I did some additional calculating, and I think I have come up with the following ESTIMATE: There is a 5% chance that my crossing could be either 66.5 days or 21 days, and a 95% chance that it will be 36 to 44 days. Here is the logic behind this estimate – please let me know if I have made any errors:

I know for sure that WiTHiN will be able to maintain an average of 7 to 8 kph for at least 12 hours per day based on my known power output capabilities are over a 24 hour period. I also know that I can expect an average surface current speed of .8 kph for 24 hours of every day. Using the simple calculations below, this was how I had estimated my record breaking 40 day finishing time:

Ocean Surface Current = .8 kph x 24 hours/day

= 19.2 km/day x

**40 days**= 768 km

12 hours of pedalling per day @ 100 watts, 8 kph

= 96 km/day x

**40 days**= 3840 km

Total distance covered = 4608 km

But what I don’t know for sure, is how wind and weather will effect my progress predictions. So, I decided to run an analysis using the 35 solo Atlantic tradewinds route (Canary Islands to West Indies) ocean rowing expeditions since 1969 from the Ocean Rowing Societies web site.

First of all, lets see if we figure out what the average speed of an ocean rowing boat is, and how that compares to actual rowing times.

The fastest solo ocean crossing (tradewinds route) in an ocean rowing boat is 42 days, the slowest is 133 days and the average of every crossing since 1969 is 82.7 days

From research of previous ocean rowers including reading archives of trip logs, the many books I have read and my communications with many of them, it seems that the average speed observed by ocean rowers while underway is about 2 knots. That converts to 3.7 kph. If we subtract the ocean surface current of .8 kph, we get an actual unassisted average speed of 2.9 kph (this compares to my unassisted average speed of 8 kph).

Let’s see if my observation of the average rowing speed works out to the total distance using an average of 12 hours of rowing per day and the ocean surface current of .8 kph:

**ROWING:**

La Gomera, Spain to Antigua = 4500 km

Ocean Surface Current = .8 kph x 24 hours/day

= 19.2 km/day x 82.7 days overall average =

**1587**km

12 hours of ROWING per day @ 2.9 kph

= 34.8 km/day x 82.7 days overall average = 2877

Total distance covered = 4464 km

So, it looks like my estimate of 2.9 kph average rowing speed without current works out to predicting the overall average time to cross the Atlantic by row boat. Therefore, my estimated crossing time of 40 days should be pretty accurate.

**Error bars**

Now lets calculate the standard deviation and error bars of all solo ocean rows and apply that to my 40 days to see what the maximum and minimum crossing time could be:

**35**

Total crossing time =

**2896 days**

Average crossing time =

**82.7 days**

Standard Deviation =

**24.11**

Error: stdev/(sqrt(count)) =

**+ – 4.076 days**

Using this standard deviation, I would expect that my crossing could take from 36 days to 44 days (approximately). Actually, that is not right… Since my average predicted crossing time is approximately 50% of the average rowing time, I think that I would need to take 50% of the rowing error which could be + – 2 days, not 4. But I’ll use 4 to be conservative.

Anyhow.. I’m a bit rusty on my stats (it’s been a while). What would the confidence level be for a 36 to 44 day crossing? 95% ??

Another way of predicting how the random chaotic nature of the weather could effect my crossing would be to simply take the maximum rowed crossing of 133 days, and divide that by my average speed compared to rowing averages (66.5 days maximum), and the fastest rowed crossing of 42 days (21 days minimum).

To summarize, we could say that there is a 5% chance that my crossing could be either 66.5 days or 21 days, and a 95% chance that it will be 36 to 44 days.

## Neville on February 26, 2008

Either way… it beats swimming! Keep it up Greg.

Neville

Chicago

## Leslie on February 26, 2008

Hello Greg,

The 95% confidence interval is for a 2-sigma standard deviation.

The normal standard deviation is a 67% confidence interval… that would be the rowing estimate value of 83±4 days.

Leaving your error at ±4 days, you effectively doubled this error as you halved the number of rowing/pedalling days estimated, so yes you've now provided a ~95% prediction. The only problem I can see with your original data set is what are the chances of a failed crossing? I.e. you forgot to include the unsucessful attempts (infinte days take to do the crossing). 😉

Like all scientists, I was quite good at statistics so these types of predictions don't mean much to me… but 10 points for including the [estimated] errors!

