Curvature of the Horizon

[ScarBest]

TL;DR Summary

I request advice on where to publish a paper on calculating the curve of the horizon seen from various altitudes.

I'm writing a paper on how to calculate the visible curvature of the horizon for any given altitude. I'm considering attempting to have it published. I wonder if anyone could suggest a publication that might be interested in such a paper. If this isn't the best forum for this request, please suggest another. Thank you.

 

[PeroK]

TL;DR Summary: I request advice on where to publish a paper on calculating the curve of the horizon seen from various altitudes.

I'm writing a paper on how to calculate the visible curvature of the horizon for any given altitude. I'm considering attempting to have it published. I wonder if anyone could suggest a publication that might be interested in such a paper. If this isn't the best forum for this request, please suggest another. Thank you.

https://en.wikipedia.org/wiki/Horizon

 

[Baluncore]

I wonder if anyone could suggest a publication that might be interested in such a paper.

The journal will depend on the application of your solution.
Have you done a literature search?
I think you will find that it has been published before.

The angular depression of the horizon, increases with altitude, and can be measured with a theodolite. The distance to the horizon is typically calculated for altitude, when range of visibility, range of radio communication, or RADAR coverage is being analysed.

When taking star sights with a sextant for astronomical navigation, angular depression of the horizon is estimated from tables.

How do you specify "curvature of the horizon"? Surely it is always a circle, 360°.
The horizon would have a radius of curvature, measured as an angle from the vertical, at the observer's altitude. Is that radius not simply 90°, minus the depression of the horizon?

 

[berkeman]

Welcome to PF.

I'm considering attempting to have it published. I wonder if anyone could suggest a publication that might be interested in such a paper.

What journals are the papers in your References from? Typically you will be publishing in a journal that you read frequently, and has similar papers in it.

 

[ScarBest]

The journal will depend on the application of your solution.
Have you done a literature search?
I think you will find that it has been published before.

The angular depression of the horizon, increases with altitude, and can be measured with a theodolite. The distance to the horizon is typically calculated for altitude, when range of visibility, range of radio communication, or RADAR coverage is being analysed.

When taking star sights with a sextant for astronomical navigation, angular depression of the horizon is estimated from tables.

How do you specify "curvature of the horizon"? Surely it is always a circle, 360°.
The horizon would have a radius of curvature, measured as an angle from the vertical, at the observer's altitude. Is that radius not simply 90°, minus the depression of the horizon?

I meant the curve that would be visible to the observer. At sea level, the horizon appears to be a straight line. From the International Space Station, there is an obvious curve visible. My paper shows how to calculate and graph the curve that would be visible to an observer at a given altitude.

 

[ScarBest]

Welcome to PF.


What journals are the papers in your References from? Typically you will be publishing in a journal that you read frequently, and has similar papers in it.

I'm not a professional scientist and I don't read journals. My calculations are a mathematical treatment that doesn't rely on the work of others, so I don't have any references.

 

[ScarBest]

https://en.wikipedia.org/wiki/Horizon

Thanks, but the Wikipedia article does not do what my paper does, which is show how the horizon would appear to be curved from the point of view of an observer at a given altitude.

 

[renormalize]

I'm not a professional scientist and I don't read journals.

But if you don't read journals and/or other references, how can you be sure your work is truly novel versus just rediscovering knowledge that is already well-known and discussed?

 

[ScarBest]

But if you don't read journals and/or other references, how can you be sure your work is truly novel versus just rediscovering knowledge that is already well-known and discussed?

I don't know this. However, I have searched online and I've found nothing like it. If I were to submit it to a journal, the editor would know and inform me.

 

[PeroK]

I don't know this. However, I have searched online and I've found nothing like it. If I were to submit it to a journal, the editor would know and inform me.

I would expect a bright high school student to be able to calculate the distance to the horizon from a given altitude. What are you doing that is more than basic trigonometry?

 

[ScarBest]

I would expect a bright high school student to be able to calculate the distance to the horizon from a given altitude. What are you doing that is more than basic trigonometry?

I'm not calculating the distance to the horizon. I'm calculating the curved shape of the horizon as it would appear to an observer. From sea level, the curve is imperceptible. From low earth orbit, it's obvious. I'm calculating and graphing the curve that would be seen from various altitudes.

