What is the Ringworld?

“Ringworld” is a Sci-Fi novel by Larry Niven.

Although there is a lot of criticism in this page, it is a tribute to an extraordinary idea which still fires the imagination so many decades after its publication.  Thank you Larry!

The story is about a world in the form of a ring around a star, built by some mysterious race.  No further spoilers! Go read the book!

The ring is constructed around a star similar to our Sun.  It has a diameter similar to that of Earth’s orbit.  Very simplistically you can think of it like this:

The ring rotates at a speed which simulates a force of gravity also similar to that felt on Earth.  Therefore it is possible to stand up on the inside (while rotating along with it of course):

To live on the Ringworld there obviously needs to be more than a strong floor to stand on:  the floor needs to be very wide, loaded with arable topsoil, there need to be rivers and lakes and above all breathable air.

In the following we refer to four layers composing the Ringworld:

• the floor, made of “scrith”
• a layer of bedrock
• a layer of topsoil
• a layer of air

The material of the floor needs to keep itself and all it supports together in a ring.

There are two more things before the Ringworld can be habitable:

• there need to be walls on the side, otherwise the layer of air would flow off into space,
• the people need to be shielded from the constant sunshine. (see chapter )

What is the Problem?

It does not matter in which sense the ring rotates, the rotation is needed to provide the centripetal force which allows objects to stand on the inner surface.

But “scrith”, the material of the floor, is therefore being pulled outwards:  it needs to be strong enough to hold the ring together.

You may have read that “the force is close to that of the strong nuclear force”.  That cannot be a correct interpretation of what we are after. The strong nuclear force is the force between quarks, it is about 10'000N, but it is a force between two individual objects.

What we are after is how much force per square metre the scrith must be able to hold.  This is its tensile strength.

The tensile strength of mild steel is about 250N/mm², that means that a wire with a cross-section of 1 square mm can hold a force of 250N before beginning to stretch and potentially break.  250N is about the force needed to lift a weight of 25kg.

If instead of steel you use a piece of string made from wool, it will need to be thicker because the tensile strength of wool is less than that of steel.  The force of the 25kg puling on it is the same however.  Note that the tensile strength of mild steel is 250×1000×1000N/m², i.e. a steel bar with a cross section of 1m² could hold up a weight of 25 million kilos.  The suspension cables of the Golden Gate bridge are 0.68m², they hold 17'000 tons each.

Why build a Ringworld?

Say you want more living space than what you can get on the surface of a planet.  Planets are balls of matter, held together by their own gravity.  The larger you make a planet, the more surface area you will have to live on, but the gravity gets stronger too.  Beyond a certain point, it becomes difficult to move objects or even just your body.

Niven’s solution to get a lot of living space is to build a world on a thin, wide ribbon, that circles its sun:  the Ringworld.

Olaf Stapledon, in “Star Maker” mentions civilisations collecting the entire output of their sun.  Freeman Dyson in his autobiographical book “Disturbing the Universe”,admits that this idea led him to propose what is now known as a “Dyson Sphere”,  not a ribbon but an entire sphere surrounding a sun.

How big is the Ringworld?

Niven’s world is big.  If watched from a point some 400 million km away from its sun, it might look like this:

Though note that in this view the wires holding the shadow squares (see ) together are visible, which they would not be (you may have to zoom in on the image).  The star in the centre is the same size as our Sun, the diameter of the ring is the same as that of the Earth’s orbit.

Dimensions

There are several parameters given in the books. The table below is a summary, “Ringworld” refers to the original novel, “Ringworld Engineers” refers to the second novel.  I assume you are familiar with numbers in scientific notation (powers of ten)

The books use a confusing mixture of imperial and metric units.  The values are not all coherent.  We’ll get to that a little later.

Note on Ringworld “gravity”: this is in fact the centripetal force the Ringworld floor has to exert on the stuff on top of it, in order to keep it circling around the sun.  The faster the ring turns, the greater this force has to be.  The ring turns at a speed such that the force is similar to the force of gravity on Earth.  But 31 feet-per-second per second is 9.4488m/s².

On Earth gravity is 1g = 9.81m/s², varying by no more than 0.7% around the globe.  Therefore the Ringworld gravity is only 0.963g, contradicting the value of 0.992g as given in the novels.

Why are there rim walls?

