Perceived Height of 38m Object at 1600m Distance

[Rutter_1980]

Hi There,

Can someone tell me if I'm standing 1600m from an object that is approx. 38m high then what will my perceived height of the object be?

Also will this remain the same if I am in an elevated position above the height of the object?
Thanks!

 

[DaveC426913]

Do you mean what angle of your vision it will subtend? This is pretty straightforward to work out with trig. Have you tried?

 

[Rutter_1980]

Do you mean what angle of your vision it will subtend? This is pretty straightforward to work out with trig. Have you tried?

Hi Dave,
Yes if I'm stood under the object it is 35m high. But if I move away it gets smaller...so how tall will it appear in my vision at 1600m away?

I can use trig to work out the height of an object from a certain distance but that gives the 'actual' height. I need the height my eye will see it having moved away from it. Not sure if that makes sense better now!

 

[Mark44]

Hi Dave,
Yes if I'm stood under the object it is 35m high. But if I move away it gets smaller

No, the object doesn't get smaller, but the size that you perceive gets smaller.

...so how tall will it appear in my vision at 1600m away?

I can use trig to work out the height of an object from a certain distance but that gives the 'actual' height. I need the height my eye will see it having moved away from it. Not sure if that makes sense better now!

Use similar triangles to find the perceived size of the object. In the larger triangle, the triangle's height represents the object's height (38 m.). In the smaller triangle, the height represents the height of the object as your eye perceives it.

 

[DaveC426913]

Hi Dave,
Yes if I'm stood under the object it is 35m high. But if I move away it gets smaller...so how tall will it appear in my vision at 1600m away?

I can use trig to work out the height of an object from a certain distance but that gives the 'actual' height. I need the height my eye will see it having moved away from it. Not sure if that makes sense better now!

You're still missing a required piece of information. Your eye does not perceive heights.

Either
- you can measure it in terms of the angle it subtends
at distance 0 it subtends 90 degrees, at distance 38m it subtends 45 degrees, at distance 1600m it subtends ? degrees
or
- you can pick a second fixed distance and use similar triangles, as Mark44 suggests, to measure its apparent height - but you must pick a second fixed distance.

e.g.: a 38m tall object at 1600m distance will appear to be 1.14cm tall if it were at 48cm distance. Which means you would be able to eclipse it with a finger at arm's length.

 

[Rutter_1980]

Yes...sorry, I realized the object doesn't physically get smaller! Hence the title subject...

I still don't know how to work this out - sorry! Do you know then what the perceived height will be if I travel 1600m from the base of the 38m high object?

 

[Rutter_1980]

You're still missing a required piece of information. Your eye does not perceive heights.

Either
- you can measure it in terms of the angle it subtends
at distance 0 it subtends 90 degrees, at distance 38m it subtends 45 degrees, at distance 1600m it subtends ? degrees
or
- you can pick a second fixed distance and use similar triangles, as Mark44 suggests, to measure its apparent height - but you must pick a second fixed distance.

e.g.: a 38m tall object at 1600m distance will appear to be 1.14cm tall if it were at 48cm distance. Which means you would be able to eclipse it with a finger at arm's length.

At the distance of 1600m the angle is 1.36°

 

[Rutter_1980]

how does this now relate to the height?

 

[DaveC426913]

That is the question we are asking you.

In a nutshell: what do you mean by "perceived height"? What are you expecting to get as an answer?

I am looking out my window at a building right now. It's six stories tall (call it 20m), and it's 200m away.
I can tell you that it subtends and angle of 5.7 degrees (sin^-1(20/200)).
I can tell you it appears to be 5cm in height at an arm's length distance of 50cm (200m:20m :: 50cm:?cm).

What would you say is its perceived height? Because I don't know what that means.

 

[Rutter_1980]

That is the question we are asking you.

In a nutshell: what do you mean by "perceived height"? What are you expecting to get as an answer?

I am looking out my window at a building right now. It's six stories tall (call it 20m), and it's 200m away.
I can tell you that it subtends and angle of 5.7 degrees (sin^-1(20/200)).
I can tell you it appears to be 5cm in height at an arm's length distance of 50cm (200m:20m :: 50cm:?cm).

What would you say is its perceived height? Because I don't know what that means.

Hi Dave...thanks for sticking with me on this one!

