Need help for calculating the height of an object

Hi guys. Lately I have seen some videos regarding Flat Earth. I thought of an experiment that could settle both sides. Even if maybe realistically it can’t be done I still belive it’s nice to entertain the idea.
Let’s say we have a boat sailing from the coast of a continent to another(the more the distance the better). Such boat has a tall pole(how tall is what I don’t know) with two motion/distance tracking “sensors” attached to it. One at the bottom and one at the very top. Due to the curvature if Earth,the distance registered by the top “sensor” will be bigger than the one registered at the bottom.
A simple experiment in theory but I have no idea if it could be reproduced in real life. I assume said pole would ideally vibrate and sway as little as possible and maybe there are other problems with this that my mind hasn’t figured out yet. So if theoretically the boat travels 3000 miles and the formula for the curvature is 8 inches/ mile(squared), how tall must the pole be in order to notice a significant distance between the “sensors”. How do I formulate all this into a mathematical equation?
Looking forward to see your replies.
I apologise in advance if I asked this question on the wrong forum.

a reply to: leopayaso1987

Do yourself a favor...

Don't watch flat earth videos for educational value...

Humor perhaps

Something to think about which will change your equation -formula ....1. Which way is the boat going ? Straight out to sea? Away from land ?.... Which boats don't typically do ....or following up the the coastline ? .... And 2. remember the earth is not exactly round . It's more Egg shaped ..and even then probabaly Not a perfect egg shape ....So which way yourBoat travels will need to Be calculated into your addition .

a reply to: leopayaso1987

I have found that the best way to calculate the height of something is to rule out everything it is not. Then after that you should have the answer you need

Or....or....Hike up a Mountain and observe the curvature of the Earth. Use Arc equation to determine circumference. Or...or...fly in a PLANE....

Even the Ancient Greek knew the Earth was round.

From Plato's Timaeus, 390BC
Wherefore he made the world in the form of a globe, round as from a lathe, having its extremes in every direction equidistant from the centre, the most perfect and the most like itself of all figures; for he considered that the like is infinitely fairer than the unlike. This he finished off, making the surface smooth all around for many reasons; in the first place, because the living being had no need of eyes when there was nothing remaining outside him to be seen; nor of ears when there was nothing to be heard; and there was no surrounding atmosphere to be breathed; nor would there have been any use of organs by the help of which he might receive his food or get rid of what he had already digested, since there was nothing which went from him or came into him: for there was nothing beside him. Of design he was created thus, his own waste providing his own food, and all that he did or suffered taking place in and by himself. For the Creator conceived that a being which was self-sufficient would be far more excellent than one which lacked anything;

a reply to: leopayaso1987

The ancient Greeks had an experiment where they measured the angle of the sun's shadow from a high object at two different locations of known distance at exactly the same time.

At one location, the Sun is directly overhead, so the shadow has zero length. At the other location, the Sun's shadow is of a measurable length. This forms a right-angled triangle. Where the Sun is directly overhead, the angle is 90 degrees. Where the Sun casts a shadow, the angle is calculated from the ratio of the height of the object to the shadow length. Scale that by the distance of the two locations, and they get the curvature of the Earth.

The ancient Greeks had an experiment where they measured the angle of the sun's shadow from a high object at two different locations of known distance at exactly the same time.

At one location, the Sun is directly overhead, so the shadow has zero length. At the other location, the Sun's shadow is of a measurable length. This forms a right-angled triangle. Where the Sun is directly overhead, the angle is 90 degrees. Where the Sun casts a shadow, the angle is calculated from the ratio of the height of the object to the shadow length. Scale that by the distance of the two locations, and they get the curvature of the Earth.

