[totallackey]

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. 

[Bobby Shafto]

Observer's eye is at toe-level.

[totallackey]

Observer's eye is at toe-level.

So?

[I_Joined_Again]

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)

[Bobby Shafto]

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.

[totallackey]

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.

[Bobby Shafto]

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.

[totallackey]

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.

[markjo]

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)?

[totallackey]

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?

[pj1]

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.

[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?

[Bobby Shafto]

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.

[totallackey]

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.

[totallackey]

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.

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.

[totallackey]

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.

[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.

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.

[Bobby Shafto]

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.

[totallackey]

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.

[totallackey]

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.