Displacement stuff - MMD .... help???

[Larks]

Michael, if you are about the place (or anyone else who knows naval architecture), this is a pretty basic statement from an assignment that I'm doing but there seems to be a bit of a "leap" that has me perplexed and because it is one of the very early parts of the chapter I'm having a struggle to get past it. (I couldn't get one of the symbols that was essentially a triangle with a cross on it to copy so have replaced it with ± ):


A body, floating freely in water will be acted upon by forces due to the water pressure at each point of its wetted surface.

The resultant of these forces will be an upward force equal to the weight of the water displaced by the body. This is the body's displacement, often denoted bythe symbol ∆. If the water density is ρ then the pressure acting upon a small area ±A will be ρh and the force will be ρh±A. By integrating over the whole of the wetted surface the total force on the body is∑ρh±A = ρV where V is the volume of the body below water (called the volume of displacement). This total force is termed the buoyancy force, or simply the buoyancy. For equilibrium this force must act vertically as any resultant horizontal force would cause the body to move in the direction of that resultant. Also for equilibrium, the buoyancy force must be equal to and oppose the weight of the body. If this were not so the body would move up or down. The two forces must act in the same line otherwise there would exist a moment causing the body to rotate.

I do understand the statement but the part that I am struggling to make sense of is:
If the water density is ρ then the pressure acting upon a small area ±A will be ρh and the force will be ρh±A. By integrating over the whole of the wetted surface the total force on the body is∑ρh±A = ρV where V is the volume of the body below water (called the volume of displacement).
Does the way this is written make sense to anyone else or should I just accept it and move on?

(Sorry about the font variations, I had to copy this from a PDF to a One Note file to convert it to word and then add symbols that didn't convert and ......oh,... you get the idea....:o)

[Paul Fitzgerald]

It makes sense but whoever wrote it needs Prozak.

[PeterSibley]

Yep ,makes sense but as Paul says , there seem to be simpler ways of stating it .

[Larks]

The whole assignment (actually a lot of the course) is written similarly and seems to be arse about face, almost like it's been translated through six languages….it's lost me completely at the moment...

[john welsford]

It seems to me to be a huge and unneccesary overcomplication. I taught marine design at college for a while, and if a student tried to put that sort of obfustication in an assignment she or he would be working midnights to rewrite it.
You could of course provide a reference to a very early work on the subject by taking a picture of you naked in hot bath with the caption underneath saying "Eureka"

JohnW

Michael, if you are about the place (or anyone else who knows naval architecture), this is a pretty basic statement from an assignment that I'm doing but there seems to be a bit of a "leap" that has me perplexed and because it is one of the very early parts of the chapter I'm having a struggle to get past it. (I couldn't get one of the symbols that was essentially a triangle with a cross on it to copy so have replaced it with ± ):


A body, floating freely in water will be acted upon by forces due to the water pressure at each point of its wetted surface.

The resultant of these forces will be an upward force equal to the weight of the water displaced by the body. This is the body's displacement, often denoted bythe symbol ∆. If the water density is ρ then the pressure acting upon a small area ±A will be ρh and the force will be ρh±A. By integrating over the whole of the wetted surface the total force on the body is∑ρh±A = ρV where V is the volume of the body below water (called the volume of displacement). This total force is termed the buoyancy force, or simply the buoyancy. For equilibrium this force must act vertically as any resultant horizontal force would cause the body to move in the direction of that resultant. Also for equilibrium, the buoyancy force must be equal to and oppose the weight of the body. If this were not so the body would move up or down. The two forces must act in the same line otherwise there would exist a moment causing the body to rotate.

I do understand the statement but the part that I am struggling to make sense of is:
If the water density is ρ then the pressure acting upon a small area ±A will be ρh and the force will be ρh±A. By integrating over the whole of the wetted surface the total force on the body is∑ρh±A = ρV where V is the volume of the body below water (called the volume of displacement).
Does the way this is written make sense to anyone else or should I just accept it and move on?

(Sorry about the font variations, I had to copy this from a PDF to a One Note file to convert it to word and then add symbols that didn't convert and ......oh,... you get the idea....:o)

[P.I. Stazzer-Newt]

∑ρh±A = ρV

True - but you have to remember that it's a vector sum - this is the usual statement of the effect where pressure at depth acts in all directions equally.

Nobody in their right mind is going to do the integration or the manual sum implied by the left hand side of the equation - the sane sensible thing to do is multiply the submerged volume by the density of the fluid.

[Larks]

Thanks everyone. John, that text is not an answer or respone to an assignment, that is the explanation of the subject of the assignment written by the lecturer. And unfortunately most of the text is written in a similar fashion.

[Hwyl]

Is it Westlawn?

[Larks]

Is it Westlawn?

No, it's not Westlawn, but I'd better not say too much more until I get marked on the assignment in case someone here knows someone who knows someone....:o:o (I'd hate to blatantly insult my marker)

[P.I. Stazzer-Newt]

See'n as yer up and about - is it a physics course or a naval arch. course?

I remember looking at the same sort of explanations about forty years ago but in a physics/engineering sci. context.

The full monty "∑ρh±A" explanation becomes relevant only in unusual cases - like immiscible fluids, or non rigid bodies, or compressible.....

[Larks]

It's actually Marine Surveying but a Naval Arch' assignment. Most of it seems pretty straight forward but I didn't do any higher level maths at school (Pure and Applied maths here in Oz) so some of this is quite foreign to me at the moment.

[Ian McColgin]

So the easier way (cheerfully glossing over mass v. weight but remembering the difference between salt water and fresh) is to note that the total volume of water displaced will equal the total weight of the boat and that the local water pressure issue is simply a function of depth below waterline. Add the area of the hole to figure how soon you'll be stepping up the the rescue craft.

