To Alaska and back (Hopefully!)

Howdy!

I know it's been a while since I've posted here but I've just been very busy.

I recently went on a voyage by ship from Washington to Alaska.

It was over 2500 miles long (about 4000km) and it took nearly a couple weeks.

I gotta say, the earth sure doesn't seem flat when you are navigating on long journeys.

(And yes I drove the boat some of the way when Captain and first mate needed a break!)

The weather has been lovely, and it never gets dark this time of year - you get a nice sunset which then turns into a nice sunrise - no night here!

Sorry to those of you who think Alaska and the north is around the outside of the flat earth - the days are just way too long here!

Anyway I'm supposed to be flying home soon (In a week or so?) - are there any particular things I should check for on the way home?

I have the theodolite app for starters. I'm also thinking of videoing ocean and land masses moving below the plane to check GPS/groundspeed to see if they match up or not.

(Of course there's only a small chance I get a window seat.)

Anyway, let me know any ideas I can do to help show the shape of the earth.

I'll try to check back here before taking to the air and see what I can learn!

Can you provide some photos and online videos prove the earth's being not flat during your travel, please?

Can you provide some photos and online videos prove the earth's being not flat during your travel, please?

to wise
What good would photos or videos be to you ? Don't all FE' discount photos as "fakes" or "photoshopped"
and of no value for evidence ?

to Tom Foolery
My suggestion :
If you are on the boat :
Take some photos and/or videos  showing the horizon and what you see when you pass out of sight or come in sight of land at various distances from the land at various distances of the height of the observer.
This would be proof of the curvature of the earth and proof the earth is not flat, but is a globe ?

Quote from: wise on June 27, 2019, 11:13:19 AM

Can you provide some photos and online videos prove the earth's being not flat during your travel, please?

to wise
What good would photos or videos be to you ? Don't all FE' discount photos as "fakes" or "photoshopped"
and of no value for evidence ?

I guess I can get an amateur real photo or video has high definition. The point here, how can he get its being round on a ship? This is BS. It is almost impossible fake the world with a live video "on the earth". Manipulation is easy because there is no space between the world and the so-called iss in an iss station. however, for an object moving in contact over the sea, this manipulation is almost impossible. I think we can easily understand that this is real or not. Any other BS suggestion?

Try to look at the stars as much as you can. Study the texture/properties of the light and shine or maybe the overall picture...any qualities you can pinpoint or stand out to you. And as many as you can. Then if you ever get south, you can compare what you've seen: if the sothern stars have the same qualities. I missed a great opportunity to do this near the equator. It seemed unrealistic when i was under those stars, but i wish i would've tried harder. You could also try taking photos, i imagine most smart phones have color filters built in, this might yield some results if compared with photos taken in the south.

Are the stars shining much at night?

Quote from: Googleotomy on June 27, 2019, 11:22:13 AM

Quote from: wise on June 27, 2019, 11:13:19 AM

Can you provide some photos and online videos prove the earth's being not flat during your travel, please?

to wise
What good would photos or videos be to you ? Don't all FE' discount photos as "fakes" or "photoshopped"
and of no value for evidence ?

I guess I can get an amateur real photo or video has high definition. The point here, how can he get its being round on a ship? This is BS. It is almost impossible fake the world with a live video "on the earth". Manipulation is easy because there is no space between the world and the so-called iss in an iss station. however, for an object moving in contact over the sea, this manipulation is almost impossible. I think we can easily understand that this is real or not. Any other BS suggestion?

The photos and/or  videos of the horizon is one of the most positive ways of  showing the evidence of the curvature of the earth.
I'm not even sure you would even see the horizon if you were on a ship at sea if the earth was flat..
I think some of Rowbotham's writings say that you would just see (Quote) '' An indistinct blur that fades away in the distance.''

Quote from: wise on June 27, 2019, 12:09:00 PM

Quote from: Googleotomy on June 27, 2019, 11:22:13 AM

Quote from: wise on June 27, 2019, 11:13:19 AM

Can you provide some photos and online videos prove the earth's being not flat during your travel, please?

to wise
What good would photos or videos be to you ? Don't all FE' discount photos as "fakes" or "photoshopped"
and of no value for evidence ?

