Float Your Boat: Newsletter Challenge (December 2012)

Decrease, very slightly.

Yes. The weight recorded at the bottom would decrease slightly, all due to Archemedes's principle.

Wjen the boat isfloating some part is above water and would weigh as much as it would in air. But when it is immersed, weight is lost equal to the volume of the boat body material, at the density of water.

So, a pressure recorder at the bottom wheere the boat sank would record a lesser weight

I think we also have to account for the altitude at which this experiment takes place, on top of Mt. Everest, pressure would be negligable due to lack of air pressure at that altitude, whereas at sea level, the pressure would be more in the floating boat than the sunk boat.

Bath's are usually filled with hot water, and it will be cooling whilst the boat sinks.

Doesn't hot water evaporate easily....

Now what would happen if somebody broke this bathtub?

Interpreting the question to allow water escaping from the tub is pushing it a bit.

I wonder if water level would drop quicker due to evaporation, or contraction as it cools ? I'll chance it, and vote for evaporation being more significant. That's based purely on my experience of steamy bathrooms, and total cowardice at the thought of trying to calculate evaporation rate from a hot bath in a sealed room.

Why, oh why, didn't they just pose the situation as being set in a bucket of water ! I'll be bashing myself with a sledgehammer unless we get the 'official' answer soon.

_{Admin - I am totally lying about the sledgehammer, so don't even bother trying it}.

Hey Kris, how did your boat do vs. surface tension?

Had a nice result with a little boat in a wineglass. Wrapped a load of foil around my pinky, then squared it off. The whole thing looked a bit messy, but it floated. Not only that, but it's inner surface looked to be a good 3 or 4 mill below water level. It was forcibly sunk, and stayed so. Not so sure it would have sunk if just toppled over. Further testing in due course...........

Splashage?

(Ed. C. knows stuff that ain't even in the dickencherry.)

I'm with you here Del. The mass of this little boat is so small compared to the water that any changes in water level from a splash in sinking the boat or even evaporation will likely obfuscate the pressure conditions.

I also dislike the imprecision of asking for the pressure at the bottom of the bathtub. The pressure differential from one location to another on the sloping bottom of this bathtub (I presume the bathtub is designed to drain properly) will change by a larger percentage than adding an aluminum foil boat. If the point one measures pressure will be a point where the boat rests on the bottom of the bathtub then there will be one plausible answer. If instead one only considers the hydrostatic pressure of the water then another plausible answer exists. If instead one considers the pressure between bathtub bottom and whatever it is resting on then another plausible answer exists.

In all of these conditions the slightest change in the quantity of water will dominate any difference measured or extrapolated.

Hey Del, who says the boat is sooooo small? Maybe they used a whole roll of foil! In any case our cat dosen't like water so its splashage is imeasureable. ( Is zero a measure? ) Gerry D

I'll go for decrease, since the water level will drop as the boat no longer displaces water.

Then again, my boat is a submarine (any good sailor will agree ) so toppling it won't make it sink. If the sub dives, then more water is displaced and pressure will increase.

Where do you obtan the variable ballast for your submarine? Pumping in some water from the tub? If so, it now displaces more water, but the difference is equal to the amount of water you have pumped into the submarine. The only change is that the surface of the air above the water is no longer flat. Air density varies non-linearly with altitude - the result is that average air pressure above the flat surface is (very marginally) lower than above the non-flat surface. The pressure will drop by an extremely small amount (unless you also increase the air pressure within the submarine).

Think again Kris! Actually, rolling a sub WILL sink it due to the loss of air in the Ballast tanks. You better have your act together when that happens! With a loss of Ballast air, it will take some serious forward speed to drive back to the surface. been there done that..full breach after emergency blow.

Darn it - I'd always figured that the ballast tanks would have some sort of valve. I sit corrected ! Cheers for the info, SubFrog.

Al = 2.70 g/mL ? That's an interesting way of putting it. Quite true, just unusual.

Here's a thought; Fold the foil to create a double hulled boat....make it such that the overall density of the formed hull is a tiny bit less than 1000 Kg/m³.....the boat will float, but when sunk it will still displace (almost) the same volume of water.

