Hollow Earth – how to measure it?

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I’m making a map for my friend – I drew a pretty neat cartographic grid with a proper map projection – and then he reminded me that his world is actually an inverted Earth with a glowing ball of concentrated magic in the middle. The sun is half black, so it creates day/night cycle by rotating.

Now, the question is, because I’m a bit stuck – how could people with roughly Renaissance technology measure the size of the planet (so they could actually draw a cartographic grid)? In our world it’s simple, just measure difference in the angle of sun rays between two position of known distance. But what for them?
Maybe there are some ways to mitigate it with small changes to the setting?

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  • 1

    @Mołot rotation of the sun – the sun is dark on its one side
    – Michał Sadowski
    Nov 29 at 16:28

  • 10

    If the sun is half-black, then you get a day/night cycle at the equator, but at all other points some part of the light side is visible 24/7, until at the poles your light level is constant throughout the day. Even at the equator, scattered light means it won’t be darker than twilight. I can answer the question this way, just be aware that you won’t actually have a full dark night this way.
    – kingledion
    Nov 29 at 16:29

  • 1

    @kingelodion if light comes as if its a point source and half-sphere of black, I can answer this too. Things get complicated if “sun” radius is big compared to the “earth” radius, and its surface that glows (hope you get what I mean).
    – Mołot
    Nov 29 at 16:35

  • 1

    David J. Lake’s novel, The Ring of Truth has a similar world, but with a sun that brightens and dims to form the diurnal cycle (and many additional complications).
    – Zeiss Ikon
    Nov 29 at 16:49

  • 3

    If the sun is a sphere with one half that emits light and one half that doesn’t, then even at the equator you can see half of the light portion of the sun at 6am, all of it at noon, half of it at 6pm, and only briefly at midnight is there a time when none of the light half is visible at all. Effectively “sunrise” and “sunset” take a full 12 hours each, and each process starts immediately after the other finishes. To get a “12 hours of light, 12 hours of darkness” set up you need something more like a point-source sun with a hemipherical shield blocking half of the light.
    – Ben
    Nov 29 at 22:59

up vote
16
down vote

favorite

3

I’m making a map for my friend – I drew a pretty neat cartographic grid with a proper map projection – and then he reminded me that his world is actually an inverted Earth with a glowing ball of concentrated magic in the middle. The sun is half black, so it creates day/night cycle by rotating.

Now, the question is, because I’m a bit stuck – how could people with roughly Renaissance technology measure the size of the planet (so they could actually draw a cartographic grid)? In our world it’s simple, just measure difference in the angle of sun rays between two position of known distance. But what for them?
Maybe there are some ways to mitigate it with small changes to the setting?

share|improve this question

  • 1

    @Mołot rotation of the sun – the sun is dark on its one side
    – Michał Sadowski
    Nov 29 at 16:28

  • 10

    If the sun is half-black, then you get a day/night cycle at the equator, but at all other points some part of the light side is visible 24/7, until at the poles your light level is constant throughout the day. Even at the equator, scattered light means it won’t be darker than twilight. I can answer the question this way, just be aware that you won’t actually have a full dark night this way.
    – kingledion
    Nov 29 at 16:29

  • 1

    @kingelodion if light comes as if its a point source and half-sphere of black, I can answer this too. Things get complicated if “sun” radius is big compared to the “earth” radius, and its surface that glows (hope you get what I mean).
    – Mołot
    Nov 29 at 16:35

  • 1

    David J. Lake’s novel, The Ring of Truth has a similar world, but with a sun that brightens and dims to form the diurnal cycle (and many additional complications).
    – Zeiss Ikon
    Nov 29 at 16:49

  • 3

    If the sun is a sphere with one half that emits light and one half that doesn’t, then even at the equator you can see half of the light portion of the sun at 6am, all of it at noon, half of it at 6pm, and only briefly at midnight is there a time when none of the light half is visible at all. Effectively “sunrise” and “sunset” take a full 12 hours each, and each process starts immediately after the other finishes. To get a “12 hours of light, 12 hours of darkness” set up you need something more like a point-source sun with a hemipherical shield blocking half of the light.
    – Ben
    Nov 29 at 22:59

up vote
16
down vote

favorite

3

up vote
16
down vote

favorite

3
3

I’m making a map for my friend – I drew a pretty neat cartographic grid with a proper map projection – and then he reminded me that his world is actually an inverted Earth with a glowing ball of concentrated magic in the middle. The sun is half black, so it creates day/night cycle by rotating.

