Short Derivation of Archie's Law

 "captain's log: supplemental"

It's been quite a while since I wrote the previous post about Archie, his wife, a crown etc.
While I was writing that, I did not want to derive Archimedes' law on my own, so I looked around on the net, to maybe find someone who already posted that, sadly, I couldn't.

Also, this morning my wife and I had an argument about this law, and guess what, she was right, I was wrong... Isn't it weird these are the arguments my wife and I have?!? I mean, seriously, how nerdy can you get? 

So for the interest of completeness (which is actually a mathematical axiom, but let's leave that be for now), I give you a short derivation of Archie's law:

Deriving the law (I AM THE LAW)

Let us consider an infinitesimal volume element of a material with density \(\rho\) , and incident area and height \(A,h\) respectively.
So now let's suppose the material is lighter than the medium surrounding it we get a force equation that looks like this:
\[\Sigma F=A\cdot P_{up} -A\cdot P_{down}-Mg\]
Where the force equivalent (sum of all forces) is positive.

Now, let's wrap the forces that are the result of pressure and name it the Buoyancy force thus:
\[F_{Buoyancy}=A\cdot\left(P_{up}-P_{down}\right)\]
and with a little massage we get:
\[A\cdot\left(P_{up}-P_{down}\right)=A\cdot h \left(\frac{P_{up}-P_{down}}{h}\right)=-V\nabla P\]

So, now, let's take a small detour to understand who \(\nabla P\) is shall we?

suppose we are dealing with an element that is filled with the same material as the medium it's in, in that case we get a mechanical equilibrium and \(\Sigma F\) is simply zero. and so we get:
\[Mg=F_{Buoyancy}=-V\nabla P\]
and breaking up the \(Mg\) element we get:
\[V\rho_{medium}\cdot g=-V\nabla P\Rightarrow \nabla P = -\rho_{medium}\cdot g\]
and so we get quite simply:
\[F_{Buoyancy}=V\cdot \rho_{medium}\cdot g\]

Proper usage

What's useful with this representation, is that we get the force outright, thus we can make force calculations directly.

An important thing to understand is that the buoyancy force is directed not UPWARDS (which is a common misconception) but against the direction of the pressure gradient . which is quite different. that means for example that in a spinning tube of air, the buoyancy force will be mostly inwards as the pressure gradient is directed outwards (in cylindrical coordinates).

That's it.

this time it was really short.

Crowns,Boats, Subs, and other things that float in water...

Abstarct:

A short story about a friend,

and things that go "ping" in the dark.

It has been some time since last I wrote something, but life, the multiverse and everything, got in the way.

By the way, if the answer to life universe and everything is 42, then for the multiverse it should be what? a Vector? a Tensor? and what are the entries? in which linear basis?
I guess I could get cute and say that its a \(42^n\) Tensor, but that's just my regular idiotic nonsense at play...

So yes, in layman terms I was simply very otherwise occupied, but today, as my wife is away, and after almost a week of sleep deprivation and other torture methods, I found the will and the way to actually start writing stuff.

Anyway the idea for this post came from a rather dubious experience, and I might have a follow-up post on the physics of water-closets, otherwise known as restrooms, bathrooms or the unsavory "latrines".

What's the connection?

And Seeing as approximately a quarter of my life was spent on boats or in the context thereof, I deem it fit to dedicate a post to the physics of these floating wonders.

What makes things float? 

Most of us at one point or the other have had a chance to hear that quite famous cry "EUREKA" but I daresay less of us actually know the origin of this cry. so here goes...

Some 2500 years ago in the golden age of the great kingdom of Crete, there lived a king and his queen, or, more probably a queen and her lackey of a king.

Now that queen was little capricious, not unlike the queen of hearts from Lewis Carroll's Alice's adventures in wonderland, and at some point in time decided that the crown allotted to her, was not fancy enough. Thus, she ordered a new crown to be made, a crown of solid gold.

