Best speed and course on a river

[mross]

Yes, excellent work.  I did suspect that there would be an optimum speed not much higher than that needed to make headway against the river current.  Can you upload your spreadsheet?

[Alan de Enfield]

10 minutes ago, Keeping Up said:

Does that clarify it

There is a flaw in your formula-Consumption is not a factor of ’speed’ but a factor of RPM. You could have a boat doing 5mph at 1000 rpm, and a boat doing 3000 rpm doing 5mph –using  your formula they would both have  the same fuel consumption.

Diesel engines use approximately 1 gal  (4.4 litres) per hour per 20 HP developed :

Fuel consumed is proportional to the power developed

Power developed is approximately proportional to the cube of the engine revs

Hence fuel consumed is proportional to the cube of the engine revs, NOT boat speed.

 Engine (example) :

 WOT = 2600 RPM

WOT = 143 HP

Fuel Consumption = 0.195 Litres/HP/Hour (equates to about 4.2 litres per 20hp)

 

Example 1 : At a 8 knot cruise at 2000 rpm that is 77% of WOT

So : 77% x 77% x 77% = 45.7%

ie 45.7% of 143hp = 65hp

 65hp @ 0.195 lt/hp/hour = 12.7 litres per hour

 

 Example 2 : At a 5 knot cruise at 1500 rpm that is 57% of WOT

So : 58% x 58% x 58% = 19.5%

ie 19.5% of 143 hp = 27.9hp

 27hp @ 0.195 lt/hp/hr = 5.4 litres per hour.

[Keeping Up]

35 minutes ago, Alan de Enfield said:

There is a flaw in your formula-Consumption is not a factor of ’speed’ but a factor of RPM. You could have a boat doing 5mph at 1000 rpm, and a boat doing 3000 rpm doing 5mph –using  your formula they would both have  the same fuel consumption.

Diesel engines use approximately 1 gal  (4.4 litres) per hour per 20 HP developed :

Fuel consumed is proportional to the power developed

Power developed is approximately proportional to the cube of the engine revs

Hence fuel consumed is proportional to the cube of the engine revs, NOT boat speed.

 etc ...

Yes but I'm not comparing one boat with another, I'm comparing figures for the same boat throughout.  To a very close approximation, on any displacement hull such as that of a narrowboat, at normal speeds the speed through the water IS directly proportional to the engine revs (there is a slip factor for the prop but it is pretty much a constant factor, assuming no cavitation or ventilation).

 

mross yes the spreadsheet is attached

River speeds.xlsx

[howardang]

2 hours ago, mross said:

Some of us find a good technical discussion fun.  You could always go and look at Facebook or Twitter if you want  to.  

Never subscribed to either, and I don't see me signing up any time soon! I am sure the so-called "technical discussion" will not be really useful to the OP, however, but no doubt most will disagree, I suspect, but if it keeps you amused, enjoy!

 

Howard

[Alan de Enfield]

5 minutes ago, Keeping Up said:

Yes but I'm not comparing one boat with another, I'm comparing figures for the same boat throughout.

Fuel consumption is not linear with an increase in RPM.

For each of my Engines

At 1000 rpm.               1.59 x 1.0³ = 1.6 litres per hour

At 1100 rpm                1.59 x 1.1³ = 2.1 litres per hour

At 1200 rpm                1.59 x 1.2³ = 2.7 litres per hour

At 1300 rpm                1.59 x 1.3³ = 3.5 litres per hour

At 1400 rpm                1.59 x 1.4³ = 4.4 litres per hour

At 1500 rpm                1.59 x 1.5³ = 5.4 litres per hour

At 1600 rpm                1.59 x 1.6³ = 6.5 litres per hour

At 1700 rpm                1.59 x 1.7³ = 7.8 litres per hour

At 1800 rpm                1.59 x 1.8³ = 9.3 litres per hour

At 1900 rpm                1.59 x 1.9³ = 10.9 litres per hour

At 2000 rpm                1.59 x 2.0³ = 12.7 litres per hour

At 2100 rpm                1.59 x 2.1³ = 14.7 litres per hour

At 2200 rpm                1.59 x 2.2³ = 16.9 litres per hour

At 2300 rpm                1.59 x 2.3³ = 19.3 litres per hour

At 2400 rpm                1.59 x 2.4³ = 22.0 litres per hour

At 2500 rpm                1.59 x 2.5³ =  24.8 litres per hour

At 2600 rpm                1.59 x 2.6³ =  28.0 litres per hour

 

 

All figures are based on my displacement boat.