A very rough approximation - Boat is a cube shape with a small hole in the top. Side of length 6, and material thickness is 1. Overall volume is 216, material volume is 216. OK the material is a bit heavier, so we reduce thickness by a factor of 2.7. Call it 0.5. If I could make a hollow cube from foil that is about 3mm each side, I'm almost in the ballpark. With enough foil, it could be scaled up and happen.

My only excuse is that I'm waiting for a parcel.

Wasn't he the villain on Johnny English? :)

Noo, you're thinking of Marcel, you sauvage. My mind was on Blaise.

Okay, first we have to ask: Why is the tub half full? Is it a Cast Iron Bathtub that someone has started breaking apart? I think we need to start a discussion about this.

If you put a pressure transducer at the bottom of the tub as shown, it will meaure a slight decrease in the water pressure once the boat has sunk. Initially, the floating boat added its weight to the weight of the water pressing on the tubs walls. Once the boat sank, its weight was transferred directly to the bottom surface of the tub, thus its weight no longer added to the pressure as measured by the transducer, and the water level dropped slightly (equal to the volume of water displaced by the boat).

Slight correction. The drop in level will correspond to the difference between the volume displaced by the boat's weight and the volume displaced by the volume of the foil used, divided by the surface area of the tub.

It will increase by a tiny amount because the water level will be slightly higher.

It wouldn't seem so, until you add surface tension into the mix...............which doesn't come into play when it is on the bottom.

I wish I could delete my last post.

Serves me right for trying to float my boat upstream. Oh well.

INCREASE

Afloat, the boat displaces its weight, sunk, it will displace the volume of the foil only. Since it sinks it must be more dense and have less volume than water so the water level and therefore the pressure at the bottom will Decrease! Gerry D.

The sum of the weights of the water and the aluminium foil does not change when the boat sinks. The area of the bottom of the bath, however that is measured, does not change either. Pressure is force/area. The pressure does not change.

Obviously A. Hero has nailed the question as generally interpreted (the water pressure), and probably as the questioner intended.

However:
The question states that the boat will end up at the bottom of the bathtub. On that definition the bottom of the bathtub is directly supporting the boat - so the pressure here will be higher....

On the other hand, without that definition there could be two slightly more interesting interpretations:
a) The air pressure on the outside bottom of the tub - yes that will change...
b) The pressure on the supports
As ever, ignore splashes, evaporation, finger-salt, variations in ambient conditions etc.
Have fun

To repeat my earlier insanity ; Could the foil be folded in such a way as to make a doble hulled boat - it's overall density being about the same as water. Dimensions calculated so that it just floats with the top being at/above the water surface ? If so (a few caveats) it could be sunk without changing the water displacement. Maybe.

Wouldn't the air inside the boat need to be incompressible? ;)

As would (pointed out by a few other sneaky folk) the foil need to be. In principle, I think that could be factored in. Bit early in the week for kitchen sink experiments, but I'll put this one up on the chalkboard. Friday night and a pint glass sounds like a good time. I'll make sure to fill the glass with water ;p

Another water displacement question! The amount of water weight therefore pressure on/in a closed body of water is equal to the weight of water displaced by the floating object. A piece of foil, be it flat, wadded into an airless ball, or formed into a boat weighs the same if it is on the surface displacing water or on the bottom. same weight, same pressure.

The pressure at the bottom of the bathtub is a function of density of water and the height difference between the surface and the bottom which is given by:

rho x g x height differential or simply rho g h.

Since g is constant and density of water does not change but height of liquid surface increases due to volume of the boat, therefore the pressure at the bottom increases accordingly.

Let's try to see how much the pressure might drop.

Making the usual simplifications - ignore splashes, evaporation, finger-salt, atmospheric pressure variation etc.

If the volume of the aluminum foil is v (m³ = or kilo-litres according to taste), and the area of the bathtub is A (sq-m):

With the boat floating, the volume of water displaced is sufficient to support the weight of the boat - 2.7*v.
With the boat sunk (additional simplification - ignore trapped air), the volume of water displaced is just v.

