Estimating VMG

It’s easy enough to estimate time of arrival when you are fetching your destination; distance divided by speed equals time. But suppose you want to figure out how long it takes to sail to San Francisco. With the prevailing summer wind, it’s going to be a beat and let’s say your boat does 6 knots through the water—just not in the right direction because you have to tack. The “ETE” or estimated time enroute, and “VMG” or velocity made good aren’t so straightforward. We heard a lot about VMG during the America’s Cup. You can figure a rough estimate, though, without knowing trigonometry or having a whole cabal of nerds running arcane computer programs for Larry Ellison.

What we’ve got when sailing into the wind is an isosceles right triangle where the hypotenuse represents the distance to your destination and the legs represent your tacks. So x + x is the distance you’ll have to sail to get from F to D via point E:

VMG

Remembering that geometry class in high school, a squared + b squared = c squared, but since the legs are the same the hypotenuse is equal to either side times the square root of two. (In this theoretical treatment it doesn’t matter whether it’s ten tacks or one.) This means if you divide the hypotenuse, which is in this case the distance from the end of Portrero Reach (F) to the City (D), by the square root of two (≈1.41) and then multiply by two, you will have the distance you need to sail to get there. And then if you divide that by your six knots of speed, you’ll have your time enroute or ETE. Divide the distance from F to D by the ETE and you’ll have your velocity made good or VMG.

Because that calculation is based on a perfect right triangle and you’re neglecting leeway, the distance is actually a little more than that so your ETE is longer and your VMG is less. So forget all that square root stuff. It turns out that just multiplying the distance from F to D by 1.5 is going to give a close enough estimate of the actual distance you’ll have to sail, assuming the destination is dead to windward. (If it isn’t, the multiplier will be less than 1.5, but of course, never less than 1.) So put those high school trig and geometry books back on the shelf next to your vinyl Pat Benatar albums. The City is roughly 7 nautical miles to windward from the channel entrance, so you’ll have to sail 7 x 1.5 or 10.5 miles at your boat speed of 6 knots. 10.5 divided by six is about one hour and 45 minutes, your ETE, giving a VMG for the seven miles of 7/1.75 = 4 knots. In fact, you can make it real easy and say your VMG into the wind is roughly 2/3 your boat speed through the water. Not so hard.

But wait. We also have to figure current, as the above assumes slack water. This can get a bit cumbersome but let’s see if we can, again, find a shortcut. Have a look at the current charts at the back of your tide book. Turn to the max ebb chart on p. 57. You’ll see 1.4 knots from behind, then a little over 2 knots of current to the right. This is on an average day at maximum current. To find what the figure is on a strong day, go to the chart on p. 48 and you’ll see the multiplier is 1.5. This means that at max ebb on a strong day, you’ll get 1.4 x 1.5, or a little over 2 knots of help halfway there, and then let’s say 2.4 x 1.5 or about 3.5 knots of being set to the right, which also helps. Averaging those, you’ll have a bit less than three knots in your favor for the whole trip. When we add this to the six knots your boat does over the water, we get a speed of nearly nine knots. The distance sailed is the same 10.5 miles, so now the trip will take about 1 hour and 10 minutes. This is at max ebb on a very strong day.

For a flood where the current is adverse, consult the chart on p. 51, and again adjusting by the chart on p. 48, we’ll multiply by 1.5 for a strong day. We get 1.5 knots of adverse current and then a little less than 2 setting us to the left, opposite of where we want. (The flood difference is smaller than the ebb difference because on average the ebb is stronger than the flood. A subject for another day.) Averaging the 2 with the 1.5, we can subtract 1.7 knots from our boat speed, bringing it to about 4.3 knots. The distance is the same 10.5 miles, so it will take a little less than 2.5 hours to get there. Again, this is against a max flood on a strong day.

The end result is we get a range of ETE from one hour, 10 minutes with a strong ebb, which yields a VMG of 6 knots; one hour, 45 minutes at slack for a VMG of 4 knots; and two hours and 25 minutes against a strong flood for a VMG of about 2.9 knots. For practice you can figure this with a boat speed of 7 knots, or you can try figuring ETE and VMG on your next trip to Drake’s Bay, which is 26 miles across the sea—to windward. The current is commonly less than a knot once you get well clear of Point Bonita, sometimes northerly in winter but usually southerly.

OK, I admit, that was a bit complicated. But you only have to figure this out once, based on the speed of your particular boat, and you’ll know at a glance how to estimate your ETE to the City or Drake’s Bay given the state of the current. All of this assumes constant wind and consistent boat handling, so your figures may vary. Not that you really care, because if you’re sailing, you’ve already arrived.

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