Earth angular size looking from the ISS

Question

Imagine you're, well, actually on the ISS so you're 395 km up.

We've all seen many photos roughly like this... however this can be deceptive;

• They might be all fisheye lenses.
• A great many of these images are simulations or comps.
• It's hard to get the real view if you don't know what lenses are used.

In short, at that height, what angle does the Earth fill horizon to horizon?

• When you're standing on a (very flat) part of the Earth it's indeed roughly 180 degrees.
• When you're on the moon it's a degree or so.

What angle, actually does the Earth take when you're at 400km?

• Does it in fact feel like the Earth is pretty flat beneath you?
• Or does it really look more like a ball?

Perhaps someone has an astronaut's comments on this or?

Again - Hollywood representations tend to fill our minds, but those could be basically quite wrong.

This would be very useful to know in making a VR where you are at that position.

I've found it difficult to get any real info.

Answer 1

As you mention, the horizon seen at ground level will appear as a plan taking up 180° of our field of view. This is known as the astronomical or sensible horizon. As soon as the observer gains altitude the horizon line moves below the observer horizontal plan, by an angle called dip angle.

The separation between the Earth and the sky now appears under an apparent angle less than 180°. This value is called the true, or geometrical or geographic horizon. This is the horizon we usually talk about in everyday life.

The visible horizon is the actual line between the sky and sea, trees, mountains, and any other obstacles. We must also be aware of the atmospheric refraction which allows us to see further than the geometrical horizon (curve in blue). But for our calculation we will just use the geometrical horizon.

The separation line itself is named the limb. The limb appears to the observer as a circle which apparent diameter decreases with observer's altitude.

Angular size of the Earth limb for an observer in altitude.

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The problem can be represented like in the image below, with distance OP being the sum of the Earth radius (OA) and the altitude of the observer.

The limb is the intersection of the sphere and the cone with vertex P and tangential to the sphere at A and B on the figure (tangential to the limb actually).

Let's assume the observer is at an altitude of 400 km.

• OA = 6371 km.
• OP = 6371 + 400 = 6771 km.

The angular size of the limb is the angle

APB = 2 * APO.

From elementary trigonometry we know that

sin APO = OA / OP = 6371 / 6771

and we can calculate APO = (about) 70°.

So APB = 140°.

It means for an observer aboard the ISS, the Earth limb appears as a disk with an apparent diameter of 140°. Of course, this apparent size will vary when the ISS altitude varies. The ISS orbit is not circular (eccentricity = .0003805), and the Earth area swept by the ISS is not a perfect sphere portion. But most significantly the ISS is subject to orbital decay between reboosts.

Note that at this altitude of 400 km, the angular size of the hidden equator is only 86°.

What does it look like from the ISS?

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ISS residents see this:

(source)

or this:

(source. Tribute to Isaac Newton)

Camera requirements to shoot the full Earth limb

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The lens required to get a full monolithic picture of the Earth must have a vertical angle of view of 140°. For a regular lens, angle of view, sensor size and focal length are linked by:

AoV = 2 * arctan ( size / (2 * f) )

For a full frame 35 mm sensor the focal length to take in 140° is 4.2 mm. For a smaller sensor, the actual focal length would be even shorter.

Such small lens are be difficult to build and locate close to the sensor (the shorter the focal, the closer to the sensor). After 110° we usually prefer a fisheye system which uses another type of projection. While a regular lens perform a rectilinear projection with a single vanishing point on the sensor, the curvilinear projection of a fisheye has five vanishing points.

The fisheye lens has a longer focal for a given angle of view, and the latter can be very large (more than 180°, which is not possible for a linear lens). There are different types of fisheyes, but the most commercialized is one using an equisolid angle projection, where equal solid angles of the 3D space are projected as equal areas on the 2D sensor. This time the relation between AoV, size and focal length is:

AoV = 4 * arcsin ( size / (4 * f) )

We can take in 140° with a lens up to 10.5 mm. This image is shot with a 8 mm, allowing to capture the limb, extra sky space and copula panels...

(source -- Focal length: 8 mm, on a full frame sensor)

The appropriate lens is not the only condition required to see the whole Earth disk, we also need a window providing a good aperture...

Is it possible to see the entire Earth disk through an ISS window?

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The largest window ever used in space is the nadir one of the ISS cupola module. it is a circular glass panel with a diameter of 80 cm.

It allows a panoramic view from the Earth-facing side (nadir) of the ISS.

(source 1 -- source 2)

Note how different the visible Earth portion is between the two pictures taken from the cupola. It's because on the first one the camera was closer to the center window. Curiously on the second image, while the camera has moved back, the visible cupola portion is smaller. It's because the angle of view of the lens is smaller in the second image (rectilinear lens) than in the first one (fisheye).

The possibility to take in the full Earth limb is discussed here: How far into space does one have to travel to see the entire sphere of earth?.

The discussion is about another spacecraft, and another window size, it says that you need to be at about 525 km altitude to see the whole limb from a 70 cm window.

Let's look at the ISS case at 400 km, with a 80 cm window, and Earth apparent size of 140°. How close from the glass the eye must be placed, so that the whole limb can be seen through the window?

(drawing taken from the linked question)

The values of d, Land θ are linked by the tangent formula:

tan ( θ / 2) = ( d / 2) / L

then:

L = (d / 2) / tan (θ / 2)
L = 14.5 cm

The eye must be at a maximum distance of 14.5 cm of the external panel.

Depending of the thickness of the window this may be possible (thinner than 14 cm) or not (thicker than 14 cm).

Answer 2

It's about 140 degrees, varying a little as the ISS altitude varies.

To get an idea of what that 140 should look like, your fist held out at arms length occupies about 10 degrees. So, look to your left, and measure two fists in. And the same on the right.

Note that this exceeds the available FoV of all VR headsets at the time of writing, apart from the HTC Vive which does 145 on the widest setting, but the edges look blurry due to the fresnel lenses.

It also exceeds the eyes natural vertical FOV, of about 135 degrees (and 120 horizontal binocular vision).

So you won't see the full Earth disk in VR, and you likely won't see it if you were in space using your real eyes either, unlike what the above "answer" claims. You will have to move your head or eyes slightly from side to side and up and down to see the full disk. And that's if the spacesuit helmet didn't get in the way (a quick google didn't show any numbers, but just by looking at a few pictures it looks nearly 180)

I was a bit surprised when I calculated how it looks from geosynchronous orbit. The Earth only occupies about 20 degrees of the sky. That's pretty high altitude.

For VR rendering you will not need to render a curved planetary surface. Unreal Engine will by default switch from stereo to mono rendering for objects further than 750m (i think). The parallax visible at that range starts to be too small to bother about, so everything further than that is rendered in mono, and could just as easily be projected onto a flat skybox, or the inside surface of a giant sphere. Obviously lighting, texturing, atmospheric effects and so on contribute to a curved look, but the curved planetary surface itself doesn't need to be explicitly rendered as one in VR. Just as a full moon looks flat to us on Earth, the difference between the distance from the astronaut to the center of Earth and the distance from the astronaut to the edge is indistinguishable.