How much longer is a beat than a run

Twister_Ken

Twister_Ken

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Having not so much been at the back of the queue when the mathematics genes were being handed out as at home in bed, I wonder if a kind person (or even an unkind one) could offer a simple formula for this.

I want to sail from A to B. The rhumb line distance is X.

If the course is dead to windward, how much further do I have to sail through the water, than X. Ignore tides, leeway and being slapped about by waves, and assume a comfortable breeze and pointing 35º to true wind direction.

Many thanks, TK
 
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I believe it will be (distance/2)/(cosine of 40 degrees) x 2 if my school maths is still strong :)
 
To explain, I split the journey into 2 halfs, each at 40' off the wind and each a right angle triangle off of the line a run would take. You know the adjacent side length (half the distance) and you know the angle. You want to find the hypotenuse so you'll need the CAH part of SOHCAHTOA. Divide the adjacent by the cos of the angle and there you have your distance for half the journey :)
Cheers
Dave
 
Twister_Ken said:
Having not so much been at the back of the queue when the mathematics genes were being handed out as at home in bed, I wonder if a kind person (or even an unkind one) could offer a simple formula for this.

I want to sail from A to B. The rhumb line distance is X.

If the course is dead to windward, how much further do I have to sail through the water, than X. Ignore tides, leeway and being slapped about by waves, and assume a comfortable breeze and pointing 35º to true wind direction.

Many thanks, TK
Click to expand...

1/cos(40) roughly 30% further. 1/cos(35) roughly 22% further.
 
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Twister_Ken said:
Having not so much been at the back of the queue when the mathematics genes were being handed out as at home in bed, I wonder if a kind person (or even an unkind one) could offer a simple formula for this.

I want to sail from A to B. The rhumb line distance is X.

If the course is dead to windward, how much further do I have to sail through the water, than X. Ignore tides, leeway and being slapped about by waves, and assume a comfortable breeze and pointing 35º to true wind direction.

Many thanks, TK
Click to expand...

It's the same, just put your engine on.

Mathematically, assuming one tack (2 legs) the extra distance is

2(X -X/cos35)

but I can't be @rsed looking up the cosine of 35deg & it is completely ignoring leeway which could add 5 deg to your angle.
 
Not sure where I got 40 degrees from, sorry all.

It doesn't matter whether you tack once or 20 times btw, the distance will be identical so always divide by 2.

I think the more interesting question is would pointing a further 5 degrees out and gaining extra speed get you there quicker?
 
a cosine is a maths function on a calculator is all you need to know for this. Google sohcahtoa and you'll find explanations of the maths.

Google cosine 35 degrees and google will do the calc for you :)

putting the X in there it is:

((X/2)/(cosine angle from wind))*2=distance over all
 
Twister_Ken said:
After posting I looked at the boat's polars. That's why I amended 40º to 35º (sorry, lustyd).

What's a cosine?

Searush - so you're saying it's X x 1.22?

PS - the Arcona doesn't really do leeway. Well, not in big dollops, anyway.
Click to expand...

Cosine of an angle is the ratio of the Adjacent side of a right angled triangle divided by the Hypotenuse. It is the ration between the tack and the direct course in your example. All sorted out by Pythagoras around 2 thousand-odd years ago.

Your answer seems about what I would expect without looking it up.

Really? Is Ancona a railway train then rather than a boat? :D
 
Twister_Ken said:
ChrisE - so you're saying it's X x 1.22?
Click to expand...

Yes, for 35' it's 1.22077459 for every 1 unit direct
40' is 1.30540729
45' is 1.41421356

Cheers
Dave

[EDIT] given these numbers it would seem one only has to achieve a water speed of 0.2 kt higher at 45' than at 35' to arrive at the same time. I suspect a lot of boats would be better than that?
 
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lustyd said:
Yes, for 35' it's 1.22077459 for every 1 unit direct
40' is 1.30540729
45' is 1.41421356

Cheers
Dave
Click to expand...

I've always used 1/3 longer, as a rough judge.

In anything other than flat water, you're not going to get 35 degrees true, and even then I find that "optimistic". So the value for 40 degrees seems to validate my guestimate!
 
lustyd said:
(snip)
[EDIT] given these numbers it would seem one only has to achieve a water speed of 0.2 kt higher at 45' than at 35' to arrive at the same time. I suspect a lot of boats would be better than that?
Click to expand...

"Keep her full & by, my son, full & by"

:D

Yup, it's all about keeping her pulling hard & well. But I'd still put the engine on or go the other way. Gentlemen don't beat to windward, doncha know?

Incidentally, many catamarans will also make faster progress if they tack downwind.
 
Searush said:
Really? Is Ancona a railway train then rather than a boat? :D
Click to expand...

No, it's a town in Italy.

The Arcona, though is a modern hull with a deep keel, and at decent sailing speeds I haven't detected measurable leeway.

As to full and bye, 35º to TW is nothing like pinching on the new boat, though it would have been on the old one.

The polars are here
 
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Searush said:
Incidentally, many catamarans will also make faster progress if they tack downwind.
Click to expand...

All boats do. DDW is slow. Broadly speaking, the faster the boat can go, the wider the angles. So for a typical cruiser racer of the Beneteau first type it will be 170 ish (in a decent breeze - much less in the light) wheras a planing sportsboat or catamaran it will be more like 140-150.

Only when it's really blowing, and you can do hull speed DDW does it pay to do that. And then, with the flow flicking directions all the time the roll tends to mean more crashes...
 
A very quick look at those polars suggests that you'd do better at 45 degrees than 35 degrees. Take the ratio of the speed at 45 deg to that at 35 deg (about 1.28 from the polars for the lowest wind speed) and compare it to the distance ratios.

Then factor in the extra time lost actually tacking, variations in tide, sea state and everything else and the theory all becomes meaningless!
 
Twister_Ken said:
No, it's a town in Italy.

The Arcona, though is a modern hull with a deep keel, and at decent sailing speeds I haven't detected measurable leeway.

As to full and bye, 35º to TW is nothing like pinching on the new boat, though it would have been on the old one.

The polars are here
Click to expand...

I note that's with a 106% jib. Are you barberhauling that? I have a feeling the polars may assume you are.

And a 100sqm kite... That's enormous! Bigger than we're carrying on a 37 footer - and we're already 20% over design size!
 
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