Twister_Ken
Well-Known Member
Please bear with me on this - just scraped through O'level maths (for the younger generation, that's what GCSEs used to be called before decimilisation and devaluation). My curiosity was raised by the question of how windward performance could be improved by forestay tension.
We want to sail dead to windward from A to B, for 10nm. Our noble vessel will sail at 45 degs to the true wind, and for the sake of argument, makes no leeway.
We set off on stbd tack, and at the halfway point (C) we tack through 90 degrees and continue on that tack until reaching B. Our track and the desired course make a triangle ABC, where AB is 10nm, and the legs AC and CB are equal, each - according to Pythagoras (the square of the hypotenuse of a right triangle is equal to the sum of the squares of the two adjacent sides) - of 7.07 nm.
We therefore need to sail 14 miles through the water to cover a 10 mile course.
If we assume we tweak the forestay and trim the sails in such a way that we can now sail at 40 degrees to the true wind, our triangle ABC is no longer a right triangle, as we can tack through 80 degress, and has shorter legs AC and CB. We need to resort to sines, etc (Mnemonic = See Old Harry Catch A Herring Trawling Off America). Using cosine 40, I get each leg of the new beat to be 6.53 nm, meaning our distance through the water is decreased by approximately one mile.
Assuming the noble vessel trogs along to windward at 5 kts, then we arrive at B 12 minutes earlier than in our untuned state.
Applied to a distance of 63 miles (Needles-Cherbourg) the tighter forestay will get us there roughly 1h 15m earlier, time to get a table in the Café de Paris before cookie goes home for the night.
Have I got that right?
We want to sail dead to windward from A to B, for 10nm. Our noble vessel will sail at 45 degs to the true wind, and for the sake of argument, makes no leeway.
We set off on stbd tack, and at the halfway point (C) we tack through 90 degrees and continue on that tack until reaching B. Our track and the desired course make a triangle ABC, where AB is 10nm, and the legs AC and CB are equal, each - according to Pythagoras (the square of the hypotenuse of a right triangle is equal to the sum of the squares of the two adjacent sides) - of 7.07 nm.
We therefore need to sail 14 miles through the water to cover a 10 mile course.
If we assume we tweak the forestay and trim the sails in such a way that we can now sail at 40 degrees to the true wind, our triangle ABC is no longer a right triangle, as we can tack through 80 degress, and has shorter legs AC and CB. We need to resort to sines, etc (Mnemonic = See Old Harry Catch A Herring Trawling Off America). Using cosine 40, I get each leg of the new beat to be 6.53 nm, meaning our distance through the water is decreased by approximately one mile.
Assuming the noble vessel trogs along to windward at 5 kts, then we arrive at B 12 minutes earlier than in our untuned state.
Applied to a distance of 63 miles (Needles-Cherbourg) the tighter forestay will get us there roughly 1h 15m earlier, time to get a table in the Café de Paris before cookie goes home for the night.
Have I got that right?