 

[Baluncore]

I'm calculating and graphing the curve that would be seen from various altitudes.

Are you calculating that for a spherical Earth, or for an oblate spheroid, such as the WGS84 standard used by GPS positioning system ?

 

[PeroK]

I'm not calculating the distance to the horizon. I'm calculating the curved shape of the horizon as it would appear to an observer. From sea level, the curve is imperceptible. From low earth orbit, it's obvious. I'm calculating and graphing the curve that would be seen from various altitudes.

It's a similar calculation. The curvature of a circle is 1/radius. It's still basic trig.

 

[PeroK]

I was thinking about what you would measure from the vantage point above the surface of the Earth. Essentially, you are looking down on a circle from some point above the centre of the circle. You could measure the angle below the horizontal to the horizon. Let's call that $\gamma$. As you scan the horizon, $\gamma$ is constant. Without measuring distances, that seems to be the key measurable quantity.

In fact, looking at the Wikipedia page, the zenith angle is defined as $\gamma + \pi/2$.

An accurate measurement of $\gamma$ would allow us to calculate the radius of the Earth, as:
$$\cos \gamma = \frac{R}{R+h}$$That's as far as high-school mathematics can take us.

However, the Wikipedia page continues with calculations involving the effect of atmospheric refraction:

https://en.wikipedia.org/wiki/Horizon#Effect_of_atmospheric_refraction

Even that would be seen by a journal editor as long-established calculations. To get something published on this, we would need something very special indeed!

 

[ScarBest]

Are you calculating that for a spherical Earth, or for an oblate spheroid, such as the WGS84 standard used by GPS positioning system ?

For a spherical earth. Calculating for an oblate spheroid would introduce considerable complexity, and since the earth is so close to being spherical, I don't think there would be any noticeable difference visually.

 

[ScarBest]

It's a similar calculation. The curvature of a circle is 1/radius. It's still basic trig.

When I talk about the curvature of the horizon, I'm not talking about the mathematical definition of curvature. I'm talking about the curved shape that the horizon would take as viewed by an observer. Since the radius of the horizon circle gets smaller the lower the altitude, your interpretation would cause the horizon to appear more curved the lower you are; the opposite is true. When an observer is close to the earth's surface, and the observer directs his attention at the point of the horizon in front of him, he is seeing a greatly foreshortened view of the horizon circle, causing it to appear as a highly eccentric ellipse, thus causing it to seem as if it had no curve at all. As you go higher, the angle to the horizon circle increases, reducing the foreshortening and increasing the amount of the circle you see in your field of vision, thus causing the horizon to appear more curved. I'm talking about the *visual* curve of the horizon - the actual shape that would be seen by an observer.

 

[Baluncore]

As you go higher, the angle to the horizon circle increases, reducing the foreshortening and increasing the amount of the circle you see in your field of vision, thus causing the horizon to appear more curved.

Who defines your angular field of vision?
How is field of vision specified?

 

[PeroK]

When I talk about the curvature of the horizon, I'm not talking about the mathematical definition of curvature. I'm talking about the curved shape that the horizon would take as viewed by an observer.

I get that:

I was thinking about what you would measure from the vantage point above the surface of the Earth. Essentially, you are looking down on a circle from some point above the centre of the circle. You could measure the angle below the horizontal to the horizon. Let's call that γ. As you scan the horizon, γ is constant. Without measuring distances, that seems to be the key measurable quantity.

In fact, looking at the Wikipedia page, the zenith angle is defined as γ+π/2.

Since the radius of the horizon circle gets smaller the lower the altitude, your interpretation would cause the horizon to appear more curved the lower you are; the opposite is true. When an observer is close to the earth's surface, and the observer directs his attention at the point of the horizon in front of him, he is seeing a greatly foreshortened view of the horizon circle, causing it to appear as a highly eccentric ellipse, thus causing it to seem as if it had no curve at all.

The horizon is a circle, although you are not at its centre, but above its centre. It can't be an ellipse, as all direction are equivalent.

As you go higher, the angle to the horizon circle increases, reducing the foreshortening and increasing the amount of the circle you see in your field of vision, thus causing the horizon to appear more curved. I'm talking about the *visual* curve of the horizon - the actual shape that would be seen by an observer.

That's all covered in my post above, and on the Wikipedia page.

 

[ScarBest]

Who defines your angular field of vision?
How is field of vision specified?