The floating boat raises the surface level by 2.7*v/A above the level in the absence of the boat
The sunken boat will only raise the surface level by v/A.

So the surface of the water will fall by (2.7-1)*v/A.
The reduction in water pressure at the bottom of the bath will be 1.7*gn*v/A

Suppose we make a large (and floppy) boat using a heavy-grade kitchen foil: say... width 0.3m, length 0.6m, thickness 22μm, and the surface area of the water in the bathtub is 0.8 sq.m.
Sinking the boat will reduce the water pressure by 1.7*9.81*0.3*0.6*0.000022/0.8 ≈ 100μP. That's about a billionth of "standard" atmospheric pressure (and equivalent to about 8mm on your super-sensitive altimeter).

I used a quarter stick of dynamite to sink the boat....unfortunately this shattered the tub, but the pressure was decreased......considerably....

The pressure would drop due to the difference between the weight and the volume of the metal, taking into account the compression of the metal due to the increased pressure at the bottom of the tub.

The weight of the metal boat displaces an equal weight of water, raising the water level when you put the boat in the water. Raising the water level raises the pressure at the bottom of the tub. Once the boat has sunk, the water at the top of the tank is no longer displaced, but the volume of the metal boat at the bottom of the tub is displaced. The difference in the amount of water displaced would be the ratio of density of the water and the metal.

Assuming that the water and the aluminum foil are both compressible (become more dense in the relatively greater pressure at the bottom of the tub), it would be the densities at the higher pressure at the bottom of the tub that would count.

"Again, since the mass of the contents hasn't changed, the bottom bears the weight of the water plus the weight of the boat Ww+Wb. Divide that force by the area and you have the integrated or average pressure, the same as for the floating boat."This assumes that there is no flexure in the tub or change in the average atmospheric pressure." This sounds the same as the weight on the feet?

However, I think it could be interesting to replace the assumption of "no change in average atmospheric pressure" with "no change in atmospheric pressure at a fixed height". This would allow you to take the effects of water compression and of air compliance into account - to predict the (entirely unmeasurable) direction of the effect on average pressure at the base of the tub.

The only thing that can change pressure in the bottom of the bathtab is a change in water head. And since the "boat" was balanced floating, the sunk part (that supports the whole boat weight) had equivalent density with water, because of the air trapped in the upper side of the "boat" shape. This, after sinking, changes, since Al is much denser than water, the sunk foil will occupy less volume than water displaced when floated, according to Archimedes, the level or head will drop, and so will the bottom pressure. S.M.

The weight of the water + the weight of the boat = \int P dA across the bottom of the tank. When you ask for the "pressure at the bottom" this is unambiguous before it sinks because the pressure is constant everywhere. After it sinks there will be different pressure where points of the boat touch and do not but the integral is the same.

To turn this into a simplified problem consider the modification of it: The boat is compressed into a solid sphere of aluminum (say 3 g/cm) that is delta thick that is then placed on the bottom. Now fill up the tank only to thickness delta. The pressure on the bottom is now P1=\rho_w*g*delta under the water and 3*P1 under the aluminum. Now fill the rest of the water to height H. There is now a pressure difference of 2*P1 depending where on the bottom you place the sensor.

In the case of a shaped boat that only rests on a few points, of negligible area, we would naturally just look at the pressure elsewhere on the bottom. The force on these points is (\rho_Al-\rho_w)*V_boat*g. The remaining force is (\rho_w)*V_boat*g+M_water*g. Thus the pressure drops by (\rho_Al-\rho_w)*V_boat*g/Area.

meant to say "disk" not "sphere" here

Being an annoyingly (so I've been told) literal thinker, I actually made a small aluminum foil boat and found that in order for it to go to and remain at the bottom of the tub one has to add ballast to it, otherwise it always floats back to the surface. Unless I crush it down to a point where no air is present (then its no longer a boat anyway) the aluminum retains its buoyancy and refuses to sink.

Of course it could just be a Boston Whaler......

Perhaps a rock would work better?