 The word spread, and a huge congregation of goldsmiths, jewelers, and their apprentices, flocked to the isle of Crete, in hopes of gaining the queen's grace and be chosen to forge the queen's new crown. 

But as is often the case with tyrannical rulers, this queen was a very suspicious being, I could theorize she had a bad experience with  goldsmiths, or might be she was simply a bitch.

In any event, she was absolutely terrified by the prospect of being swindled and as a result she wanted to make sure the crown was actually made of solid gold. 

But here's the pickle... suppose she WASN'T swindled and she melted the crown to check if it's solid gold, she's now left with a pot of molten gold, while still having to pay the goldsmith.

Thus a method had to be devised to check the crown without damaging it.

Enter stage left: Archimedes.

Archie, our friend happened to be in the vicinity and having a reputation of the genius he was, was charged with the daunting task of finding that method.

Well, Archie thought hard, maybe losing some weight (why can't I?) and some hair (why am I?) at the prospect of failing the queen and subsequently failing to breath, and at the end of 3 excruciating days, his wife decided she would have none of it anymore, "You stink!" shrilled the shrew "Go have a bath or it's the couch tonight, for you!".

Archimedes, being the thoughtful husband that he was, climbed into a warm bath, and noticed, that when he submerged more of his body, the water level rose and spilled over the sides of the tub. 

"EUREKA"

he shouted then, followed by his wife's "Shut up already you git! you'll wake the baby and then you'll have to deal with it!!" 

Anyway, I will leave you wondering as for how the story ends, did old Archie indeed wake the baby, how long was spent in the dog-house, and whether or not a goldsmith found his premature demise.

Now for the physics:

Suppose an object with a volume V is partially submerged in water - the elevation force is due to pressure differences, the partially exposed part of the object experiences just the atmospheric pressure, but the underside experiences the upward pressure from the water, so let's see what that pressure is:
Let's consider a column of water and a thin strip \(\Delta Z\) thick
\[\Sigma F=0 \Rightarrow s\cdot\left(P(z+\Delta z)-P(z)\right)-s\cdot\Delta z \cdot \rho_{water} g \\
\text{or in other words} \frac{\partial P}{\partial z}=\rho_{water}\cdot g\Rightarrow P(z)=\rho_{water}\cdot g\cdot z+ P_{atm}\]

Where z is the depth of water.

So, we have to consider \[mg=s\cdot\rho_{water}\cdot g\cdot z \Rightarrow h\cdot\rho_{object}=\rho_{water}\cdot z \]

In other words, the depth of immersion is given by the height and relative density of the object and the fluid.

Now, there's an easy way to see that this dynamic is correct, simply take a piece of wood , and see that it submerges deeper when you hold it length up, than when it's laying flat on the water.

So that takes care of boats, we just have to make sure the average density of the boat is lesser than the density of the water, mind you we're talking the average density of the space occupied by the boat meaning also the air inside the boat, unless you start to take on water, and then guess what? your downward bound.

That also might provide a hint why we sometime encounter "unflushables"...

That also might provide an insight as to how submarines stay submerged at a constant depth:
When the sub is at "bubble up" state, basically the density of the sub is lower than the surrounding water thus the sub tends to float up. to hasten the process the sub might or might not apply it's propeller or other means of propulsion.

When the sub is at "bubble down" state, the average density of the sub is higher than that of the surrounding water making the sub "heavier" and thus sinks down, again applying propulsion or not is at the captain's discretion.

By the way, subs mostly have compressed air tanks, which they discharge into  ballast sections, to change the average density of the vessel, and then use compressors to re-compress said air to the tanks, evacuate the ballast air ballast sections to increase the average density (water then flood the ballast sections).

So it seems fairly simple right? WRONG, we actually took the water's density to be constant where it really isn't, cold water is denser than hot water, and deep water is a tad denser than shallow, so what's the deal? 

Well the physics for this is fairly complicated in terms of the math involved, but the IDEA is fairly simple, water density is a product of the mutual forces between water molecules, that are essentially electric in nature, and so external pressure is somewhat involved in this, but even more so, temperature.

