The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Investigations => Topic started by: totallackey on August 03, 2018, 02:00:18 PM
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Observer witnesses a distance measured along the base to an object is 1700 miles.
Object top is an unknown distance above the earth but can be seen.
A ten foot tall pole (as measured from the baseline to the top of the pole) is situated between the observer so the top of the object can be seen just above the pole. Observer is three feet away from the pole.
Object computes to have a height of roughly 5600 miles above the surface.
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Observer's eye is at toe-level.
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Observer's eye is at toe-level.
So?
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Based on the assumption that the pole extends 10ft above the horizontal eye level of the viewer your maths appears correct. The hypotenuse would lay at 73.3 degrees above horizontal. [ arctan(10/3) = 73.3 degrees ]
For height h
arctan(10 / 3) = arctan(h / 1700) = 73.3 degrees
Therefore 10 / 3 = h / 1700
h = 1700 x 10 / 3
h = 5666.6 miles
(10 feet divided by 3 feet is a ratio , the ratio is multiplied by 1700 miles and gives an answer in miles)
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Observer's eye is at toe-level.
So?
So, as long as you're aware of that and that's the vantage point for aligning the top of the pole with the top of the distant object, no problem.
But if someone is standing 3' from the 10' pole and sighting along the top of the pole to align with the distant object from that standing position, then he needs to account for the height of his eye because it changes the angle and thus the calculation of the height of the distant object.
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Observer's eye is at toe-level.
So?
So, as long as you're aware of that and that's the vantage point for aligning the top of the pole with the top of the distant object, no problem.
But if someone is standing 3' from the 10' pole and sighting along the top of the pole to align with the distant object from that standing position, then he needs to account for the height of his eye because it changes the angle and thus the calculation of the height of the distant object.
Given the requirement the top of the object and the top of the pole must be visible to the observer, your point about the angle of the hypotenuse is moot.
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The angles of the hypotenuse are anything but moot, and calculating the height of the distant object depends on getting the angle right.
Knowing the height of eyelevel when 3' away from the 10' pole (the bottom of which we assume is level with the base of the distant object) when the tops are aligned is key to calculating that angle.
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The angles of the hypotenuse are anything but moot, and calculating the height of the distant object depends on getting the angle right.
Knowing the height of eyelevel when 3' away from the 10' pole (the bottom of which we assume is level with the base of the distant object) when the tops are aligned is key to calculating that angle.
Tell me how the angle of the hypotenuse even figures in the calculation?
The calculation is?
Here is mine...
1700 miles = 8976000 ft
10 foot pole
3 foot distance between pole and observer
8976000/3 = x/10
x=5667 miles in height.
Hypotenuse of the angle c is irrelevant.
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Observer witnesses a distance measured along the base to an object is 1700 miles.
Object top is an unknown distance above the earth but can be seen.
A ten foot tall pole (as measured from the baseline to the top of the pole) is situated between the observer so the top of the object can be seen just above the pole. Observer is three feet away from the pole.
Object computes to have a height of roughly 5600 miles above the surface.
I'm sorry, but what's the question again? The math obviously works, even if it is pretty much a nonsense scenario.
So, what's the point to this little exercise (other than to make fun of anyone who responds)?
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Observer witnesses a distance measured along the base to an object is 1700 miles.
Object top is an unknown distance above the earth but can be seen.
A ten foot tall pole (as measured from the baseline to the top of the pole) is situated between the observer so the top of the object can be seen just above the pole. Observer is three feet away from the pole.
Object computes to have a height of roughly 5600 miles above the surface.
I'm sorry, but what's the question again? The math obviously works, even if it is pretty much a nonsense scenario.
So, what's the point to this little exercise (other than to make fun of anyone who responds)?
The exercise concerns measuring altitude of an observed point above the surface of the earth.
Where do you see or read or otherwise detect ridicule taking place here in this thread?
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Where do you see or read or otherwise detect ridicule taking place here in this thread?
Probably just the general condescension in most of your replies. Also the arrogance of expecting people to spend time responding without any apparent point or purpose.
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The exercise concerns measuring altitude of an observed point above the surface of the earth.
But you've already given the altitude of the observed object in your OP (5600 miles).
So, once more, what's the point?
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The angles of the hypotenuse are anything but moot, and calculating the height of the distant object depends on getting the angle right.
Knowing the height of eyelevel when 3' away from the 10' pole (the bottom of which we assume is level with the base of the distant object) when the tops are aligned is key to calculating that angle.
Tell me how the angle of the hypotenuse even figures in the calculation?
The calculation is?
Here is mine...
1700 miles = 8976000 ft
10 foot pole
3 foot distance between pole and observer
8976000/3 = x/10
x=5667 miles in height.
Hypotenuse of the angle c is irrelevant.
Because your ratio approach assumes the sighting point (eye) is level with the base of the pole. If it's not, then that ratio doesn't work. If you're looking from 3' away but from a height of 5', you need a new ratio because the sides of the triangle are not 3' and 10' anymore. The angle of the hypotenuse changes.
(http://oi67.tinypic.com/2n0kj0k.jpg)
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Where do you see or read or otherwise detect ridicule taking place here in this thread?
Probably just the general condescension in most of your replies. Also the arrogance of expecting people to spend time responding without any apparent point or purpose.
You choose to label my replies as condescending?
Arrogance of expecting people to respond?
I am going to refer you to AR for my full response to your post here.
I made an OP and it is apparent you have ZERO meaningful content to add to the OP.
