What made Pilot Cutter so fast? Just rumors?

[Lee Jerry]

Originally Posted by NewAlexandria

I don't have white paper just yet. I am struggling with calculations and I was not able to prove it to myself that I'm even correct. I need more time. I was only able to calculate that naval architects (mostly Marchaj) are wrong. So for now in order to be on the same page I will define the correct units in physics and what I will use in my calculations. This is very important, because naval architects never transferred physics correctly to naval architecture, and their math, and concepts all fail during calculations and real examples.

Below are the units and simple examples that calculate displacements correctly. I wonder if you will agree with the math. If you agree I will show you that weight and mass are completely 2 different things when it comes to naval architecture. Combining Weight/Mass into same thing works only on LAND, because it is proportional. Once you put an object in water and confuse weight with mass or vise versa you are a magician, but no longer a naval architect. The second an object is in water, or under water, or in space, or in air, or on another planet, or anywhere else in universe your calculations must be accurate. They must observe every force acting upon an object. You can no longer calculate things on land and apply them to water.

If you agree with the above I will introduce a new concept into naval architecture, but not new to physics: APPARENT WEIGHT. One of the concepts that never transferred over into naval architecture, but possibly the most important one. For now, let me post how displacement is calculated in Metric and US/British/Imperial standards.


m = mass
w = weight
p = density
g = gravity
Fb = force of buoyancy
aw = apparent weight
aw = w - Fb



METRIC:
mass in kg
weight in Newtons
water density p = 1000kg/m3

EXAMPLE:
1000k g boat displaces: 1 cubit meter

CALC:
1000*9.8 = 9800N
9.8*1000V = 9800
V = 9800/9800
V = 1



US/BRITISH/IMPERIAL:
mass in slugs
weight in pounds
1 slug = 32.17405lb
water density p = 1.94 slug/ft3
1000kg = 68.5218 slug

EXAMPLE:
68.5218 slug boat displaces: 35.3205 cubit feet

CALC:
68.5218*32.17405 = 2204.6238lbf
32.17405*1.94V = 2204.6238
V = 2204.6238/62.417657
V = 35.320515155


FINAL
1 cubic meter = 35.3205 cubic feet


I wrote the above, because I noticed no one knows how to calculate displacement on the US/BRITISH side. No book has it defined, or I missed it, but I looked for it, and can't find it.

Wow! So much wrong here. I'm not sure I can address / find / remember them all, but let's get started.

What EXACTLY are naval architects (mostly Marchaj) wrong about? Is this just a units thing?

Originally Posted by NewAlexandria

...because naval architects never transferred physics correctly to naval architecture, and their math, and concepts all fail during calculations and real examples.

So is that what happened to the Titanic?

Originally Posted by NewAlexandria

...I will show you that weight and mass are completely 2 different things when it comes to naval architecture.

Yes AND no.

Weight is the force that acts on a mass, so they are not the same. F = ma is not a particularly new concept. But here on Earth, the force of gravity, especially at the surface, varies very little. Therefore, we (generally everyone who is not you) use a standard value for it. In SI units that is 9.80665 m/s^2 or 9.81 m/s^2, and in US customary 32.174 ft/s^2 or 32.2 ft/s^2. Therefore the two (weight and mass) are proportional and that proportion doesn't change.

Furthermore, the slight variation (~0.5%) in the force of gravity from the equator to the poles would apply to both the vessel's weight force and the buoyant force, so the net effect would (likely, I've never actually looked at or even thought at it) be even less. In other words, negligible.

Originally Posted by NewAlexandria

Combining Weight/Mass into same thing works only on LAND, because it is proportional. Once you put an object in water and confuse weight with mass or vise versa you are a magician, but no longer a naval architect. The second an object is in water, or under water, or in space, or in air, or on another planet, or anywhere else in universe your calculations must be accurate. They must observe every force acting upon an object. You can no longer calculate things on land and apply them to water.

Well, we're not in space or on another planet, so as mentioned, the changes are negligible and that amount of accuracy isn't needed on land or water.