So with pressure \(\rho_{water}\) rises linearly at first but pretty quick stabilizes to a constant.

Like so:

Courtesy Windows To the Universe (NESTA)

With temperature the change is much more pronounced, but still pretty much the same applies, A linear rise in density when temperature drops, and then exponential decay to a constant.
I suspect somewhere in the middle there's actually a point where it all turns to ice...

That remind me of the cool Thermometer where there are different glass bells with different nifty colored liquids in a glass water tube, and when the water in the tube is in thermal equilibrium with the area, some bells float up, some sink down, and the one left in the middle shows the right temperature on it.... pretty cool if you ask me...

Told you water density changes with temperature!

So anyway making the calculations needed to predict the density at a certain depth and temperature is a pretty nasty undertaking thus usually subs employ feedback loop mechanisms to apply the right density. either that or they do it by hand and eye i.e. "bubble up"\"bubble down" mechanism.

By the way, remember the couple of pictures in the beginning? well there you go:

Baby Ruth is swimming pool - not quite what you think...

I could go on and on about this, about weighing ships, (as opposed to sheep), and using partially submerged sonar buoys, Thermocline, and using different water densities to mislead enemy vessels as to your true location etc. etc. But I'm pretty sure if you read Clancy's "The hunt for Red October" you'd learn all this and have great time doing so...

Physics of Lenses and Idiots (part II)

 Abstract:

Burning ships, perfect lenses, Red shirts

and weird Al

Hello again folks...

Just for spite. this post is just for spite.
well, not really but it's fun to say it is.

As some of you might remember I wrote an inhumanely long post, and my liege-wife told me to cut it short.
so I did, but no good deed goes unpunished or as theorized by the great-but-not-overly-sane Newton "To every action there is always opposed an equal reaction", and so it is with great yet perverse pleasure, that I give you the second part of the Lenses & Idiots Trilogy(!!!)

Not quite the shilling, though he probably minted some shillings...

 
Anyway, last time I was reminded of two nice stories....

The first about Archimedes and the Siege of Syracuse, where supposedly the defending forces used about 300 round bronze shields, polished to a shine, in order to concentrate sunlight, and burn the marauding fleet to a crisp.did anyone say he loved the smell of napalm in the morning?
well let's put that theory to the test:

groceries:
1. 300 bronze body shields of area 1.3m x 0.7 m, polished to an "\(\varepsilon\)" sheen.
2. the sun - Earth's solar constant. - 1360 \(\frac{J}{sec\cdot m^2}\)
3. auto-combustion temperature of wood - at the most \(~ 450^{\circ}c\)
4. a  distance estimate say at 500 meter.

and now:

300 shields yield an incident area of about 273 square meters. thus according to our previous calculations we simply substitute the respective areas and get the temperature at a 1 square meter of surface on the wood to be about  1600K at  \(\varepsilon=1\), which means this should burn like kindle wood but experimental results here assert differently... so what went wrong?

well, a couple of things:

first off, bronze is not a mirror, meaning for starters, \(\varepsilon\neq1\) in fact not even close.
secondly, we assumed the image area of the shields on the side of the boat is at the same size of the shield. this is miserably wrong.

Consider a piece of shiny metal or a wrist-watch you use to blind the lecturer at a "Mechanics and Special Relativity" course, it is pretty obvious that the area of light traced by the image of the watch's surface is larger than the actual surface of the watch right?

