How much Push or Pull do you need ?

[Canada Bob]

Hi,

Hope you guys with far more knowledge that I have don't mind me asking what's probably a very basic question.

I'm trying to figure our how much "push or pull" you need to exert on a vessel to make it move.

To keep things simple and to present my curiousity in graphic terms lets say we have a canoe {simple/standard form hull} that weighs in at 1,000 lbs {yea, I know, that's one heavy canoe, but I'm just trying to round off numbers}.

How do you calculate {without getting into various hull shapes} what effort would be required to start a vessel moving, and to keep it moving.

What effect would a towing line of say 10lbs have on the boat ? or a force of say 10 lbs/sq inch pushing the boat have ? all in flat calm waters this is.

[Submarine Tom]

CB,

Surprisingly little!

-Tom

[Canada Bob]

I'd never be able to figure out the thrust needed by a propeller but I might be able to get my head around how they figured out how much paddle they needed in the water to move the Mississippi {style} paddle boats.

How did they know how many square inches they needed to push the boat ? and how much pressure per sq inch to move it along ?

[Submarine Tom]

Put a strain gauge on your paddle and start paddling. Or, put a fish scale in a tow line and start towing, you can then measure drag at different velocities.

-Tom

[Canada Bob]

There must be some sort of calculation though Tom ? must be something better than trial and error even 150 years ago, how did they know how much paddle to have in the water.

I guess like in rowing the greater the effort/lbs per sq inch combined with the speed of the stroke, the faster you go. Maybe they start out knowing what energy is available human strength or horse power and then calculate the optimum sq inches in the water to propel the boat. There has to be a basic calculation though to give some idea of how much energy, {lbs / sq inch} that a "paddle" needs to put into the water to move a 1,000 boat.

[Mr Efficiency]

Move at what speed, though ? In the absence of wind or tide, it takes only a tiny force to move ( however imperceptibly) a very large vessel. There is only inertia to overcome, not an initial resistance as you would find with a vehicle on a flat road.

[Canada Bob]

I'm not looking for high speeds, 5 knots or 10 knots would be fine, on calm waters with no current to overcome.

[Canada Bob]

I guess that as you move a boat through water you get a build up of pressure at the bow ? would that increase as the boat went faster ?

[Mr Efficiency]

The strain gauge idea should give a guide as to the resistance involved, which for your purposes will increase with speed. Tow it at a distance behind a boat with an accurate GPS, to see how it varies with speed.. Or put the canoe in the shallows of a running stream and you hold the line with the strain gauge. You will need to judge the current speed.

[Canada Bob]

Understood Mr E, but trial and error / observation is one thing, but there has to be an equation to show how much force / lbs per sq inch would be needed to move a boat ?

Re your comment above, if this theoretical 1,000 lb canoe was towed at a speed of 5 knots with a strain of {let's say} 50 lbs, does that mean to say that if we applied a 50 lbs push at the stern that we would get 5 knots out of it ?

Speaking of that 50 lbs would that be 50 lbs per sq inch ? or 5 lbs per sq inch over equally over 10 sq inches ?

[Mr Efficiency]

The 50 lbs would consist of the wave-making resistance and skin friction combined, two distinct retarding factors, Plus a tiny component of air resistance, it is not a question of psi. And yes, if the strain gauge or spring fish scale is showing 50 lbs, a 50 lbs force applied to the stern of the canoe should cancel that out.

[Canada Bob]

Thanks for the feedback, appreciated...

Leaving out wave and air resistance, just to get to grips with the basic principle and to keep it at a level that I can relate to

What I'm trying to figure out is, how many lbs per sq inch this canoe thing of mine would need applied to the stern to push it along ? at a couple of knots.

Would 50 lbs applied to 1 sq inch be the same {propelling force} as 5 lbs per sq inch spread over 10 sq inches ?

[Canada Bob]

OK, here's another way to look at it, lets say I was considering a water jet propulsion system for this canoe, if the water came out of 1 sq inch at 50 lbs / sq inch, would it have the same propulsion if it came out of a 10 inch square outlet at 5lbs per square inch ???

[Mr Efficiency]

You could test it with an electric outboard, they are rated in pounds of thrust. I think it would be a case of suck it and see, maybe googling 'canoes with electric outboard' will get you some figures to compare, obviously it is dependent on the shape and size and displacement, different boats will have different characteristics, including resistance.

[Canada Bob]

Thanks, thinking out aloud at this end, plus the feedback from you guys has brought me a mental step further in a little project that so far is just pen and paper stuff.

It's just the principle {not the fine arts of hull shapes and wave forms} at this stage, but it has helped me considering what you guys have had to say, it's valued and appreciated.