Not "every force acting upon an object" is always needed. Small effects are often "ignored." When you step on the scale in the morning do you correct the reading for the atmospheric pressure that day/time, because the buoyant force provided by the air changes slightly?

Originally Posted by NewAlexandria

If you agree with the above I will introduce a new concept into naval architecture, but not new to physics: APPARENT WEIGHT. One of the concepts that never transferred over into naval architecture, but possibly the most important one.

Apparent weight is only applicable to a submerged object, like an anchor or a submarine. For an object floating on the surface, like a boat or ship, the apparent weight is zero.

Originally Posted by NewAlexandria

For now, let me post how displacement is calculated in Metric and US/British/Imperial standards.

You don't/didn't "post" (I assume you mean something like "show" or "demonstrate") how displacement is calculated. You just stated what the mass is only "post" (poorly and sloppily) how to get a weight force from that mass and then into a volume. Again, F = ma. Groundbreaking stuff there...

Furthermore, you seem to be confusing some terms. "Displacement" is a force. It is usually measured / reported in kilograms-force (often shortened to just kilograms) or metric tons, but Newtons are of course acceptable too, in the SI system or pound-force (often shortened to just pounds) or long tons in Imperial system. "Displaced volume" is just that, a volume measurement usually using cubic meters or cubic feet, respectively.

There are a few ways to "calculate" the displacement. If it is a small boat, you could just weigh it (since weight = displacement), but not really a "calculation" I suppose. For larger vessels, you could take freeboard readings and then use the lines plan to get the displaced volume and combine with the specific gravity of the water (it changes with temperature and salinity) to calculate the displacement. Or you could get the displaced volume by placing the vessel in a closed tank or similar and measure the increase in height of the water, the way Archimedes did a couple thousand years ago (although he did with a submerged object) and do the same thing with specific gravity to get to displacement.

Originally Posted by NewAlexandria

m = mass
w = weight
p = density
g = gravity
Fb = force of buoyancy
aw = apparent weight
aw = w - Fb



METRIC:
mass in kg
weight in Newtons
water density p = 1000kg/m3

EXAMPLE:
1000k g boat displaces: 1 cubit meter

CALC:
1000*9.8 = 9800N
9.8*1000V = 9800
V = 9800/9800
V = 1

What's a "cubit meter?" I know (roughly) what a cubit is and I know what a meter is, but not a cubit meter.

What is "N" in the highlighted line? If it's supposed to be the units of Newtons, then why don't the other values have units given for them?

What is "V?" It isn't in your list of abbreviations.

Originally Posted by NewAlexandria

US/BRITISH/IMPERIAL:
mass in slugs
weight in pounds
1 slug = 32.17405lb
water density p = 1.94 slug/ft3
1000kg = 68.5218 slug

EXAMPLE:
68.5218 slug boat displaces: 35.3205 cubit feet

CALC:
68.5218*32.17405 = 2204.6238lbf
32.17405*1.94V = 2204.6238
V = 2204.6238/62.417657
V = 35.320515155

How come the SI calculations only got two significant digits but the Imperial ones get up to seven?

Originally Posted by NewAlexandria

FINAL
1 cubic meter = 35.3205 cubic feet

No.

1 cubic meter = 35.3146667 cubic ft

Originally Posted by NewAlexandria

I wrote the above, because I noticed no one knows how to calculate displacement on the US/BRITISH side. No book has it defined, or I missed it, but I looked for it, and can't find it.

Then you didn't look in the right place(s). I could give you some recommendations.

[Lee Jerry]

Originally Posted by NewAlexandria

You do understand the guy is correct right? Except he does not know how to word it correctly and prove it mathematically. I can, but I'm not starting that discussion until I have it all tested and calculated for everything I need to in order to make this reasonable, and not breakable by any other laws, or formulas. I simply need to make this bullet proof and simple. Otherwise you guys will behave like that pope few hundreds years ago trying to burn that guy on stakes for showing him what revolves around what.

I'm assuming you meant to say "of" not "if". Not picking on you, but for technical purposes.