Now, the question is, because I’m a bit stuck – how could people with roughly Renaissance technology measure the size of the planet (so they could actually draw a cartographic grid)? In our world it’s simple, just measure difference in the angle of sun rays between two position of known distance. But what for them?
Maybe there are some ways to mitigate it with small changes to the setting?

share|improve this question

I’m making a map for my friend – I drew a pretty neat cartographic grid with a proper map projection – and then he reminded me that his world is actually an inverted Earth with a glowing ball of concentrated magic in the middle. The sun is half black, so it creates day/night cycle by rotating.

Now, the question is, because I’m a bit stuck – how could people with roughly Renaissance technology measure the size of the planet (so they could actually draw a cartographic grid)? In our world it’s simple, just measure difference in the angle of sun rays between two position of known distance. But what for them?
Maybe there are some ways to mitigate it with small changes to the setting?

map-making sunlight renaissance hollow-earth

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asked Nov 29 at 16:17

Michał Sadowski

18315

18315

  • 1

    @Mołot rotation of the sun – the sun is dark on its one side
    – Michał Sadowski
    Nov 29 at 16:28

  • 10

    If the sun is half-black, then you get a day/night cycle at the equator, but at all other points some part of the light side is visible 24/7, until at the poles your light level is constant throughout the day. Even at the equator, scattered light means it won’t be darker than twilight. I can answer the question this way, just be aware that you won’t actually have a full dark night this way.
    – kingledion
    Nov 29 at 16:29

  • 1

    @kingelodion if light comes as if its a point source and half-sphere of black, I can answer this too. Things get complicated if “sun” radius is big compared to the “earth” radius, and its surface that glows (hope you get what I mean).
    – Mołot
    Nov 29 at 16:35

  • 1

    David J. Lake’s novel, The Ring of Truth has a similar world, but with a sun that brightens and dims to form the diurnal cycle (and many additional complications).
    – Zeiss Ikon
    Nov 29 at 16:49

  • 3

    If the sun is a sphere with one half that emits light and one half that doesn’t, then even at the equator you can see half of the light portion of the sun at 6am, all of it at noon, half of it at 6pm, and only briefly at midnight is there a time when none of the light half is visible at all. Effectively “sunrise” and “sunset” take a full 12 hours each, and each process starts immediately after the other finishes. To get a “12 hours of light, 12 hours of darkness” set up you need something more like a point-source sun with a hemipherical shield blocking half of the light.
    – Ben
    Nov 29 at 22:59

  • 1

    @Mołot rotation of the sun – the sun is dark on its one side
    – Michał Sadowski
    Nov 29 at 16:28

  • 10

    If the sun is half-black, then you get a day/night cycle at the equator, but at all other points some part of the light side is visible 24/7, until at the poles your light level is constant throughout the day. Even at the equator, scattered light means it won’t be darker than twilight. I can answer the question this way, just be aware that you won’t actually have a full dark night this way.
    – kingledion
    Nov 29 at 16:29

  • 1

    @kingelodion if light comes as if its a point source and half-sphere of black, I can answer this too. Things get complicated if “sun” radius is big compared to the “earth” radius, and its surface that glows (hope you get what I mean).
    – Mołot
    Nov 29 at 16:35

  • 1

    David J. Lake’s novel, The Ring of Truth has a similar world, but with a sun that brightens and dims to form the diurnal cycle (and many additional complications).
    – Zeiss Ikon
    Nov 29 at 16:49

  • 3

    If the sun is a sphere with one half that emits light and one half that doesn’t, then even at the equator you can see half of the light portion of the sun at 6am, all of it at noon, half of it at 6pm, and only briefly at midnight is there a time when none of the light half is visible at all. Effectively “sunrise” and “sunset” take a full 12 hours each, and each process starts immediately after the other finishes. To get a “12 hours of light, 12 hours of darkness” set up you need something more like a point-source sun with a hemipherical shield blocking half of the light.
    – Ben
    Nov 29 at 22:59

1

1

@Mołot rotation of the sun – the sun is dark on its one side
– Michał Sadowski
Nov 29 at 16:28

@Mołot rotation of the sun – the sun is dark on its one side
– Michał Sadowski
Nov 29 at 16:28