So do yourself a favor and put me on ignore, okay moran?
That way, you do not need to respond to posts (particularly mine) where you are expected to be able to perform basic math.
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The angles of the hypotenuse are anything but moot, and calculating the height of the distant object depends on getting the angle right.
Knowing the height of eyelevel when 3' away from the 10' pole (the bottom of which we assume is level with the base of the distant object) when the tops are aligned is key to calculating that angle.
Tell me how the angle of the hypotenuse even figures in the calculation?
The calculation is?
Here is mine...
1700 miles = 8976000 ft
10 foot pole
3 foot distance between pole and observer
8976000/3 = x/10
x=5667 miles in height.
Hypotenuse of the angle c is irrelevant.
Because your ratio approach assumes the sighting point (eye) is level with the base of the pole. If it's not, then that ratio doesn't work. If you're looking from 3' away but from a height of 5', you need a new ratio because the sides of the triangle are not 3' and 10' anymore. The angle of the hypotenuse changes.
(http://oi67.tinypic.com/2n0kj0k.jpg)
Once more with clarity.
You take an object in the sky that has an unknown altitude above the surface of the earth, but that object has a known distance as measured along side B (base of the triangle) from the observer to the object.
I gave FIXED and KNOWN measurements.
Why do you insist on changing the FIXED, KNOWN, and OBSERVED parameters.
You cannot demonstrate a single instance of the ratio approach NOT working.
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The exercise concerns measuring altitude of an observed point above the surface of the earth.
But you've already given the altitude of the observed object in your OP (5600 miles).
So, once more, what's the point?
The point is the title of the OP.
If you do not wish to answer the title of the OP, fine.
Go play with the cat clods in your sandbox.
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The exercise concerns measuring altitude of an observed point above the surface of the earth.
But you've already given the altitude of the observed object in your OP (5600 miles).
So, once more, what's the point?
The point is the title of the OP.
If you do not wish to answer the title of the OP, fine.
My answer is that your OP is very poorly presented.
You present 1700 miles as the base of a right triangle. Then you present an observer with an unknown eye level 3 feet away from a 10 foot pole that is the very tip of that 1700 mile right triangle. I contend that you have not provided enough information to properly calcualte that the object in the sky is 5600 miles high.
The object, the top of the pole and the eye line of the observer must line up along the hypotenuse of the right triangle. You must either provide the height of the observer's eye level or completely remove the observer in order to make the problem solvable.
Where do you see or read or otherwise detect ridicule taking place here in this thread?
Well, right here, for one.
Go play with the cat clods in your sandbox.
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I gave FIXED and KNOWN measurements.
Why do you insist on changing the FIXED, KNOWN, and OBSERVED parameters.
I didn't change any values. I provides variables. You can fill in with the fixed, known values.
I thought d was 1700 miles but with your last response this might be how you could fill them in:
d = 1700 miles - y
y = 3 feet
pole height = 10 feet
Sorry, I failed to label pole height.
The ratio you use works for the triangle where observers' eye is at the base of the triangle. It keeps working if you raise that eye but also proportionately reduce distance to the pole.
But if you rise the eye up while keeping distance to pole fixed, you've got a new triangle and new ratio even though all your known values remained fixed.
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The exercise concerns measuring altitude of an observed point above the surface of the earth.
But you've already given the altitude of the observed object in your OP (5600 miles).
So, once more, what's the point?
The point is the title of the OP.
If you do not wish to answer the title of the OP, fine.
My answer is that your OP is very poorly presented.
Okay.
You present 1700 miles as the base of a right triangle.
Discerned that from a poorly written OP.
Then you present an observer with an unknown eye level 3 feet away from a 10 foot pole that is the very tip of that 1700 mile right triangle.
Eye level does not matter.
The height of the pole (10 feet) and the fact the top of the object is visible despite the pole is the only data in question.
I contend that you have not provided enough information to properly calcualte that the object in the sky is 5600 miles high.
Your contention is wrong.
The object, the top of the pole and the eye line of the observer must line up along the hypotenuse of the right triangle. You must either provide the height of the observer's eye level or completely remove the observer in order to make the problem solvable.
Wrong.
All that is necessary is for one to know the top of the pole still allows for the top of the sighted object to remain visible.
Where do you see or read or otherwise detect ridicule taking place here in this thread?
Well, right here, for one.
Go play with the cat clods in your sandbox.
That was written to you AFTER the post you made, repeating your meaningless off-topic and senseless objections to an OP you know provides correct math and a way for anyone to calculate the true altitude of the sun.
Now cease with the off-topic banter and pure, senseless objections to the OP.
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I gave FIXED and KNOWN measurements.
Why do you insist on changing the FIXED, KNOWN, and OBSERVED parameters.
I didn't change any values. I provides variables. You can fill in with the fixed, known values.
I thought d was 1700 miles but with your last response this might be how you could fill them in:
d = 1700 miles - y
y = 3 feet
pole height = 10 feet
Sorry, I failed to label pole height.
The ratio you use works for the triangle where observers' eye is at the base of the triangle. It keeps working if you raise that eye but also proportionately reduce distance to the pole.
But if you rise the eye up while keeping distance to pole fixed, you've got a new triangle and new ratio even though all your known values remained fixed.
If you rise the eye up, then the pole obscure LESS of the object.
The only way for one to accurately measure for an unknown height of a distant object where the base line distance to the base of the object is known is to place a pole in between the line of sight of the TOP of the object and have that pole cover the remainder of the object.
