A simple navigation question

Greenheart

Greenheart

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Quick navigation question, re. beating/close reaching.

I’ve forgotten the mathematical factors by which a straight-line distance increases, when you have to tack towards your destination. A little while spent with paper and protractor and I could work it out, but I recall somewhere seeing a handy sliding-scale…

...which calculated the approximate increase of total distance, as one bears away. Such that if one was tacking through 45º, the increase over a straight line would be 50%, or whatever it may be…and the flatter the tack, the greater the overall distance.

Does anyone know of a link, to a page detailing the figures?

I recognise that many other factors affect distance & journey time...I was thinking about the basic passage-planning.
 
****EDIT sorry the previous message was wrong, this should be ok**


Supposing your destination is exactly up/downwind and alfa is your tacking angle, then the increase in distance is dist/cos(alfa)
For example with 45° tacking angle the distance will be multiplied by about 1.4, 14 miles instead of 10miles.


These are the correct increases (it's divided by cos(alfa))
30 1.156519
35 1.223496345
40 1.309396167
45 1.42001263
50 1.564181229



edit
"tacking angle" above to be read "true wind angle", half the tacking angle actually
 
Last edited:
dancrane said:
Quick navigation question, re. beating/close reaching.

I’ve forgotten the mathematical factors by which a straight-line distance increases, when you have to tack towards your destination. A little while spent with paper and protractor and I could work it out, but I recall somewhere seeing a handy sliding-scale…

...which calculated the approximate increase of total distance, as one bears away. Such that if one was tacking through 45º, the increase over a straight line would be 50%, or whatever it may be…and the flatter the tack, the greater the overall distance.

Does anyone know of a link, to a page detailing the figures?

I recognise that many other factors affect distance & journey time...I was thinking about the basic passage-planning.
Click to expand...

Tacking angle is zero (straight line) no increase
Tacking angle is 120 degrees then 100% increase, i.e. double the distance. (You are sailing two sides of an equlateral triangle).

Surely a straight line proportionally between is good enough for a rough estimate.

If so then the answer is tacking angle divided by 120.

If your boat tacks over 120 degrees then you should sell it!
 
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Roberto said:
Supposing your destination is exactly up/downwind and alfa is your tacking angle, then the increase in distance is 2*cos(alfa)
For example with 45° tacking angle the distance will be multiplied by about 1.4, 14 miles instead of 10miles. More exactly by sqrt(2) :)

alfa 2cos(a)
30 1.729327404
35 1.634659562
40 1.527421609
45 1.408438177
50 1.278624217


edit
"tacking angle" above to be read "true wind angle", half the tacking angle actually
Click to expand...

Edit again Roberto! It's 2/cos(a) where a is the angle between the initial CMG ( ie away from the direct line) and the answer will be something like 1.646 to make up a number. The percent increas is that number minus 1 and multiplied by 100.

This assumes you make the same angle on each tack. And it's nothing to do with the wind angle
 
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Edit original question: PLEASE BEAR IN MIND, I'M THICK!

Gentlemen, I thank you for your replies...sort of. The subject, which was hazy, is now a full-on pea-souper. :mad:

Is it the mathematicians' code of defensive obscurity - to make fairly simple formulae, more complex than the question? :D :rolleyes:

For the sake of sanity, I'll do it the slow way...I'm going to draw a 10 cm wide/high right-angle on an A4 page, then use protractor & ruler to determine how much longer the track is, from page-bottom to page-top, according to each extra degree the course is, from vertical.

Simpler than 2*cos(alfa), methinks!

Thanks anyway. In another life, I might have understood your musings perfectly well. ;)
 
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wotayottie said:
Edit again Roberto! It's 1/(1/(2sin(a)) +1/(2cos(a)) where a is the angle between the initial CMG ( ie away from the direct line) and the answer will be something like 1.646 to make up a number. Multiply that number by the direct distance.

This assumes you make as much distance on each tack. And it's nothing to do with the wind angle
Click to expand...


I corrected mine again sorry.

I think you are referring to a boat which for example is already going say at 45° to the true wind and bears away to 50°, this increase maybe ?

I was answering to the question: you are at point A, destination D is exactly upwind of you, you sail two legs with the same true wind angle.
In this case you sail AD/cos(CAB)

bitbit.jpg





dancran, I have a ruler for you let me dig it out :)
 
If you are tacking 45degrees to the true wind, you are sailing the long side of rightangle triangle, the other two sides are equal.
Pythagoras is your best friend.
The long side is sqrt(2) or 1.414 times the short side.

On a calculator
45
cos
1/x
 
As an aside, it's worth knocking up a little spreadsheet and looking at your downwind angles, to see how little your speed has to increase to make it pay to head up a few degrees from a dead run!
 
