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I have the following situation, two objects A and B, at a distance of x from each other. Both objects have their own 2d heading Ah and Bh and their own speeds As and Bs. I'm trying to determine the rate that A would perceive B to be moving, i.e. if I were standing on the back of A how quickly would I have to turn my head to keep my eye on B? I've been barking up the angular velocity / transverse velocity tree but got myself totally confused applying the theory to the context. I'm sure this is pretty simple stuff but I've got my wires so crossed now I can't think straight!

A layman's explanation of how to calculate / express this would be appreciated. I guess I'm looking at how many degrees/rads I'd need to turn my head per second??

Many thanks!

Raven
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1 Answers1

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Here is the way it can be done.

Use $Ah$ as the $x$ axis. Call $\theta$ the angle between $Ah$ and $Bh$.

You have the coordinates of each point by having

$x_A(t)=x_A(0)+As \times t$, $y_A(t)=0$ (by construction)

$x_B(t)=x_B(0)+Bs \times t \times \cos \theta$

$y_B(t)=y_B(0)+Bs \times t \times \sin \theta$

Then $x_{AB}(t)=x_B(0)-x_A(0)+Bs \times t \times \cos \theta-As \times t$

and $y_{AB}(t)=y_B(0)+Bs \times t \times \sin \theta$

Call $\phi$ the angular angle of the vector $AB$

Then $\tan \phi(t)=\dfrac{y_B(0)+Bs \times t \times \sin \theta}{x_B(0)-x_A(0)+Bs \times t \times \cos \theta-As \times t}$

Or more simply $\phi(t)=\arctan \dfrac{\alpha +\beta t}{\gamma + \delta t}$

You can derive this to have the angular speed.

Martigan
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