Apart from the topsoil and water, on top of some rocky layer, there also needs to be air.  The air would flow off the sides, unless there were sufficiently high walls along the rim of the floor.  These are the rim walls.

Meteorite impact shield

Although I did not find a direct reference to it, there is a meteorite impact shield, made from scrith foam.  This layer of foam is on the outside of the floor, and presumably also on the outside of the rim walls.

Cross section

The cross section of the Ringworld then roughly (and very oversimplified) looks like this:

The scrith bottom is not supposed to be flat, it is said to be sculpted to provide relief:  moutains, lakes, rivers.

And the sun is directly overhead.  It shines all the time.  Or does it?

Shadow squares

There are ancillary problems such as coping with the erosion caused by flowing water, making relief (mountains, lakes, rivers), and so on.  The most important ancillary problem however is shielding Ringworld life from the sunshine:  the sun is always exactly overhead.  Without periodic shielding from the sun, life would be burned to death.  Niven’s solution is a second ring of “shadow squares”, which also turn around the sun, but are closer to it. They are spaced, held together by (strong) wires.  The shadow squares cast shadows on the ring, and if they turn at a different speed than the Ringworld, then their shadows move along the ring, providing a simulation of day and night.  Simplistically:

(the wires holding the shadow squares are not shown)

The shadow squares are actually not squares, but that’s what they are called in the novel.  They make the views of night and day somewhat strage.  In the first image at the beginning, you can already see the shadows on the ring as it arches up from the horizon.  If you look up about 45º, you would see this:

If you looked at the sun, directly above you:

At “night”, when the shadow of a shadow square falls on the area where you stand, it would be somewhat like this perhaps:

And if you looked straight up at the sun at “night”:

Directions

While standing on this fast revolving ribbon, we will call downward the direction outward from the sun, upward the direction towards the sun.  Two other useful directions are spinward and antispinward:  in the direction of the rotation and against it.  Finally, the direction to the rim wall, while facing spinward, is called port and the other direction is starboard.  Directions are not important for this page.

Summary

Physics and Astronomy

Before we try to calculate the tensile strength of scrith, we must understand some basic physics, mechanics and astronomy.  Below are some useful notions.

We will (of course) use the metric system.  One of the two basic ideas behind the metric system was to use only factors of 1 or powers of ten.  A mile is 1'760 yards, but a kilometer is 1'000 metres.  An inch is 0.08333 feet, but a centimetre is 0.01 metres.

Physics

Let us go over these line by line:

A length or distance s is measured in metres, unit:  m. A door is 2m high, a normal person is around 1.8m tall.  Standard ceiling height is 2.5m.

A mass is measured in kilograms, unit:  kg.  A liter of water, (approximately also milk, beer etc.) is 1kg.  Mass is NOT weight!

A span of time is measured in seconds.  Note:  there are 60 seconds in a minute, 60 minutes in an hour and 24 hours in a day;  therefore 24×60×60=86400 seconds in a day.  Originally the metric time unit was 1/100'000 of a day, but in the end the classic second was kept, and therefore the second is not strictly speaking a metric unit.

An angle is measured by the length of a circle’s radius around its circumference.  Therefore there are 2π radians to go around a full circle or 360º.  A radian is about 57.3º.

After the units of the metric system were defined to very high precision and reproducibility, the old Imperial units were defined in terms of the metric units:  the inch is defined as 25.4mm, from which the lenghts of the foot and mile follow.

All other mechanical units are combinations of the three fundamental units of length, time and mass.  For example, speed is measured in metres per second.  Kilometres per hour is NOT a metric unit, because the hour is 3600 seconds, and 3600 is not a power of 10.

Further physics involving electrical phenomena need a unit related to electric charge.  That is the Ampère, but we don’t need electric stuff here.

Acceleration is the increase of speed over time, hence is measured in metres-per-second per second.  This shows s² or “square seconds” which is a little strange at first.

The unit of force is N or Newton (in honour of Isaac Newton) and it is the force necessary to give a mass of 1kg an acceleration of 1 metre-per-second per second.  That force is about the same force as the weight of 100grams of mass.  At Earth’s surface, the weight of a kilogram of something is a force of 9.81N.  This is close to 10N and is how you should remember the value of a Newton of force:  about a tenth of the weight of a kg or a litre, at the Earth’s surface.

Angular speed, the rate of turning of a spinning object, is measured in radians per second.