 

 

[Keeping Up]

2 minutes ago, Alan de Enfield said:

Fuel consumption is not linear with an increase in RPM.

 

Indeed, I never said it was.

I said the speed through the water is linear with an increase in RPM.

Fuel consumption is proportional to the power developed (as you said)

Power required is proportional to the cube of the speed, ie proportional to the cube of the RPM.

Thus fuel consumption (per hour) is proportional to the cube of the speed through the water and to the cube of the RPM.

 

 

[mross]

I put Alan's figures into a spreadsheet and added a trendline.  The best fit was a second-order polynomial.  This shows that the fuel consumption is not rising with the cube of RPM but partly linearly and partly with the square of the RPM (the first term in the formula above)

[Keeping Up]

1 hour ago, mross said:

I put Alan's figures into a spreadsheet and added a trendline.  The best fit was a second-order polynomial.  This shows that the fuel consumption is not rising with the cube of RPM but partly linearly and partly with the square of the RPM (the first term in the formula above)

I'm puzzled by this. Surely Alan's figures are by definition 1.59 X (RPM cubed) where RPM is specified in thousands of revolutions pee minute

[Sea Dog]

10 hours ago, Keeping Up said:

 

I said the speed through the water is linear with an increase in RPM.

Surely not? Not my field, but I'd have expected increasing prop slip as speed through the water increases.

[mross]

28 minutes ago, Keeping Up said:

I'm puzzled by this. Surely Alan's figures are by definition 1.59 X (RPM cubed) where RPM is specified in thousands of revolutions pee minute

I did not look at the middle column!  I thought he had taken figures from the engine manufacturer.

[Tacet]

12 hours ago, Alan de Enfield said:

 

Power developed is approximately proportional to the cube of the engine revs

 

Surely this can only be case over a very small part of the power curve.

Like Keeping up, I had a few trial and error attempts on a spreadsheet and rather than his 1.5 factor, came close to something to do with the root of 2, which felt plausible.  But it wasn't spot on, so the theory fell apart. 

[Alan de Enfield]

20 hours ago, Keeping Up said:

Thus fuel consumption (per hour) is proportional to the cube of the speed through the water and to the cube of the RPM.

 

I understand where you are ’coming from’ but cannot agree 100% with your theory.

A slightly modified steel skip is not very aqua-dynamic and as the speed increases the resistance will increase such that a 1mph increase in speed, at say, 5mph to 6mph, will require much more power (RPM) than an increase of 1mph at lower speeds as it is pushing an ever increasing bow-wave.

As the speed starts to approach the maximum hull speed (probably around 7 to 8mph) however much you increase the RPM/Power you will see a very, small increase in speed, until you get no increase in speed irrespective of how much power you apply.

My boats max hull speed is a tad over 7knts (8mph)  and is achieved at 1800 rpm.

At 1800 rpm fuel consumption is 7.3 litres per hour.

If I increase the rpm to 2600 rpm, consumption is now 22 litres per hour and the speed remains at 7 knots.

RPM may well be closely linked to a displacement boats speed at low speeds, but is certainly not linear throughout its speed range.

[Keeping Up]

I agree with you 100% but I don't think it changes anything. I was deliberately concentrating on speeds less than the maximum hull speed of say 7 knots. I certainly agree that once you reach that speed it doesn't matter how much extra power you apply the boat won't go any faster.

At lower speeds I have always been told the speed is directly proportional to RPM as the slip factor is a constant percentage. I too find it counter intuitive but I don't know of any source material that states otherwise.

As you say a displacement hull is not very hydrodynamic. That is why an increase from say 5 to 6 mph needs a lot more power; that is where the cubed factor comes from.

[Tacet]

The answer is, I am told, 1.5 times the flow.  Only on the basis of Keeping Up's cube rule, of course.

Maths is attached - with apologies for poor quality images.  It is reasonably intelligible until one needs to differentiate

DS³/s-a

to identify when a small change in S (speed through water) results in no change to F (fuel) consumed.  i,e dF/dS=0.  I am reliably informed that the resulting graph is truly horrible but can be straightened out by making various adjustments such as D (distance) not being zero or negative etc.

It all ends up with S=3a/2 when F is at the minimum.  With "a" being the flow of the stream. Which is so strikingly simple that perhaps some easier method of calculation ought to be available.

Either way, I won't be taking questions on the details for  errr.....   my own reasons.

 

 

Edited December 14, 2017 by Tacet

[mross]

11 hours ago, Tacet said:

The answer is, I am told, 1.5 times the flow.  Only on the basis of Keeping Up's cube rule, of course.