I guess I can get an amateur real photo or video has high definition. The point here, how can he get its being round on a ship? This is BS. It is almost impossible fake the world with a live video "on the earth". Manipulation is easy because there is no space between the world and the so-called iss in an iss station. however, for an object moving in contact over the sea, this manipulation is almost impossible. I think we can easily understand that this is real or not. Any other BS suggestion?

The photos and/or  videos of the horizon is one of the most positive ways of  showing the evidence of the curvature of the earth.
I'm not even sure you would even see the horizon if you were on a ship at sea if the earth was flat..
I think some of Rowbotham's writings say that you would just see (Quote) '' An indistinct blur that fades away in the distance.''

the disappearance of an object behind the waves at sea is a simple perspective problem and the distant object disappears behind the nearby waves. this has nothing to do with the so called curvature of the earth.

I guess I can get an amateur real photo or video has high definition

And the problem with that is you have no intention of ever accepting a photo or video that shows you are wrong to be real.
Instead you will just dismiss it as fake.

this manipulation is almost impossible.

Or, it isn't manipulation at all, because Earth isn't flat.
You are yet to show any manipulation.

the disappearance of an object behind the waves at sea is a simple perspective problem

Only if you are below the waves. If you aren't, then it isn't a perspective problem.
The vast majority of cases presented are where both the viewer and the object are well above the waves.
This means that perspective can't be the reason. The only reason left is the curvature of Earth.

Quote from: Googleotomy on June 27, 2019, 12:45:24 PM

Quote from: wise on June 27, 2019, 12:09:00 PM

Quote from: Googleotomy on June 27, 2019, 11:22:13 AM

Quote from: wise on June 27, 2019, 11:13:19 AM

Can you provide some photos and online videos prove the earth's being not flat during your travel, please?

to wise
What good would photos or videos be to you ? Don't all FE' discount photos as "fakes" or "photoshopped"
and of no value for evidence ?

I guess I can get an amateur real photo or video has high definition. The point here, how can he get its being round on a ship? This is BS. It is almost impossible fake the world with a live video "on the earth". Manipulation is easy because there is no space between the world and the so-called iss in an iss station. however, for an object moving in contact over the sea, this manipulation is almost impossible. I think we can easily understand that this is real or not. Any other BS suggestion?

The photos and/or  videos of the horizon is one of the most positive ways of  showing the evidence of the curvature of the earth.
I'm not even sure you would even see the horizon if you were on a ship at sea if the earth was flat..
I think some of Rowbotham's writings say that you would just see (Quote) '' An indistinct blur that fades away in the distance.''

the disappearance of an object behind the waves at sea is a simple perspective problem and the distant object disappears behind the nearby waves. this has nothing to do with the so called curvature of the earth.

How high are waves?

Quote from: Googleotomy on June 27, 2019, 12:45:24 PM

Quote from: wise on June 27, 2019, 12:09:00 PM

Quote from: Googleotomy on June 27, 2019, 11:22:13 AM

Quote from: wise on June 27, 2019, 11:13:19 AM

Can you provide some photos and online videos prove the earth's being not flat during your travel, please?

to wise
What good would photos or videos be to you ? Don't all FE' discount photos as "fakes" or "photoshopped"
and of no value for evidence ?

I guess I can get an amateur real photo or video has high definition. The point here, how can he get its being round on a ship? This is BS. It is almost impossible fake the world with a live video "on the earth". Manipulation is easy because there is no space between the world and the so-called iss in an iss station. however, for an object moving in contact over the sea, this manipulation is almost impossible. I think we can easily understand that this is real or not. Any other BS suggestion?

The photos and/or  videos of the horizon is one of the most positive ways of  showing the evidence of the curvature of the earth.
I'm not even sure you would even see the horizon if you were on a ship at sea if the earth was flat..
I think some of Rowbotham's writings say that you would just see (Quote) '' An indistinct blur that fades away in the distance.''

the disappearance of an object behind the waves at sea is a simple perspective problem and the distant object disappears behind the nearby waves. this has nothing to do with the so called curvature of the earth.

I guess the problem with flat earthers is:
If they accepted the reality that the earth wasn't really a flat disc it would be the end of their life.
Sad.
I  guess you probably  have the problem that you never had the opportunity to have ever been to sea or have ever been on a beach and looked out to sea you would know what the horizon really was like.
If you don't believe what I say maybe you should talk to someone else who has ever been to sea or been down to the shore about the horizon.
I still like to think that  you should take this up with some one who could get you straightened out.
Or maybe you just have a warped sense of humor  and are just maybe making up all this nonsense you keep spouting and thinking it's funny ?
Well, most of it is so much nonsense that  maybe some people might think it is funny to them ?