Mechanics

Let us again go over these line by line.  Circular motion is the movement of an object along a circle:

The object moves at a speed v, which is depicted as the blue vector (arrow) in the figure at the left.  If there were no centripetal force F_c, the object would just continue straight along its speed vector.  The force needed to keep it going round the circle must be perpendicular to the speed vector v.

Because it wants to go straight, while moving around the circle the object subjectively still feels an urge to fly away from the centre of the circle, and this is why we as persons feel a “centrifugal force”.

But what keeps us going around the circle needs to be a force that pulls towards the centre.  Because F=ma we know the acceleration (change in speed) needed, and this change is not one of size but of direction only, which is why the centripetal acceleration is perpendicular to the speed vector.

We will not derive the formula here, but the centripetal force is proportional to the mass of the object and proportional to the square of the angular speed.  Conversely, for a given force, mass and radius of orbit, this gives the angular speed.

Two other formulae give the acceleration as a function of the orbital speed and the orbital speed as a function of the acceleration;  we will need them later.

Gravitation is the attraction between two masses.  It is a force that is proportional to the mass of each object, times a very small number called the gravitational constant.  The gravitational force diminishes with the distance d between the masses:  at twice the distance the force goes down by a quarter.

There is a simple explanation for this inverse-square law:  the gravitational influence of an object spreads through space, but as it goes further out, it spreads over a larger area.

The figure on the right shows that the area increases by a factor four when the distance increases by a factor two:  an area grows as the square of its sides.

Now suppose we have an object of mass m, e.g. some planet, that orbits around an object of mass M, say our Sun.  In reality they will both turn around a point in between, which is the centre of gravity of both masses. But if M is very much larger than m, this point is very close to the centre of the Sun.  So we can think of the planet as turning around a stationary Sun.

To achieve an orbit in free space, no strings attached between the Sun and the planet, we only have the gravitational attraction F_g of the Sun as centripetal force F_c on the planet.  Thus F_c = F_g and since we know both, we can find the orbital speed for a given distance r from the Sun. This is Kepler’s Third Law of planetary motion.  If the planet comes closer to the Sun, it has to move faster to prevent falling into the Sun, if it moves further away, it has to move slower or it will fly off into deep space.

Kepler’s First Law is that the orbit is an ellipse.  But since a circle is a special case of an ellipse.  Since most planets  move in orbits that are close to a circle and since the Ringworld is circular, we will only consider circular orbits here.

Ringworld Physics

The Solar System

In the novels Niven compares the Ringworld to objects in our solar system, therefore we give a few measures of the Solar system here:

The mass of the Sun is enormous, it is a thousand times that of the planet Jupiter.

The Ringworld has air, and to calculate the tensile strength of scrith, we need to know its mass, since it also “stands” on the ring.  We assume that the atmospheric pressur at ground level on the Ringworld is approximately the same as on Earth.  On Earth the pressure at sea level is about 10N/cm² and this corresponds to a mass of 1kg weighing down on 1cm² or ten tons per square metre.  These ten tons of air are spread over the entire height of the atmosphere, with the density gradually diminishing as one goes higher and higher.  At sea level the density of air is about 1.2kg/m³.

The Ringworld System

As shown above, Niven gives dimensions in the first and second novel, using a mix of metric and imperial units.  The numbers are summarised here again, in metric, so we can calculate more easily:

Remaining uncertainties

Some numbers in the novels are vague.  We will have to make some decisions, but fortunately the result of the tensile strength calculation does not depend significantly on them.

The length of the day is given as 30 hours, but there is no definite indication that these are exactly Earth hours.  We will assume so, and also that a day is divided into 15 hours of sunshine and 15 hours of night.

The ring mass is a very round number, but it is said to be close to the mass of our planet Jupiter.  We will stick to the round number.

We have no other indication about the sun than that it is “barely smaller and cooler” than the Sun.  “Barely” as in “barely any users upgraded to Windows11” may be less than 1% (as mentioned by one on-line article at the time I write this in 2022).  We shall assume 0%. i.e. the Ringworld sun has properties identical to ours.

The radius of the ring is close to but slightly larger than the radius of Earth’s orbit.  It puts the ring’s surface at the same distance from its sun as the Earth is, but this configuration gives the inhabitants about the same amount of radiation as at the Earth’s equator.  The day is 30 hours, assuming it is divided equally between daylight and night, that exposes the inhabitants to 15 hours of sun directly above.  Can they stand that?  But the night may be longer.

