Calling any mathematician's (kind of boaty)

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

Gezzz

And there was me hoping for (x) PSI.

So what did you agree on 500N or 1000N?

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

Geoffatstanpit is quite right. C.A.Marchaj covers this in his book "Aero-Hydrodynamics of sailing". Unfortunately, his formula for the force q=0.994 x V x V requires velocity in ft/sec and gives force in pounds/square ft. Your water flowing at 2 knots is covering 2 x 6080 ft/hour which is 3.38 ft/sec. Your 1 square metre plate has an area of 10.7 square feet. Therefore q=0.994 x 3.38 x 3.38 = 11.35pounds /sq ft. The force on the 1 square metre plate is 11.35 x 10.7 =121.5 pounds. Of course, we don't express force in pounds nowadays. 121.5 pounds is the same as 55kg which gives us a force of 550 newtons.

By the way, the formula would apply to a plate fully immersed in the water, so that some of the water flows over the top edge as well as round the sides and underneath. Your pontoon float is evidently floating, so the formula would not give quite the right answer in this case, but it would be a first approximation.
 
Re: Calling any mathematician\'s (kind of boaty)

Oh dear, sorry, I forgot the divide by 2 in the formula. That brings the calculated force down to 275 newtons
 
Re: Calling any mathematician\'s (kind of boaty)

Some early morning thoughts:

Not much help, but not as easy as considering a flat plate:

(As an aside: Marchaj says drag coeff flat plate is 0.004 to 0.008 for flow parallel to plate, & 1.9 for perpendicular to plate)

But you don't have a flat plate you have a cube (or a box anyway).

Also the cube is at the interface between air & water so you will be making waves. This will increase drag. (as per above post just seen it!)

In other words the drag coefficients for flat plates wont be relevant.

Googled this:
http://kathrein-scala.com/tech_bulletins/FlatPlate.pdf
(talks about air, but equations work for any fluid, just use the properties of salt water)
Says drag coefficient for cube, face forwards is 1.07.

This doesn't include effects of waves, and maybe the submerged portion of your pontoon isnt a cube exactly, but getting closer!

I would say, get your pontoon and tie it to something with a spring balance tied into the mooring rope and read what it says!
 
Re: Calling any mathematician\'s (kind of boaty)

... and also if a boat is tied to the pontoon with a stong wind blowing then that's going to affect you calculations as well
 
Re: Calling any mathematician\'s (kind of boaty)

[ QUOTE ]
(As an aside: Marchaj says drag coeff flat plate is 0.004 to 0.008 for flow parallel to plate, & 1.9 for perpendicular to plate)

[/ QUOTE ]
Drag coefficient of 1.9 makes it 1004.8N. Can anyone cite any reference to Cd being higher than 2?
 
Re: Calling any mathematician\'s (kind of boaty)

Also...

I would guess that the maximum force wont be in the steady state anyway, a shock load like when subjected to a swell/wash etc & snubbing (cf anchoring) would give much higher peak loads & would mean that you would have to bung on a large safety factor after doing your basic calculation.

What is the reason for doing the calc?

How will you fix the pontoon?

eg ropes/chain with catenary (spelling?!) will absorb shocks much more than fixed to post.
 
Re: Calling any mathematician\'s (kind of boaty)

There is a good forum at www.eng-tips.com , may be try there.

My guess on it - imagine the pontoon/barge is free floating and you are on a fixed structure holding it with a rope, do you think you could hold it? I reckon yes, quite easily - 400-600N. now put a boat on it, now apply a gale of wind etc. Pontoon go bye bye!
 
You must log in or register to reply here.

Share this page

Top