If you go to 8.37 on this video you will see someone filming the Boston skyline from 4.7 miles away...he zooms in and can read the Prudential Building sign. And then using the same camera he films the moon.....200,000+ miles away and then he zooms in with the same camera from the same position.
Can anyone explain what I am seeing?

a reply to: stormcell

Could you help me understand how the ancient greeks measured the sun at the 'exact time'....what instrument did they use to accomplish this task?

a reply to: leopayaso1987

the polite answer is that the spurious movement from the pitch and roll of the vessel - would be infiniltly greater than the difference in true motion between the two

hint the eath has a diameter of 12740km

use the formulat pi D - to get circumferance

and try again adding 100m to the radius

OS - why are you not prepared to just accept all the other evvidences of a spheroid earth

your " plan " is theoretically superficially plausible - but in reality - it just is utterly impractical - and has greater error margins than the " difference " you is attemppting to measure

I dont think it will work.

The waves would have those sensors all over the place.
However, if you set up a GPS antenna on that pole and sailed from the northern hemisphere to the equator, you could plot out your traverse and see in AUTOCAD that your journey was curved.

Or try this trick from your home:

Download RTKLIB of Emlid Reach website.

Then find a city around the equator and search for its local CORS network.
Do the same thing with a Northern Hemisphere city and find its CORS network data, download both.

The CORS station youve chosen will be base station broadcast data.
So its important that the times chosen for the CORS data from both the equator and northern city be relative to each other.
Use GPS time, that will make it easier, because there wont be any time zone conversions.
for right now the GPS Time is:

GPS week: 2017 GPS week mod 1024: 993 GPS seconds of the week: 38027
Ya its a weird time method but its accurate as eff.

Once you get the data, open up RTKLIB and run RTKPOST
Bring in the observation data (.obs file) for either city,
then bring in the observation and navigation files from the other city (.obs and .nav)

The program will process the 2 base stations by analyzing the time it took each base stations broadcast to hit each satellite. It will probably take 5 minutes to fully processes.

Once it does hit PLOT
then from the drop down menu select sat vis and boom!

you will see the path of each satellite that received each stations broadcast times and a plot of their path in 3D.
Meaning you will see how they followed the curvature of the earth.
Problem solved, you proved every flat earther to be a wacko.

Make a youtube video and you will be the spokesperson for round earth because you visually shown these buffoons that the earth is round.

Guys,relax. I never said I belived in Flat Earth. Just wanted to know if this was possible and how tall the pole has to be.

a reply to: Meldionne1

Let’s assume it travels in a straight line from Australia to New Zealand. Let’s just assume for practical reason that Earth is a sphere.

cheers for that

a reply to: ignorant_ape
Who said I didn’t accept the curvature. Nowhere do I mention I belived in Flat Earth.

a reply to: Macenroe82
That was way to complicated for me to understand. But thank you.

a reply to: pointr97

How about communication with light (mirrors)?

a reply to: leopayaso1987

Too answer the question,
you wouldn't need a large pole.

And you would only need 1 RTK sensor (Real Time Kinematic)
You could plot that sensors position in a CAD program after the journey, and see the curvature it traveled over in 3D

a reply to: leopayaso1987

Well your experiment, using a boat's mast, wouldn't reveal much. The mast isn't really high enough to detect an appeciable distance difference. However, your conceptual question is valid.

Here's a better example...

At Sea Level the perceived horizon (by a 6' tall person) is approximately 12 miles away (the actual horizon is only 3 miles away, but we won't get into this geometry), meaning a 50' tall object will disappear over the horizon at 12 miles. However, from a jetliner traveling at 37,000 ASL (Above Sea Level) the horizon is 235 miles away. So, that same 50' tall object would disappear over the horizon at 247 miles away.

HERE is a calculator to help you out.

originally posted by: Meldionne1
Something to think about which will change your equation -formula ....1. Which way is the boat going ? Straight out to sea? Away from land ?.... Which boats don't typically do ....or following up the the coastline ? .... And 2. remember the earth is not exactly round . It's more Egg shaped ..and even then probabaly Not a perfect egg shape ....So which way yourBoat travels will need to Be calculated into your addition .

If the earth is egg shaped why do photos depict it as round?

In response to the OP, the concept sounds fine but I think their would be way to many uncontrollable variables to get a definitive outcome.