10

10

If the sun is half-black, then you get a day/night cycle at the equator, but at all other points some part of the light side is visible 24/7, until at the poles your light level is constant throughout the day. Even at the equator, scattered light means it won’t be darker than twilight. I can answer the question this way, just be aware that you won’t actually have a full dark night this way.
– kingledion
Nov 29 at 16:29

If the sun is half-black, then you get a day/night cycle at the equator, but at all other points some part of the light side is visible 24/7, until at the poles your light level is constant throughout the day. Even at the equator, scattered light means it won’t be darker than twilight. I can answer the question this way, just be aware that you won’t actually have a full dark night this way.
– kingledion
Nov 29 at 16:29

1

1

@kingelodion if light comes as if its a point source and half-sphere of black, I can answer this too. Things get complicated if “sun” radius is big compared to the “earth” radius, and its surface that glows (hope you get what I mean).
– Mołot
Nov 29 at 16:35

@kingelodion if light comes as if its a point source and half-sphere of black, I can answer this too. Things get complicated if “sun” radius is big compared to the “earth” radius, and its surface that glows (hope you get what I mean).
– Mołot
Nov 29 at 16:35

1

1

David J. Lake’s novel, The Ring of Truth has a similar world, but with a sun that brightens and dims to form the diurnal cycle (and many additional complications).
– Zeiss Ikon
Nov 29 at 16:49

David J. Lake’s novel, The Ring of Truth has a similar world, but with a sun that brightens and dims to form the diurnal cycle (and many additional complications).
– Zeiss Ikon
Nov 29 at 16:49

3

3

If the sun is a sphere with one half that emits light and one half that doesn’t, then even at the equator you can see half of the light portion of the sun at 6am, all of it at noon, half of it at 6pm, and only briefly at midnight is there a time when none of the light half is visible at all. Effectively “sunrise” and “sunset” take a full 12 hours each, and each process starts immediately after the other finishes. To get a “12 hours of light, 12 hours of darkness” set up you need something more like a point-source sun with a hemipherical shield blocking half of the light.
– Ben
Nov 29 at 22:59

If the sun is a sphere with one half that emits light and one half that doesn’t, then even at the equator you can see half of the light portion of the sun at 6am, all of it at noon, half of it at 6pm, and only briefly at midnight is there a time when none of the light half is visible at all. Effectively “sunrise” and “sunset” take a full 12 hours each, and each process starts immediately after the other finishes. To get a “12 hours of light, 12 hours of darkness” set up you need something more like a point-source sun with a hemipherical shield blocking half of the light.
– Ben
Nov 29 at 22:59

4 Answers
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This actually is pretty easy using the similar methods to what you mention.

The “sun” is rotating, meaning that if you look up at night when the shaded side is toward you, you can see the inside of the sphere where it is still daylight.

You can figure out how long it takes for the sun to rotate once. Lets say 24 hours just for fun.

Then you figure out the distance between two points somewhere along the sphere where you can see them at night, and see how long it takes for the terminator line to pass between them.

share|improve this answer

    up vote
    13
    down vote

    What light looks like from the surface

    The sun in your world is half-bright, half-dark. Assume that it is a perfect sphere with any radius greater than zero. Therefore, for any point on the surface, the percentage of the sun that is facing that point at any given time is

    $$cos^2phicos^22theta + frac{1}{2}sin^2phi.$$

    Here are three plots of what that looks like at the equator:

    enter image description here

    At a point in the temperate zone at about 40 degrees latitude:

    enter image description here

    And at the poles:

    enter image description here

    How to measure latitude between two points

    Without sunrise over a horizon or a pole star, it is much harder to measure latitude. The best you can do is to measure the brightness of the sun at various points in the day, and to compare with the trigonometeric properties of its brightness from the last section.

    The the best of my knowledge, there is no way to scientifically measure brightness until you have a photographic plate. That will not be available to people of Renaissance technology. However, you can at the very least dead-eye reckon brightness, so we will assume some sort of brightness metric. One potential way to measure brightness is that brightness is proportional to the ‘bright’ portion of the sphere of the sun that is visible to you from your location. If you look at the sun through a dark lens, you may be able to measure this, depending on how large the radius of the sun is.