That's a good question. How far one can see to the side when looking ahead is more a question of psychology than math or physics. I can calculate the shape of the curve at all peripheral angles up to a chosen limit. I've been using 45 degrees.

 

[PeroK]

That's a good question. How far one can see to the side when looking ahead is more a question of psychology than math or physics.

It's a question of angles. Unless you can measure the distance, angles are all you have. The zenith angle is the same in all directions.

I can calculate the shape of the curve at all peripheral angles up to a chosen limit. I've been using 45 degrees.

It's always a circle, viewed from a point above its centre.

 

[russ_watters]

At the surface the horizon looks flat because it covers half the sphere of vision; 180 degrees, so if you want radius it's 90 degrees. Once you calculate $\gamma$ then the radius is just 90 - $\gamma$

This is just a more complex case of calculating the angular diameter of a celestial object where its 3d (sphere) size matters instead of just it's 2d diameter. $\gamma$ is the harder part but as said, not that hard.

 

[jedishrfu]

Ignoring atmospheric effects, one point you may be missing is that at sea level, the horizon appears as a straight line unless you realize that it surrounds you as you follow the line. Hence, you are in the center of a circle.

The same is true as you go higher in altitude, but it becomes slightly more curved until you begin to see a circle.

 

[russ_watters]

I don't think OP is missing that, I think it's what they want to somehow quantify (which the math given so far pretty much already does).

 

[renormalize]

I don't know this. However, I have searched online and I've found nothing like it. If I were to submit it to a journal, the editor would know and inform me.

That attitude greatly disrespects journal editors and their referees. They aren't schoolteachers who critique and grade the solution to a student's homework problem. Editors rightly expect a submitter to be expert in their field and its literature. Otherwise, how can the submitter even judge whether their own work is novel and useful?

 

[PeroK]

That attitude greatly disrespects journal editors and their referees. They aren't schoolteachers who critique and grade the solution to a student's homework problem. Editors rightly expect a submitter to be expert in their field and its literature. Otherwise, how can the submitter even judge whether their own work is novel and useful?

The problem would fit better in the physics or mathematics homework forums on here.

 

[OmCheeto]

Apologies if this has already been answered

TL;DR Summary:
...visible curvature...

Has anyone determined what this means, mathematically?

By visible, does that limit us to human eyes and brain power?: 180° field of view, & resolution.
If not, then mathematically, we should be able to determine that the earth is curved from anything above zero altitude. A somewhat trivial problem which I solved several years ago.

If we are limited to human eyes and brains, then google claims we have to be at an altitude of 35000 feet to 'visually' see it.

In any event, I'm confused how to measure 'visible curvature' in other than in binary terms: 'Yes' it's curved, and 'No' it's not.

A quick search finds that wiki's circle definition of curvature yields 'curvature = 1/radius'
which indicates 3 terms:
1/(radius --> 0) = ∞
1/(0 > radius > ∞) = curved
1/∞ = not curved

Which strikes me as stupid, as changing how one measures the radius changes the curvature.

 

[Baluncore]

The observer is at the apex of the tangent cone surface. The radius of the horizon is the half-angle of the cone, measured by the observer, as the angle from the axis of the cone to the tangent horizon. The curvature of the horizon is (1 / radius angle), specified in units of inverse angle.

 

[OmCheeto]

The observer is at the apex of the tangent cone surface. The radius of the horizon is the half-angle of the cone, measured by the observer, as the angle from the axis of the cone to the tangent horizon. The curvature of the horizon is (1 / radius angle), specified in units of inverse angle.

66 years around the sun, and I have zero recollection of a single conversation regarding anything involving an 'inverse angle'.

 

[DaveC426913]

Has anyone determined what this means, mathematically?

By visible, does that limit us to human eyes and brain power?: 180° field of view, & resolution.
If not, then mathematically, we should be able to determine that the earth is curved from anything above zero altitude. A somewhat trivial problem which I solved several years ago.

If we are limited to human eyes and brains, then google claims we have to be at an altitude of 35000 feet to 'visually' see it.

Joining a little late but it seems to me the OP is talking about how an observer would see a segment of the horizon in a limited view - as if projected on a plane perpendicular to his line-of-sight. It would still be a circle but, being a different field-of view, it seems the apparent size would be different. (But I may be talking out of my hat. Dammit Jim! I'm a photographer, not a geometer).