Inconsistencies

From the mechanics of circular motion we know the relation between the orbital speed and the centripetal acceleration for a given radius:
a_c = v²/r
v = √(a_cr)
We can calculate one from the other.

Niven gives the rotation speed as 770 miles per second, which needs an acceleration of 10.04m/s² but he gives the acceleration as 0.992g (9.73m/s²), which gives a speed of 1'220km/s or 758 miles per second.  He also states the gravity is 31 feet/s² which is 0.9449m/s² or 0.9632g and gives a speed of 1202km/s or 747 miles per second.  These numbers are actually quite close to each other but still inconsistent.  A table compares them:

We will take Niven’s acceleration of 9.45

Ringworld Composition

The necessary centripetal acceleration will be the decisive number for calculating the tensile strength.

In holding things together the scrith is helped a little by the fact that the sun’s gravitation attracts the ring and thus already provides a fraction of the centripetal force needed.  How much is that help?  At ringworld distance from the sun (which, as we decided above, has the same mass as our Sun), a kg of mass is attracted by a force of 0.0057N (F=GmM/r²).  Which, by virtue of the metric system’s use of factors that are only 1 or powers of 10, is also a centripetal acceleration of 0.0057m/s².  But we need an acceleration close to 9.8m/s².  Even at the lowest value of 9.2m/s² from the table above, the help we get from the sun’s gravitation is 0.062%, less than a tenth of a percent!  And we err on the positive side, since the sun is smaller than our Sun (though “barely”), hence helps even less.  We will neglect that help.

The stuff on top of the scrith

That leaves us with first finding out how much stuff the scrith needs to hold together:  recall the cross section from earlier.

I found an illustration by John Hewitt, presumably from his book “Ringworld Companion”:

Hopefully John will not be unhappy that I mention him and his book;  I could find no easy way to contact him.  Much in the table below is derived from the novels and John’s illustration.

(*) I do know that “a column of air on a m²” on Earth is slightly tapering outward because of the Earth’s curvature, and slightly tapering inward on the Ringworld for the same reason, but since the significant layer of air is only about 20km thick, this tiny difference can safely be ignored.

Go through this table row by row again.

The mass of air bearing down is assumed the same per m² as on Earth.

Take an average of 15m topsoil (and water). The density of topsoil varies, but a good average seems to be 1.5.

Take also an average thickness of the bedrock of 15m.  We are told there are no minerals, so the density is about 2.5, average for rocks.

I call structural scrith the part that bears all the load of holding the ring together.  The thickness of the structural scrith is said to be 50ft “if it were hull metal”.  By hull metal is meant the material used by another alien race to build their space ships.  That thickness is only a bit more than 15m, but John Hewitt’s drawing gives an average value of 35m.  That permits a smaller tensile strength.

The total thickness is then 65m, which is 3 times what the novel says, but maybe Niven meant metres not feet when he wrote the number.

There is a meteorite impact shield, made of scrith foam.  I found industrial foam is around 10% of the density of the solid material, but there is much less dense and much denser foam.  10% is a guess.  I assume that this foam does not help hold the ring together.  However it counts for the mass of stuff the floor needs to hold together.

There is no explicit mention (that I could find) of the thickness of the foam.  I decided on an average of 800m, given John’s drawing.  There is also no mention of foam on the sides of the rim walls, but in the simplistic cross section it is shown and has the same thickness as on the bottom.  What happens to meteorites impacting at an angle but coming in above the rim wall is unclear.  Most small ones would burn up in the atmosphere, as they do on Earth.

From these dimensions the cross section of the foam can be computed and from that the total foam volume.  Only 10% of that is equivalent to structural scrith, and with the volume of structural scrith we get the total volume of scrith used to build the ring floor.

Computing the amount of scrith

Still using the above table:  we know the total mass.  If we subtract the mass of the stuff on top, we have the mass of the scrith (structural and foam).

On each square metre of ring there is air, topsoil and bedrock and that gives a total of 70'097kg.  70 tons of stuff per square metre.  Sounds right.  Multiply by the surface of the ring and we have the mass of stuff.

Subtract from the total and we have the mass of the scrith: 1.892×10²⁷kg.  It turns out that the useful stuff is negligible compared to the mass of scrith!  Not really a surprise:  the mass of a house is usually far greater than that of the stuff you put in it;  the mass of a bridge bigger than that of the cars crossing it.

