/math]

 

Running the formula with some actual numbers, the orders of magnitude look right - for example, if I have a 3,000 meter circle and a T1 of .004, V needs to be greater than 12. But something feels off to me, so I thought I'd ask for help.

Edited by Greg H.
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The weapon attached to T can be considered to be mounted in a turret with a fixed rotation speed of T1. Sorry, that wasn't clear.

 

If it helps, imagine a tank with no tracks, but the turret still spins, being orbited by an airplane at a fixed distance. Given the fixed rotation speed of the turret, and the fixed radius of the orbit, find a minimum value of V such that the tank can't track the plane and shoot it down.

Edited by Greg H.
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Sorry - Using more familiar terms. A quick idea

 

[latex]\omega = \frac{d \phi}{d t}[/latex]

 

[latex]v_{tangential} = \omega \cdot \frac{d \phi}{dt} = r \cdot \omega[/latex]

 

You want omega_tank to be less than omega_plane

 

[latex]\omega_{plane}=\frac{v_{plane}}{r}[/latex]

[latex]\omega_{tank}=given[/latex]


NB

 

forgot to give terms

v= velocity, tangential in this case [metres/sec]

omega = angular velocity [rad/sec]

r = radius [metres]

phi = angle [radian]

t=time [seconds]

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A stopped clock is exactly correct twice a day.

 

It takes one circuit by A for T to aquire A's angular velocity or cycle time.

 

T can then fire at the appropriate point of A's second circuit to score a hit, whatever the angular velocity of A.

Edited by studiot
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A stopped clock is exactly correct twice a day.

 

It takes one circuit by A for T to aquire A's angular velocity or cycle time.

 

T can then fire at the appropriate point of A's second circuit to score a hit, whatever the angular velocity of A.

While in real life, I would agree with you, the simulation I'm using isn't that smart.

 

@imatfaal: Thanks for the explanation. It confirms that I was on the right track, just in a terrible way. :)

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