It may be no consolation, but I think I'm going to have to try this as well ! Will go check the kitchen for tinfoil stocks......

by putting the Alufoil into the water, the pressure at the bottom of the bathtub increases (part of a million or something like that), if you float the boat with water of outside the bathtub, the pressure increases too (by parts of a million); if you float the boat with water from inside the bathtub the pressure at the bottom of the bathtub will not change!

if the alu-foil touches the bottom of the bathtub, then the points where the alufoil touches the bottom have another pressure then the environmental water!

the pressure inside the water to the alufoil depends only from the height of the water at the bottom or in between bottom and surface.

because the density of the alufoil is greater then the density of the water the touching points from the alufoil at the bottom have an higher pressure at the bottom then the water pressure at the bottom is.

this is the reason why the boat sinks into the mud at the bottom of the sea/lake or something else!

Thanks for coming down to chat.

increase (very slightly!) the foil the boat is built from willtake up space in the water, increasing the depth in the bathtub by a small amount, this will in turn increase the water pressure.

When floating, the Al boat displaces an amount of water equal to its weight. Water column height increases in tub.

When sunk, the Al boat displaces an amount of water equal to its volume.

Al has a density 2.7 x that of water, meaning when it is sunk, it will displace 1/(2.7), let's round, 1/3 its weight it water.

Thus, the water column height will drop (relative to that of the floating boat), therefore, the pressure will drop at the bottom of the tub.

The pressure of the fluid as measured at the bottom of the standard home bathtub will drop ever so slightly. This is due to the marginal reduction in the level as the fluid in the tub that was displaced by the mass of the floating boat is now only displaced by the volume of the aluminum itself. Unless the measurement device is incredibly accurate, you will probably not see a change in pressure.

**This is assuming that the floating boat had enough freeboard to displace the necessary volume of water to support the boats mass.**

a very simple question in fact

The total weight of water + boat is supported by the bottom of the bath tube. Nothing else!

This can be done either :

by uniform pressure distribution on the bottom when any object in the water float (or before they reach the bottom while sinking).

When the "non floating anymore" object reach the bottom a part of the weight (depending on density of the object) is directly applied to the bottom (on a neglectible surface area) the rest is still applying uniform pressure distribution.

If you want to consider it from pressure only point of view : before the pressure in uniformly spread on the surface and is equal to weight of (water + boat) / bottom area.

Once the boat touch the bottom you have some aera supporting the boat weigth. The pressure at those point is higher (if not the water won't be gone !)

As the global weight and surface of bottom is constant the average pressure should remain constant, and because some points have higher pressuree the rest should have less.

This is to say if you neglect water evaporation, atmosphéric pressure change, ground movement, apparent gravity variation (tidal effect), temperature and dilatation change during the test, and chemical réactions (like oxydation of the foil)...that could be, in fact, more than neglectable !

I know this question for very long because my great uncle challenged me we a similar one when I was only 11 or 12...

"When the ice cube melt in my glass of whisky, does the level of liquid change?"...the answer is VERY difficult (alcool and water are not neutral liquids, atmospheric moisture can't be neglected...) and I still don't know for sure. As a very good side effect, I HATE whisky and any form of alcool since that time!

So, are you saying the pressure at the points where the boat does not touch stays the same?

Your analysis of average pressure at the bottom is not quite complete - even if we also assume that the bathtub iteself is perfectly rigid and water is incompressible. The difference is due compressibility of air, as the water+boat surface is non-uniform initially.

Air pressure on the surface of boat+ water (or only water after wreckage!) is supossed to be constant and should be removed from the balance.

Of course we are only dealing with the vertical component of all forces as all horizontal are balanced so no lateral movement occurs.

I say water pressure decrease a bit, while aluminium foil to bath tube contact create small aera were the pressure (of aluminium on btahtube bottom) is a bit higher

So the overall effort on the bath tube bottom remain constant. As it surface remain constant the AVERAGE pressure (aluminium AND water together) remain constant TOO.

"Air pressure on the surface of boat+ water (or only water after wreckage!) is supossed to be constant and should be removed from the balance."
This would require a change in the ambient atmospheric pressure; to my mind this seems to be out of kilter with the usual form of such constraints, which is that every external environmental factor remains constant. For simplicity, I would specify that the atmospheric pressure at some fixed location remains constant - say at the top of the bathtub. The reason for the sugestion was to introduce some interest to a problem that otherwise appears trivial.