Since we have the volume of the scrith (including the foam when collapsed back to solid) we find that the density of scrith is a mere 10'650kg per cubic metre.  The chemical element closest in density is silver, at 10.5.  Is there something wrong?

Our vertical column of 1m² carried 70 tons of stuff.  With the 35m of scrith and 800m of foam, it comes to 1'295 tons.  The foam is the largest contribution.  None of this includes the rim walls.

Computing the Forces

Finally we have everything to compute the necessary tensile strength.  We use an engineering trick:  cut the ring on both sides and think of how much you would have to pull to provide the centripetal forces.

The large forces F on either side are equal and represent the sum of the components of the small ones in their direction.  The forces directly adjacent are at right angles and don’t contribute anything, but as we go along the ring, they contribute more and more, until at the middle the contribution is 100%.  The integral is the same as if there were centripetal forces all along the straight diameter between both Fs(*).  Thus F is the same as the centripetal force totalled over a radius.

(*) if you don’t believe this, think of going along the diameter from left to right while looking “up” at the arch: in your first steps you will see a lot of small forces pointing to the right but only a little down, as you go on the number of little forces gets smaller but they point more downward, and when you reach the middle there is only one put it points straight down.  At each step the amount of downward pointing is the same.  You should of course work this out using calculus.

In the table we have the force needed on a m², multiply that by the width of the ring, and then by the radius.

Then we divide it again over the area of load bearing scrith.

Finally we find that the tensile strength necessary is about 500 million tons per square mm.

No known material has this strength at that density. (euphemism of the page)

Other Considerations

The Shadow Squares

The ring turns once in 7.5 days and a day is 30 hours.

There are 20 shadow squares, therefore at any time there are 20 days and nights marked along the ring.  The shadow squares are linked by wires and themselves form a ring.

The ring of shadow squares itself must turn to keep the wires taut, and therefore at least slightly faster than the orbital speed.  But how fast do they have to turn?

Any point on the ring must pass through 7.5 dayzones in one ring turn.  There are only two possibilities:

1. the squares turn in the same sense as the ring
2. the squares turn in the other sense

If they turn in the same sense:

1. if they turn faster than the ring, then in one turn of the ring the squares must be 7.5 segments ahead or 7.5/20 turns ahead per ring turn. The squares angular speed is then (20+7.5)/20 = 27.5/20 times the ring’s angular speed.
2. if they turn slower, they must be 7.5 segments of lagging per turn, and the angular speed is then 12.5/20 of the ring’s angular speed.

If they turn in the other sense:

if the shadow squares stood still, there would be 20 days per turn of the ring, hence they cannot turn in the other sense because there would be even more days per turn. Therefore they turn in the same sense.

There is no reason to make them go faster than ring speed:  it would require more energy at set-up and stronger wires. Therefore they probably turn slower, in which case dawn comes from spinward. (this is confirmed in … ?) Thus the angular speed is 12.5/20×7.76E-06 rad/s which is 4.85E-6 rad/s.

At any distance r of a sun of mass M, the orbital angular speed is ω = √(GM/r3) therefore for any given ω the radius for which this is orbital speed is r = 3√(GM/ω2).  We know the desired orbital speed, the corresponding radius is then 17.8 million km.

If the squares are closer to the sun than this limit, they will have to turn faster than 4.85E-06 rad/s in order not to fall into the sun.  If they are further away, their orbital speed is slower therefore if they do turn faster they will have to be held together by wires.  The shadow squares are therefore slightly further from the sun than the 17.8Mkm limit, but not necessarily by much.

At that distance the sun’s radiation on the squares is 73 times more intense than it is on the ring ( (152/18)² ). The solar constant is 1.3kW/m2, on the shadow squares it would therefore be 95kW/m2

If 95kW/m2 can be handled, then the rectangles should be put close to the limit since further away would require more material and stronger wires.

Why they are probably not square

Suppose half of a day is in the shadow, the other half in the sun, and the shadow squares are indeed square.

Each shadow square throws a shadow that is 1/40 of the ring’s circumference, or 24 million km, long.  This is much more than the width of the ring.  If a shadow square really is a square, it will then cast shadow laterally that extends far over the width of the ring.  Seen from far away, this would make the sun blink like a lighthouse over a fair sector of space:

This might attract undesired attention. The shadow squares should be rectangles.

