Power/Speed/Displacement Calculation

[NB Willawaw]

I am involved with a new boat design which has the following underwater/under waterline criteria:

Bow: Triangular Section 3.50m long x 2.08m beam x 0.66m draft
Mid: Rectangular Section 10.31m long x 2.08m beam x 0.66m draft
Stern: Triangular Section 2.90m long x 2.08m beam x 0.66m draft

I estimate this corresponds to a displacement of 18.5 metric tonnes approx in fresh water (boat will only be used in fresh water).

How do I calculate the power that will be needed to drive this hull at 6mph ignoring wind and tide ?

Can anybody help me with some formulas please ??

Mark

[gonzo]

First of all, the displacement doesn't change from fresh to salt water, or out of the water.
Calculations are for speed on the water, tides go up and down anyway.
With triangular sections, the wave making is going to absorb enourmous amount of power and make the boat difficult to handle. Is there a reason to not streamline the hull?

[NB Willawaw]

Not convinced about the first. Draft will vary slightly due to salinity therefore displacement must also vary slightly. Its irrelevant really as thats not my question and the boat will only be used on fresh water.

Tides do go up and down, but in confined spaces like estuaries, you get flow. We have a 4-6 Knot flow in the River Thames caused by spring tides.

The boat doesnt have triangular sections - it is streamlined.
I have calculated the shape as a combination of simple shapes i.e rectangle and triangles to simplify the maths.

Do you know the answer to the question ?

Originally Posted by gonzo

First of all, the displacement doesn't change from fresh to salt water, or out of the water.
Calculations are for speed on the water, tides go up and down anyway.
With triangular sections, the wave making is going to absorb enourmous amount of power and make the boat difficult to handle. Is there a reason to not streamline the hull?

[Guillermo]

Originally Posted by NB Willawaw

Not convinced about the first. Draft will vary slightly due to salinity therefore displacement must also vary slightly...

Well, it depends on what you understand as 'displacement'. Most usual assumption for displacement is the weight of the boat, so it's irrelevant where the boat is floating, fresh water, salt water or melted lead. But if you mean submerged volume, then yes, it varies.

Regarding your question, a quick estimative suggests an engine power not in excess of 50 HP, but to be more accurate Lwl and Cp are needed.

[Robjl]

In your first post you state;
".......needed to drive this hull at 6mph ignoring wind and tide ?"
In your second;
"Tides do go up and down, but in confined spaces like estuaries, you get flow. We have a 4-6 Knot flow in the River Thames caused by spring tides."
I'm not sure what speed you need to make?
Is it a flat 6mph or is it 6mph into a 6 knot flow?
There is a lot of difference!
I'm guessing about a 16m waterline, is this correct?

My estimate for power needed;
20Hp continuous to do 6 knots in flat water
120Hp cont to do 11 knots in flat water
My suggestion;
I agree with the 50hp but....
if the tide turns... drop anchor.

Displacement is the weight of water displaced by the boat, the density of the water makes no difference. It is a matter of using terminology correctly or as is the norm in this area. Just as we usually refer to speed in knots rather than mph.
Hope this is useful.

[terhohalme]

For a displacement boat you need approx. 4 kW/metric tonne. At 18 tonnes it makes 72 kW and you can achieve closely 10 knots while your LWL is 16.7 metres. This power includes some reserve against wind and waves.

For a canal boat halv the power is enough, say 36 kW.

For future purposes have a basic book of boats like Gerr's Nature of Boats.

[NB Willawaw]

For the purposes of clarification:

1. Ignore wind and tide. I need the boat to do 6mph on a non-tidal inland waterway - if you need Knots thats 5.2 Knots.

2. The length/beam dimensions given are at the waterline so LWL is 16.71m

I would like to understand how the required power is calculated so can somebody actually explain the formula to me please.

Originally Posted by Robjl

In your first post you state;
".......needed to drive this hull at 6mph ignoring wind and tide ?"
In your second;
"Tides do go up and down, but in confined spaces like estuaries, you get flow. We have a 4-6 Knot flow in the River Thames caused by spring tides."
I'm not sure what speed you need to make?
Is it a flat 6mph or is it 6mph into a 6 knot flow?
There is a lot of difference!
I'm guessing about a 16m waterline, is this correct?

My estimate for power needed;
20Hp continuous to do 6 knots in flat water
120Hp cont to do 11 knots in flat water
My suggestion;
I agree with the 50hp but....
if the tide turns... drop anchor.

Displacement is the weight of water displaced by the boat, the density of the water makes no difference. It is a matter of using terminology correctly or as is the norm in this area. Just as we usually refer to speed in knots rather than mph.
Hope this is useful.

[terhohalme]

OK, just for xmas...

While the speed is so low, you can ignore wave making. Friction resistance can be estimated by Froude's friction formula:

RF=161*(0.00871+ 0.053/(8.8+LWL))*WSA*V^1.825

where
RF is friction resistance in Newtons
LWL is waterline length in metres
WSA is wetted surface area in square meters
V is speed in meters/second

Effective power (Pe) in Watts is

Pe = V * RF

Installed power is twice that much (propeller efficiency!).

Study BOOKS!!! This goup is not a free design office!

[NB Willawaw]

Thanks for your help.

Firstly, I'm not a naval architect.
Secondly, this is a hobby not a paid job
Thirdly, if I could get easy access to the books that provide this information I wouldn't need to ask the question.

Lastly, if you can't provide assistance to a fellow boater with good cheer, you shouldn't reply.

Thanks for your help.

Have a good Christmas.

Originally Posted by terhohalme

OK, just for xmas...

While the speed is so low, you can ignore wave making. Friction resistance can be estimated by Froude's friction formula:

RF=161*(0.00871+ 0.053/(8.8+LWL))*WSA*V^1.825

where
RF is friction resistance in Newtons
LWL is waterline length in metres
WSA is wetted surface area in square meters
V is speed in meters/second

Effective power (Pe) in Watts is

Pe = V * RF

Installed power is twice that much (propeller efficiency!).

Study BOOKS!!! This goup is not a free design office!

[Robjl]

This spreadsheet is on another thread...can't remember where...but it seems to be accurate.
Merry Xmas.

[Raggi_Thor]

NB, if you read you own postings carefully, you may see that the tone in your writing is a little aggressively demanding, even if that was not your intention. You did get a lot of questions answered, I think :-)

[gonzo]

You can't simplify the math by ignoring the shape of the boat. It will make a significant difference. You are asking a question without giving the proper information and demanding the "right" answer. Post the lines of the hull for a better understanding of your needs.

[wdnboatbuilder]

I would just like to say thank you to all the guru's on this and all forums
thank you