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The exercise concerns measuring altitude of an observed point above the surface of the earth.
But you've already given the altitude of the observed object in your OP (5600 miles).
So, once more, what's the point?
The point is the title of the OP.
If you do not wish to answer the title of the OP, fine.
My answer is that your OP is very poorly presented.
Okay.
You present 1700 miles as the base of a right triangle.
Discerned that from a poorly written OP.
Then you present an observer with an unknown eye level 3 feet away from a 10 foot pole that is the very tip of that 1700 mile right triangle.
Eye level does not matter.
The height of the pole (10 feet) and the fact the top of the object is visible despite the pole is the only data in question.
If you put the observer a specific distance (3 feet) away from the 10 foot pole, then the eye level of the observer absolutely matters because the eye line of the observer, the top of the pole and the object must all line up. If you had not give that 3 foot distance to the observer, then the problem becomes much more reasonable and we could move on to your actual point, whatever that might be.
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If you rise the eye up, then the pole obscure LESS of the object.
But rise up from what height? That's the part you didn't include in the what was known or fixed. When I mentioned it in my first post you asked why it mattered, so that's what everything's been about since: why does it matter how high the observer's eye is when 3' behind the pole. I'm trying to explain to you that different heights will provide different angles for the hypotenuse (the line connecting the top of the pole with the top of the object), and thus different ratios.
The only way for one to accurately measure for an unknown height of a distant object where the base line distance to the base of the object is known is to place a pole in between the line of sight of the TOP of the object and have that pole cover the remainder of the object.
That means placing your eye level with the base of the pole. And to get the tops to line up your eye is 3' behind the pole, then your ratios work.
But it's not the only way. You can stand and work in your height. And to keep the ratios the same in the same scenario, you'll be closer than 3' behind the pole.
But you didn't specify whether eye level was at toe level or standing eye level. It matters. If you're 3' behind the pole and your eye is 5' high when aligning the pole's top with the distant object, the angle will be shallower and your calculation of ratios different.
(http://oi68.tinypic.com/2efruc0.jpg)
You can't be just at any height if a fixed 3' behind the pole and use the ratios you cited. They only work from one elevation 3' behind the 10' pole, and that's with your eye at ground level.
If that's what you were trying to depict in the first place. Fine. And I said if true, no problem. But you asked why it matters, and that's what I've been telling you. It matters to the triangle/ratio whether your standing or lying on the ground, given the "KNOW, FIXED and OBSERVED" figures you provided.
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If you put the observer a specific distance (3 feet) away from the 10 foot pole, then the eye level of the observer absolutely matters because the eye line of the observer, the top of the pole and the object must all line up. If you had not give that 3 foot distance to the observer, then the problem becomes much more reasonable and we could move on to your actual point, whatever that might be.
Without actual supporting evidence contrary to my OP?
Of course everything must line up (I STATED THIS IN THE OP).
This latest reply from you clearly indicates you really have nothing to contribute.
Why not try and state why this is not an accurate way to measure the altitude/height of any object?
I think I already know why you will not, but you try anyway.
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If you rise the eye up, then the pole obscure LESS of the object.
But rise up from what height? That's the part you didn't include in the what was known or fixed. When I mentioned it in my first post you asked why it mattered, so that's what everything's been about since: why does it matter how high the observer's eye is when 3' behind the pole. I'm trying to explain to you that different heights will provide different angles for the hypotenuse (the line connecting the top of the pole with the top of the object), and thus different ratios.
Yeah, and since then I have attempted to write exactly why you are wrong but:
A) You are incapable of understanding it; or,
II) You do understand why your point is meaningless and bears not on the issue at hand and are being purposefully obtuse.
The only way for one to accurately measure for an unknown height of a distant object where the base line distance to the base of the object is known is to place a pole in between the line of sight of the TOP of the object and have that pole cover the remainder of the object.
That means placing your eye level with the base of the pole. And to get the tops to line up your eye is 3' behind the pole, then your ratios work.
But it's not the only way. You can stand and work in your height. And to keep the ratios the same in the same scenario, you'll be closer than 3' behind the pole.
But you didn't specify whether eye level was at toe level or standing eye level. It matters. If you're 3' behind the pole and your eye is 5' high when aligning the pole's top with the distant object, the angle will be shallower and your calculation of ratios different.
(http://oi68.tinypic.com/2efruc0.jpg)
You can't be just at any height if a fixed 3' behind the pole and use the ratios you cited. They only work from one elevation 3' behind the 10' pole, and that's with your eye at ground level.
If that's what you were trying to depict in the first place. Fine. And I said if true, no problem. But you asked why it matters, and that's what I've been telling you. It matters to the triangle/ratio whether your standing or lying on the ground, given the "KNOW, FIXED and OBSERVED" figures you provided.
[/quote]
No, it does not.
If one were to stand further away, of course the fixed values would remain the same.
The variable in the measurement will always remain x.
The other numbers (the height of pole, distance to object from the observer as measured along the base, distance from pole to observer) can have any value. That would determine the angle of the hypotenuse.
But the angle of the hypotenuse is not necessary to make the calculation.
You know that so everything you have written here has been purposeless.
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Why not try and state why this is not an accurate way to measure the altitude/height of any object?
If you assume a flat earth and give the relevant information, then it is a perfectly valid and accurate way to measure the altitude/height of an object. In fact, it's used all the time by surveyors.
