How much HP to push a barge??

[JamesG]

How do you calculate the horse power needed for a boat to push a barge?

The barge will be totally square, all right angles, so basically a box.

I need a formula or graph that lets me input the weight of the barge and the dimentions as variables.

Thanks a lot!
-james

PS- Is torque a major factor or is it mainly HP?

[Landlubber]

2 hp will move it, but how fast do you want to move it?

....it would help greatly if you would consider telling us just how big it is and what it displaces.....

[JamesG]

I don't know how big it will be, but it will probably be somewhere in this range:

It might be as small as 10'X30' with a weight of 13,500lbs and a draft of about 14".
Or as large as 50'X100' with a weight of 200,000lbs and a draft of about 14".

These are all really rough numbers.

-james

[Landlubber]

These are all really rough numbers.


...yep....

surely if you are for real you have some idea of just what you are designing mate?

[Submarine Tom]

James,

A box, a square, or a cube?

10 X 30 and 50 X 100 are not squares.

I think you may be in real trouble here?

But, if you insist on a "formula", here's one:

1 HP per 10 square feet.

Good luck!

-Tom

P.S. It's torque.

[PAR]

Everyone's a naval architect. Some concept of what you're doing would be a reasonable idea JamesG, don't you think? Are you familiar with Archimedes?

How hard could it be, right? Lets say you've "designed" up a nice, basic 50'x100' barge. You've got it lightly loaded and she's sunk down and now draws about 12" of water. This little barge is displacing about 311,000 pounds or about 141 ton. It's PPI is about 26,000 pounds or about 12 ton, for each inch you further sink her down. Well my boy, this is a fair bit of engineering to handle all this weight, on a lightly loaded barge, so I'll ask again, are you up to the engineering necessary to keep this puppy from folding in half, when you load it up?

[JamesG]

All i'm asking for is a formula for the amount of horsepower needed to push a barge at a certain speed. If there is no formula then maybe there's a graph because this is probably a hydrodynamic/fluid dynamic equation.

The inputs would probably be:
Weight of the barge
Displacement of the barge OR dimensions of the barge
Drag coefficient (it is a square, blunt, hull in this case (imagine a shoebox))
And Speed

The output would be :
Horse Power of an engine.

My apologies for not being clear enough last time. Let me know if you need more info.

[JamesG]

Originally Posted by PAR

Everyone's a naval architect. Some concept of what you're doing would be a reasonable idea JamesG, don't you think? Are you familiar with Archimedes?

How hard could it be, right? Lets say you've "designed" up a nice, basic 50'x100' barge. You've got it lightly loaded and she's sunk down and now draws about 12" of water. This little barge is displacing about 311,000 pounds or about 141 ton. It's PPI is about 26,000 pounds or about 12 ton, for each inch you further sink her down. Well my boy, this is a fair bit of engineering to handle all this weight, on a lightly loaded barge, so I'll ask again, are you up to the engineering necessary to keep this puppy from folding in half, when you load it up?

Who says I wanna keep it from folding in half?

[haru]

We have many skeptics here so until you come clear help comes slow.

This is physics 101 you could just have asked in a physics forum.
Anyway, here's the formula simplified:
P = 0.00067 * v^3 * cos(alpha) * rv * A
P: power in HP
v: relative velocity
rv: river flow
alpha: angle to river flow; 0 = same direction
A: area

You need to know torque in terms of pressure from the pushing boat and the transported loads. Pushing gently will do.

[gonzo]

There is also wind resistance. In a lightly loaded barge it is often more than the wave and skin friction resistance. If you fold it in half, it depends if it is lengthwise or sidewise. You need to ask the right questions to get useful answers. That means studying the basics first.

[jonr]

HP will be the major factor. One can get whatever torque desired with gearing.

[Submarine Tom]

Originally Posted by JamesG

All i'm asking for is a formula for the amount of horsepower needed to push a barge at a certain speed. If there is no formula then maybe there's a graph because this is probably a hydrodynamic/fluid dynamic equation.

The inputs would probably be:
Weight of the barge
Displacement of the barge OR dimensions of the barge
Drag coefficient (it is a square, blunt, hull in this case (imagine a shoebox))
And Speed

The output would be :
Horse Power of an engine.

My apologies for not being clear enough last time. Let me know if you need more info.

James,

In post #5 above I gave you a formula.

Why are you asking for one again?

-Tom

[JamesG]

Originally Posted by Submarine Tom

James,

In post #5 above I gave you a formula.

Why are you asking for one again?

-Tom

Tom,

Thank you for the formula. I am actually am using it along with that other one.


-james

[JamesG]

Originally Posted by haru

We have many skeptics here so until you come clear help comes slow.

This is physics 101 you could just have asked in a physics forum.
Anyway, here's the formula simplified:
P = 0.00067 * v^3 * cos(alpha) * rv * A
P: power in HP
v: relative velocity
rv: river flow
alpha: angle to river flow; 0 = same direction
A: area

You need to know torque in terms of pressure from the pushing boat and the transported loads. Pushing gently will do.

Thanks Huru!

Just curious. Are you sure rv stands for "river flow" or does it stand for "relative velocity"? I just wanna make sure i use it correctly.

And is "relative velocity" the barge's speed through the water, not the ground speed? Kind of like with airplanes, "airspeed" vs "ground speed"?

thanks,
james

[haru]

I made rv up. :lol: since I converted it to your particular case.
Cos(alpha) * rv = velocity vector of the river
v is relative to the water so it's like airspeed

As gonzo said, no wind is accounted. Also side way movement isn't really accounted either so it will sidetrack, but it should get to the other side.
Having more HP certainly won't hurt at least it won't strain the engine.