# Seeing something from only one angle means you have only seen (what?)% of its surface area at most?

Is there a logical/mathematical way to derive what the very maximum percentage of surface area you can see from one angle of any physical object?

For instance, if I look at the broad side of a piece of paper, I know I have only seen 50% of its surface area (minus the surface area of the very thin sides). Is 50% always the maximum amount of surface area you can see of any object from one angle?

Assumptions: This is assuming we aren't considering transparent/semi-transparent objects or the extra parts we can see with the help of mirrors. Just looking at the very surface of an object from one particular angle.

• There was a Vsauce video recently on a related topic How much of the earth can you see from space Aug 7, 2017 at 19:42
• A more interesting question: given a surface $f(x,y,z)=0$, find the point in space from which we can see the maximum percentage. Aug 11, 2017 at 19:32
• Clarification - are you using a single point of vision or normal human binocular vision? I can see approximately 100% of a toothpick's outer surface (less whatever part is obscured by my fingers as I hold it) pointing directly out of the bridge of my nose. Each eye sees approximately half, and the sharp end pointing away is pointy enough that it comes to a point. However this relies on binocular vision and ignores the "one angle" requirement. Aug 12, 2017 at 4:51
• Think of looking at a screw from the head (something like i.imgur.com/DiKIj19.jpg). Since you can't see the length of the screw from this viewpoint, its surface area can grow without bounds. Aug 12, 2017 at 10:17
• I assume that we talk about "seeing" as one point as opposed to putting the paper between your eyes and seeing both sides with one eye each. Aug 15, 2017 at 9:01

There is no such upper bound.

As a simple counter-example, consider a thin right-angled solid cone of base radius $r$ and height $h$, observed on-axis from some large(ish) distance $z$ away from the cone tip. You then observe the tilted sides, of area $\pi r\sqrt{r^2+h^2}$, and you don't observe the area of the base, $\pi r^2$, so you observe a fraction \begin{align} q &=\frac{\pi r\sqrt{r^2+h^2}}{\pi r^2+ \pi r\sqrt{r^2+h^2}} \\ &= \frac{\sqrt{1+r^2/h^2}}{r/h+\sqrt{1+r^2/h^2}} \\ &\approx 1- \frac rh \end{align} of the surface, in the limit where $r/h\ll 1$, and this can be arbitrarily close to $1$ so long as the cone is thin enough and long enough.

• Aug 9, 2017 at 1:45

As a completely tangential type of answer. Consider a neutron star; due to the General Relativistic bending of light in curved space we are not bounded by the dull constraints of Euclidean geometry!

If the radius falls below 1.76 times the Schwarzschild radius for its mass$$^1$$, then then all of the surface is visible, when viewed from any direction (e.g. Pechenik et al. 1983; Kraus 1998). Light rays emitted tangentially to the surface from a point on the neutron star that is opposite a (distant) observer would be bent through 90 degrees whilst travelling to the observer. According to that observer, the antipodal point would actually form the circular boundary of the observable disk and hence they could "see" 100% of the surface.

For neutron stars with radii even less than this (as a fraction of the Schwarzschild radius), then there are multiple paths for light to reach the observer for some parts of the surface and they could be seen more than once on the visible disk.

e.g. a neutron star just below the critical $$R/R_s$$ value where the entire surface can be seen. (From http://www.spacetimetravel.org/ssm/ssm.html )

$$^1$$ @SeleneRoutley asks whether such objects are theoretically possible or stable? A full answer to this is contained within my response to this Physics SE question, but to summarise:

GR does impose a limit on the mass/radius ratio of neutron stars, but the exact limit depends on the equation of state governing the structure. The diagram below shows the theoretical mass-radius plane, adapted from Demorest et al. (2010). The shaded part top-left and marked "Causality" is outlawed by GR and any equation of state where the sound speed is less than the speed of light (as it must be).

I have added a thick red line which marks where the radius equals 1.76 times the Schwarzschild radius. Any neutron star above this line would have a completely visible surface from any viewing direction.