1/cos alpha ( yes this cleverdick got it wrong last time :o ) gives:

30 degree tacking 15.5% longer
40 degree 30.5%
45 degree 41% longer
50 degree 55.5%
60 degree 100%
70 degree 190% longer
 
this first one is for angles which might be useful in sailing downwind: what speed you need to achieve to justify a change from direct course.

Example: if you are sailing at 6 knots while pointing directly at destination (number 6 on the horizontal axis), if you change the course by 20° you need to sail at least at 6.5 knots to compensate for the additional distance: the vertical line starting from 6 on the horizontal crosses the diagonal line marked 20° at a point where the circles indicate 6.5kt. This supposing equal legs.
Basically: read your direct speed on the horizontal, go up in vertical until you cross the diagonal with the degrees of deviation you want, then back down *along the circles* ti read the speed.
Hope it is clear ?


Allung_uk.jpg


Have another one for tacking angles nearer to 40-50°
 
Roberto said:
this first one is for angles which might be useful in sailing downwind: what speed you need to achieve to justify a change from direct course.

Hope it is clear ?


Allung_uk.jpg


Have another one for tacking angles nearer to 40-50°
Click to expand...

Much clearer, many thanks. :)
 
dancrane said:
Much clearer, many thanks. :)
Click to expand...

How does knowing this help other than intellectual curiosity? Surely the most important thing to appreciate is VMG to the wind direction? If you sail too close to the wind, your speed drops off and so does the VMG. As you bear aware, your VMG goes up and then as you bear away further, it drops off again. Even a few degrees seems to make a huge difference to the VMG.

FWIW, I often sail with the VMG diplayed on the multi display just to prove the point to others that sailing as close as possible is not necessarily the fastest option to get to where we want to go if it is upwind

Having said all the above, the graph is very neat and does answer the question of the OP, so I am not being overly critical

TudorSailor
 
tudorsailor said:
How does knowing this help other than intellectual curiosity? Surely the most important thing to appreciate is VMG to the wind direction? If you sail too close to the wind, your speed drops off and so does the VMG.
Click to expand...

The OP is a cruiser and therefore will have the engine on all the time anyway. He wants to know how much longer it will take him if he maintains his 6.25 knots at 45 degs to the wind with the main up pretending to sail ("motorsailing") as opposed to being honest and just motoring there directly. 1.41 times as long, is the answer.
 
The trouble with VMG displays is that they mislead the helmsman, because the boat takes a while to change speed after changing course.
If your boat is sailing optimally at say 45 degrees to the true wind, then you head up, the VMG indicated will rise due to the better heading then slowly decrease.
Also because the unit's idea of true wind is calculated from 3 quantities boatspeed, wind speed and apparent wind angle, all of which have errors and averaging delays, the VMG displayed is often nonsense.
Being able to work out whether heading up 2 degrees and losing 0.1 of a knot is good or bad can be very useful.
 
Nico said:
The OP is a cruiser and therefore will have the engine on all the time anyway. He wants to know how much longer it will take him if he maintains his 6.25 knots at 45 degs to the wind with the main up pretending to sail ("motorsailing") as opposed to being honest and just motoring there directly. 1.41 times as long, is the answer.
Click to expand...

What a mysteriously presumptuous, caustically critical thing to say! I don't have any boat right now, less still an engine!

While I'm looking at different boats (with very differing sailing characteristics, re. favoured points of sailing and best course-compromises), I just wanted to know the extent to which the course distance increases, as one bears away. Pretty sure I never mentioned speed... :rolleyes:
 
This could get complicated :)

The speed your boat is making through the water and distances covered is not in itself truly relevant. It is the speed towards the destination (VMC) that becomes important in passage planning.

When sailing on a direct course to the destination then I agree that the speed you are doing in the direction of the objective is what you need (in order to calculate something like your arrival time).

When tacking.. it is the VMC (not necessarily VMG unless your target is directly upwind) that is important.

If for example the target is 5 miles directly upwind and you are doing 5 knots VMG... then it will take an hour.. Whether you do 6 or so knots and point at 45 degrees.. or 7 knots and point at 50 degrees.. The calculation would be simplified by just considering the speed you are making upwind and forgetting the distance covered.

My tuppence.. it's another way of looking at the question.

Note - By checking polars for various boats, you will see that their VMG (speed made to windward) peaks at a certain speed in a fairly light wind. However much stronger the wind gets, they don't make it to windward any quicker.
 
As an extra.. You might find it interesting to note that to sail the quickest path to a destination that is not directly to windward of where you start is to not necessarily sail best VMG on each tack!!!

Example here

Then the maths gets you into Wallying where in a shifty wind, you don't luff nor bear away to the full amout in each shift. It's all very mathematical.
 
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