When the boat sinks the pressure at the bottom drops so there is a small change in surface height. Since air pressure falls off exponentially (with scale height of about 10 kilometers) it is an extraordinarily tiny correction. Of course, if your boat sinks in a broad volume of water the change from this is also very tiny (but it will always massively dominate).

Actually, taking Legray's assumptions and assuming vertical sides to the bath, the total volume of the solid and liquid materials filling the bath will not change - so the average height of the surface (including the depression due to the boat) will not change. The only difference is due to changes in air density - which is going to be absolutely miniscule. The direction of the change is nevertheless calculable.

I don't agree : I never talked about VOLUME. I said weight! In fact the boat "forces" some air "into" the water so it displaces more weight of water than it aluminium volume and thus float !

When the boat sink, some water gets IN, in fact the volume of water getting IN the boat is higher than the volume of boat getting IN water.

Globally the levels decrease a bit.

As far atmospheric pressure, the small decrease of water level is compensated by some air comming "over" the bath surface which in turn apply some pressure.

If taken into account this cause about 0.13% correction to the pressure decrease computed.

Agreed, you never talked about volume.

Unfortunately, this sort of problem can best be fully addressed by including volume in the considerations. Specifically:
Globally (under the asssumptions), we can see that the total volume of air inside the tub (taking the plane of the rim as the top of the tub) will not change. We take it that the pressure at the top of the tub is constant. Thus, the remaining variable that affects the average pressure on the bottom of the tub is the average density of the air within the tub. So the question is how this might change when we sink the boat.

Now that you mention it, I have to correct my view :

If you consider the bath as closed (put a virtual lid on it) BOTH mass and volume will stay constant. So whatever the order of constent the weight of them on the bottom is constant.

When the boat floats the global weight of AIR+Aluminum+Water is uniformly spread on the bath bottom causing a given pressure. When the boat reaches the bottom part of the weight of aluminium is then supported directly by the bath bottom. the water pressure therefore decrease accordingly. The level of water doesn't make any difference. The shape of the bath will changes how much the pressure change at the bottom but is not making significant difference and will not reverse the result.

Air will make absolutly no diference and has no impact on the result.

I think the original post stated it was the pressure at the bottom, not on the bottom. I interpret that as pressure of the water at a specified depth.

Agreed - but wasn't that problem rather too easy?

:)

Let us try an extreme thought experiment:
Imagine that the water sits on top of the air (conceptually we could use a weightless rigid divider with friction-free edge seals). As the volume of air is constant, the air pressure is unchanged - which means that the pressure on the bottom of the bath is substantially reduced. This is theoretically possible because the pressure on the top surface of the water is reduced (due to your ideal virtual lid).
Now open the lid: the air under the water will compress. The base of the bath is now supporting the original air, the water, and the additional air that comes in as the water compresses. So (with vertical sides) the pressure on the base of the bath will increase.

There are similar effects (due to air supporting air) when you change the shape of the surface of the water.

So try a second relatively extreme thought experiment:
Confine the water to one side of the bath (open top), using a vertical divider (so the divider itself provides no vertical forces. You have removed some of the volume available for air near the top of the bath, and replaced it with air neat the bottom of the bath. The air pressure in the new location will be higher than in the old location, so the air will be more dense.

The floating boat will have similar (but smaller) effect.

The situation with the top closed (as you proposed) is more difficult - so I leave it as an exercise for the reader ;-)

I believe that the answer "the average pressure on the bottom of the bath stays constant" would only be correct if the surface area of the water in the tub did not change with the fill level. It may be that this is true of some tiled tubs that were constructed in situ; however, I think it is more reasonable to assume the situation that applies to all pre-fabricated tubs - that the surface area increases with increasing depth of water...

BTW, my response to the water in whisky question would have been different from yours (i.e. an excuse for at-least-daily experimentation with a good single malt); however, (unfortunately) I have always felt much the same about placing ice in whisky as I do about placing it in red wine...

