[markjo]

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.

[Bobby Shafto]

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.



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.

[totallackey]

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.

[totallackey]

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.



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.

[markjo]

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?

However, if you assume a round earth, then you must adjust for the curvature of the earth, but it can still valid and accurate.

[Bobby Shafto]

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?

[totallackey]

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'

[totallackey]

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.

[Rama Set]

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?

[totallackey]

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.

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.

[markjo]

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

[Rama Set]

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.

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’. 

[totallackey]

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.

[Rama Set]

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.

[markjo]

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?

[Bobby Shafto]

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'

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.



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.

[Rama Set]

The angle of the hypotenuse doesn’t matter, per se. what matters is that similar triangles are compared.

[totallackey]

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'

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.



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.

[Rama Set]

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?

[totallackey]

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.