P.S.It's not because of "perspective." It's because the earth is "round." And take that up with someone else  who  is in  the Navy or ever has been . LOL

Quote from: wise on June 27, 2019, 01:22:21 PM

Quote from: Googleotomy on June 27, 2019, 12:45:24 PM

Quote from: wise on June 27, 2019, 12:09:00 PM

Quote from: Googleotomy on June 27, 2019, 11:22:13 AM

Quote from: wise on June 27, 2019, 11:13:19 AM

Can you provide some photos and online videos prove the earth's being not flat during your travel, please?

to wise
What good would photos or videos be to you ? Don't all FE' discount photos as "fakes" or "photoshopped"
and of no value for evidence ?

I guess I can get an amateur real photo or video has high definition. The point here, how can he get its being round on a ship? This is BS. It is almost impossible fake the world with a live video "on the earth". Manipulation is easy because there is no space between the world and the so-called iss in an iss station. however, for an object moving in contact over the sea, this manipulation is almost impossible. I think we can easily understand that this is real or not. Any other BS suggestion?

The photos and/or  videos of the horizon is one of the most positive ways of  showing the evidence of the curvature of the earth.
I'm not even sure you would even see the horizon if you were on a ship at sea if the earth was flat..
I think some of Rowbotham's writings say that you would just see (Quote) '' An indistinct blur that fades away in the distance.''

the disappearance of an object behind the waves at sea is a simple perspective problem and the distant object disappears behind the nearby waves. this has nothing to do with the so called curvature of the earth.

How high are waves?

Those waves would have to be mighty high to make those ships disappear.
Just another indication that someone has never been to sea.

the disappearance of an object behind the waves at sea is a simple perspective problem and the distant object disappears behind the nearby waves. this has nothing to do with the so called curvature of the earth.

Not if the eyes or camera are higher than the highest wave.

Quote from: wise on June 27, 2019, 01:22:21 PM

the disappearance of an object behind the waves at sea is a simple perspective problem and the distant object disappears behind the nearby waves. this has nothing to do with the so called curvature of the earth.

Not if the eyes or camera are higher than the highest wave.

Nope. Since the angular size isn't linear, although you stay the point heigher than the heighest wave, even so you can't see the objects behind the waves. You should remember impossible problems of geometry but I can remind it. and jackblack has agreed angular size being a reverse trigonometric function. It is not linear you know. You can not see! If you claim you can see so prove it geometrically by chosing heighnesses close and distances far.

Crossing the north is the real deal.

If you go northwards continouesly are you gonna reach North Russia? Or reach the darkness, even get blocked by a solid material AKA Dome?
I believe the latter.

Prove me wrong. 

Correct me if I'm wrong.
But......
According the most accepted '' Flat Earth Model ''
(Which is really the Unipolar or North Polar Azimuthal Equidistant Projection of the Globe.)
It looks as if you would go north, around Alaska and Siberia to go by ocean from the west coast  of the USA to Japan ?
Example: San Diego, California, USA to Yokosuka , Japan

Crossing the north is the real deal.

If you go northwards continouesly are you gonna reach North Russia? Or reach the darkness, even get blocked by a solid material AKA Dome?
I believe the latter.

Prove me wrong. 

Prove yourself right.

Quote from: wise on June 27, 2019, 01:22:21 PM

the disappearance of an object behind the waves at sea is a simple perspective problem and the distant object disappears behind the nearby waves. this has nothing to do with the so called curvature of the earth.

Not if the eyes or camera are higher than the highest wave.

On a clear calm day at sea the radar or a lookout in a crow's nest might be more than at least 50 to 100  feet above the sea and still see ships passing out of view as has been shown as they pass over and beyond the horizon.

Crossing the north is the real deal.

If you go northwards continouesly are you gonna reach North Russia? Or reach the darkness, even get blocked by a solid material AKA Dome?
I believe the latter.

Prove me wrong. 

Ever been there yourself? These people have and I'd trust their navigation more than your map:

Who Was the First Person to Reach the North Pole? | National Geographic


First Flight Over North Pole (1926), British Pathé


Solo Flight over the North Pole, Harry Anderson.