Be very patient in this thread. This is not about me. I will leave you off with data, charts, theories that will change it all for all of us. I'm not kid of the street.

No, he's not right.

I assume you meant to say "off" not "of." Not picking on you, but for technical purposes.

Originally Posted by NewAlexandria

I know exactly what I'm doing.

So do we...being a troll. And not a particularly good one.

[NewAlexandria]

Originally Posted by Lee Jerry

Wow! So much wrong here.

If you read your own reply you would realize there is nothing wrong. You did not find anything wrong except for missing units in one and not the other and not defining V. My cubic feet to 1 meter was from my abbreviated calculations. I don't look up online defaults if I don't have to. If you looked carefully you would realize I calculated it myself, hence I'm off by 0.0058333. It has no meaning in grand scale of things. You are simply trying to find a problem where there isn't one. You can keep doing it, you are wasting your own time.

Your other assumptions are incorrect btw. NO. Apparent weight of objects in water submerged fully or partially IS NOT ZERO. Apparent Weight is always there as long as there is BUOYANCY. You have incorrect understanding of physics. You just trying to look smart by picking minor issues that have no meaning.

You also keep putting words in my mouth. I never talked about minor differences due to temperature. And knowing that you referred to temperature differences I can tell you simply have no idea how forces work upon objects in air, in water, on land, on Moon, etc.

Apparent Weight is nothing more than actual weight minus buoyancy force. The second there is buoyancy force acting upwards, you have apparent weight ,because your object is no longer without a force acting upon it. Don't believe me? Take a heavy object and slowly submerge it in your bathtub, and tell me if it get's lighter as you keep putting it in and heavier as you remove it, or it is only light when you fully submerge it.

If you can't feel the difference in the example above, go to the beach, stand in water up to your belly and raise one leg. Does she feel lighter than on land? How come she is lighter if you are not fully submerged ? Your upper body is above water. According to you your leg should be as heavy as on land.

And if you can't do both, take a scale, go to the elevator. Put the scale down, read your weight. Press 5th floor, go up and read your weight. You will see what apparent weight means. It has nothing to do with water, fully submerged, or partially.

As far as letters, spellings, etc. I suffer from dyslexia. I speak flew languages, but I don't have a first language. Can't spell anything, but it has no meaning here.

Good night.

[Adelie]

Originally Posted by NewAlexandria

Yes, proving they are wrong was easy, but it has no meaning to me without redefining it correctly. It's easy to destroy it, but rebuilding takes time.

What specifically were they wrong about? Everything?

If you can prove that they are wrong then that should give you hints about how to proceed with proving what is right.

[NewAlexandria]

Originally Posted by Adelie

What specifically were they wrong about? Everything?

If you can prove that they are wrong then that should give you hints about how to proceed with proving what is right.

That is exactly what I'm doing, but the problem with naval architecture for me at least is that first I actually need to learn it myself. Even the wrong way. I knew nothing about it when I started this. So I have some understanding of it's physics now. To the point where I'm able to spot the mistakes. The mistakes happened to be fundamental, and the entire architecture is build on those mistakes. So in order to straighten it all out I need to explain everything from beginning to the end.

I would not be able to do this before Marchaj's book, because other authors were very vague and they did not expand on topics. Marchaj did. He went crazy with numbers and ideas, and that's where I started to see the errors. By page 100 of the book I knew what is wrong. But what I know and what I can proof is 2 different things. So I'm writing out bunch of examples, charts, calculations in order to make this black and white simple. That is hard. I had to restart the calculations 3 times. I got it wrong. I'm doing it on big grid paper manually with a pencil.

I think the problem for me and for everyone is the "water". Boats sit in 2 different densities, with forces acting upon them, but naval architects use calculations from "land" and apply them to water throwing everything off. It's super hard to prove that this way of thinking is wrong, because we all deal with objects on land except for airspace and space missions.