A surveyor standing 100 meters from a bridge. She determines that the angle of the elevation to the top of the bridge is 35°. The surveyor's eye level is 1.45 meters above ground. What is the height of the bridge?
(https://d2gne97vdumgn3.cloudfront.net/api/file/8q9cWVgrTviJfQKGoEXT)
However, if you assume a round earth, then you must adjust for the curvature of the earth, but it can still valid and accurate.
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No, it does not.
If one were to stand further away, of course the fixed values would remain the same.
The variable in the measurement will always remain x.
The other numbers (the height of pole, distance to object from the observer as measured along the base, distance from pole to observer) can have any value. That would determine the angle of the hypotenuse.
But the angle of the hypotenuse is not necessary to make the calculation.
You know that so everything you have written here has been purposeless.
Let's try it your way. New problem:
3 fixed, known values:
Distance to object from observer as measured along the base: 1383'
Height of pole: 10'
Distance from pole to observer: 3'
1 unknown value to be determined:
Height of the object: x
What's the value of x?
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No, it does not.
If one were to stand further away, of course the fixed values would remain the same.
The variable in the measurement will always remain x.
The other numbers (the height of pole, distance to object from the observer as measured along the base, distance from pole to observer) can have any value. That would determine the angle of the hypotenuse.
But the angle of the hypotenuse is not necessary to make the calculation.
You know that so everything you have written here has been purposeless.
Let's try it your way. New problem:
3 fixed, known values:
Distance to object from observer as measured along the base: 1383'
Height of pole: 10'
Distance from pole to observer: 3'
1 unknown value to be determined:
Height of the object: x
What's the value of x?
Well, you point out the distance between observer and object as measured along the base is 1383' without the need to add the 3'.
So:
1383/3 = x/10
461*10 = 4610'
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However, if you assume a round earth, then you must adjust for the curvature of the earth, but it can still valid and accurate.
Feel free to point out why anyone would make such a foolish assumption when:
A) Such a thing (curvature) is not visually apparent; and,
II) The supposed amount of arc between the object base and the observer in this case would not yield any appreciable difference in the result.
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No, it does not.
If one were to stand further away, of course the fixed values would remain the same.
The variable in the measurement will always remain x.
The other numbers (the height of pole, distance to object from the observer as measured along the base, distance from pole to observer) can have any value. That would determine the angle of the hypotenuse.
But the angle of the hypotenuse is not necessary to make the calculation.
You know that so everything you have written here has been purposeless.
Let's try it your way. New problem:
3 fixed, known values:
Distance to object from observer as measured along the base: 1383'
Height of pole: 10'
Distance from pole to observer: 3'
1 unknown value to be determined:
Height of the object: x
What's the value of x?
Well, you point out the distance between observer and object as measured along the base is 1383' without the need to add the 3'.
So:
1383/3 = x/10
461*10 = 4610'
Can you explain why you set these ratios as equivalent? And what does the number 3 represent here?
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No, it does not.
If one were to stand further away, of course the fixed values would remain the same.
The variable in the measurement will always remain x.
The other numbers (the height of pole, distance to object from the observer as measured along the base, distance from pole to observer) can have any value. That would determine the angle of the hypotenuse.
But the angle of the hypotenuse is not necessary to make the calculation.
You know that so everything you have written here has been purposeless.
Let's try it your way. New problem:
3 fixed, known values:
Distance to object from observer as measured along the base: 1383'
Height of pole: 10'
Distance from pole to observer: 3'
1 unknown value to be determined:
Height of the object: x
What's the value of x?
Well, you point out the distance between observer and object as measured along the base is 1383' without the need to add the 3'.
So:
1383/3 = x/10
461*10 = 4610'
Can you explain why you set these ratios as equivalent? And what does the number 3 represent here?
Yep...I can.
(https://ibb.co/icNLOz)
The diamond in the upper left of the right triangle (A) represents the object.
The base of the triangle (measured from angleB to angleC is the total distance from the intersecting line of the object to the surface of the earth. Note the pole represented by the line touching line BC and line AC .
The angles are equivalent so the ratios would be equivalent.
3' represents the distance between observer and pole.
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However, if you assume a round earth, then you must adjust for the curvature of the earth, but it can still valid and accurate.
Feel free to point out why anyone would make such a foolish assumption when:
A) Such a thing (curvature) is not visually apparent; and,
II) The supposed amount of arc between the object base and the observer in this case would not yield any appreciable difference in the result.
A) The curvature of the round earth was established over 2000 years ago and has worked out quite well in real world applications ever since, so it seems like a pretty safe assumption.
II) The circumference of the round earth is 24,900 miles. That works out to 69.166666... miles per degree. That means that 1700 miles covers a little over 25.5 degrees of arc. I would contend that 25.5 degrees of arc in the base of a triangle would be pretty significant to the result
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No, it does not.
If one were to stand further away, of course the fixed values would remain the same.
The variable in the measurement will always remain x.
The other numbers (the height of pole, distance to object from the observer as measured along the base, distance from pole to observer) can have any value. That would determine the angle of the hypotenuse.
But the angle of the hypotenuse is not necessary to make the calculation.
You know that so everything you have written here has been purposeless.
Let's try it your way. New problem:
3 fixed, known values:
Distance to object from observer as measured along the base: 1383'
Height of pole: 10'
Distance from pole to observer: 3'
1 unknown value to be determined:
Height of the object: x
What's the value of x?
Well, you point out the distance between observer and object as measured along the base is 1383' without the need to add the 3'.