The added length of a line image is given by
\[L_{image-line}=L_{original-line}\left(1+\sin(\alpha)r\right)\]
so basically the area increases like
\[A=A_{original}\left(1+2\sin(\alpha)r+\sin^2(\alpha)r^2\right)\]
So even though the flux is multiplied 300-fold, by all the shields reflecting the sun at the same site, still the area of incident is also multiplied, and even if we take only the linear approximation and not the whole deal we STILL get  the incident area to grow like \(1+cr\) thus:
\[\Phi_{boat}=1360\cdot 300 \cdot \frac{1}{1+c500}\]
And even taking the  angle to be fairly small let's say \(15^{\circ}\) we get the temperature to be about 410K which is about \(110^{\circ}c\) and that with an \(\varepsilon\) of 1!! its enough that we assume \(\varepsilon\) to be 50% which is  of course a gross over estimate, we get a temperature of about \(50^{\circ}c\).
That kind of a temperature is not enough to boil water, much less  trigger wood auto-combustion ,in fact, even if we were talking about perfect reflective surfaces at vacuum conditions this just ain't enough or differently put: "I just can't do it captain, I don't have the power" - Hmm... maybe if the wood was treated with a combustion agent first...

So the obvious conclusion here is that the people of Syracuse had an inside agent!!!
They had a traitor in their midst!!!!

Or most likely this never actually happened....

Warning- computer geek humor below:

<Computer geek humor>

try{kill_Redshirts();}

Throw Exception("attempted divide by zero.");

By the way, did you ever notice that while redshirts are being slaughtered by the dozens in the original star trek series, Scotty actually wears a red shirt, but is immune to the redshirt-death-rule?
hmm... makes me want to throw an exception....
</Computer geek humor>

Anyway...

Let's see what happens when we apply a lens at the 1 square meter target area, with such accuracy as to concentrate the rays at an area no bigger than 1 square centimeter, we get with the same type of calculations, even taking into account the \(\frac{1}{r}\) factor, a temperature of about 4100K, and if we have for example a solar tower, surrounded by perfectly reflective parabolic mirrors, that span an area of,say, 500 square meters, all directed at a Zeiss Parabolic lens of a perfect nature, with the longest mirror to lens distance of about 200 meters, and an \(\varepsilon\) factor of 70% we get, about 5400K.

Again what melts at 5400K?
hmm a short list that includes about everything...
a short list of things that actually BOIL at 5400K includes among others:
Carbon, Platinum, Rhodium, Titanium, Silicon, Palladium, Cobalt, Nickel, Iron etc...

For a reference the surface temperature of the SUN is evaluated at about 5800K.


But, interestingly enough, a star-ship that tries to show-off and make a run near the sun will have to withstand the ludicrously high temperature of about  5 million Kelvin of the sun's corona, about 3 orders of magnitude higher then the surface temperature - thus while bathing in the sun's surface sea of fire might be enjoyable, getting there could prove messy.
Of course this goes against star-trek TNG's episode  "redemption II" story-line, when a Klingon Bird Of Prey survived the corona only to explode on the surface...

A Klingon BOP taking a nice warm bath in a sun.

The second story goes something like this:
I have a friend who's father is the head of solar energy research at a notable institute.
He told me the story of acquiring an almost perfect Zeiss lens. and it goes something like this:
One day, he and one of his colleagues were wondering the streets of Dresden Germany,  biding their time in between lectures, when they saw a group of kinder-garden age kids, playing around with a large lens, having fun with reflections and images.
This physics professor immediately recognized the lens to be of tremendous quality, and approached the kinder-garden staff with an offer to buy the lens.
They really didn't know what they had in their hands, or simply didn't care too much, but they pretty much GAVE the lens away to that professor, with a simple demand - give us a lens that will do what this one know how to do, so the kids will be able to continue playing. He gave them an ordinary, albeit good quality lens, and basically got this magnificent lens for a song...
It is that lens that is still on the top of the solar tower at his laboratory, and through which they get temperatures high enough to burn through steel as if it was butter.

Now as promised a short mass on Fresnel Lenses vs. parabolic lenses:
In short - a parabolic lens has the interesting feature where every light beam coming straight from infinity hits the same focal point, and vise-verse, if you put a light source at the focal point of a perfect parabolic lens, you can be damn sure that all the light goes straight ahead - in fact that is how the high beams on your car works, in other words - know the mechanism of the bastard that blinds you!!!