Just read the conversation above, even simple apparent weight is giving people problems. They assume there is a magic line where only submerged items have apparent weight, otherwise it's ZERO. It's like saying airplanes can't take off, because their apparent weight is negative only when they do take off. But before they take off they go from apparent weight being equal their true weight, and then they gradually "lose" the apparent weight to the point the lift force is stronger than that and they fly. Or it's like saying the second you open that door of the capsule on the Moon your weight is different.

Obviously that's not how this works, but explaining that to others is difficult. That's what I'm struggling with. I need simple concrete examples that finish this topic once and for all. Also, what is in those books I'm reading does not define everything, so I actually need to define it. Stability is one. Most important one.

[Lee Jerry]

Originally Posted by NewAlexandria

If you read your own reply you would realize there is nothing wrong. You did not find anything wrong except for missing units in one and not the other and not defining V.

Riiiiight...

Originally Posted by NewAlexandria

My cubic feet to 1 meter was from my abbreviated calculations.

That's a convoluted way to convert cubic meter to cubic feet. It's much easier to simply do the linear conversion cubed - divide m^3 by 0.3048^3 to get ft^3 (or multiply to go the other way).

Originally Posted by NewAlexandria

I don't look up online defaults if I don't have to. If you looked carefully you would realize I calculated it myself, hence I'm off by 0.0058333. It has no meaning in grand scale of things. You are simply trying to find a problem where there isn't one. You can keep doing it, you are wasting your own time.

I'm not trying to look smart. I'm just trying to help you out. Obviously, you neither see nor desire that. But you are probably right, that I am wasting my time.

Originally Posted by NewAlexandria

Your other assumptions are incorrect btw. NO. Apparent weight of objects in water submerged fully or partially IS NOT ZERO.

That's not what I said.

What I said was the apparent weight of a boat floating on the surface is zero. The weight (or displacement) equals the buoyant force. Look at your definition of apparent weight. This thread is about sailboats, isn't it?

Originally Posted by NewAlexandria

Apparent Weight is always there as long as there is BUOYANCY. You have incorrect understanding of physics. You just trying to look smart by picking minor issues that have no meaning.

Apparent weight is always there, whether there is any buoyancy or not. If there is no buoyancy (and no other forces acting), then the apparent weight is simply equal to the weight.

Again, since this is a thread purportedly about sailboats, I'll point you to the analogy of apparent wind.

Originally Posted by NewAlexandria

You also keep putting words in my mouth. I never talked about minor differences due to temperature. And knowing that you referred to temperature differences I can tell you simply have no idea how forces work upon objects in air, in water, on land, on Moon, etc.

I didn't put any words in your mouth. I was simply explaining the conversion from weight to volume of water, and that it is dependent on salinity and temperature. There's nothing there about you saying it. I don't see what the big deal is about a true parenthetical statement.

Originally Posted by NewAlexandria

Apparent Weight is nothing more than actual weight minus buoyancy force.

Well, not exactly. What's the buoyant force in your elevator example below?

Originally Posted by NewAlexandria

The second there is buoyancy force acting upwards, you have apparent weight ,because your object is no longer without a force acting upon it.

Your object was never without a force acting on it. There was at least the force of gravity acting. (There could be others too.) The buoyant force is an additional force, not the first one.

If you're going to prove that a century or more of modern naval architecture is wrong, you should probably try to be more precise.

Originally Posted by NewAlexandria

Don't believe me? Take a heavy object and slowly submerge it in your bathtub, and tell me if it get's lighter as you keep putting it in and heavier as you remove it, or it is only light when you fully submerge it.

If you can't feel the difference in the example above, go to the beach, stand in water up to your belly and raise one leg. Does she feel lighter than on land? How come she is lighter if you are not fully submerged ? Your upper body is above water. According to you your leg should be as heavy as on land.

And if you can't do both, take a scale, go to the elevator. Put the scale down, read your weight. Press 5th floor, go up and read your weight. You will see what apparent weight means. It has nothing to do with water, fully submerged, or partially.

Yes. And none of this has anything to do with a sailboat floating on the surface of the water.