Wide-angle shots versus human-eye-esque fields-of-view tend to greatly distort the proportions of objects.

In any event, I'm confused how to measure 'visible curvature' in other than in binary terms: 'Yes' it's curved, and 'No' it's not.

It would be quantifed as a subtended angle of the observer's view, just like any other object.
"At altitude X the horizon appears as a segment of a circle whose radius/diameter subtends Y degrees in my field of view."

Because of the limited field of view, and the oblique angle of observation I am not sure that the circle in diagram B (above) would appear to the observer to be the same radius as it would in diagram A.

 

[ScarBest]

Joining a little late but it seems to me the OP is talking about how an observer would see a segment of the horizon in a limited view - as if projected on a plane perpendicular to his line-of-sight.

Yes, exactly, thank you. However, the horizon circle would not be at the angle shown in your diagram, but rather at the angle that I've added in yellow. When the observer is looking at the central point of the horizon (from his point of view), he will be viewing the horizon circle at an inclined angle, so the circular arc of the horizon will be seen as an elliptical arc, due to foreshortening of the horizon circle arc in the direction in which the observer is looking.

 

[PeroK]

Joining a little late but it seems to me the OP is talking about how an observer would see a segment of the horizon in a limited view - as if projected on a plane perpendicular to his line-of-sight. It would still be a circle but, being a different field-of view, it seems the apparent size would be different. (But I may be talking out of my hat. Dammit Jim! I'm a photographer, not a geometer).

Wide-angle shots versus human-eye-esque fields-of-view tend to greatly distort the proportions of objects.
View attachment 356738

In the upper diagram, which you've marked with an X, the red line represents the horizon. In the second diagram the red line does not represent the horizon. The red line below the apex would be hidden behind the horizon. The horizon must be symmetrical in all directions. If it's an ellipse, then what defines the major and minor axes?

 

[Baluncore]

When the observer is looking at the central point of the horizon (from his point of view), he will be viewing the horizon circle at an inclined angle, so the circular arc of the horizon will be seen as an elliptical arc, due to foreshortening of the horizon circle arc in the direction in which the observer is looking.

I do not believe that. The observer is at the apex of a cone, the horizon is tangent with the sphere. The distance to the horizon, from the observer, is the same in any direction. The horizon will always be part of a circle, never an ellipse. To get an ellipse, you must cut a diagonal plane through a cone, but the horizon is a cut perpendicular to the axis, so must remain a circle, when viewed from the apex.

 

[DaveC426913]

In the upper diagram, which you've marked with an X, the red line represents the horizon. In the second diagram the red line does not represent the horizon.

Diagram not-to-scale.

The red line below the apex would be hidden behind the horizon. The horizon must be symmetrical in all directions. If it's an ellipse, then what defines the major and minor axes?

The point I am trying to draw attention to is the axis of the observer's line of sight. (The centre, dotted red arrow), and the observed angle of the arc of the horizon is perpendicular to that.

A camera would show this by projecting the curve onto its focal plane, which is perpendicular. Closing one eye would also achieve the same effect - by removing any clues about distance and foreshortening. @ScarBest: This is why I think your yellow line annotation is incorrect. But I may be misinterpreting you.


In diagram A, the observer's line of sight coincides with the centre of the circle - the horizon equidistant in all directions. And the viewing angle of the horizon is highly acute (magenta) not perpendicular.

In diagram B, it lines up with the horizon's near edge, and is perpendicular to it (magenta). Only a portion of the horizon is visible, most of it being off the bottom of the observer's field-of-view.

 

[PeroK]

Diagram not-to-scale.


The point I am trying to draw attention to is the axis of the observer's line of sight. (The centre, dotted red arrow), and the observed angle of the arc of the horizon is perpendicular to that.

A camera would show this by projecting the curve onto its focal plane, which is perpendicular. Closing one eye would also achieve the same effect - by removing any clues about distance and foreshortening. @ScarBest: This is why I think your yellow line annotation is incorrect. But I may be misinterpreting you.


In diagram A, the observer's line of sight coincides with the centre of the circle - the horizon equidistant in all directions. And the viewing angle of the horizon is highly acute (magenta) not perpendicular.

