The natural wind is from 360 degrees at 10 knots. The desired relative wind down the angled deck is 10 degrees left of the ship’s heading at 30 knots.
What heading and speed must the carrier go?
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2$\begingroup$ Is it not 10 degrees and 20 knots? $\endgroup$– Ron BeyerCommented Nov 17, 2018 at 3:53
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$\begingroup$ Ha! Yeah, I think you’re right. Thet’d give you a bit less than 30 kts, but pretty durn close. I was making it waaay too hard. Thanks. $\endgroup$– Tad ChamberlainCommented Nov 17, 2018 at 8:10
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$\begingroup$ As I think about this further, I see it’s not right. Heading the ship 10 right of the natual wind would put wind down the angle only when the ship was stopped. As the ship speed increases the relative wind direction will move closer and closer to the ship’s heading. $\endgroup$– Tad ChamberlainCommented Nov 18, 2018 at 0:08
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$\begingroup$ If the ship has to steam forward to make its own wind, it will never be right down the angled deck. Just the limitations of the layout. $\endgroup$– Michael HallCommented Nov 19, 2018 at 2:44
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1$\begingroup$ Where are we going to find a carrier to test the answer? :) $\endgroup$– Juan JimenezCommented Nov 19, 2018 at 13:42
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2 Answers
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A carrier heading of 010 degrees at a speed of 20 knots would give you an acceptable solution. You could perform some vector addition to compute the resultant wind and perhaps optimize this somewhat, but in reality the winds are generally not constant enough to make minor adjustments worth chasing.
Even if the winds were calm and the ship was making all 30 knots the crosswind component is almost negligible, and well within limits for any carrier based aircraft. This happens a lot. There is a bigger negative effect from turbulence caused by the island structure when the ship has to make wind than any effect of increased crosswind.
answered Nov 19, 2018 at 18:55
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Let’s suppose that the runway is angled by 10º to the left. To start with, let’s assume there is no wind, and that we wish 30 kts of relative wind parallel to the runway. The speed of the carrier (with any heading) will be 30 kts/cos 10º = 30,46 kts.
Now, we have a wind with a speed of 10 knots in relation to the sea. We change the heading of the carrier in order to take the maximum advantage of that wind, so that it may be parallel to the runway. So the carrier steers to a heading exactly 10º to the right of the wind direction. Now we have too much wind, 30 + 10 = 40 kts, and we should reduce the ship’s speed in proportion to that excess of 40/30. In other words, the carrier should steam at 30,46 x 30/40 = 22,85 kts.
Hence, the carrier should have a heading of 10º to the right of the north, and a speed of 22,85 kts...
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$\begingroup$ The conditions given in the original question was 10 knots of natural wind from due north. $\endgroup$ Commented Nov 26, 2018 at 16:11
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