    In this case, you can calculate latitude relative to the equator easily. Maximum brightness at latitude $phi$ ($max(B_phi)$) is $1 – 1/2sin^2phi$ times that at the equator ($max(B_{eq})$), solve backwards for $phi$:

    $$phi = sin^{-1}sqrt{2left(1 – frac{max(B_phi)}{max(B_{eq})}right)}.$$

    It is possible that maximum brightness at the equator is a well known standard value in your scientific community, even for those who have never been there. You have to know, or at least be able to estimate, the maximum brightness to be able to calculate latitude. As an alternate measurement, not2 that equatorial max brightness is twice the constant brightness at the poles. You can re-work everything in terms of that value as
    $$max(B_{eq}) = 2B_{pole}.$$

    Now there are two ways to measure radius

    Assuming the hollow world is a perfect sphere, there are two ways to measure the radius. you can compare latitude of two points to the distance between them, or you can compare the time it takes the terminator to travel between two points.

    The terminator line

    The terminator is important here, because with no stars in the background, the only way to ensure that two points are at exactly the same longitude is to visually signal when the terminator passes by the points. This can be done using one lighthouse to signal another point within visual range of that lighthouse. Since your world is hollow and thus concave from the point of view of someone on the surface, the line of sight of a lighthouse is actually very long, limited only by atmospheric attenuation of light (due to water vapor, or whatever).

    The terminator can be exactly identified by looking at the sun through a telescope with a dark lens. As soon as there is no bright patch visible, the terminator line has passed.

    Comparing latitude method

    The latitude way is take two points that you know are at the same longitude, and calculate their latitude using the above method and distance ($d$) between them, using what ever method.

    If the latitude delta is $alpha$, then the polar radius of the hollow Earth is $$r_{pole} = d/alpha$$ in radians.

    Timing the terminator method

    To time the terminator, you will need to get two points at the same latitude (confirmed using the methods above), and measure the distance ($d$) between them and the time it takes the terminator to pass between them ($t$). You also need to know the length of the day $t_{day}$.

    Then, the equatorial radius of the hollow Earth is $$r_{eq} = 2pi dfrac{ t_{day}}{t}.$$ Note this only works if the points are less than $pi$ radians apart on the surface, that is, in the same hemisphere.

    These numbers might not be the same!

    These two methods could give you different answers, polar and equatorial radius. If you have a perfect sphere, the two calculated radii should be the same, but if your planet is inscribed within an oblate spheroid with an equatorial bulge(like our Earth), or even a polar bulge, then the two numbers will not be the same.

    Conclusion

    It is actually pretty difficult to time astronomical pheomena with only one available astronomical object. But, you can use the unique configuration of the sun’s surface to do this.

    This does require some photometric skills more advanced than what was available in the Renaissance, which I’m not sure how to replicate with Renaissance technology, but I have faith that the Galileo’s and Newton’s of the world could figure it out. Also, there needs to be some sort of standard measurement of equatorial brightness to make these calculations, but given this number’s importance in observational ‘heliometry,’ I would expect this number to be a well established topic in a Hollow Earth’s scientific community.

    share|improve this answer

    • 1

      I have serious doubts about people being able to judge latitude based on brightness – the human eye is very bad at judging between subtle changes in brightness. Would it be possible to use something different, like using the dark lens to see how much light is visible at the top/bottom of the sun during night (or darkness during day)?
      – Rob Watts
      Nov 29 at 18:15

    • 1

      @RobWatts A good idea! I specifically left the brightness measurement assumed. As i mention in the conclusion, given the importance of equatorial max brightness in astronomy, I assume that a Galileo or Newton would find a good measure, even if I can’t right now.
      – kingledion
      Nov 29 at 18:39

    • With latitude: $-pi/2 le theta le pi/2$, longitude: $-pi le phi le pi$ I got i.stack.imgur.com/cTq0e.png (python pastebin.com/VDijE4WN) which is a pretty intriguing pattern! Is this right? Is there a source or derivation for your expression somewhere? Thanks!
      – uhoh
      Nov 30 at 2:49

    • 1

      @uhoh That looks right, the key finding is that for any point on the interior of the hollow sphere, the light intensity integrated over a full rotation is the same, 0.5! So for the derivation, I did a 2-d integration over the surface of the sphere with respect to $theta$ and $phi$. I did it myself, so not 100% that it is right. My answer is already too long, I’ll do it up in latex and share with you if you ping me in chat, if you want.
      – kingledion
      Nov 30 at 4:21

    • For equipment I wonder if you could use something like a pinhole camera. it would let you see which parts of the light source are occluded by the shade, though it wouldn’t really help with light levels except by the most skilled observers.
      – AndyD273
      Nov 30 at 18:00

    up vote
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    Two points, a known distance apart, will have a straight line-of-sight between them that will not be parallel to the curving surface. Stand a pole at the centre between the two points. Where the sight line intersects the pole will give you the height (er – depth) of the arc. That will allow them to calculate the radius of their world. Even if they don’t have the trig to calculate it, they can always just make a scale diagram and measure it.