So:
1383/3 = x/10
461*10 = 4610'
Can you explain why you set these ratios as equivalent? And what does the number 3 represent here?
Yep...I can.
(https://ibb.co/icNLOz)
The diamond in the upper left of the right triangle (A) represents the object.
The base of the triangle (measured from angleB to angleC is the total distance from the intersecting line of the object to the surface of the earth. Note the pole represented by the line touching line BC and line AC .
The angles are equivalent so the ratios would be equivalent.
3' represents the distance between observer and pole.
Ok, I agree with that calculation. However, that is for a person whose eyes are at a point 3’ from the base of the 10’ pole. If the time person is standing, and their eyes are at a height of 6’, and they are standing 3’ from the base of the pole, then the object is approximately 1,840’.
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However, if you assume a round earth, then you must adjust for the curvature of the earth, but it can still valid and accurate.
Feel free to point out why anyone would make such a foolish assumption when:
A) Such a thing (curvature) is not visually apparent; and,
II) The supposed amount of arc between the object base and the observer in this case would not yield any appreciable difference in the result.
A) The curvature of the round earth was established over 2000 years ago and has worked out quite well in real world applications ever since, so it seems like a pretty safe assumption.
II) The circumference of the round earth is 24,900 miles. That works out to 69.166666... miles per degree. That means that 1700 miles covers a little over 25.5 degrees of arc. I would contend that 25.5 degrees of arc in the base of a triangle would be pretty significant to the result
Again, your answer to A is personally UNKNOWN to you and simply a personal assumption.
You may choose to hold on to it and trumpet it is the gospel truth but that makes you nothing more than a religious zealot.
Your answer to II is accompanied by absolutely ZERO math and is simply a contention made without substance.
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So just to summarize what some of the people have been saying here. The OP is correct as long as the observers eyes are at the base of the triangle.
Mathematically, the angle of the hypotenuse is not directly relevant to comparing similar triangles as long as both triangles have the same hypotenuse.
If the observer’s eyes are above ground level, then the OPs calculation needs to be reworked because you are no longer comparing similar triangles. I am going to make a drawing demonstrating this in half an hour or so.
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However, if you assume a round earth, then you must adjust for the curvature of the earth, but it can still valid and accurate.
Feel free to point out why anyone would make such a foolish assumption when:
A) Such a thing (curvature) is not visually apparent; and,
II) The supposed amount of arc between the object base and the observer in this case would not yield any appreciable difference in the result.
A) The curvature of the round earth was established over 2000 years ago and has worked out quite well in real world applications ever since, so it seems like a pretty safe assumption.
II) The circumference of the round earth is 24,900 miles. That works out to 69.166666... miles per degree. That means that 1700 miles covers a little over 25.5 degrees of arc. I would contend that 25.5 degrees of arc in the base of a triangle would be pretty significant to the result
Again, your answer to A is personally UNKNOWN to you and simply a personal assumption.
No, it's a well established fact. It's also irrelevant seeing as I said that the method works for a flat earth as well as a round earth, as long as the appropriate adjustments are made.
Your answer to II is accompanied by absolutely ZERO math and is simply a contention made without substance.
Seriously? Do I really need to show how to divide 29,400 miles by 360 degrees to get 69.1666... miles/degree and then divide 1700 miles by 69.1666... miles/degree to get 24.587 degrees?
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Let's try it your way. New problem:
3 fixed, known values:
Distance to object from observer as measured along the base: 1383'
Height of pole: 10'
Distance from pole to observer: 3'
1 unknown value to be determined:
Height of the object: x
What's the value of x?
Well, you point out the distance between observer and object as measured along the base is 1383' without the need to add the 3'.
So:
1383/3 = x/10
461*10 = 4610'
(http://oi65.tinypic.com/33x8qqw.jpg)
The diamond in the upper left of the right triangle (A) represents the object.
The base of the triangle (measured from angleB to angleC is the total distance from the intersecting line of the object to the surface of the earth. Note the pole represented by the line touching line BC and line AC .
The angles are equivalent so the ratios would be equivalent.
3' represents the distance between observer and pole.
I left out a crucial piece of information. Your triangle assumes the observer has zero height.
Actually, the observer in my configuration has an eye-level of 5.75' and if standing 3' behind the 10' pole the tops of the pole and the object are aligned, the angle formed is 54.78° and not the steeper 73.3° of your triangle.
(http://oi65.tinypic.com/5mrg4.jpg)
The height to the top of the object, then isn't 4610' but rather 1965'.
Height of the observer's eye matters.
The angle of the hypotenuse matters.
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The angle of the hypotenuse doesn’t matter, per se. what matters is that similar triangles are compared.
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Let's try it your way. New problem:
3 fixed, known values:
Distance to object from observer as measured along the base: 1383'
Height of pole: 10'
Distance from pole to observer: 3'
1 unknown value to be determined:
Height of the object: x
What's the value of x?
Well, you point out the distance between observer and object as measured along the base is 1383' without the need to add the 3'.
So:
1383/3 = x/10
461*10 = 4610'
(http://oi65.tinypic.com/33x8qqw.jpg)
The diamond in the upper left of the right triangle (A) represents the object.
The base of the triangle (measured from angleB to angleC is the total distance from the intersecting line of the object to the surface of the earth. Note the pole represented by the line touching line BC and line AC .
The angles are equivalent so the ratios would be equivalent.
3' represents the distance between observer and pole.