Very Rare Three Submarines Surfaced In The Arctic At Once For ICEX 2018 (and a Canadian Ski-do ) The Joint Forces Channel


Voyage to North Pole on Nuclear Icebreaker '50 Years of Victory'/50 Лет Победы by tletter

Nope.

Angular size wont save you here.
If both you and the object are above the waves then they can't get in the way of your view.
To claim otherwise is akin to claiming you would never be able to see anything because your big toe is larger in angular size than the objects you are looking at as it is so close to you.

You are the one making the insane claims, so the burden of proof is on you.
Explain how something below you obstructs the view to something above it.

If you really want some images, I made some quite a while ago, and no, I'm not going to waste my time uploading them on some site of your choosing just for you to dismiss them.

This is an image of a small wave in front of you, but below you. Notice that it only obstructs a small area, with you being able to see the water far enough behind it.
It isn't going to be able to block the bottom of a distant object.
Here is another example:

On the left shows what it is in reality, with a 1m tall wave, with its crest 2 m below you, with a 10 m tall building at 10 km.
On the right you have what it looks like in terms of angles. Notice the angle measurement used is quite small, only an arcsecond. The wave starts off quite well below you, and its angular height shrinks as it gets further and further away.
This only hides the section of the building which is actually at the wave or below.

In order to have the waves obstruct the view to a distant object you would need something like this:

This is where the angular size magically remains constant and instead the waves magically grow to be 10 m tall to obstruct the 10 m tall building.

Again, angular size being an inverse trigonometric function will not help you.
It can explain why putting your thumb just in front of your eye will obstruct your view.
It doesn't explain how a wave well below you can obstruct a distant object nor why objects disappear from the bottom up.

Nope.

Angular size destroyed rabinoz. You won't save rabinoz.

Are you claiming rabinoz does not get mathematic at all?

Quote from: JackBlack on June 27, 2019, 07:31:58 PM

Nope.

Angular size destroyed rabinoz.

No it didn't!  And now read JackBlack's post again!

Quote from: wise on June 27, 2019, 03:32:07 PM

Nope.

Angular size wont save you here.
If both you and the object are above the waves then they can't get in the way of your view.
To claim otherwise is akin to claiming you would never be able to see anything because your big toe is larger in angular size than the objects you are looking at as it is so close to you.

You are the one making the insane claims, so the burden of proof is on you.
Explain how something below you obstructs the view to something above it.

If you really want some images, I made some quite a while ago, and no.

This is an image of a small wave in front of you, but below you. Notice that it only obstructs a small area, with you being able to see the water far enough behind it.
It isn't going to be able to block the bottom of a distant object.
Here is another example:

On the left shows what it is in reality, with a 1m tall wave, with its crest 2 m below you, with a 10 m tall building at 10 km.
On the right you have what it looks like in terms of angles. Notice the angle measurement used is quite small, only an arcsecond. The wave starts off quite well below you, and its angular height shrinks as it gets further and further away.
This only hides the section of the building which is actually at the wave or below.

In order to have the waves obstruct the view to a distant object you would need something like this:

This is where the angular size magically remains constant and instead the waves magically grow to be 10 m tall to obstruct the 10 m tall building.

Again, angular size being an inverse trigonometric function will not help you.
It can explain why putting your thumb just in front of your eye will obstruct your view.
It doesn't explain how a wave well below you can obstruct a distant object nor why objects disappear from the bottom up.

Now you have no excuse for your failure to provide a sensible answer!

Quote from: wise on June 27, 2019, 10:50:45 PM

Quote from: JackBlack on June 27, 2019, 07:31:58 PM

Nope.

Angular size destroyed rabinoz.

No it didn't!  And now read JackBlack's post again!

Quote from: JackBlack on June 27, 2019, 07:31:58 PM

<fakery>

Jackblack's post does not magically save you. So you have admitted you don't know anything about angular size but accept and repeat what jackblack says, right? So why do you write me but don't return your home and leva here to jackblack? Are you leading him and he does all the technical fakery?

And even according to these images it is clear that waves hide the behind building. Try again with adding the the point of observer. Remember it? Where is observer? We are still on this point. See, jackblack can not magically save you.