Maths is attached - with apologies for poor quality images.  It is reasonably intelligible until one needs to differentiate

DS³/s-a

to identify when a small change in S (speed through water) results in no change to F (fuel) consumed.  i,e dF/dS=0.  I am reliably informed that the resulting graph is truly horrible but can be straightened out by making various adjustments such as D (distance) not being zero or negative etc.

It all ends up with S=3a/2 when F is at the minimum.  With "a" being the flow of the stream. Which is so strikingly simple that perhaps some easier method of calculation ought to be available.

Either way, I won't be taking questions on the details for  errr.....   my own reasons.

 

 

PLEASE try again and in focus!  It looks interesting but gives be a headache trying to read it 

[Sea Dog]

So, set revs that your engine and cooling system is happy at and also gives reasonable progress, then enjoy the cruise. After you arrive, either: a. measure exactly how much fuel you've used, put your figures into a spreadsheet, compare it with a load of graphs and solve a number of equations, or b. go to the pub. I recommend course of action a because there's money to be saved - potentially a tiny amount on fuel, but a much larger amount by doing maths whilst everyone else is at the pub.

[Tacet]

2 hours ago, mross said:

PLEASE try again and in focus!  It looks interesting but gives be a headache trying to read it 

Sorry, I can't - as I don't have the hard copy.  I will ask for a better image though. 

[Alan de Enfield]

1 hour ago, Sea Dog said:

So, set revs that your engine and cooling system is happy at and also gives reasonable progress, then enjoy the cruise. After you arrive, either: a. measure exactly how much fuel you've used, put your figures into a spreadsheet, compare it with a load of graphs and solve a number of equations, or b. go to the pub. I recommend course of action a because there's money to be saved - potentially a tiny amount on fuel, but a much larger amount by doing maths whilst everyone else is at the pub.

It is more the stimulation and something 'brain testing' rather than the ultimate answer.

You don't even need to believe in the stance you are taking, its called debate, and, you can even change sides during it.

Long dark nights - anyone can sit in the pub and vegetate. 

 

[rusty69]

1 minute ago, Alan de Enfield said:

 

You don't even need to believe in the stance you are taking, its called debate, and, you can even change sides during it.

Long dark nights - anyone can sit in the pub and vegetate. 

 

Does it have anything to do with  Mass? Not recommended in the pub. 

[Keeping Up]

19 minutes ago, Alan de Enfield said:

It is more the stimulation and something 'brain testing' rather than the ultimate answer.

You don't even need to believe in the stance you are taking, its called debate, and, you can even change sides during it.

Long dark nights - anyone can sit in the pub and vegetate. 

 

Exactly

If I'd needed the answer urgently I'd have posted during the summer. In fact I've enjoyed thinking about the problem, and posting it here helped me to arrive at my own solution - which may or may not be correct but it looks easible and suggests that for most of the time, when the river was flowing very slowly this summer (probbaly 1mph), I was travelling marginally too fast at 3mph and could have saved a little more money by going slower, which would have left me with more coins to spend in the pub cogitating and vegetating.

Nobody has yet answered my other questionusefully though. When travelling downstream, what are the conditions that make it quicker to travel rund the outside of the bend where the current is greater, as opposed to travelling round the inside of the bend which is a shorter distance.

[Naughty Cal]

30 minutes ago, Keeping Up said:

Exactly

If I'd needed the answer urgently I'd have posted during the summer. In fact I've enjoyed thinking about the problem, and posting it here helped me to arrive at my own solution - which may or may not be correct but it looks easible and suggests that for most of the time, when the river was flowing very slowly this summer (probbaly 1mph), I was travelling marginally too fast at 3mph and could have saved a little more money by going slower, which would have left me with more coins to spend in the pub cogitating and vegetating.

Nobody has yet answered my other questionusefully though. When travelling downstream, what are the conditions that make it quicker to travel rund the outside of the bend where the current is greater, as opposed to travelling round the inside of the bend which is a shorter distance.

Is that because you wouldn't have got to the pub as quick?

[Sea Dog]

1 hour ago, Keeping Up said:

Nobody has yet answered my other questionusefully though. When travelling downstream, what are the conditions that make it quicker to travel rund the outside of the bend where the current is greater, as opposed to travelling round the inside of the bend which is a shorter distance.

The conditions are that if you choose the shorter distance by cutting the inside of a river bend, sooner or later you'll go aground - at best, this will make your journey take longer.  This is because the flow tends to scour out the outside of a bend and the inside can be very shallow. It doesn't make any difference whether you do it downstream or upstream.