Ok so the height of my object is 38m high. As I move away from it the object 'appears' to get smaller and smaller until eventually it will vanish (even though it is still 38m high).
So at a distance of 1600m away what height will the object 'appear' to be to my eye?
I get the angle it subtends to is 1.36° @1600m but how does this relate to the height the object now 'appears' to be in my vision?
I thought there may be some easy relationship that for every 1m you move away from an object it 'appears' to get 1cm smaller or something!?
OR
How do I now get my 1.36° to tell me roughly what size the object now 'appears' to be to me
I can't check this in reality as this 35m object is being built 1600m from my home and I want to try and work out what it will 'appear' to look like once it's there...
Thanks.

 

[elektrokokke]

As Dave says, you can only relate the "perceived" height to another object at another given distance. E.g., at 1600 m distance, the 38 m building will appear as tall as a 19 m building which is 800 m away.

You could measure the "perceived" size of the object in terms of centimeters on a ruler you are holding, but the measurement will depend on your distance from the ruler. E.g., if you hold the ruler at a distance of 32 cm, the building should measure 7.6 mm "on the ruler". This is the simple "rule of proportion".

One absolute scale you could relate to an object's size at a distance would be the size of the object on your retina, but that cannot be easily determined, as it depends on many factors. Also, your brain is totally ignorant of this, so you couldn't call it "perceived size".

On another note, the brain DOES get a notion of absolute size from what it sees. A big role plays the distance estimated by stereo vision, but even with only one eye the brain is quite good at measuring sizes, e.g., by experience, focus, color, etc. In this sense, a 38 m tall building will still be perceived as being more or less 38 m tall by a trained viewer, even if it is 1600 m away.

 

[DaveC426913]

I'd like you to try an experiment. You tell us how large the perceived 38m tower is to you. Doesn't matter if your answer is inaccurate, just tell us how you deduced the answer.

 

[Rutter_1980]

I'd like you to try an experiment. You tell us how large the perceived 38m tower is to you. Doesn't matter if your answer is inaccurate, just tell us how you deduced the answer.

Is that directed to me Dave?

 

[DaveC426913]

Yes.

 

[Rutter_1980]

Yes.

Well that's kind of why I asked! I have no real idea how to do it. At first I thought it was a simple case of working out the angle your viewing it from at the distance1600m but I don't know where to go from there...things 'appear' smaller to us as we move away from them and I hoped someone could tell me some easy relationship that you could apply but it seems far more tricky than I thought!

The only thing left for me to do is to drive 2 hours to where one of these 38m high pylons is sighted - drive 1600m away from it and view it from there to see how it looks! Guess that's the only real way of knowing!

 

[DaveC426913]

Well that's kind of why I asked! I have no real idea how to do it.

What I'm after here is to find out what kind of answer you're expecting.

Were you going to say 'at this distance the tower seems to be about one metre tall'? That only makes sense if you conclude that the tower, rather than being 1600m distant, is perceived as being (1600/38*1) 42 metres away. But why there? Why wouldn't you perceive the tower as being 1 metre away (thus only 2.4cm tall), or 10,000 metres away (thus 237 metres tall)?

The lesson here is that it doesn't make sense to try to assign an absolute height of an apparent image of an object. All you can do is assign an angle that it subtends.

The angle an object subtends is the inverse tan of its height divided by its distance:
θ = tan^-1(38/1600) = 1.36°

The only thing left for me to do is to drive 2 hours to where one of these 38m high pylons is sighted - drive 1600m away from it and view it from there to see how it looks! Guess that's the only real way of knowing!

I can tell you how it'll look. It will subtend a 1.36° angle.
At arm's length*, you will be able to pinch it between your finger and thumb when held 1.9cm apart (1600m : 38m :: 80cm : 1.9cm).
*assuming your fingers are 80cm from your eye

Just pick something nearby, like I did with the (fictional) 20m tall building, 200m away.
Work out the principles, then apply the formula to your 38m tower.

This is not arm's length:
http://travelswithoutpants.com/blog/wp-content/uploads/2014/04/DSCN0471-1024x768.jpg

 

[DaveC426913]

BTW, you might also find this star-gazing technique useful.

So, your tower should be almost entirely eclipsed by the width of your pinkie.

I use this technique both for star-gazing and for sailing, when I need to judge the size of distant objects, or the distance to objects of known size.

You can practice by trying to gauge the altitude of commercial planes passing overhead (once you know the model and thus length of a given plane).