Quote from: rabinoz on June 27, 2019, 11:23:06 PM

Quote from: wise on June 27, 2019, 10:50:45 PM

Quote from: JackBlack on June 27, 2019, 07:31:58 PM

Nope.

No it didn't!  And now read JackBlack's post again!

Quote from: JackBlack on June 27, 2019, 07:31:58 PM

<fakery>

So you have admitted you don't know anything about angular size.

I admitted no such thing. I just can't be bothered wasting time when you refuse to make any useful post.

Now you post your sensible answer, please. I guess that you are unable to!

PS Angular size for distant objects is close enough to (object height)/(distance to object) in radians, of course!

Quote from: wise on June 27, 2019, 10:50:45 PM

Quote

Nope.

Angular size destroyed rabinoz.

No it didn't!  And now read JackBlack's post again!

Quote from: JackBlack on June 27, 2019, 07:31:58 PM

YES IT DID! Look at here! Don't hide behind jacblack! Come here and fight for your belief!

Quote from: wise on June 28, 2019, 12:40:30 AM

Quote from: rabinoz on June 27, 2019, 11:23:06 PM

Quote from: wise on June 27, 2019, 10:50:45 PM

Quote from: JackBlack on June 27, 2019, 07:31:58 PM

Nope.

No it didn't!  And now read JackBlack's post again!

Quote from: JackBlack on June 27, 2019, 07:31:58 PM

<fakery>

So you have admitted you don't know anything about angular size.

I admitted no such thing. I just can't be bothered wasting time when you refuse to make any useful post.

Now you post your sensible answer, please. I guess that you are unable to!

PS Angular size for distant objects is close enough to (object height)/(distance to object) in radians, of course!

Your telling "close enough" does not magically make them close enough. It mathematically has hidden behind the wave. Ship is like you and jackblack is like wave. ship hidden behind wave and you hidden behind jack. 

Source: Impossible problems of geometry









rabinoz please prove your objections here.

Are you leading him and he does all the technical fakery?

You not being able to comprehend something doesn't make it fakery.
You not liking something because it shows you are wrong doesn't make it fakery.
Me responding when you were addressing someone else doesn't make it fakery.

You have provided absolutely no reason for anyone to think it is fakery.
This is based upon the very real facts of how angular size and angular positions works.

And even according to these images it is clear that waves hide the behind building.

No. Not even close.
The only time that it showed the building obscured by the waves was when the waves magically grew to the size of the building.
The point of the observer is quite clear in my graphs. That is the point where height w.r.t. eye=0, and where the angular position=0.

Appealing to angular size will not save you. It cannot explain why objects disappear from the bottom up.

YES IT DID! Look at here! Don't hide behind jacblack! Come here and fight for your belief!

No it didn't.
Again, refer to any of the images I provided.
It is somewhat difficult to tell exactly where you have gone wrong due to the many errors.
You seem to be trying to show the physical size while also trying to show the angular size.
In effect, it seems like you are determining the angular size of the object, and then pretending that is the physical size to then pretend you can't see the object.

You can either entirely work on physical size, or entirely work in angular size and position. You can't just use both.
If you want to use the former, that 10 m tall building needs to be 10 m tall, not a tiny fraction of that.
If you want to use the latter, then you need to ensure you have your reference for angular position correct.
It isn't the ground.
It is eye level (which can technically be in any direction, but for simplicity, level with the ground, or on a line directly to your target works).
That means the ground isn't shown well below your feet remaining there.
It starts at a position of 90 degrees down, and then rapidly comes up to be quite close to 0 and getting closer and closer to 0 as it gets further away (assuming Earth is flat).

That is what was shown in this image:

You have an observer at a height of 3 m above the bottom of the wave and 2 m above the top of the wave (so the wave height is 1 m), and you have a building that is 10 m tall.
As the wave remains below the eye, it remains at a negative angular position.
There is no way for it to obstruct the view to the building with a positive angular position.
The best you get is the bottom 1 m of the building being obstructed by a wave right on it.

The only one being destroyed here is you.

(and there is nothing impossible about your geometry problem, just mistakes by you and you not bothering to figure out integrals).

<BS>

Writing same BS does not magically prove anything. Look mister; I draw it with the language you can get, but rabinoz can not. 



Here:

J: Your lenght is J. I was thinking R but there is a possiblity you confuse on radius, so J is better.
X: Your distance to a wave
h: Height of wave
L: Distance to ship
H: Heigt of ship.