I haven't had time to make that phone call to the port authorities by the way…

Les.

## Leslie on February 26, 2008

The 95% confidence interval is for a 2-sigma standard deviation.

The normal standard deviation is a 67% confidence interval… that would be the rowing estimate value of 83±4 days.

Leaving your error at ±4 days, you effectively doubled this error as you halved the number of rowing/pedalling days estimated, so yes you've now provided a ~95% prediction. The only problem I can see with your original data set is what are the chances of a failed crossing? I.e. you forgot to include the unsucessful attempts (infinte days taken to do the crossing). 😉

Like all scientists, I was quite good at statistics so these types of predictions don't mean much to me… but 10 points for including the [estimated] errors!

I haven't had time to make that phone call to the port authorities by the way…

Les.

## Adventures of Greg on February 27, 2008

Les:

yes – you are right about including the unsuccessful attempts – didn't think of that. I think that may fall into a separate study – first, if a successful attempt, then what is the predition, and second, what are the chances of a failed attempt – first time out, second time out, etc.

I need to consider a failed attempt – these are common. I would say without looking at the sata – probably a 25% chance of a failed attempt. And since I am going with a support boat, any failed attempt would probably mean postponing until the year following unless I were able to quick-find another support boat (unlikely).

I'll do a follow up and point out my standard deviation error as well as consider the chances of a failed attempt

gk

## Anonymous on February 27, 2008

Go for it Greg! If there were a way to monitor nature forces like wind and current so you can keep them at your back, should easily have a new record. At the least you don't want to buck them as that means wasted time and energy. River rats do this all the time by studying eddies and piggybacking on them. Sometimes even able to make unusual headway UPSTREAM using these techniques. Hope this helps.

## Anonymous on February 27, 2008

Hi Greg,

I can not check your confidence level like leslie, but there are a few other things I see:

1. You are rowing in a certain period.

Then you could calculate the expected bad wheather and obviously you need some statistical info from the period you are rowing.

Let:

P(i|n) be the chance of zero days of weather against you, given that the journey will be $n$ days.

If you could get this info, you could calculate all kinds of stuff. Like "How many days will I expect to have bad weather?"

This should be something like

the sum of j x P(j|n) for j goes from 1 to n.

Anyway, this is probably too much. maybe you could just find the number of days you expect bad weather in the statistics…. (forget the other stuff).

2. I was wondering how much effect a series of "bad" days will have on the subsequent days. Think of:

1. In bad weather you'll probably not sleep as well or as long.

2. In bad weather, you'll spend more energy for less speed.

3. What will this take off your energy output of the days following the bad days?

You might be able to prepare yourself for bad weather by (and I am fantasizing this, cause I have no idea how it will be at sea):

1. Conciously spend less effort with bad weather, as it will have a lot less effect (does this make sense?)

2. carry some kind of sleep assisting medicine, that helps, but doesn't knock you out.

3. carry something to add to your energy levels, especially following "bad" days (but dont use it all the time). This might not be medically sound advice …..

Whatever you do, the event is probably an enourmously memorable one, however it goes.

And also remember, no statistic, no matter how cleverly made, is going to replace the real thing:

You are going to add to statistics.

You are going to make history, mate!!

Guus Bonnema, Leiden, The Netherlands.

## David Tangye on March 29, 2008

That looks like a good initial approach. The factor that need much better refinement is your estimate of 8kph. You need to know what is your average wind profile in % by direction and speed, and average your speed by running 12 hour tests in each representative condition. Eg what can you sustain over 12 hours in 10, 15, 20, & 30knots upwind?. (Say 4kph, 3kph, 2kph & 0kph respectively). Then the same across and downwind.

The only way to know this is to do a heap of coastal/offshore trials. I am guessing that your average will be much closer to 5kph if that.

Then get the figures for the expected average wind direction and speed for each section of the voyage, eg coastal departure, mid sections broken by region eg current/stream, coastal arrival region.