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Need help for calculating the height of an object

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posted on Sep, 2 2018 @ 07:15 AM
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a reply to: pointessa

Because at the distance you have in order to be to see the Earth being round it is difficult to detect the oval shape. In other words, the Earth is far more 'round' than it is 'oval'.

Think about it this way, if you had a giant square box with rounded corners, up close it would look like a box with rounded corners, but if you got dozens of miles away from this box it would look square and you wouldn't be able to perceive the rounded corners.


edit on 9/2/2018 by Flyingclaydisk because: (no reason given)




posted on Sep, 2 2018 @ 07:18 AM
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Use a ruler? Hope it helps.



posted on Sep, 2 2018 @ 07:20 AM
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originally posted by: leopayaso1987

...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.



The results you'd expect to see in your experiment can be measured today by studying tall bridges, for example;

Verrazano-Narrows Bridge


Because of the height of the towers (693 ft or 211 m) and their distance apart (4,260 ft or 1,298 m), the curvature of the Earth's surface had to be taken into account when designing the bridge—the towers are 1 5⁄8 in (41.275 mm) farther apart at their tops than at their bases; they are not parallel to each other. The bridge's two towers are the tallest structures in New York City outside of Manhattan.





posted on Sep, 2 2018 @ 07:47 AM
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a reply to: leopayaso1987

Have you taken into account that light bends around heavy gravity objects.. like earth



posted on Sep, 2 2018 @ 07:50 AM
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a reply to: Spacespider

The Earth doesn't have enough mass to appreciably bend light. Any object with mass affects light, but not in a way that humans can perceive unless we're talking about incredibly massive objects like stars.



posted on Sep, 2 2018 @ 07:56 AM
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a reply to: weirdguy

A ruler is actually a pretty good example...

If you take a ruler and set it atop a round beach ball and look at it from the side, the ruler only touches the beach ball at one point (the point of tangency). The left end of the ruler is the height of someone's eyes looking at the horizon, and the point of tangency is the horizon. Anywhere along the right side of the ruler to the right of the point of tangency is the measure of how far something is below the horizon.



posted on Sep, 2 2018 @ 07:58 AM
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a reply to: Spacespider
Why would I take that in consideration for my experiment?



posted on Sep, 2 2018 @ 08:01 AM
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a reply to: Flyingclaydisk
I understand what you’re saying. I just wanted to know,in theory how to calculate the height.



posted on Sep, 2 2018 @ 08:05 AM
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a reply to: pointessa
I think you are right,I know realistically the variables would be one to many to get a good result. I still want to know how to theoretically claculate the height



posted on Sep, 2 2018 @ 08:17 AM
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a reply to: leopayaso1987

I provided you a calculator to do so.

It's in the link "HERE"



posted on Sep, 2 2018 @ 09:15 AM
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a reply to: leopayaso1987

This was already solved using a laser sight and a pole on a boat. 1 mile was something like 6' difference and 3 miles was something like 18-24'. I think it was on the show made by the secondary cast mythbusters.




posted on Sep, 2 2018 @ 09:30 AM
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just for entertyainment ,

the diffrential of distance travvellled for a complete circumnavigation of the globe for :

bottom sensor @ 5m above the water line

and " top sensor " @

100m = 314m

for top sensor @ 500m

its : 1570m

BTY this is BASIC maths

i cannot think of a sensor system [ independant ] that can measure to that accuracy while being sent on a circumnavvigation of the globe

EDIT [ for clarification ] my reference to circumnavigation = a THEORETICAL path that ignores land masses . in reality - the longest " leg " possible is AFAIK a partial circumpolar transit - from the ice shelf off antactica - travvelling due north along longitide 170w untill you hit ice in the bering straits [ that should be about 16000 km ]
edit on 2-9-2018 by ignorant_ape because: (no reason given)



posted on Sep, 2 2018 @ 09:32 AM
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a reply to: leopayaso1987

this is your second thread - thats JAQing off - regards a flat earth - you dont post many threads - so with 2 - it raises my suspicions



posted on Sep, 2 2018 @ 09:39 AM
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a reply to: ignorant_ape
First of all thank you for your calculations. I bet it’s basic math,I just wouldn’t know in my head how to formulate that. I’ve been a member for a number of years now. I just don’t post a lot of things,that’s all. I don’t understand why you’d consider me suspicious.



posted on Sep, 2 2018 @ 09:41 AM
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a reply to: ignorant_ape
Could you post a pic of where you calculated this,I would really like to understand what you’re saying but my limited mind just can’t do it.



posted on Sep, 2 2018 @ 11:06 AM
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a reply to: leopayaso1987

It's simple math! Pi x D = Circumference.

Pi ~= 3.1415927 (which is close enough for what you're doing).

D = 41,804,400 (feet) or 7917 .5 (miles) nominaly. Using the equation above, the circumference (C1) = 131,332,397.9 (feet)

Change the D by adding how tall you want your mast to be (say 1,000 feet) and you get (D) = 41,805,400.

Now run the equation (above) again...

C2 = 131,335,539.5

C2 - C1 = 3,141.6 (feet)






edit on 9/2/2018 by Flyingclaydisk because: (no reason given)



posted on Sep, 2 2018 @ 01:57 PM
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a reply to: Flyingclaydisk

And now please explain how you get the diameter. Because as correct as your math is, it does not help him in his quest.



posted on Sep, 2 2018 @ 04:50 PM
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a reply to: verschickter
Thanks,as much as you guys are helping I am also a bit confused. Where does the curvature come into acount in the calculus(does that word exist?,sorry not English,I’m trying). I mean,I can just Google the distances and all of that in order to get the numers. Can someone explain it to me in terms like an equation where we have only one unknown that being what I would like to find out: the height of the pole. When I thought of this experiment,in my mind I assumed that the greater the distance travelled the shorter the mast would have to be. I thought something like this: So,I have the distance(doesn’t have to be a real route,we can think of any given one, and then I had the rate of curvature,and all other numbers if I needed I could Google or something. So I visualized something along these lines: 1+x=4 where “x” is the height of the mast.
Relax,I know 1+x=4 is not the equation,just trying to make you guys understand how I thought this through.



posted on Sep, 2 2018 @ 04:52 PM
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a reply to: verschickter
You perfectly understood my struggle



posted on Sep, 3 2018 @ 11:39 AM
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a reply to: leopayaso1987
Look into Eratosthenes and what he did in egypt.

-That´s how you can work out the diameter.
-If you know the diameter you can calculate the circumference.
-Stick a pole into the ground at the shore line for example 2m high.
-drive out and measure the distanceit takes you to barely see the pole anymore from 2m above sea level.

The line of sight is your tangent. From there it´s simple math.

Hardest part will be finding a relative flat plane. You could take a barometer to get the height above sea level and use it as an adjusting parameter in your calculation. But then you need to get to the point where your line of sight tangents (is that the correct english word??) the earth surface and also take a measurement there.

Because height above ground on land is relative to the ground and not sea level.

Either way, you have to work it out step by step and you need to know the distance between you and the thing you observe while increasing the distance.







 
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