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What Steve was trying to say is that if the plane were to be flying a steady descent from 9 to 6 it would be consuming ALL the airspace from 9 to 6, not just the diagonal vector. So they cannot vector a plane at 7k across that path because it's *possible* that the plane will descend from 9 to 6 in less than a mile vs. the hoped for gentle descent.
5Z Yes, exactly Tom. The further out that they can get an airliner low, the less airspace they will consume getting him down in close to the vicinity of the terminal where the airspace is more precious. |
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