How much longer is a beat than a run

TK

are you happy with those polars ? They have already been optimised for best speed directly to windward (VMG) for the different wind speeds. The diamond gives you the best VMG; so at 3 m/s windspeed it's best to head about 42 degrees off, and for 10 m/s, about 32. (That's VMG to windward)

In the first case, your boat speed will shoot up from 4.1 kts at 32 degrees off the wind to 5.3 kts at 42 degrees, and is more than enough to make up for the extra distance sailed.


Now you can start worrying about the effect of lee-bowing on the equation !

And VMC...
 
flaming said:
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...
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Strangely, experiments on my boat show that DDW is faster than tacking downwind. It's probably down to the rig which doesn't lose as much efficiency as a rig with sheeting angles restricted by stays. Easy to work it out if you have a GPS or integrated wind instrument. Just steer to maximise VMG.

To get the ratio of distance travelled when tacking upwind, divide water speed by VMG.
 
lustyd said:
I'd have thought it would be quicker again Gybing downwind...:D
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FWIW I agree, but there are some on here who say "tack downwind" to mean gybe downwind. I suppose they could also beat downwind if they wanted to.
 
images
 
Twister_Ken said:
What's a cosine?
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When you are going up or down a hill, the cosine of the angle is the ratio of the horizontal distance travelled to the distance you walked on the slope. Its maximum is 1 (when the ground is flat every bit of every step takes you forward) and its minimum is zero (when you climb up a vertical cliff you don't go forward at all, no matter how high you climb).
 
flaming said:
I note that's with a 106% jib. Are you barberhauling that? I have a feeling the polars may assume you are.

YES

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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I have an North G2 A sail (new this season, not yet flown) which is 78 sqm, so not quite so enormous! Flies from the masthead.
 
sarabande said:
TK

are you happy with those polars ? They have already been optimised for best speed directly to windward (VMG) for the different wind speeds. The diamond gives you the best VMG; so at 3 m/s windspeed it's best to head about 42 degrees off, and for 10 m/s, about 32. .
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Yes - I was assuming about F4 which is approx 6 - 7 m/s, putting the diamonds at about 35º.
 
I often wonder about these polars.

These I see come out of a velocity prediction program, and so I cynically believe are not the result of empirical observation. Which would be difficult and expensive in terms of time therefore money.
 
lustyd said:
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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ER, No! (Sorry if I get accused of contradicting ;))
If you do 1.2 knots at 35o, then you only need to do 0.2kt better at 45o.
But if you do 2.4kt at 35o, you need to do 2.8 at 45o.
If you do 4.8kt at 35o, then you need to do 5.6 at 45o
etc.

At least that's the case in the world of polars. In the real world...
 
timbartlett said:
ER, No! (Sorry if I get accused of contradicting ;))
If you do 1.2 knots at 35o, then you only need to do 0.2kt better at 45o.
But if you do 2.4kt at 35o, you need to do 2.8 at 45o.
If you do 4.8kt at 35o, then you need to do 5.6 at 45o
etc.

At least that's the case in the world of polars. In the real world...
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Ah very good point! I think my boat can go faster than 1.2kt, I'll have to test!
 
KellysEye said:
>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.

If you don't take into account tide/current and waves your calculations will be signifcantly out.
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I do rather know that, and I know how to allow for tides. All I was interested in was a factor to multiply the rhumb line distance by. Thanks to various scholars of this parish, I now have it.
 
>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.

If you don't take into account tide/current,waves and leeway your calculations will be signifcantly out. For example we sailed two five hundred mile passages (Curacao to the Caribbean island chain) upwind with a two to three knot current from the east and apparent wind twenty to twenty five knots. We sailed eight hundred miles to cover five hundred miles. Your angle would give six hundred and seventy five miles.

Tide is a different matter. If the tide is against you you will be pushed down, if it's with you you will be leebowed and pushed higher. If you sail both tide cycles there is still a small loss because most boats are quicker and point higher on one tack. You also still have to allow for being pushed back by waves. Occasionally we have come to a complete stop and that's in a heavy displacement boat.

What we did over a number of passages was work out the average apparent wind speeds, the average tide, the course and distance over ground made good and the wind angle on each tack. Do that over various wind speed and tides and you will be well set for planning passage time.
 
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flaming said:
But looking at the wavelets they aren't sailing DDW!
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Looking at the sails, you don't get much more downwind. They are being wrapped around the shrouds and spreaders. I can't sail downwind as I have a fractional rig.
Mentalpause was with me and we bemoaned the fact that they ran off away from us as we couldn't follow them.
 
Back to the question.

Sail for 7 miles in a staight line and then tack through 90 deg and sail another 7 miles.

The 90 degree tac for the calculation allows for a bit of leeway when sailing at, say, 40 degrees off the wind.

You have now sailed 14 miles.

The distance between your start and end points is, as I remember, the square root of the sum of the squares of the two legs you have just sailed.

We thus need the square root of 98, which to me is pretty close to the square root of 100. We have thus travelled a straight line distance of 10 miles, having sailed 14 miles.

Not super accurate but near enough to give you some idea, and you don't need a calculator or trig tables!

Or am I being simplistic?
 
steve48 said:
Back to the question.

Sail for 7 miles in a staight line and then tack through 90 deg and sail another 7 miles.

The 90 degree tac for the calculation allows for a bit of leeway when sailing at, say, 40 degrees off the wind.

You have now sailed 14 miles.

The distance between your start and end points is, as I remember, the square root of the sum of the squares of the two legs you have just sailed.

We thus need the square root of 98, which to me is pretty close to the square root of 100. We have thus travelled a straight line distance of 10 miles, having sailed 14 miles.

Not super accurate but near enough to give you some idea, and you don't need a calculator or trig tables!

Or am I being simplistic?
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But I'm not tacking through 90º, but through 70º.
 
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