Is there anything you can not understand? If you verify it is not so we can continue.

Lets assign them some values. You claim during you are heing enough than wave so you see an object heigher than wave, right? Inother say you claim this:



Lets make an example overlap with this shape.

J: 2 metres
X: 10 metres
h: 1,90 metres
L: 10 000 metres
H: 2,00 metres.

Can you see this normally?



Yes, you can.

So; lets calculate:

Appearent size of ship: h: 2 metre, L: 10 000 metre >> a= arctan 2/10.000 = 0.01145916 degrees

Appearent size of wave, ie here:



Appearent size of wave ~= arctan (1,90/10) = 10,7579

Apperant distance to wave: arctan (2 / 10) = 11,30 degrees;



apperant distance of ship from wave to ship, ie this one:



Sum of apperant distance, ie here:



Sum of apperant distance: arccotan (2/10000) = 90-arctan (2/10000) = 90- 0.01145916 = 89,986 degrees

appearent distance from wave to ship: 89,986 - 11,30 = 78,69 degrees

The perception of the length of an object is perceived in relation to the angle, not the actual length. so let's write angle values ​​instead of lengths.

So; lets show appearent sizes in drawing:



As we see that, we need to be on a point more and more higher than the wave between we and ship to see the further ship.

we can calculate the angular equivalent of this height in proportion. when you increase the height, this causes minor changes in angles, and you must always go higher than the middle wave to see the distant object. the higher the object, the higher the height you need to go.

We have calculated the appearent sizes depend on you stay 2 metre high. Actually it has nothing in this drawing. It just an affect reduces or increases the angular sizes. But it is not so much greater affect. The important points here the waves hiding the ship.

I know you're surprised how I put together different concepts in this way. This is for you to understand. I hope you understand.

Get it rabinoz?

Quote from: rabinoz on June 27, 2019, 11:23:06 PM

Quote from: wise on June 27, 2019, 10:50:45 PM

So you have admitted you don't know anything about angular size.

I admitted no such thing. I just can't be bothered wasting time when you refuse to make any useful post.

Now you post your sensible answer, please. I guess that you are unable to!

PS Angular size for distant objects is close enough to (object height)/(distance to object) in radians, of course!

rabinoz please prove your objections here.

That diagram of yours is completely incorrect!
In tnat the side view the wave and ship should retain their original size. Your diagram has applied perspective twice.

Read what Samuel Birley Rowbotham wrote on angular size:

Zetetic Astronomy, by 'Parallax' at sacred-texts.com CHAPTER XIV.
"The smallest angle under which an object can be seen is upon an average, for different sights, the sixtieth part of a degree, or one minute in space; so that when an object is removed from the eye 3000 times its own diameter, it will only just be distinguishable; consequently the greatest distance at which we can behold an object like a shilling of an inch in diameter, is 3000 inches or 250 feet."

The above may be called the law of perspective. It may be given in more formal language, as the following:. when any object or any part thereof is so far removed that its greatest diameter subtends at the eye of the observer, an angle of one minute or less of a degree, it is no longer visible.

The angular height of an object is simply (height of object)/(distance to object) radians - take it or leave it.

It is just as shown in JackBlack's diagram:

That means the ground isn't shown well below your feet remaining there.
It starts at a position of 90 degrees down, and then rapidly comes up to be quite close to 0 and getting closer and closer to 0 as it gets further away (assuming Earth is flat).

That is what was shown in this image:

You have an observer at a height of 3 m above the bottom of the wave and 2 m above the top of the wave (so the wave height is 1 m), and you have a building that is 10 m tall.
As the wave remains below the eye, it remains at a negative angular position.
There is no way for it to obstruct the view to the building with a positive angular position.
The best you get is the bottom 1 m of the building being obstructed by a wave right on it.

The only one being destroyed here is you.

(and there is nothing impossible about your geometry problem, just mistakes by you and you not bothering to figure out integrals).

If you don't like it tough, but you are still wrong.

Writing same BS does not magically prove anything. Look mister; I draw it with the language you can get, but rabinoz can not. 

Yes, that is right. Writing the same BS doesn't magically prove anything.
You posting the same BS won't magically make you correct,

Again, you either deal with angular positions, or real positions.
Pick one and make your diagram.