This is only possible is the density of the boat material exactly equalled that of water. It could then kind of "sink" (by just placing it there) but not put any pressure on the bottom.

Are you saying that the average pressure on the bottom will only stay constant if the vessel becomes neutrally buoyant when sunk? If so I agree (for the usual tapered tub).

Oui

Actually, I think I was somewhat in error. The boat fills with water and is now partly supported by the buoyancy of the material itself. This causes the water level to drop. In the limit of a very broad tank this is pretty good. I now remember the Archimedes Paradox. This basically says you can float a battleship in a tablespoon of water if you just shape the tank closely to the shape of the boat. If such a boat sinks the water level now decreases by a large height (of order of magnitude of the height of the ship's waterline). This can give a massive decrease in the pressure at the bottom where the water actually sits (which is now inside the boat). To avoid this one has to assume the volume of water is more than the volume of the boat and the shape of the boat can completely submerge beneath it.

Basically Archimedes principle is kind of wrong the way it is usually stated but it is true in the limit of a massive tank.

A fun way to think of this is to lower the boat down on a string. When it floats the force on the string is zero. When it is suspended at depth the string exerts a force of Vol(rho_b-rho_w)g. We can now see how the pressure on the bottom has dropped and less mass is displaced so the water drops.

A more realistic example is a boat in a tiny pond. The slope and curvature of the pond edge determines how much the level raises when I drop a boat in it. It also affects how much it will actually drop when I sink the boat. This lets the boat completely submerge but the pressure still varies because the area of the bottom varies and is not all at the same height.

The pressure still drops but how much depends on the geometry of the system not just the volume and density of the boat material.

'Basically Archimedes principle is kind of wrong the way it is usually stated'

I.e. "Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object." ?

I think the issue is the definition of the term "displaced". If you take an etymological meaning for "displaced" equal to the net amount of water that is moved it is indeed wrong - so long as any part of the object is above the level of the surface prior to the immersion. The "get-out" for this is a technical definition of the term "displaced".
Maybe something was gained in translation?

You can ask what it means for fluid to be "displaced." If you draw a line across the boat at the waterline, then the volume of the boat below this is the relevant volume to compute the force. The problem is that for bodies of size large enough relative to volume of water (or area large relative to its area), this is not the volume of water shifted by lowering the boat into the water.

I think that means the same as my contribution. As someone (presumably you?) marked my commment off topic, would you (or they) care to clarify why this might have been done?

I pressed the post comment by mistake before editing...
It should have read

You have provided the "technical definition" that I had omitted from my contribution (because I assumed everyone knew it).
Was my assumption of the reason for your writing 'Basically Archimedes principle is kind of wrong the way it is usually stated' incorrect, or can you think of a different reason that someone marked my commment off topic?

I thought your comment was really a less specific rehash of what I posted earlier but I didn't post it as off topic. You seem to know more about what is going on that the vast majority of people here.

I think the best way to teach this stuff is by using \int PdA to give the force and then demonstrate that external boundary constraints (fixed density, volume of material and vessel shape) alter the final equilibrium solution height. The height then is altered by two factors. The air "displaced" at the surface before sinking and the shape of the boundary over the change in height of the fluid fix the result.

Teaching via \int PdA ??
I may be misinterpreting, as this is not a notation I use...
'Presumably dA' here is the area subtended onto a horizontal plane rather than the area of the boat's surface! [Otherwise you need to multiply by cos(theta) - usual definitions]
I rather fear this method of teaching would cause even more confusion.

I believe that it is better to stick with an Archimedes-style explanation on this: i.e.
. that if we swap the boat for the "diplaced" water (i.e water that that comes level with the surface with the boat present), the water would be in equilibrium
. that when we make the swap, the forces at the surface do not change
. therefore the force on the boat is just sufficient to support the displaced water.
I've probably not expressed it perfectly, and even the "proper" version may not look mathematical at first sight; however, I believe it to be rigorous

To the extent that gibberish can be rigorous, you have expressed it perfectly.

