@Scott T-Bird - I think the key is that you're thinking purely about sailing straight upstream, but that's not how to do this. Let me go through your points and show how I see them...I'm not sure I understand your argument correctly.
Yes, exactly. Current is 10 knots from the west, so True wind is 10 knots from the east.That would mean that True wind is whatever the current is.
Well, you wouldn't have the apparent wind until you start moving, but yes, let's say you start sailing 5 kts through the water, 45 degrees down stream (north east). Now you have the 10 knots of true wind from the east from before, plus 5 knots of boat-generated wind from the north east. The total apparent wind is now 14 kts from the east north east.I agree that if you pointed the bow of the boat down current at a 45 degree angle, you would have 10 knot True Wind (in their case) and an apparent wind with vector that is a little bit more aft that would actually give you lift.
Don't think of tje ground yet, or the actual wind relative to the ground. Think of the boat's motion through the water, in the 10 knots of true wind - like you said above. Since the true wind is 10 knots, you can sail away from that wind with a VMG greater than 10 if you have a foiling catamaran. This means that relative to the water's surface, you're sailing west with a VMG more than 10. Since the water is moving east relative to the land at 10 knots, and you're moving west relative to the water faster than that, you're sailing west relative to the land.It is an interesting question if you can actually "sail" downwind at higher SOG than the current when the actual wind is 0
But the idea is to never just sit in the water with boat speed equal to current speed. Again, you're not going straight upstream or downstream - you're always reaching. For this next part you might want to sketch out the vectors on some paper... I had to... Yes, say at some point you will have a VMG upstream that is 10 knots. But at that time you'll be broad reaching, so you're also moving north across the river. That motion across the river generates an apparent wind from the north. So your boat is pointing West North West (35 degrees north of the oncoming current). Through the water, you're going 12.2 knots. The component of your speed up-river is 10 knots, which cancels the true wind. But you also have a component of your speed, 7 knots, directly north. That northward movement, at 7 knots, generates a 7 knot northerly apparent wind. That apparent wind hits your sails at a 55 degrees off the bow (a close reach). The key is to have a boat that's fast enough that it can still accelerate to a higher speed when it is going 12.2 knots through the water, with an apparent wind of 7 knots from 55 degrees. The AC boats are efficient enough to do that. Therefore their speed through the water increases. The increase in STW changes means you have VMG upstream.#1: True wind goes to zero when your boat speed (in the water) equals the current (when SOG = 0).
No, definitely not DDW. I agree that you can't sail DDW at the wind speed, which is why you can't go straight up the river. You (well, not you, but a foiling cat) can, however, sail on a broad reach at a greater that VMG than the TWS (see their polars and the data from the AC races). That's all that's happening here - Sailing with a greater downwind VMG than the 10 knots of True Wind.Think about what happens when you sail DDW
Edit - Sorry, I adjusted the numbers in my example because I realized the triangle calculator I was using was changing the triangle legs around.... The concepts hold and now the numbers should work.
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