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Ground Speed? How is it calculated?

2K views 18 replies 10 participants last post by  tenuki 
#1 ·
I know if you’re going 4 knots and the current is 3 knots against you, your only making 1 knot ground speed, or if the current is 3 knots with you, you’re at 7 knots ground speed. But is there a calculation for ground speed when you are going cross current, say, 4 knots across a 3 knot current? I’m guessing you’d be at about 5 or 6 knots ground speed, but that’s just a guess.
 
#2 ·
I suppose you can use trigonometry to caluclate. Draw a triangle 1 side 3kts length another 4kts length at right-angle to each other, the 3rd side length is your resultant speed and angle of your track (bearing other influence like wind).
 
#3 ·
You can calculate it mathematically—it's called trigonometry... or you could use the GPS to show you your SOG. ;) Or you could draw the speed of the boat and the current graphically, and add the vectors to show you your SOG
 
#4 · (Edited)
What your looking for is called the "vector sum". The two velocities are at right angles to each other, the vector sum can be determined using the Pythagorean theorem.

vresultant2 = (4kt)2 + (3kts)2

vresultant = Sqrt[ (4kt)2 + (3kts)2 ]

vresultant = 5 knots

or....

4 knots squared (16) + 3knots squared (9) = 25, and the square root of 25 = 5 knots!

But the real issue isn't usually speed, but how far off course the current is pushing you. To calculate this you need to calculate how long (time) it will take you to get a certain distance (say 100 nautical miles).

t = d/v

Once you know the time, then you miltiply that by the current (3 knots) to give you how many miles you'll be off course at the end of the 100 miles, and correct your heading from there.

Destination = v x t

So...

100 nautical miles at 5 vector sum, knots per hour = 20hrs, multiplied by 3 knots current = 60 miles that you will be off course for every 100 miles traveled.

Edit: Sorry about all the math
 
#5 · (Edited)
Umm.

Knotaloud/trantor-

The current isn't always perpendicular to the wind... in fact 99% of the time it won't be, so the pythagorean theorum falls on its face. Also, knotaloud's equations should have been written

vresultant^2 = (4kt)^2 + (3kts)^2
rather than what you had, since the ^ is often used to express raising something to an exponent. Merely having the number outside the parentheses is usually assumed to mean multiplication.

Also, the current is not always going to be the only thing affecting your course. Leeway also affects your course, and may, in many cases, offset the set of the current to some degree, so basing your calculations on just the current's speed, which also may not be perpendicular to the line of travel, is less than accurate.

BTW, "T=D/V" is Distance = Velocity * Time, not Destination.

Finally, I generally recommend doing the calculations graphically, since it is far more difficult to screw them up by doing them graphically. Also, it can be done right on the chart. Any calculations for set and drift are going to be estimates...and a graphic solution generally will get you in the ball park, without a lot of math.... and the extra precision of the mathematics-based solutions is generally both unnecessary and inaccurate-it is like measuring with a micrometer and then marking with a piece of chalk... why bother-doing the calculations graphically is simpler and faster as a general rule...

And often, like in many other things where time is of the essence, a good answer in navigation right away is better than a perfect answer that takes four times as long... since there are often times you need to have idea of what to do immediately...and the extra time the equations take can put you closer to danger by delaying any decision you make.
 
#11 ·
Tea-Rex, you still have one of those things? I haven't seen one of those since I sold it in 1972.

Ground speed is calculated by dropping a bouhyied line and timing how many knots you've tied into the line at 6 fathom intervals transverse through your hands during a sandglass timed interval of 3 minutes. This is a modern feature added as an extra on new Catalinas, Beneteaus and Hunters. All the rest of us (particularly at my age) don't really care how fast we're going, just that we're still going:D
 
#12 ·
Ian-

IIRC, using a chip knot log... will tell you the boat speed relative to the water, but not the speed over ground, since it doesn't account for whatever current set and drift may be occurring at the time.

ianhlnd said:
Ground speed is calculated by dropping a bouhyied line and timing how many knots you've tied into the line at 6 fathom intervals transverse through your hands during a sandglass timed interval of 3 minutes. This is a modern feature added as an extra on new Catalinas, Beneteaus and Hunters. All the rest of us (particularly at my age) don't really care how fast we're going, just that we're still going:D
 
#14 ·
LOL... very true Ian...but the OP was asking specifically for SOG...
 
#16 ·
Plotting frequent positions on the chart, ascertained by the most reliable methods available, is the best method of determining set and drift. It is a navigational axiom that one should do upon the chart as much as possible. Using formulas and calculations one is apt to make the dreaded math error. Plotting on the chart is much less error prone as each conclusion must "look right". Besides, doing all that math, when a simple fix will do, keeps one too long in the chart room and not on watch where one belongs.
 
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