I left out a crucial piece of information. Your triangle assumes the observer has zero height.
Actually, the observer in my configuration has an eye-level of 5.75' and if standing 3' behind the 10' pole the tops of the pole and the object are aligned, the angle formed is 54.78° and not the steeper 73.3° of your triangle.
(http://oi65.tinypic.com/5mrg4.jpg)
The height to the top of the object, then isn't 4610' but rather 1965'.
Height of the observer's eye matters.
The angle of the hypotenuse matters.
Your entire diagram depicts an entirely different set of circumstances to that which you wrote.
Moving goalposts.
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To be fair, you never really had a goal. You presented a mathematical statement which was true, and nobody disagreed with. It was pointed out that depending on the real world application, it might need to be tweaked. That is it. Do you agree that your OP is not correct if the observer is standing 3ft from the sighting pole?
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To be fair, you never really had a goal. You presented a mathematical statement which was true, and nobody disagreed with. It was pointed out that depending on the real world application, it might need to be tweaked. That is it. Do you agree that your OP is not correct if the observer is standing 3ft from the sighting pole?
My OP is correct.
The stated method is a legitimate means for determining the altitude of an object above the earth if one knows the baseline distance to the object in question.
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To be fair, you never really had a goal. You presented a mathematical statement which was true, and nobody disagreed with. It was pointed out that depending on the real world application, it might need to be tweaked. That is it. Do you agree that your OP is not correct if the observer is standing 3ft from the sighting pole?
My OP is correct.
The stated method is a legitimate means for determining the altitude of an object above the earth if one knows the baseline distance to the object in question.
The only things that were really ever in question about your OP were the location of the observer's eye level and your inability to understand its relevance.
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To be fair, you never really had a goal. You presented a mathematical statement which was true, and nobody disagreed with. It was pointed out that depending on the real world application, it might need to be tweaked. That is it. Do you agree that your OP is not correct if the observer is standing 3ft from the sighting pole?
My OP is correct.
The stated method is a legitimate means for determining the altitude of an object above the earth if one knows the baseline distance to the object in question.
If, and only if, the base of the triangles being compared are flat.
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To be fair, you never really had a goal. You presented a mathematical statement which was true, and nobody disagreed with. It was pointed out that depending on the real world application, it might need to be tweaked. That is it. Do you agree that your OP is not correct if the observer is standing 3ft from the sighting pole?
My OP is correct.
The stated method is a legitimate means for determining the altitude of an object above the earth if one knows the baseline distance to the object in question.
If, and only if, the base of the triangles being compared are flat.
Durrr...
You got something else you want to offer relevant to the OP?
Why would the base of the triangles not be flat?
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To be fair, you never really had a goal. You presented a mathematical statement which was true, and nobody disagreed with. It was pointed out that depending on the real world application, it might need to be tweaked. That is it. Do you agree that your OP is not correct if the observer is standing 3ft from the sighting pole?
My OP is correct.
The stated method is a legitimate means for determining the altitude of an object above the earth if one knows the baseline distance to the object in question.
The only things that were really ever in question about your OP were the location of the observer's eye level and your inability to understand its relevance.
Since it is irrelevant to the stated method, then I suggest you are the one with the disability in this thread.
You cannot provide evidence of your position I have such inability.
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To be fair, you never really had a goal. You presented a mathematical statement which was true, and nobody disagreed with. It was pointed out that depending on the real world application, it might need to be tweaked. That is it. Do you agree that your OP is not correct if the observer is standing 3ft from the sighting pole?
My OP is correct.
The stated method is a legitimate means for determining the altitude of an object above the earth if one knows the baseline distance to the object in question.
If, and only if, the base of the triangles being compared are flat.
Durrr...
You got something else you want to offer relevant to the OP?
The OP doesn't have a point, really. We are just trying to flesh out an idea here. So, this being the Flat Earth Society, the posters are relating your incredibly vague and pointless OP to the ongoing debate over the shape of the Earth. If you decide to rock up with a photo that shows a ten foot pole, with the Willis Tower's top matching up with the top of the pole, and write "Hey! Look! This doesn't match properly! The Earth is Flat!" we would already have a strong basis to discuss the truth of that statement.
Why would the base of the triangles not be flat?
It wouldn't be flat if it was curved, obviously.
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Your entire diagram depicts an entirely different set of circumstances to that which you wrote.
Moving goalposts.
It does? I did?
Let's check. Here's the scenario:
3 fixed, known values:
Distance to object from observer as measured along the base: 1383'
Height of pole: 10'
Distance from pole to observer: 3'
1 unknown value to be determined:
Height of the object: x
And here's the diagram:
(http://oi65.tinypic.com/5mrg4.jpg)
Distance to object from observer as measured along the base: 1380' + 3' = 1383' : CHECK
Height of pole = 10' : CHECK
Distance from pole to observer = 3' : CHECK
What did I change from setting the problem to depicting the diagram? What goalpost did I move?
The goal was to figure the height of the distant object. My diagram includes all of the KNOWN, FiXED values for the attributes I provided. I didn't change any of them.
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To be fair, you never really had a goal. You presented a mathematical statement which was true, and nobody disagreed with. It was pointed out that depending on the real world application, it might need to be tweaked. That is it. Do you agree that your OP is not correct if the observer is standing 3ft from the sighting pole?
My OP is correct.
The stated method is a legitimate means for determining the altitude of an object above the earth if one knows the baseline distance to the object in question.