    enter image description here

    share|improve this answer

    • 1

      You’d need a tower, not a pole, because the local curvature of the Earth can be quite different from the large scale curvature, for example, on a hillside.
      – kingledion
      Nov 29 at 22:51

    • 1

      Don’t try to measure the planet while on a hillside. 😐
      – DaveC426913
      Nov 30 at 2:30

    • You don’t really need a pole/tower. Angles could be measured quite accurately since ancient times. Just measure the angles between the straight line and the sun and use the fact that internal angles of a triangle add up to half circle to determine the difference in longitude/lattitude.
      – Jan Hudec
      Nov 30 at 9:09

    • 1

      You’ll end up with a triangle that is n miles at the base and 4000 miles tall. You’ll be working with an angle that is 0.15 degrees from square. A lot of room for error there.
      – DaveC426913
      Nov 30 at 16:15

    up vote
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    Providing long distance travel is practical, his could be done with pure geometry ( or more easily, assuming the natives had advanced to usable trigonometry). One could measure the distance between two points on the inner surface that are visible from the same location, then travel to one point and directly measure the distance to the other point — or, lacking trigonometry, one could select two points that are sixty degrees apart (therefore forming an equilateral triangle between the two and your observation point), and know that the distance to either one is exactly one third of the circumference of the world.

    Angle bisection could be used, along with long distance signaling (with flashing mirrors, perhaps), to shorten the distance requiring measurement, at the cost of some loss of precision.

    If there’s not access to the antipodes, one could measure a distance across water (perhaps by surveying around the shore of a large lake or inland sea), and directly measure how much the surface dips below the sight line; this would allow deriving the radius of the sphere directly via similar triangles (same method as is used to calculate horizon distance on the outside of the Earth).

    share|improve this answer

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      4 Answers
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      4 Answers
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      accepted

      This actually is pretty easy using the similar methods to what you mention.

      The “sun” is rotating, meaning that if you look up at night when the shaded side is toward you, you can see the inside of the sphere where it is still daylight.

      You can figure out how long it takes for the sun to rotate once. Lets say 24 hours just for fun.

      Then you figure out the distance between two points somewhere along the sphere where you can see them at night, and see how long it takes for the terminator line to pass between them.

      share|improve this answer

        up vote
        26
        down vote

        accepted

        This actually is pretty easy using the similar methods to what you mention.

        The “sun” is rotating, meaning that if you look up at night when the shaded side is toward you, you can see the inside of the sphere where it is still daylight.

        You can figure out how long it takes for the sun to rotate once. Lets say 24 hours just for fun.

        Then you figure out the distance between two points somewhere along the sphere where you can see them at night, and see how long it takes for the terminator line to pass between them.

        share|improve this answer

          up vote
          26
          down vote

          accepted

          up vote
          26
          down vote

          accepted

          This actually is pretty easy using the similar methods to what you mention.

          The “sun” is rotating, meaning that if you look up at night when the shaded side is toward you, you can see the inside of the sphere where it is still daylight.

          You can figure out how long it takes for the sun to rotate once. Lets say 24 hours just for fun.

          Then you figure out the distance between two points somewhere along the sphere where you can see them at night, and see how long it takes for the terminator line to pass between them.

          share|improve this answer

          This actually is pretty easy using the similar methods to what you mention.

          The “sun” is rotating, meaning that if you look up at night when the shaded side is toward you, you can see the inside of the sphere where it is still daylight.

          You can figure out how long it takes for the sun to rotate once. Lets say 24 hours just for fun.

          Then you figure out the distance between two points somewhere along the sphere where you can see them at night, and see how long it takes for the terminator line to pass between them.

          share|improve this answer

          share|improve this answer

          share|improve this answer

          answered Nov 29 at 16:31

          AndyD273

          30.5k255132

          30.5k255132

              up vote
              13
              down vote

              What light looks like from the surface

              The sun in your world is half-bright, half-dark. Assume that it is a perfect sphere with any radius greater than zero. Therefore, for any point on the surface, the percentage of the sun that is facing that point at any given time is

              $$cos^2phicos^22theta + frac{1}{2}sin^2phi.$$

              Here are three plots of what that looks like at the equator:

              enter image description here

              At a point in the temperate zone at about 40 degrees latitude:

              enter image description here

              And at the poles:

              enter image description here

              How to measure latitude between two points

              Without sunrise over a horizon or a pole star, it is much harder to measure latitude. The best you can do is to measure the brightness of the sun at various points in the day, and to compare with the trigonometeric properties of its brightness from the last section.