If you want to go through the math you are making it much harder than it needs to be with all the extra angles (and likely loads of mistakes thrown in).
All you need is their angular position and angular heights.

You have your eye-level, this is your reference.
The important thing is how far below or above something is to your eye-height (both physical and angular height).
So with your eye height at 2 m, everything else needs to drop down 2 m.
This means the relevant height of the wave isn't 1.9 m. It is 0.1 m below your eye level, i.e. h-J.
At a distance of 10 m this results in an angular height of atan((h-J)/X)=atan(-0.1/10)=-0.5729 degrees.
This appears quite small, but as a comparison, the sun is roughly 0.5 degrees.

We can then easily equate this angular height with a physical height much further away.
For example, at a distance of L, it would be L*tan(-0.572938698), so for L=10 000 m, you would be able to see all the way down to ~100 m below eye level.
This can also be done with similar triangles, or by using slopes.
If over 10 m, it drops 0.1 m, then over 10000 m it will drop 10000*(0.1/10) m=100 m
This means this wave would block only object 100 m below eye level.

Or to put it in perspective of ground level/sea level, it would be 98 m below that. So the ground nearer the object will be far more important.

Again, this matches what I have already provided.

If you really need to get all the angles in, then the size of the wave you have calculated is only an approximation.
To do it correctly, you would need to find a difference in angles or use the cosine rule with additional lengths.
The angular size of the wave would be atan(2/10)-atan(0.1/10). This is 10.7370 degrees, not the 10.7579 you show.
To find the angular distance to the wave (from straight down), you don't use atan(2/10). You use atan(10/2). This is 78.6901 degrees.
atan(2/10) will give you the distance to the bottom, from straight out. That would be the 11.3099 degrees you calculated. So placing it at the bottom is either wrong or very deceptive.
This means the top of the wave would be 89.4271 degrees from straight down or 0.5729 degrees down from straight out.

You did get the apparent height of the ship correct. That is because you actually have the right angle triangle needed.
So it would be atan(2/10000)=0.0115 degrees.
But you got the angular distance to the ship wrong from the wave. Completely wrong.
The angle you started finding (the sum) is the angle from straight down out to the ship.
That means when it comes time to subtract the angle, you need to subtract the one that is from straight down for the wave as well.
So that means what you actually want is (90-atan(2/10000))-atan(10/2)=11.2985 degrees.
This is larger than the angular size of the wave.
That means the bottom of the distant ship will be able the wave.

Even using the numbers you provided you end up with the bottom of the ship above the wave.
So no, what we actually see is that the wave does not block the view to the ship. The ship should still be clearly visible, even with us only 10 cm above the top of the wave.

And no, our height of 2 m is VERY important. It is important because it places us above the wave.

Also, it is the height of the wave which determines how high you need to go, not the height of the object.

Edit: fixed a typo.

That diagram of yours is completely incorrect!

Your childishly claiming its being wrong is completely a wrong estimation. I want to remind the issue and continue with calculation:



This is the ship we want to see. Logically we have to see it. But practically we can not.

Short of calculations:

Appearent size of ship: h: 2 metre, L: 10 000 metre >> a= arctan 2/10.000 = 0.01145916 degrees
Appearent size of wave ~= arctan (1,90/10) = 10,7579
Apperant distance of wave: arctan (2 / 10) = 11,30 degrees;
Sum of apperant distance: arccotan (2/10000) = 90-arctan (2/10000) = 90- 0.01145916 = 89,986 degrees
appearent distance from wave to ship: 89,986 - 11,30 = 78,69 degrees

Now we have arrived this drawing:



We have to assign an angular size to our altitude comperative to waves angular size. Lets do it:

Firsty, we stay about paralel to wave and our angular size have to just a bit bigger than it. So, bigger part of angular size of our altitude is the angular size of wave meanwhile, ie 10,7579.



We can caltulate the remained part as follow;

Now we have to use real distances to calculate the angle shown in diagram:



Now. Lets calculate our total angular size comperatively the angular size of wave: 10,7579 + 0,572 = 11,32

Lets put it in the angular size drawing:



Now let's draw this shape in scaled on autocad. and so let's see if we can see a ship at this height and distance.



As we see that, we can not see the ship from this heigh. It stays hidden behind the wave.

Get it rabinoz? The name of Rowbotham can not magically save you.