Your position in the river should allow you to pass any oncoming vessels port side to port side. In other words you should be able to keep, or be able to manoeuvre yourself in plenty of time, far enough to the right of the safe channel to avoid a collision. This may sometime mean giving up the most favorable or quickest course to ensure safe navigation. 

2 hours ago, Alan de Enfield said:

It is more the stimulation and something 'brain testing' rather than the ultimate answer.

You don't even need to believe in the stance you are taking, its called debate, and, you can even change sides during it.

Long dark nights - anyone can sit in the pub and vegetate. 

 

I did recommend course of action a: course of action b was the pub!

[Tacet]

1 hour ago, Keeping Up said:

Exactly

If I'd needed the answer urgently I'd have posted during the summer. In fact I've enjoyed thinking about the problem, and posting it here helped me to arrive at my own solution - which may or may not be correct but it looks easible and suggests that for most of the time, when the river was flowing very slowly this summer (probbaly 1mph), I was travelling marginally too fast at 3mph and could have saved a little more money by going slower, which would have left me with more coins to spend in the pub cogitating and vegetating.

Nobody has yet answered my other questionusefully though. When travelling downstream, what are the conditions that make it quicker to travel rund the outside of the bend where the current is greater, as opposed to travelling round the inside of the bend which is a shorter distance.

In the summer, you should have been going much slower to minimize fuel consumed (on the cube rule basis) that is 1.5mph through the water and only 0.5mph over the ground. 

I don't see you are going to get a mathematical answer to the without lots of assumptions.  Obviously one need to know the distance saved and reduction if river speed by taking thee short cut. If you were just drifting, my instincts are that you would do best by not losing any momentum (i.e. not taking to water flowing more slowly than the main channel) as it may well take a while to regain the speed.  This may continue to apply at low, powered speeds through the water but could diminish as the differential increase.

[Alan de Enfield]

17 minutes ago, Tacet said:

In the summer, you should have been going much slower to minimize fuel consumed (on the cube rule basis) that is 1.5mph through the water and only 0.5mph over the ground.

Surely the most economic routes and speeds must encompass more than just  fuel usage.

Time must play an important part in the calculation :

If (for example) you are only achieving 0.5mph SoG and you have 20 miles to go to your destination, then it will take you 40 hours (2 days), this could then involve having to take on extra crew for the 'night shift', or paying for moorings for at least one night (maybe 2 nights), it could also mean you losing a days wages if you arrive back a day later than expected.

If (for example) you are achieving 4 mph SoG and you have 20 miles to go to your destination, then it will take you 5 hours. It may use a considerable amount more fuel, but you save additional crew costs, mooring costs and loss of wages.

So, which is the most economical speed ?

[Keeping Up]

2 hours ago, Sea Dog said:

The conditions are that if you choose the shorter distance by cutting the inside of a river bend, sooner or later you'll go aground - at best, this will make your journey take longer.  This is because the flow tends to scour out the outside of a bend and the inside can be very shallow. It doesn't make any difference whether you do it downstream or upstream.

Your position in the river should allow you to pass any oncoming vessels port side to port side. In other words you should be able to keep, or be able to manoeuvre yourself in plenty of time, far enough to the right of the safe channel to avoid a collision. This may sometime mean giving up the most favorable or quickest course to ensure safe navigation. 

I did recommend course of action a: course of action b was the pub!

I agree entirely, but consider the case of a river that is about 200ft wide and is very deep. Cutting the corner to within say 20ft of the bank will reduce the distance significantly without any risk of grounding; and in these circumstances it is perfectly possible that your view will be sufficiently good for a great distance to determine that it is safe to occupy the 'wrong' side of the river. If necessary you could always show a Blue Board but of course in this country most people would not know what it meant.

Yes the pub option is definitely a good one

 

2 hours ago, Alan de Enfield said:

Surely the most economic routes and speeds must encompass more than just  fuel usage.

Time must play an important part in the calculation :

If (for example) you are only achieving 0.5mph SoG and you have 20 miles to go to your destination, then it will take you 40 hours (2 days), this could then involve having to take on extra crew for the 'night shift', or paying for moorings for at least one night (maybe 2 nights), it could also mean you losing a days wages if you arrive back a day later than expected.

If (for example) you are achieving 4 mph SoG and you have 20 miles to go to your destination, then it will take you 5 hours. It may use a considerable amount more fuel, but you save additional crew costs, mooring costs and loss of wages.

So, which is the most economical speed ?

These calculations take time and fuel cost into account, but as you say not the costs of any crew's wages, nor the fact that if you spend less time in the pub you will probably drink less.