If, and only if, the base of the triangles being compared are flat.
Durrr...
You got something else you want to offer relevant to the OP?
The OP doesn't have a point, really. We are just trying to flesh out an idea here. So, this being the Flat Earth Society, the posters are relating your incredibly vague and pointless OP to the ongoing debate over the shape of the Earth. If you decide to rock up with a photo that shows a ten foot pole, with the Willis Tower's top matching up with the top of the pole, and write "Hey! Look! This doesn't match properly! The Earth is Flat!" we would already have a strong basis to discuss the truth of that statement.
The OP point is the altitude of the sun over the flat earth plane can be measured.
That has always been the point.
The longer you participate in the thread, the more apparent it becomes you are simply trying to fluff it off.
Why would the base of the triangles not be flat?
It wouldn't be flat if it was curved, obviously.
Since I have offered an OP that makes no assumptions concerning curvature as there is no observed curvature between the observer (C) and point B (the point where line AB join), then your injection of curvature is nonsensical.
Of course, feel free to demonstrate your supposed math...and the difference the supposed arc would make in the final answer to an appropriate equation.
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Your entire diagram depicts an entirely different set of circumstances to that which you wrote.
Moving goalposts.
It does? I did?
Let's check. Here's the scenario:
3 fixed, known values:
Distance to object from observer as measured along the base: 1383'
Height of pole: 10'
Distance from pole to observer: 3'
1 unknown value to be determined:
Height of the object: x
And here's the diagram:
(http://oi65.tinypic.com/5mrg4.jpg)
Distance to object from observer as measured along the base: 1380' + 3' = 1383' : CHECK
Height of pole = 10' : CHECK
Distance from pole to observer = 3' : CHECK
What did I change from setting the problem to depicting the diagram? What goalpost did I move?
The goal was to figure the height of the distant object. My diagram includes all of the KNOWN, FiXED values for the attributes I provided. I didn't change any of them.
Bobby, simply check with any surveyor and ask them about how to solve for an altitude of any object if the baseline distance from observer to bottom of object is known.
They will CLEARLY STATE to you they diagram it in the way I PRESENTED, not your method.
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Bobby, simply check with any surveyor and ask them about how to solve for an altitude of any object if the baseline distance from observer to bottom of object is known.
They will CLEARLY STATE to you they diagram it in the way I PRESENTED, not your method.
It's not the method that's the issue. You just need to make sure you've got the right triangle and thus right ratios.
There's nothing wrong with your opening post calculation, as long as the sighting point that's 3' behind the 10' pole is at the vertex of the triangle giving you the ratio.
If the base of the 10' pole is at the same level as the instrument or eye making the alignment sighting, then you've got the correct triangle and you can use the values of the sides of the triangle using your ratio method. But if you're standing 3' behind the pole but you calculate as if the triangle corner is at your shoes, then you've got the wrong triangle and your ratios using that wrong triangle will give you an incorrect answer. The vertex of the correct triangle will be further along the baseline behind you and the angle to the distant object that you've aligned with your eye from 3' behind the pole will be shallower than the angle coming from your feet.
The problem isn't your method the problem. It's that you said height of the aligning eye doesn't matter. It does. It matters to surveyors.
It's been stimulating, but I can't make it any clearer. If we're still not on the same wavelength, good luck to you. I have to move on.
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Since I have offered an OP that makes no assumptions concerning curvature as there is no observed curvature between the observer (C) and point B (the point where line AB join), then your injection of curvature is nonsensical.
Of course, feel free to demonstrate your supposed math...and the difference the supposed arc would make in the final answer to an appropriate equation.
If you want to apply this to the real world and calculate the altitude of the sun, then you will first have to ensure that the base of your triangle is not curved.
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Since I have offered an OP that makes no assumptions concerning curvature as there is no observed curvature between the observer (C) and point B (the point where line AB join), then your injection of curvature is nonsensical.
Of course, feel free to demonstrate your supposed math...and the difference the supposed arc would make in the final answer to an appropriate equation.
If you want to apply this to the real world and calculate the altitude of the sun, then you will first have to ensure that the base of your triangle is not curved.
Given you would not presume to state this same sentence to a surveyor, I will take it you have no real objection.
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Since I have offered an OP that makes no assumptions concerning curvature as there is no observed curvature between the observer (C) and point B (the point where line AB join), then your injection of curvature is nonsensical.
Of course, feel free to demonstrate your supposed math...and the difference the supposed arc would make in the final answer to an appropriate equation.
If you want to apply this to the real world and calculate the altitude of the sun, then you will first have to ensure that the base of your triangle is not curved.
Given you would not presume to state this same sentence to a surveyor, I will take it you have no real objection.
Uhhh... what? Why wouldn't I do that? If you have a curved base to your triangle, you must determine how material that curve is to the calculation. Since you could only level a straw man against my objection, I can only ask you to try again.
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Since I have offered an OP that makes no assumptions concerning curvature as there is no observed curvature between the observer (C) and point B (the point where line AB join), then your injection of curvature is nonsensical.
Of course, feel free to demonstrate your supposed math...and the difference the supposed arc would make in the final answer to an appropriate equation.
If you want to apply this to the real world and calculate the altitude of the sun, then you will first have to ensure that the base of your triangle is not curved.
Given you would not presume to state this same sentence to a surveyor, I will take it you have no real objection.
Uhhh... what? Why wouldn't I do that? If you have a curved base to your triangle, you must determine how material that curve is to the calculation. Since you could only level a straw man against my objection, I can only ask you to try again.