In diagram B, it lines up with the horizon's near edge, and is perpendicular to it (magenta). Only a portion of the horizon is visible, most of it being off the bottom of the observer's field-of-view.




View attachment 356757

Unless you get into the optics of an eye or camera lens, the horizon is a horizontal cut of a cone and is hence a circle. Even projected to an intermediate horizontal plane, you get a circle.

The right angle in the second diagram makes no sense to me in terms of points on the horizon.

 

[DaveC426913]

Unless you get into the optics of an eye or camera lens,

Exactly so. The OP mentions at least once or post that he is talking about what the observer sees visually.

If the observer looked out the window of his craft, with the horizon vertically centred, and the window perpendicular to his line-of-sight, he could draw an arc right on the window with a sharpie.

This arc would be of a portion circle/ellipse that would subtend a measurable amount of his view (at that specific distance from the window).

I doubt the radius of that curve would be the same as the actual horizon if he were looking straight down on the Earth, because, looking out the porthole like this, the rest of the horizon wouldn't be below his feet, it would be literally behind him, almost viewable out the opposite porthole.

The right angle in the second diagram makes no sense to me in terms of points on the horizon.

It's no longer points on a horizon; It's a curve on a plane perpendicular to the observer. That curve is what the observer sees, absent any 3-dimensional/foreshortening clues.

Maybe I've got it wrong. If the OP can't seem to nail down a description of what he's doing, I'm not going to do much better.

 

[PeroK]

This arc would be of a portion circle/ellipse that would subtend a measurable amount of his view (at that specific distance from the window).

It's a circle. It can't be an ellipse.

I doubt the radius of that curve would be the same as the actual horizon if he were looking straight down on the Earth, because, looking out the porthole like this, the rest of the horizon wouldn't be below his feet, it would be literally behind him, almost viewable out the opposite porthole.


It's no longer points on a horizon; It's a curve on a plane perpendicular to the observer. That curve is what the observer sees, absent any 3-dimensional/foreshortening clues.

Maybe I've got it wrong. If the OP can't seem to nail down a description of what he's doing, I'm not going to do much better.

You can't see a defined radius, you can only measure a radius. A ball held at a certain distance will appear to have the same radius as the Moon. You need to calculate the distance before you can say how big something is.

 

[DaveC426913]

It's a circle. It can't be an ellipse.

A circle, seen at an oblique angle is observed to be an ellipse.

The oblique angle is due to the fact that the plane of his image is in front of him.

If he used that sharpie to extend the arc of the horizon as seen through the porthole, he could see the entire curve on the wall under the porthole at his feet.

You can't see a defined radius, you can only measure a radius. A ball held at a certain distance will appear to have the same radius as the Moon. You need to calculate the distance before you can say how big something is.

A red herring. We're not talking about how big something is, we're talking about a curve, as seen by the observer that subtends a portion of his vision.


We are clearly talking past each other. I don't know what else I can say. Best I can do is have you re-examine these diagrams. (I have eliminated the red lines representing the horizon - that is causing confusion.)

One has the line-of dight aligned with the centre of the Earth; its plane of imaging tangential with the Earth's surface, and the entire horizon visible.

The other has line-of-sight aligned with the horizon; its plane of imaging perpendicular to the Earth surface, and only a portion of the horizon in his field of view (much of it is outside his line of sight - even below/behind him).

Would the two arcs - draw with a blue sharpie - on the planar surface - have the same radius?

 

[PeroK]

A circle, seen at an oblique angle is observed to be an ellipse.

The oblique angle is due to the fact that the plane of his image is in front of him.

Sorry, this is still wrong:

https://physics.stackexchange.com/q...izon-visually-appear-curved-at-high-altitudes

 

[DaveC426913]

Sorry, this is still wrong:

https://physics.stackexchange.com/q...izon-visually-appear-curved-at-high-altitudes

It is not wrong. A circle, seen at an oblique angle is, in fact, an ellipse. Your reference is a red herring. (It doesn't mean you don't have a valid point, but you haven't made that case with what you wrote.)



A circle the size of a horizon can only be seen if you have a completely undistorted 360 view from edge-to-edge, which humans don't have when looking horizontally at one part of the horizon.

If human is looking at the horizon from the ISS, they will see a curve - even though the other edge of the curve is literally behind him. But he can still sketch out the curve he sees on the plane that is perpendicular to his line of sight - with the caveat that, at some, point, his observed curve will deviate from the true circle due to perspective distortion. I guess what he would draw is a parabola, with the ends reaching the floor.