How wide are those rectangles?  Let s be the radius of the sun, r the distance of the shadow squares from the sun centre, R the radius of the Ringworld and W its width. What is the width w of the shadow squares such that no sunlight can fall on the ring but no shadow extends beyond the ring?

Note that W>2s (though not by much) and that the sun is not a point light source.

Consider the upper half of this figure, in an exaggerated configuration so that the problems become more obvious:

C is the sun centre, M the middle of the ring, R is the radius of the ring, Z half of the ring width and z half of the shadow square width. Q is at a rim wall, line QP touches the sun in P.  Extend QP to where it intersects the extension of MC at O.

Let d=OC and q=PQ.

In the right triangles CPQ and CQM we have

q²+s² = CQ = Z²+R²

whence

q² = Z²+R²-s²

which gives us q.  q has to be larger than R though not by much.

The right triangles OPC and OQM are similar (they share the angle POC), thus we have:

s/d = Z/OQ = Z/(q+OP)  and  s/OP = Z/(d+R)

From the second follows:

sd+sR = ZOP

or:

OP = (sd+sR)/Z

From the first follows:

sq+sOP = Zd

Substituting OP in the first equation:

sq+s(sd+sR)/Z = Zd

sqZ+s²d+sR = Z²d

d(Z²−s²)=sqZ+sR

d=(sqZ+sR)/(Z²−s²)

which gives us d.  But the value of z is proportional to that of Z:

z/(r+d) = Z/(R+d)

z=Z(r+d)/(R+d)

This relies on Z>s; if Z=s then d=∞

For a radius of 17.8Mkm of the shadow square ring, this gives a value for z of 644Mm or w=2z=1'288'751km

The shadow squares are not much less wide than the ring.  This is to be expected, since the ring is only slightly wider than the sun.

Shadow Square Ring Stability

The Ring’s orbit is unstable:  the slightest perturbation would start making it go like a hoola-hoop around the sun and it would eventually touch the sun.  In the sequel novel “Ringworld Engineers” there is discussion of attitude jets, mounted on the rim walls and tangential to the floor.

The same applies to the ring of shadow squares.  But that ring also has to be kept in the same plane as the Ringworld, therefore that ring not only must have atttitude jets, but their controls must be more sohpisticated.

The Ring’s attitude jets are so-called ramscoop engines, making their “fuel” from scooped up hydrogen in interstellar space.  This scooping requires a certain speed, and therefore is another factor in determining how large the shadow square ring has to be.  It may turn out that the shadow square ring must be ominously close to the habitable ring!

Issues

These are some questions I came up with:

What is the density/strength ratio to make a minimum ring?  I.e. if the strength is low but the density high, then any thickness of floor will break under its own “weight”.  Would a carbon fibre ring be possible?

Salt:  In one encounter Louis Wu wonders if the species would tolerate salt, since he reckons that there is no salt available except in the two large salt oceans.  However, salt from the oceans must via flup recycling slowly distribute everywhere.

Scrith must be flexible and malleable, otherwise it would have been impossible for “Fist of God” mountain to form.  But if it is, then it would get stretched flat by the enormous tensile force and there would be no contoured bottom, it would be impossible to “sculpt” the landscape.

At several hundred metres of scrith foam covering, there would be no detailed contours visible from below, only larger lakes and mountains. No way to see small rivers or deltas, as is claimed.

If the width is 1.6 million km, then it takes more than 5 seconds for a radio signal to cross the width.  There is no way to control the lander from the rim wall.

Will the atmosphere “slosh” over the rim walls?  On a scale model with width 1 metre the rim walls are only 1mm high.  In this drawing they appear still 4 times higher than they are said to be:

Have you tried to walk with a very shallow tray filled with water?

The meteor defence system cannot work by exciting solar flares: it takes 8 minutes for a signal to travel to the sun and another 8 minutes back.

There is no need for two enormous salty oceans counterbalancing each other by being placed at diametrically  opposite positions. Earth’s oceans are at most 10km deep, and 20'000km across, therefore there is ample room for many such “tiny lakes” on a ring that is 1'600'000km across! If the floor is bent, the arrow alone is over 2'000km, an ocean of 10km deep is peanuts in comparison.  Placed between the rim walls, the Pacific Ocean would look like this tiny blue spot:

Put as many salty oceans as you like…

What if a flup pipe gets blocked?