Now there are possible equations of state (marked as labelled loci in this diagram) that allow high-mass neutron stars to exist above the red line. However it is currently not possible to measure the mass and radius of any neutron star with enough accuracy to place them unequivocally above this line, but the measurements for some are consistent with this.

The phenomena discussed above are "routinely" taken account of when inferring spatial structures associated with the pulse profiles of rotating neutron stars at X-ray wavelengths (Sotani & Miyamoto 2018).

• Awesome! I presume neutron stars this dense / massive can survive without collapsing further? Aug 11, 2017 at 13:30
• @WetSavannaAnimalakaRodVance The 1.76 S.R. limit is ok for lots of equations of state. Smaller neutron stars may not be stable, depending on the EOS. There is a hard GR limit at about 1.2 S.R. Aug 12, 2017 at 11:11
• Thanks, Rob, my expectation was that you would have qualified the question if there was a caveat about stability, but just wanted to make sure. I was aware of the hard GR limit, but I don't have many knowledges of real equations of state, so my comment was more about them. It's a wonderful, out of left field answer whatever the caveats, though! Aug 12, 2017 at 11:26
• This is so far the only answer considering physics. All the others assume Euclidean geometry and don't consider what is physically possible. Aug 13, 2017 at 17:38

Consider a thin piece of paper, then as you said you can see only half of it's surface. Now, take an arbitrary point in the middle of the paper, and extrude the sheet from this point, ie. create a pyramid. You can increase the apparent surface as much as you want (just by increasing the height of the pyramid), so you can see from 50% (included) to 100% (not included) of the surface area of this object. Looking at the other side of the pyramid, can see 0% to 50% of the object.

• Wouldn't the concave side still be ~50% of the paper's surface area, though? Aug 8, 2017 at 17:56
• What do you mean by the concave side ? Aug 8, 2017 at 21:13
• @user1717828: I am sure it is meant to be a filled pyramid, with no concave part. Aug 9, 2017 at 3:42

There are not bounds in either direction.

## Invisible majority:

Take a rod (a solid cylinder with circular cross section, and disks for end caps) of radius $r$, axis of revolution running along the $z$-axis, lower end cap at $z=1$ and upper endcap at $z = u$. Place the observer at the origin. The lower endcap is the only portion of the cylinder that is visible, so the visible fraction is \begin{align*} &(\pi r^2) / (2 \pi r^2 + (u-1) \cdot 2 \pi r) \\ &= 1 / (2 + 2(u-1)/r) \text{.} \end{align*} By increasing $u$ or decreasing $r$, we can make this last expression as small as we like.

The basic idea here is that we can arrange for the object to have a small "shield" close to the observer, which occults a huge surface.

## Invisible minority:

Take the spherical shell of inner radius $1$ and outer radius $2$. (This is the set of points with $1 \leq \text{radius} \leq 2$.) Drill $n$ cylindrical holes of radius $r$ along rays from the origin. For the same reason that the curved wall of the cylinder in the above case is invisible, the curved wall of these drillings is visible. Suppose we choose $n$ so that half of the inner sphere's surface has been drilled through. (We could do better. We can easily exceed 50% by close packing the spherical caps corresponding to the drillings on the surface of the inner sphere.) Then the ratio of visible to all area is \begin{align*} &( (1/2) \pi + n \cdot 2 \pi r ) / ( (1/2) \pi + n \cdot 2 \pi r + (7/2) \pi) \\ &= 1 / (1 + 7/(1 + 4 n r)) \text{.} \end{align*} Notice that if we reduce $r$ by a factor of $2$, we increase $n$ by a factor of $4$, so by decreasing $r$, we may increase $1 + 4 n r$ as much as we like. That is, we can make this ratio as close to $1$ as we like.

The basic idea here is to make most of the surface area barely converge toward the rays of vision so that only a small amount of the surface is occulted. Then fill in the volume "behind" these convolutions so that the resulting back face has a small area.