 

[PeroK]

It is not wrong. A circle, seen at an oblique angle is, in fact, an ellipse. Your reference is a red herring. (It doesn't mean you don't have a valid point, but you haven't made that case with what you wrote.)

Please calculate the eccentricity of the ellipse as a function of altitude.

 

[DaveC426913]

I ask you look again at the two diagrams in post 37. In the second diagram, the plane on which the curve is drawn is not perpendicular to a line drawn to the centre of the sphere - it is oblique.

Instead, it is perpendicular to the top edge of the horizon.

Therefore, the image on the plane's oblique 2-dimensional surface will be distorted from a circle.

 

[DaveC426913]

Please calculate the eccentricity of the ellipse as a function of altitude.

That is exactly what the OP is claiming he can do. If I had solved that, I would have surpassed the OP.

 

[Baluncore]

Next thing, someone will claim that a rainbow does not form a perfect circle, because you are looking at a small part of it, or part is underground. No part of a rainbow is ever an ellipse, it is always a circle to the observer.

 

[PeroK]

View attachment 356769
I ask you look again at the two diagrams in post 37. In the second diagram, the plane on which the curve is drawn is not perpendicular to a line drawn to the centre of the sphere - it is oblique.

Instead, it is perpendicular to the top edge of the horizon.

Therefore, the image on the plane's oblique 2-dimensional surface will be distorted from a circle.

That circle is NOT part of the horizon. You are always looking at the horizon from the conical centre, never obliquely. Oblique would be off centre. As drawn erroneously in that diagram, imagining that different points on the horizon are different distances away from the observer.

This is sheer nonsense now.

 

[DaveC426913]

All right. I concede. I thought I knew where the OP was going, but it is possible I have over-interpreted his idea. Thanks guys for being patient with me.

 

[DaveC426913]

Next thing, someone will claim that a rainbow does not form a perfect circle, because you are looking at a small part of it, or part is underground. No part of a rainbow is ever an ellipse, it is always a circle to the observer.

It's an interesting analogy; one I tried to employ, actually, but abandoned. But it kind of highlights where I've been going. A rainbow will always fit with an observer's point of view, with negligible distortion, and can therefore be treated as a circle.

The Earth's horizon, OTOH, can easily extend well-outside the observer's field of view, necessarily distorting it from a circle - as the observer sees it in the 2D plane of his vision.

 

[PeroK]

The Earth's horizon, OTOH, can easily extend well-outside the observer's field of view, necessarily distorting it from a circle - as the observer sees it in the 2D plane of his vision.

At some point, depending on your measurement apparatus, you measure only the arc of a circle. Then the circle is constructed from many of these homogeneous arcs.

 

[Baluncore]

The Earth's horizon, OTOH, can easily extend well-outside the observer's field of view, necessarily distorting it from a circle - as the observer sees it in the 2D plane of his vision.

Those two dimensions are azimuth and elevation in spherical coordinates, it is not a flat x-y plane. The observer, at the centre of those spherical coordinates, is located on axis, at the apex of the tangent cone. The eye is not a flat sheet like an image sensor, or a flat photograph.

The only time the circle of the horizon, (or a rainbow), can be an ellipse, is when a diagram is drawn from the viewpoint of a third person, (an artist not on the conical axis), picturing an observer at the centre of the observer's horizon, with the horizon and rays sketched as seen by the primary observer.

 

[russ_watters]

All right. I concede. I thought I knew where the OP was going, but it is possible I have over-interpreted his idea. Thanks guys for being patient with me.

Honestly, I think you're 95% of the way there, you just got a small detail slightly wrong (the angle in your second photo, which the OP pointed out in the next post). Otherwise and more importantly the main point does seem to be what the OP is after: The horizon is flat when you look at it level at zero elevation and curved downwards when you look at it from higher elevation.

I was hoping my planetarium software would show this, but unfortunately it gives a limited and cartoonish view.

 

[DaveC426913]

Honestly, I think you're 95% of the way there, you just got a small detail slightly wrong (the angle in your second photo, which the OP pointed out in the next post).

Yes. I regret even putting that line in. It wasn't meant to be to-scale. It was a sketch, not meant to be nit-picked.