What is the material of the shadow squares?  Does it reflect anything at all?  If so, their sunward sides are visible from the ring through the gaps.

In the map room Teela and Speaker-to-Animals use data to draw a map which they take with them.  But they draw with what and on what?  Pen and paper?

Why is the world already full of people?  Or does that indicate it takes only a few million years?

How does one make scrith foam?

If you were a Pak Protector, would you trust the Ringworld?  You lose everything if the ring ever breaks.

In one place Niven mentions the “gravity” in m/s but should be m/s²

With the same radiation as at our equator and a 15 hour part of the day with the sun at the zenith, will people get roasted?  Our Sun rises and sets at the horizon, giving much less radiation over the day.

Think of applying an engineering safety factor of 2.

The speed at the rim is sufficient for the ramscoop attitude engines to get their interstellar hydrogen fuel.  How does this hydrogen affect the Ringworld’s atmosphere?  What if the sun and therefore the world, move with respect to that interstellar gas?  The attitude engines would not get a constant flow.   Our Sun and the Earth move in space, but we have no noticeable influence.

Is the floor flat or bent?  The width of the  Ringworld is sufficiently large that the direction of the sun is noticeably different from one rim wall to the other if the floor is flat.  Consider a slice from a sphere:

This is exaggerated, but shows that the floor could well be bent instead of being flat.  How much?

The width W of the floor, seen at distance r from the sun, subtends an angle α, where αr=W therefore α=0.601º (for comparison, our Sun subtends about 0.5º, and the Ringworld’s width is somewhat larger than the Sun’s diameter, so that value must be right)

If bent, the arrow of the curve, r(1−cos(α/2)), would be 2105km.  Seemingly a lot, but not much compared to the 1.6 million km width.  If the floor is flat, then the centripetal force is perpendicular to the floor everywhere.  But the sun is then off the vertical by 0.3º when standing near one rim wall, and by the same amount but in the other sense at the other rim wall.  Buildings not in the centre cast a small shadow.  Only in the centre of the floor will the sun be exactly at the zenith.  If the floor is slightly bent so that the sun is directly overhead everywhere, then the “gravity” force will be off by the same amount, i.e. a plumb line does not meet the floor at 90º except at the centre again.  But I’m convinced the floor is flat.

Gravity from the stuff on the floor:  standing near a rim wall, there are over a million km worth of stuff exerting a gravitational attraction away from the wall.  Is this noticeable?  It turns out to be too little stuff, because it is only a thin layer and it gets distant fast.  Consider an area of 1000km around a point on the rim wall.  The stuff in that area is surprisingly little:  e.g. I live near the highest part of the Jura mountains,  along a 10km long block from it, at 1700m high, and only a few km from my house, attracts with a gravitational force almost 1000 times more than the stuff in that whole 1000km area.

Inspiration

I mentioned that “Ringworld” still inpires us today.  A ringworld appears in a recent episode (2022) of a Star Wars spin-off:

Another one, much smaller, is “Elysium” in the movie of the same title:

No way those walls can keep enough air in for any pressure at the surface.

Making the Illustrations

Given the unknowns, it was not easy to make illustrations that give a reasonable impression of what one might see when standing on the Ringworld.

I used a 3D modelling and rendering program to make a model with as much as possible the correct relative dimensions.  Thus, if the radius of the ring is 158 units, the width of the ring is only 1.6 units and the height of the rim walls is insignificant.

The atmosphere was the most difficult part to model:  I found no off-the-shelf volumetric textures that simulate our air, which is a light-scattering gas that thins out with altitude.  It is not surprising that such a texture is not available, since skies used in animated movies are normal skies, lit by a distant Sun, whereas the one of the Ringworld curves up and is not limited by a horizon.  One person suggested using an image editing program to overlay a blueish layer with alpha-channel gradients.  While any post-processing of this kind would work for the production of a sci-fi image, in this particular case I did not want a pretty, imaginative picture, I wanted a correct simulation of the physics.

What you see in the illustrations is close to that, or so I believe, but it is not entirely correct.  Studies have been made of rendering air, but the mathematics are horrendously complex.  Look for: “Real-Time Rendering of Planets with Atmospheres”, or authors T. Nishita, M. Falk, T. Ertl et al..