Originally Posted by NewAlexandria

As far as letters, spellings, etc. I suffer from dyslexia. I speak flew languages, but I don't have a first language. Can't spell anything, but it has no meaning here.

That's cool. But you're making my point, which is that maybe you aren't the best person to be commenting on others typos.

[NewAlexandria]

Originally Posted by Lee Jerry

That's not what I said.

Originally Posted by Lee Jerry

Apparent weight is only applicable to a submerged object, like an anchor or a submarine. For an object floating on the surface, like a boat or ship, the apparent weight is zero.

You said a boat or a ship has apparent weight of ZERO. It says above, you wrote that. That is WRONG. Fundamentally wrong. Even authors of the books I read don't assume that.

You also said: submerged objects like anchor or submarine. Those 2 don't float partially that's why you used this as an example.

You need to own your words. You did not understood apparent weights and forces until I explained them to you. You understand them now, which is very good, because I was able to solve the last remaining calculations and I can prove everything with math, physics and disprove important naval architecture assumptions.

It would be good if you could stop distracting, because you are injecting confusion, by being wrong, then saying you meant something else. Just read your own quotes above.

[Lee Jerry]

Originally Posted by NewAlexandria

You said a boat or a ship has apparent weight of ZERO. It says above, you wrote that. That is WRONG. Fundamentally wrong. Even authors of the books I read don't assume that.

Yes, and I'll say it a third time: A boat floating on the surface of the water has zero apparent weight because the weight force is exactly equal to the buoyant force.

If you think this is wrong, then prove it, show it, demonstrate it, explain it - anything but simply state it.


(I expect I know where you are going with this. It has to do with mixing, confusing, conflating statics with dynamics.)

Originally Posted by NewAlexandria

You also said: submerged objects like anchor or submarine. Those 2 don't float partially that's why you used this as an example.

There is no such thing as "float partially." This is (probably) the crux of your confusion. It's like "sort of pregnant." The object either floats or sinks*.


(* There is a third option, where the object is held partially submerged in the water but not yet floating. Like when a boat is launched with a crane. But this is, as they say, an uninteresting case. Or all of those silly examples you listed in your prior post. Therefore, I ignored this case. But I look forward to you showing that it somehow applies to a sailboat.)

Originally Posted by NewAlexandria

You need to own your words. You did not understood apparent weights and forces until I explained them to you. You understand them now, which is very good, because I was able to solve the last remaining calculations and I can prove everything with math, physics and disprove important naval architecture assumptions.

Just because I didn't explain to you chapter and verse at the first mention of it doesn't mean I didn't understand it.

Please stop just saying what you can prove and DO IT.

Originally Posted by NewAlexandria

It would be good if you could stop distracting, because you are injecting confusion, by being wrong, then saying you meant something else. Just read your own quotes above.

Please show me where I was wrong.




And way to ignore everything else I said.

[Bowdrie]

Originally Posted by NewAlexandria

You said a boat or a ship has apparent weight of ZERO. It says above, you wrote that. That is WRONG. Fundamentally wrong. Even authors of the books I read don't assume that.

You're beating a dead horse.
Write/call/text/email, 100 yacht/ship/boat design offices.
Ask them how much "apparent weight" means in their calculations.
The ones who might reply would probably say something to the effect of a line by Rhett Butler, "Frankly my dear, I don't give a d*mn".
You said you want to learn Naval Architecture, if so, sign up at Westlawn or MIT.
If you refuse to believe Archimedes, they'll have good reason to throw you out on your a**.
Do you really believe that the principles discovered/learned/written down about floating objects > advancing to "Naval Architecture" over a couple thousand years are all wrong?
That you can do better?
"Apparent weight, apparent weight, apparent weight", give it a break, it's a nothingburger.
I'll tell you the most important/first and foremost principle of naval architecture.
When all is boiled down to the zero baseline it only takes four words.
The. Ship. Must. Float.

[NewAlexandria]

Archimedes was probably correct, you guys twisted it all wrong. Sit tight I'm preparing a reply to Lee Jerry. It's not my fault you guys all learned it wrong, and now are defending it through words, this is not a dead horse at all. This is trivial, and no advanced concept of naval architecture can be calculated without understanding this. Worse, stability of objects in water is depended on this. That's why this is important to me.