To show that, we take a simple parabola - \(f(x)=\alpha x^2\), and we will take a beam that comes from straight up and hits the parabola at \(x=x_0\).
the tangent to the graph at that point is given by \(y=2\alpha x_0 x+b\), and the beam that hits the parabola at that point is diverted to a line that looks like \(y=-\frac{1-(2\alpha x_0)^2}{4\alpha x_0}x +c\) Where calculating \(c\) yields \(c=\frac{1}{4\alpha}\).

The fact that where the diverted light ray cuts the "y" axis, doesn't include \(x_0\) as an argument already shows that all rays meet at the same place i.e. at the same focal point at \(\left(0,\frac{1}{4\alpha}\right)\).

A Fresnel lens is a neat way to manufacture a closely approximated parabolic lens, while reducing physical bulk and bill of materials. What you want to do is take a regular parabolic lens, slice it in regions, and piece together what you got. I found a nice illustration of this on the net, here is the picture...

Fresnel Vs. Parabolic lens, notice the corresponding regions.

Thus it is pretty clear, that while Parabolic lenses capture all light rays, it is heavier and bulkier than the corresponding Fresnel lens.
On the other hand Fresnel lenses capture MOST light ray, but not all, so while being lighter and more compact, it creates some distortion in the image, depending of course on the quality of material used, and the "resolution" or the density of "cut&paste" done to the original Parabolic lens to get the Fresnel lens.

Of course this is a very low concern for lighthouses, and other navigational signs and lights, like for instance masthead, port and starboard lights and also port entry and hazard lights.

And so we see that using Fresnel lenses is more efficient in the long run, because it takes up more energy to rotate or move a bulky and cumbersome lens then a lightweight one, and of course reducing weight and energy immediately corresponds to corrosion and mechanical faults ratio reduction.

So why not use a parabolic reflective surface? this could be lightweight as well as cheap AND efficient in terms of light reflection?
Well I think today most mechanisms where we have enough space to imbed that surface, actually DO use reflective parabolic surfaces, another good example of this would probably be the lenses and mirrors inside a telescope.

As for why NOT to use parabolic lenses and mirrors? - mainly physical space considerations probably, but also, in older parabolic mirrors I THINK there might have been significant energy losses via a nifty little mechanism called the "skin effect" but that is all for now, I may discuss this effect in a different post.


Wow, this was long, I hope you enjoyed this...

If indeed you have, you have the making of a true nerd. Oh and sorry I didn't get to almost die in this post, I have plenty of cases in which I did, and I hope they will always continue to be of the "almost" type in order for me to continue recounting them...

a true nerd...

By the way that's Time Independent Schrodinger Equation (a.k.a TISE) for a particle in central E&M potential given by a point charge plastered there behind weird Al. It's used ,for instance, for calculations regarding the Hydrogen atom .... - there I proved I'm a nerd, even though I'm not fluent neither in JavaScript or Klingon, and I hate mayonnaise. 

As always, next time : oh... why bother...

Physics of Lenses and Idiots (part I)

Abstract:

Found a magnificent Lens, Got the story.

and thoughts about Idiocy in organizations

This time I didn't lie!

How's about me keeping y'all on your toes ha?

Well, as promised this post will deal with lenses, and idiots.
Following our recent brush with hypoxia, my lovely wife and myself proceeded to tour the coastline of California, we drove through San Francisco, met with a distant relative, and all in all had a great time.

At some point we drove through Cambria, which is a quaint little town somewhere on the coastal road between Carmel-by-the-sea and L.A.

WARNING: next is a tacky description fitting of a less-than-mediocre dungeon master:

"It was a cool morning. the air was still. thick fog covered the valley and wayward sounds made their eery way to our ears. the cold ate at our bones, and it was difficult to see anything further then the tip of your nose. 

Somewhere to the left you hear the crack of dried bones, and you think you might have felt something brushing against your boot, or maybe it was your imagination...."

Ok, sorry, got carried away there just a tad....
Anyway, it was cold that morning so  the town was shrouded in mist. we took a little stroll down to the waterfront, where it was rather stinky as the sea washed ludicrous amounts of kelp to the shore and it just lay there rotting.