You are the one supporting a curve exists.
I support no curve because one is not visible to me in performing the observation and the resulting calculations.
But go ahead, noble RE-er...
Provide us the math taking into account the supposed sphericity found in 1700 miles of the mythical ball earth surface...
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You are the one supporting a curve exists.
Feel free to quote in this thread where I did this. I can save you the time though, I never did.
I support no curve because one is not visible to me in performing the observation and the resulting calculations.
So if it is not visible then it does not exist?
But go ahead, noble RE-er...
Provide us the math taking into account the supposed sphericity found in 1700 miles of the mythical ball earth surface...
I am merely pointing out a fault in your method. If your method does not take verify the geometry of the base of your triangle, then you are introducing a source of error that you are not accounting for.
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To be fair, you never really had a goal. You presented a mathematical statement which was true, and nobody disagreed with. It was pointed out that depending on the real world application, it might need to be tweaked. That is it. Do you agree that your OP is not correct if the observer is standing 3ft from the sighting pole?
My OP is correct.
The stated method is a legitimate means for determining the altitude of an object above the earth if one knows the baseline distance to the object in question.
The only things that were really ever in question about your OP were the location of the observer's eye level and your inability to understand its relevance.
Since it is irrelevant to the stated method, then I suggest you are the one with the disability in this thread.
You cannot provide evidence of your position I have such inability.
Your inability to show the relevance is shown quite clearly here:
Then you present an observer with an unknown eye level 3 feet away from a 10 foot pole that is the very tip of that 1700 mile right triangle.
Eye level does not matter.
Eye level matters if you want it to line up with the object and the top of the 10' pole.
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You are the one supporting a curve exists.
Feel free to quote in this thread where I did this. I can save you the time though, I never did.
I support no curve because one is not visible to me in performing the observation and the resulting calculations.
So if it is not visible then it does not exist?
But go ahead, noble RE-er...
Provide us the math taking into account the supposed sphericity found in 1700 miles of the mythical ball earth surface...
I am merely pointing out a fault in your method. If your method does not take verify the geometry of the base of your triangle, then you are introducing a source of error that you are not accounting for.
OF course its verified by observation.
Unless you have visible evidence to the contrary...
And of course, you will offer something unsubstantiated by personal observation.
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So it’s ok to appeal to an authority when it’s surveying methods, but not ok when it’s geodesy and topography. Wow you have shitty standards. Regardless, this is about your methodology and if you aren’t measuring the datum for your observation to ensure that your triangles are indeed similar, you are introducing sources of error, so your method can’t be used in RL circumstances.
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So it’s ok to appeal to an authority when it’s surveying methods, but not ok when it’s geodesy and topography. Wow you have shitty standards. Regardless, this is about your methodology and if you aren’t measuring the datum for your observation to ensure that your triangles are indeed similar, you are introducing sources of error, so your method can’t be used in RL circumstances.
Real life?
Funny, I did this just the other day.
I checked.
I am real and I was indeed alive at the time.
I have made no appeal to authority. I stated a fact when I wrote that surveyors use this method all the time when solving for unknown elevations and they do not, REPEATING DO NOT take into account any supposed sphericity.
But okay, okay, go ahead and provide the method utilized by your standards.
Demonstrate how the result would be drastically different.
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So it’s ok to appeal to an authority when it’s surveying methods, but not ok when it’s geodesy and topography. Wow you have shitty standards. Regardless, this is about your methodology and if you aren’t measuring the datum for your observation to ensure that your triangles are indeed similar, you are introducing sources of error, so your method can’t be used in RL circumstances.
Real life?
Funny, I did this just the other day.
I checked.
I am real and I was indeed alive at the time.
I have made no appeal to authority. I stated a fact when I wrote that surveyors use this method all the time when solving for unknown elevations and they do not, REPEATING DO NOT take into account any supposed sphericity.
Which surveyors? They don’t account for the curve of the Earth over any distance? If so, what is your source?
But okay, okay, go ahead and provide the method utilized by your standards.
I don’t know the method for geodetic surveying but surely you are smart enough to know that triangulation doesn’t work on a non-Euclidean plane?
Don’t derail the thread.
Demonstrate how the result would be drastically different.
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So it’s ok to appeal to an authority when it’s surveying methods, but not ok when it’s geodesy and topography. Wow you have shitty standards. Regardless, this is about your methodology and if you aren’t measuring the datum for your observation to ensure that your triangles are indeed similar, you are introducing sources of error, so your method can’t be used in RL circumstances.
Real life?
Funny, I did this just the other day.
I checked.
I am real and I was indeed alive at the time.
I have made no appeal to authority. I stated a fact when I wrote that surveyors use this method all the time when solving for unknown elevations and they do not, REPEATING DO NOT take into account any supposed sphericity.
Which surveyors? They don’t account for the curve of the Earth over any distance? If so, what is your source?
Don't try and switch this around...
You brought up the issue so name one that has or does.
But okay, okay, go ahead and provide the method utilized by your standards.
I don’t know the method for geodetic surveying but surely you are smart enough to know that triangulation doesn’t work on a non-Euclidean plane?
Don’t derail the thread.
I'm not derailing the thread.
I started the thread.
You are the one derailing it with your mythical sphericity claims.
Demonstrate how the result would be drastically different.
Still waiting for you to do it...
You are the one subscribing to the issue of sphericity so go ahead and figure it out.