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The capture velocity is the velocity that the particle must travel to fall a distance 'x' (see figure above) in the same amount of time that a particle traveling at

{latex}\large\[V_\alpha\]{latex}

{latex}\large\[L + b\tan (\alpha )\]{latex}

Now, we know that one way to determine the travel time of a particle is to divide the distance travelled by the particle by the particle velocity. Therefore, we can equate the following travel times:

{latex}\large\[\frac{d}{{V_c \cos (\alpha )}} = \frac{{L + d\tan(\alpha )}}{{V_\alpha  }} = \frac{{L\sin (\alpha ) + \frac{b}{{\cos (\alpha )}}}}{{V_{up} }}\]{latex}

Solving for the capture velocity by equating the first and third expressions, we find that:

{latex}\large\[V_c  = \frac{{dV_{up} }}{{L_{tube} \sin (\alpha )\cos(\alpha ) + d}}\]{latex}

Where L is the length of the tube, d is the inner diameter of the tube, Vup is the vertical velocity through one tube, and

{latex}\large\[\alpha \]{latex}

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