Well we needed to refuel so we pulled over, and while we were in the station, after grabbing a cup of coffee something caught my eye.

by that time the sun was about half way between horizon and zenith, and the glint of a glass house - the kind you should not throw stones from - was pulling at me.

We went to investigate only to find something incredible.
there is was, encased in a glass gazebo, welded shut and otherwise unapproachable, like a queen enthroned in a crystalline palace - The most wonderful Fresnel Lens I have ever laid eyes upon....

Piedras Blancas lighthouse Fresnel Lens at Cambria

I was psyched.
We went to find someone who could maybe open the locked carousel, so I could take a closer look at it, and found an elderly fellow who was kind enough to tell us the wonderful story of this lense.

In his crackled yet fiery voice he told the story of the lens....  

It seems that ever since a couple of decades ago, as the US coastguard decommissioned several lighthouses along the US coastline, this lens was all but forgotten by all, laying embedded in a non-active lighthouse's torch.

Imagine this untouched treasure trove, laying partly in the sand, covered by rotting kelp, the sand and salt slowly eating away at the tempered glass, scratching it, scarring it....
oh the humanity!!! 

The good people of Cambria led by local coastguard veterans secured the consent of the US coastguard, took the lens, fixed it, re-polished it, and enshrined it in the glass carousel in town,
by the veterans center.

Now as it turns out, the US coastguard, found out that this kind of lens is not manufactured anymore. kind of like the precise German lenses and mirrors used in WWII to blind British air-fighters on their runs over bombarded Berlin. it just costs too damn much to manufacture these kind of lenses especially when there are "better" alternatives to be found, more on that later...

That no one manufacture these anymore is a nicer way of saying these babies cost a small fortune each!!! and now, the US coastguard found out about that and wanted to put their hands on this lens.... snatch it away from the land lubbers' keep.

After a long and epic struggle, almost as epic as Gandalf's struggle with the Balrog at the bridge of Khazad-dum (and later falling, and in the waters, and on mountains... sheesh he must've been really tired that night), or the no-less epic struggle of scraping myself out of bed each morning....
the land lubbers won the lens and it can still be seen lit on special events and festivals of Cambria

An epic struggle

An equally epic struggle

Another nice story about this lens is this:

When first they brought the lens to town, they put a regular old incandescent light bulb at the focus of this lens, only to find out that the housing for the light bulb melted completely!!!

it took them a while (and several light bulbs and housings) to understand that at the focal point things get heated. so much so in fact that the whole thing simply melted.


Warning: physics ahead-

Let's try to understand how hot it gets at the focal point shall we?

Let's say the incident area of the lens is about 1 square meter.
assuming the light bulb is approximately round with a radius of say 5 cm the incident area it presents is about \(\frac{\pi}{4}\cdot 10^{-3}\,m^2 \).

Now the light flux that goes through the lens goes to the focal point with a fractional portion accounting for flux losses due to diffraction, interference and other energy losses (crepuscular rays for instance) so basically:
\[\Phi_{bulb}=\varepsilon\Phi_{lens}\]

Now the flux per area averaged over all earth's incident area, over all emission wave-lengths, and over night-day at the mean earth-sun distance is called the earth's solar constant, and the observed value is \(S_{\oplus}\approx1360\, \frac{J}{sec\cdot m^2} \) , so:
\[J_{bulb}=\frac{\Phi_{bulb}}{A_{bulb}}=\varepsilon\frac{\Phi_{lens}}{A_{bulb}}=\varepsilon S_{\oplus}\frac{A_{lens}}{A_{bulb}}\]

Where \(J\) denotes the energy flux per area unit, which is proportional to the temperature in the 4th power by Stefan-Boltzman's law:
\[J=\sigma T^4\]
In which \(\sigma\) is the Stefan-Boltzman constant which is given in SI units by \(\sigma=5.67\cdot 10^{-8} \frac{Jouls}{sec\times m^{2}K^{4}}\)
And so if we substitute all the constant and data we get that:
\[T_{bulb}=\sqrt[4]{\varepsilon 3.05\cdot 10^{12}}\]

Where providing no energy loss whatsoever (\(\varepsilon=100\%\)) we get the temperature there to be about 1300K or about \(1000^{\circ}c\).
Let's say the situation is pretty bad and we have \(50\%\) energy loss we then get "only" 1100K
which is about \(800^{\circ}c\)... you get the picture.