Thank you so much for your kind vote of confidence. The physical principle as addressed is correct, so maybe I should try another tack:

"Teacher": What would happen if I replaced the boat with fluid that came up to the same level as the surrounding fluid?
"Pupil": The system would be in equilibrium
Teacher: What would the net force on the bottom of the new fluid be?
Pupil: Exactly sufficient to support it - therefore equal to the weight of the fluid that "replaced" the boat.
Archimedes: So what would the force be on the boat if I now substitute the boat for this fluid?
Pupil: the same as the force on the fluid - so equal to the weight of the fluid.

Better?

Of course you have to include the angle corrections to get the net force on a rigid body. Landau and Lifshitz is not my favorite but pretty standard physics treatment for hydro. Your treatment is a good one. If you break it up into vertical columns, it becomes the differential equivalent of the one I gave.

Depending on the origins, I believe that the evaluation of dot-products (or tensor arrangements) using cosines or projections is either
. treated as axiomatic or
. it is derived using the concepts of pressure and the equilibrium of (for example) a fully immersed right-triangle right-prism in a gravitation-free enviroment.

The axiomatic arrangement relies on a combination of self-consistency and consistency with observation. We can base calculations on this, but I would not regard it as explanatory.

To my mind, calculations based on the physical derivation are even more prone to confusion than the direct method - but clearly we need a better wording than the one I dragged out.
I think Archimedes would have used the Aristotle-style (Q&A) technique - I'll have another try..

P.S. I see you've been off-topicked now as well - it seems that someone thinks that knowing what constitutes displaced water is off topic

Somebody with Tourette's is on an off-topic frenzy. Maybe it is time they justify their choices. Details of the magnitude of the buoyancy is relevant to the question. Different perspectives on deriving the answer are also such. It is interesting that the wrong answers above are not off-topiced.

The floating boat displaces water to the extent of its weight, thereby raising the water level. When sunk, it displaces water according to its volume. Being denser than water, the sunken aluminum will displace less water than the floating one, thereby lowering the water level, and thus the pressure at the bottom of the vessel. Note that the WEIGHT of the tank with its sunken aluminum foil boat will increase sightly due to the lowered center of gravity, putting the center of mass closer to the center of the earth.

Good thinking. However...
For a 200-mm drop that I think that the weight increase would be about one part in ten-million of the weight of the aluminium.
On the other hand, if the inside of the boat that is below water level is on average 1.4-mm below the water surface, the expansion of the air that provided the buoyancy would be roughly the same. This is due to reduction in the air pressure of the the air previously in the boat, as it is raised to fill the difference between the old and new surface heights.
That would make it too close to call - unless of course a significant amount of air was sunk with the boat.

Hmmm. Thanks. Didn't think about the air column. So, it seems we would need some numbers: initial water depth (it was a bathtub, I think, so we could figure that approximately), whether we are at STP, altitude. But I'd like to do this without arithmetic, a la W. W. Sawyers little puzzle with the wine and the water. P.S. We live on a sailboat, so it's all the more interesting.

A canvas sail sounds complex - it will doubtless entrap air and also change its residual density as it sinks (all to be avoided in your case).

On the other hand, any superstructure above water level will further increase the effective height change of the air.

since this is an open-system, pressure will remain the same

The pressure will decrease because of lower water level.

The floating boat displaces its own weight of water, so when submerged any increase or decrease in water volume in receptical will be equivelent to the mass of the material used in the boats construction. This is given away by the evidence of the sinking boat which indicates a higher specific gravity than water so the volume displaced will relate to that higher specific gravity, when submerged the volume of that mass would displace a reduced amount of water due to its higher density per unit of volume.

So the answer is the water level would decrease and so the pressure at the bottom of the tub would decrease

Assume that the pressure of the water is evenly distributed over the full bottom of the bathtub i.e. the depth is a constant. The weight of the water plus the foil boat remains the same whether the boat is floating or has sunk. So therefor the pressure at the bottom of the bathtub will remain constant.

What I don't know is if a high pressure weather system moves in will this increase in atmospheric pressure increase the pressure at the bottom of the tub. My gut feel is that it won't as the atmospheric pressure will be equal in every which direction.

Who knows?