@Lee Jerry, before I make a reply with calculations, you do realize, the answer is in your own words right? You said this:

"weight force is exactly equal to the buoyant force." meaning: Fb = w

Well if that is the case can you tell me what would happen if I added 1 more pound to the boat?
In other words, if I'm adding weight to the boat, am I growing her Buoyancy? Very curious.


I'm going to help a bit:
I have a boat that weights: 464 lbs, her water line volume is 7.44ft3.
According to you her buoyancy force is 464lbs, but the formula for force of buoyancy says something else: Fb = pVg which is 1.94 * 36 * 32.2 = 2248 lbs. So all the sudden our buoyancy is not equal our weight. Why?

[NewAlexandria]

Sorry, I forgot to say that 36 is entire volume of my boat.

You can look at it this simple way looking at waterline:

1.94*7.44*32.2 = 464 <------- this is what you are saying, this is what I agree with as well, but there is a problem, 70% of the boat above that is unaccounted for. How come my weight of the boat weights as much as 30% of my boat all the sudden? Do you see this already or no?

Archimedes would say I never talked about boats, I was talking about objects in water (fully submerged). And if they are not fully submerged you need to do your own calculations.

[Bowdrie]

I'm not trying to be snarky, but is this what you're trying to wrap your mind around concerning "Apparent weight"?
I'll present a scenario; it could be done with 2 boats or any number, I'll chose 4.
Let's imagine that we made 4 fiberglass hulls from the same mold and each one of them weighed exactly 2,900 lbs., as weighed in the shop on the same scale.
With hull #1 we place a 100 lb. block of steel inside the hull.
With hull #2 we glue the 100 lb. block of steel on the outside bottom of the hull, (with weightless glue).
With hull #3 we placed a 100 lb. block of Balsa wood in the hull.
And on hull #4 we glue a 100lb. block of Balsa wood on the outside bottom of the hull, (with weightless glue).
So, set upon the shop scale all 4 hulls weigh the same now at 3,000 lbs.
Now we'll set all the hulls in our giant tank of water.
Will all 4 hulls float at the same waterline?

[NewAlexandria]

I'm not trying to wrap my mind around it, I was distracted, but I liked the idea of solving that also.

To answer your question:

V = weight / mass * density
V = 3000/62.4

48 cubic feet of boat under water. If your ballast on the "outside" takes volume, then waterlines will be different. This was already solved above I think. It does not really define apparent weight problem we were talking about. Underwater, only density of water maters, Balsa or iron has no meaning. That 3000lbs of buoyancy is up to the water line.

Fb = 48 * 32.2 * 1.94
Fb = 3000

You see this? All the sudden the boat weights 3000 pounds like on your scale, but this time only up to the waterline. Where is the rest? Can you calculate it? Or at least acknowledge a problem.

[Bowdrie]

Yes, hulls #1 and 3 will float at the same waterline.
Hull #2 will float a very small amount higher than #1 or#3.
Hull #4 will float a substantial amount higher than the other 3.
In a perverse sort of way, one might say that what was done to hull #4 was to glue-on "buoyancy".
Can I do the calculations, you mean like using Simpsons rules on a lines drawing?
Only with extremely tedious effort.
Should I want to design a boat I would be tempted to do what others have done for centuries, carve a scale half-model and take the lines off.
A mold taken from a half-model at its design waterline, and then weighing the water needed to fill the mold can give a pretty good approximation of the displacement of the full-size boat at that waterline.
Of course, the accuracy of scale counts for much.
One who has a good streak of artistry/vizualation and facile with ships curves and a planimeter is starting from a different position.

[NewAlexandria]

Don't do any calculations. Read my post again. See if you can spot the problem.

Later I will post the drawings and how come the buoyancy force equal boat's weight is only in the waterline section. Fb = pVg.

Is anyone else here going to give this a shot and figure it out before I post?