Just for proper comparison purposes:
Aluminum melts at about 900K, Copper at about 1350K, Lead and Zinc  at 600K and 700K respectively and Iron at about 1800K.
So obviously if the bulb housing was made with aluminum wiring it is fairly easy to see the whole thing would simply melt even in relatively cold days, not to mention the plastic in the housing...
after a fairly short research it seems most plastics melt at a temperature no higher than about \(200^{\circ}c\), while even temperature and fire resistant brands do not fair well above \(400^{\circ}c\) to say the least. so really the lens could be in a horrendous state or simply our assumption for \(\varepsilon\) could be way off, since even at \(\varepsilon=5\%\) it turns out the plastic housing would have melted...

Ok, so what they did in the end is lower the bulb housing by maybe ten centimeters, so now, while the lighthouse does not yield such illumination as it used to, it is now operable.

So my lovely wife went through this post and had several remarks:
1. This post is too long.
2. I am a complete and utter nerd, but she loves me still.
3. it is late and I should go to bed if I prefer hell not to be risen....

Thus I am at this point apologizing, and cutting this post in two.

Let's conclude for now by deriving an interim equation for the ambient stupidity in an organization...
 
It is obvious that  the bigger an amount of money is involved in a project, the more foreign consideration (i.e. bribes, conflicting interests etc. )  will come into play.
Also there is a pretty known principle where a chain or for our purposes a net is only strongest as it's weakest link, this also applies to teams, where a team is only strongest as its weakest member.
and for our purposes a project team is only as capable as it's dumbest member.
Also we want to make an observation that not only the above is true, but as team grows larger it becomes more sluggish and bureaucracy bound, now the only question is what are the relations, so , obviously when a lot of people work on the same project with no proper guidance (i.e. tyranny or otherwise simple dictatorship) the average work tend to scatter in all directions equally thus yielding zero work.On the other hand it is a well known fact that given a strong and small team, work is done in greater efficiency.
so I propose the next relations:
\[eff \propto Ne^{-\frac{N}{N_0}}\]
Where N is the number of people working on the same project, and \(N_0\) is there for normalization purposes, and \(eff\) denotes the efficiency.

also it is a well known fact that you need a team of people to fix one persons stupid mistake, and if I am VERY lenient I will only take a quadratic decrease in efficiency in respect to the amount of money, thus I propose a refinement:

\[eff\propto \frac{B}{\beta B_s}\frac{N}{$^2}e^{-\frac{N}{N_0}} \]

Where \(B\) is the average brains of the team, \(B_s\) is the brains possessed by the dumbest person on team, \(\beta\) is the coefficient that represents how many other team members it takes to fix one dumb mistake...

Keeping in mind that idiocy is the reciprocal of efficiency we get the ambient idiocy of a team or organization to be:

\[I_{diocy}\propto \frac{\beta B_s}{B}\frac{$^2}{N}e^{\frac{N}{N_0}}\]

This could very well account for the coastguard behavior in that instance, albeit in military organizations this dynamic is really not enough to account for all the stupid s@#% one encounters there, and I have rich experience of such encounters...  

Also - basically whichever organization it was that hired me, probably raised it's idiocy bar by several orders of magnitude... :p

That's all for now folks, part II of this discussion will be very interesting, with an explanation of why, for instance,it's highly unreasonable that the Klingon ship captained by Kurn son of mogh, with Worf on board out maneuvered two Klingon birds of prey by flying into a star's corona, and while Kurn's ship was able to escape and break orbit, the pursuing ships fell into the star's sea of lava and consequently exploded...