[Larks]

Hi Ian, thanks, it's the formulas and how they are applied to calculate the volumes of displacement, waterplane area and coefficient, Midship coefficient, horizontal prismatic coefficient, vertical prismatic coefficient etc etc etc that I'm struggling with....but getting there...slowly

[Ed Burnett]

Which formulae specifically?

The various volumes, areas and coefficients you mention are tools that describe the size and shape of the hull - they don't really say how or why it floats.

[Ian McColgin]

When I took Organic Chemisty ('67) my math and computer skills were not up to handling the calculus. Especially as lowly undergrads we could only get at the computer (Remember when a lucky campus had but one computer that was about as big as a four bedroom house?) at ohfriggingdarkthirty and I just could not get all the integrations done in a reasonable time. So, on discovering that K&E graph paper was of fantasticly consistant quality I just graphed the reaction (very fine line #4HHH and latex gloves to keep any moisture from my hands from migrating into the paper), cut it out, and weighed it on the triple arm analytic beam balance. Faster and both more accurate (absolute truth) and more precise (replicablity) than the various mathematical approxomations available at the time.

When I got playing at boat design I faced a similar problem and I couldn't afford a planimeter. Remember planimeters? But if you carefully cut out the boat's sections and find for any design the constant you need for length, you have a pile of volumes close enough that you can cut along any water plane for any heel or lading, just stack them on the balance and bazingo.

Side variation for navigators seeking the better noon sight: Graphing the sights either side of local noon will make a nice parabola. In theory you can derive a parabola from three points but in practice there is some experimental error in the sights so even with a computer deriving the parabola well enough to do better than fudge the local noon apex is really hard. Turn the graph so's it hangs and use something nice and lank like a length of fine link gold chain. Dangle it from the ends till the chain's curve best averages all the sights, discarding any that are cleary way off the norm of observations.

Who says the theories of gravitation are not a matter of design?

[bott]

When I took Organic Chemisty ('67) my math and computer skills were not up to handling the calculus. Especially as lowly undergrads we could only get at the computer (Remember when a lucky campus had but one computer that was about as big as a four bedroom house?) at ohfriggingdarkthirty and I just could not get all the integrations done in a reasonable time. So, on discovering that K&E graph paper was of fantasticly consistant quality I just graphed the reaction (very fine line #4HHH and latex gloves to keep any moisture from my hands from migrating into the paper), cut it out, and weighed it on the triple arm analytic beam balance. Faster and both more accurate (absolute truth) and more precise (replicablity) than the various mathematical approxomations available at the time.

When I got playing at boat design I faced a similar problem and I couldn't afford a planimeter. Remember planimeters? But if you carefully cut out the boat's sections and find for any design the constant you need for length, you have a pile of volumes close enough that you can cut along any water plane for any heel or lading, just stack them on the balance and bazingo.

Side variation for navigators seeking the better noon sight: Graphing the sights either side of local noon will make a nice parabola. In theory you can derive a parabola from three points but in practice there is some experimental error in the sights so even with a computer deriving the parabola well enough to do better than fudge the local noon apex is really hard. Turn the graph so's it hangs and use something nice and lank like a length of fine link gold chain. Dangle it from the ends till the chain's curve best averages all the sights, discarding any that are cleary way off the norm of observations.

Who says the theories of gravitation are not a matter of design?

Indeed, I just love it when "reality" can be used for analytical understanding. One of my physics professors in undergrad had a challenge for "eyeballing" a linear fit and calculating the deviation and comparing it to a computer fit. The differences were often negligible, or at least far within error. I often wish that I had been trained on a slide rule for basic math for the physical, tangible connection to the numbers if brings...

Just to be truely nerdy, the chain would make a cantenary (or hyperbolic cosine), not a parabola. But very close, and likely plenty precise when compared with the significant error inthe sights themselves.

[mmd]

vini, vidi.... vamoosed!

I see that the question seems to have been well answered by others, but thanks for thinking of me.

Busy with clients today.

[Larks]

Which formulae specifically?

The various volumes, areas and coefficients you mention are tools that describe the size and shape of the hull - they don't really say how or why it floats.

Hi Ed, my problem was really just trying to get past that original sentence, which I simply couldn't decipher and as it was right at the start of the assignment text I was starting to worry if it would mean the difference between understanding the rest of the text or not.

I understandthe theory so in the end, after reading the comments here, I moved on and am now working through the coefficients of fineness (though in one context the author refers to coefficients of fitness) and Moments.

[Larks]

Well would you believe that after a morning of scratching my head I've just realised that at least half of my problems in understanding this stuff are due to the way the course notes, which are presented as a PDF file, have been typed up and the obscurity of the symbols that they have used?

All the symbols have come out a bit squonk, but there is one symbol in particular being used throughout that looks like a triangle with a cross on the top of it which I showed above as ±. The further I get into the notes the more confusion this is causing because it is everywhere and I simply could not find a reference anywhere as to what it is supposed to represent. As a result I have been stalled for hours going nuts.

I have stared at it and stared at it, walked away and played here and gone back and suddenly realised that it is supposed to be sigma δ……doh!!!

[PeterSibley]

When I got playing at boat design I faced a similar problem and I couldn't afford a planimeter. Remember planimeters? But if you carefully cut out the boat's sections and find for any design the constant you need for length, you have a pile of volumes close enough that you can cut along any water plane for any heel or lading, just stack them on the balance and bazingo.



Thank you Ian ,talk about a timely piece of information ! :):)