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 May 16, 201510 yr by Greg H.

What do you mean by track, if T1 is fixed?

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 May 16, 201510 yr by Greg H.

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]

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 May 17, 201510 yr by studiot

Apart from Studiot's astute observation above

 

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

 

[latex] v_{plane}> r \cdot \omega_{tank} [/latex]

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.

I would have said you are into modular arithmetic and that A can acquire T, what ever the angular speeds, by this method.

 

If this is so then there is no safe speed for T.