Calling any mathematician's (kind of boaty)

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TheBoatman

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Calling any mathematician\'s (kind of boaty)

I would like to know how much force is applied to a 1 metre square flat face if it was subjected to a 2 knot tide.

The reason for asking is I want to know how much force is being applied to a pontoon float that is at right angles to the tide?

Peter.
 
Re: Calling any mathematician\'s (kind of boaty)

I think this is engineering rather than pure maths. I am neither mathematician nor engineer but I am an empiricist and can visualise an experiment to measure the force in a rough and ready way.
 
Re: Calling any mathematician\'s (kind of boaty)

that's the eqivalent of a ton of water being pushed against it per second , but obviously theres fluid mechanics involved so erm erm erm have'nt a clue
 
Re: Calling any mathematician\'s (kind of boaty)

[ QUOTE ]
Could you assume one ton of water is being brought to a stop in one second, from two knots? That would be 1000N.

[/ QUOTE ]
i suspect not, given that it doesn't actually stop but flows around, that's the dynamics bit. I'm sure someone who actually knows something about this will be along soon.
 
Re: Calling any mathematician\'s (kind of boaty)

You can't assume that at all.

Unfortunately I didn't pay enough attention in fluids lectures to know what you do have to do, but I know you don't do that!
 
Re: Calling any mathematician\'s (kind of boaty)

Won't there be some pressure on the other side to take into account as well. Also depends how thick the water is
 
Re: Calling any mathematician\'s (kind of boaty)

Just make sure the chain or mooring rope is strong enough!
 
Re: Calling any mathematician\'s (kind of boaty)

It will be a half of the density of the water times the square of the velocity times the area. Simple arithmetic. The trick is to get all the right units /forums/images/graemlins/confused.gif
 
Re: Calling any mathematician\'s (kind of boaty)

[ QUOTE ]
[ QUOTE ]
I'm not sure .. its blinkin complex
see http://en.wikipedia.org/wiki/Pressure

[/ QUOTE ]

This is a bit more relevant
http://en.wikipedia.org/wiki/Drag_coefficient

Discounting viscosity and assuming a Cd of 1, that would make it 500N.

[/ QUOTE ]

Don't think that helps cos the drag coefficient can be greater than 1, and a flat surface a right angles to the flow will have a maximum drag coefficient
 
Re: Calling any mathematician\'s (kind of boaty)

It is easy. First you have to define pontoon. What pontoon area is in water (Length X draught). Then you have to define Force on pontoon from Tide. It is continuous force on every part of pontoon area. This continuous force then have to be approximate in Force which will be positioned in the middle of pontoon underwater area. After that you have made same calculation for upper part of the pontoon regarding wind ( 200 km/h).
Those 2 Forces will make Moment in the connection with ground.
When you have Moment in that point (Force X Length of pontoon/2 ), it is easy to dimension bolts for connection with shore. You can find this calculation in " Strength of materials" lectures ( Wikipedia !?)
I know that you can not make calculation from this shirt explanation but can help you to find way.
 
Re: Calling any mathematician\'s (kind of boaty)

you mahave part of the answer .. after calling in outside help from another forum where the following helpful response was posted

Do a search for rudder design calcs - Kemps Eng Yearbook (the bible!) gives:
F = kAv2T
Where
F = force
k = constant (between 15.5 and 18 for rudders behind twin or single screws - I'd go for 18 for a worst case).
A = Area in m squared
v2 = velocity squared (m/s)
T = Angle to flow in degrees.

It should give an indication of force, but it will only be rough.
Hope it helps.



this gives 18*1*1*2*90 = 3240N
 
Re: Calling any mathematician\'s (kind of boaty)

[ QUOTE ]
[ QUOTE ]
I'm not sure .. its blinkin complex
see http://en.wikipedia.org/wiki/Pressure

[/ QUOTE ]

This is a bit more relevant
http://en.wikipedia.org/wiki/Drag_coefficient

Discounting viscosity and assuming a Cd of 1, that would make it 500N.

[/ QUOTE ]


another reply on the same forum backs you up

temp and what the surface is will make a vast difference but here goes.

flow velocity v = 1.02889 m/s
length l = 1 m (assuming square)
viscosity of sea water @ 10C u = 1.346 x 10E-3 Pa.s
density of sea water @ 10C p = 1028 kg/m3
drag coefficient Cd = 1.17 (square plate)

Reynolds Number = pvl/u = 7.86 x 10E5 therefore the flow will probably be turbulent so we can apply the standard drag equation:

Force = .5p*v^2*Cd*A = 637 Newtons


Jings I'm glad its a quiet afternoon at work!
 
Re: Calling any mathematician\'s (kind of boaty)

Looking at it from first principles, I think I'll stick with my initial 1000N. Or accepting your figures for v and p I'll amend that to 1057.69N. That's the force that would be generated by completely stopping the flow - ie all water is brought to a stop before changing course to pass around the obstacle. The real force will be less than that figure, but can't be more.
 
Re: Calling any mathematician\'s (kind of boaty)

What you really need is described here
Calculations of Nonlinear Free Surface Flows
Around Submerged and Floating Sea Caches

Hamn-Ching Chen and Tuanjie Liu
Department of Civil Engineering, Texas A&M University
College Station, Texas, USA

Erick T. Huang
Naval Facilities Engineering Service Center
Port Hueneme, California, USA

'Viscous flow simulations were performed for each sea-cache
configuration at 3 different positions: (a) anchored on the
seafloor; (b) tethered at mid-height; and (c) floating on the surface.' The quote perhaps indicates the extent of the problem. The shape of the pontoon itself may reduce or increase the resulting forces.
 
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