Sailing problem

[RogerG]

Can you help please. I am trying to construct a trig formula to solve this problem:

I am sailing a straight line (side c) to get to my destination which is 5.42 nautical miles away. My boat speed is 3 knots. The tide is flowing at 2.5 knots (side a) at an angle of 3 degrees (Angle B). I need to know how long this will take or in other words the length of side b of the triangle. I know that the figures I am using mean that it is possible to say that it will take a little less than an hour but I am looking for a formula that will work with variations of data - for example if Angle B was 120 degrees.

Is this possible please?

 

[mmm4444bot]

how long this will take or in other words the length of side b of the triangle.

It seems to me that the length of side b is a distance, not a time.

I'm also thinking that you will end up at vertex C instead of vertex B (unless you've already included a course correction to account for the tide).

I'm probably misinterpreting the posted information. Can you post a diagram?

 

[RogerG]

Apologies for the delay - night time over here.

In the example I gave I meant to convey that the current was against the boat. But I am trying to construct a formula that would apply in all circumstances - whether the current was with, against or varying degrees of across the track.

So using the data I specified:

B|
 |
 | A-B =3.27nm Boat Speed 4 kts
 |
 |
 | Angle C = 120
A \
    \
 1.5   \
         \
           \C

So in my diagram, I am starting at point A intening to go to Point B. If I do nothing, the tide will take me to Point C. I intend to average 4 knots and obviously I will in due course steer line C-B to get to my destination.

Ah, having posted I see the forum software removes the spaces that set line AC 120 degrees from line AB. BUt does that make it clearer?

Grateful for the assistance.

 

[mmm4444bot]

B|
 |
 |
 |
 | A-B =3.27nm Boat Speed 4 kts
 |
 |
 |
 |
 |
 | Angle C = 120
A \
     \
       \
         \
           \
             \
               \
         1.5   \
                  \
                   \ C

To prevent the suppression of extra word spaces, we can use the Code tags.

Okay, I see my confusion. I was thinking of a triangle.

Per your diagram, the tide is flowing at a 60-degree angle to the vertical. We need to calculate the vertical component of the tide, to know by how much the speed of the boat is reduced.

[attachment=0:51vs0jvm]Knots.JPG[/attachment:51vs0jvm]

I added point D, to form a right-triangle with the tide vector as its hypotenuse.

Angle CAD is 60 degrees. The tide vector runs from A to C, so the vertical component of the tide runs from A to D.

The distance AD is the adjacent side of the 60-degree angle, so its length is 1.5 times the cosine of 60 degrees.

AD = 1.5 cos(60 degrees) = 0.75.

If we consider the positive direction as running from A to B, then the vertical component of the tide (from A to D) is negative because it's the opposite direction.

In other words, the vertical force of the tide reduces the boat speed by 0.75 knots.

4 - 0.75 = 3.25

The net vertical speed of the boat is 3.25 knots.

time = distance/speed

time = 3.27/3.25 = 1.0062 (rounded)

So, I'm thinking that it will take 1 hour 22 seconds to move from A to B.

When the tide changes direction, the new vertical component will need to be calculated. When angle CAD exceeds 90 degrees, the tide will increase the speed of the boat because the vertical component of the tide will be positive. If angle CAD = 90 degrees (i.e., the tide is running perpendicular to the boat's direction), then the vertical speed of the boat is unaffected.

Better wait for somebody else to check my reasoning, before you set sail across the Atlantic, though.

 

[RogerG]

gosh thanks. I follow the adverse tide logic completely and as I understand you when there is a positive tide (ie the angle is less than 90 degrees), I can do a mirror image of your calculation to find the positive component to add to the boat speed to find the speed over the ground made good.

I am just having a little trouble with the assertion that a tide at 90 degrees to the track has no negative effect. Logic says you are right of course. In practice I would have to sail at 45 degrees to the track to stay on course to counteract the tide and there would be some friction losses I guess. But for this purpose I can ignore that.

Thank you very much. This is appreciated.

Regards

Roger

 

[mmm4444bot]

having a little trouble with the assertion that a tide at 90 degrees to the track has no negative effect.

I'm thinking that a perpendicular tide would have no vertical effect on the boat's speed.

Again, I'm not 100% confident that I properly reasoned this exercise. My concern is that the boat is not headed from A towards B for the entire trip because a correction for the horizontal component of the tide must be made, at some point.

We need a second opinion.