              The the best of my knowledge, there is no way to scientifically measure brightness until you have a photographic plate. That will not be available to people of Renaissance technology. However, you can at the very least dead-eye reckon brightness, so we will assume some sort of brightness metric. One potential way to measure brightness is that brightness is proportional to the ‘bright’ portion of the sphere of the sun that is visible to you from your location. If you look at the sun through a dark lens, you may be able to measure this, depending on how large the radius of the sun is.

              In this case, you can calculate latitude relative to the equator easily. Maximum brightness at latitude $phi$ ($max(B_phi)$) is $1 – 1/2sin^2phi$ times that at the equator ($max(B_{eq})$), solve backwards for $phi$:

              $$phi = sin^{-1}sqrt{2left(1 – frac{max(B_phi)}{max(B_{eq})}right)}.$$

              It is possible that maximum brightness at the equator is a well known standard value in your scientific community, even for those who have never been there. You have to know, or at least be able to estimate, the maximum brightness to be able to calculate latitude. As an alternate measurement, not2 that equatorial max brightness is twice the constant brightness at the poles. You can re-work everything in terms of that value as
              $$max(B_{eq}) = 2B_{pole}.$$

              Now there are two ways to measure radius

              Assuming the hollow world is a perfect sphere, there are two ways to measure the radius. you can compare latitude of two points to the distance between them, or you can compare the time it takes the terminator to travel between two points.

              The terminator line

              The terminator is important here, because with no stars in the background, the only way to ensure that two points are at exactly the same longitude is to visually signal when the terminator passes by the points. This can be done using one lighthouse to signal another point within visual range of that lighthouse. Since your world is hollow and thus concave from the point of view of someone on the surface, the line of sight of a lighthouse is actually very long, limited only by atmospheric attenuation of light (due to water vapor, or whatever).

              The terminator can be exactly identified by looking at the sun through a telescope with a dark lens. As soon as there is no bright patch visible, the terminator line has passed.

              Comparing latitude method

              The latitude way is take two points that you know are at the same longitude, and calculate their latitude using the above method and distance ($d$) between them, using what ever method.

              If the latitude delta is $alpha$, then the polar radius of the hollow Earth is $$r_{pole} = d/alpha$$ in radians.

              Timing the terminator method

              To time the terminator, you will need to get two points at the same latitude (confirmed using the methods above), and measure the distance ($d$) between them and the time it takes the terminator to pass between them ($t$). You also need to know the length of the day $t_{day}$.

              Then, the equatorial radius of the hollow Earth is $$r_{eq} = 2pi dfrac{ t_{day}}{t}.$$ Note this only works if the points are less than $pi$ radians apart on the surface, that is, in the same hemisphere.

              These numbers might not be the same!

              These two methods could give you different answers, polar and equatorial radius. If you have a perfect sphere, the two calculated radii should be the same, but if your planet is inscribed within an oblate spheroid with an equatorial bulge(like our Earth), or even a polar bulge, then the two numbers will not be the same.

              Conclusion

              It is actually pretty difficult to time astronomical pheomena with only one available astronomical object. But, you can use the unique configuration of the sun’s surface to do this.

              This does require some photometric skills more advanced than what was available in the Renaissance, which I’m not sure how to replicate with Renaissance technology, but I have faith that the Galileo’s and Newton’s of the world could figure it out. Also, there needs to be some sort of standard measurement of equatorial brightness to make these calculations, but given this number’s importance in observational ‘heliometry,’ I would expect this number to be a well established topic in a Hollow Earth’s scientific community.

              share|improve this answer

              • 1

                I have serious doubts about people being able to judge latitude based on brightness – the human eye is very bad at judging between subtle changes in brightness. Would it be possible to use something different, like using the dark lens to see how much light is visible at the top/bottom of the sun during night (or darkness during day)?
                – Rob Watts
                Nov 29 at 18:15

              • 1

                @RobWatts A good idea! I specifically left the brightness measurement assumed. As i mention in the conclusion, given the importance of equatorial max brightness in astronomy, I assume that a Galileo or Newton would find a good measure, even if I can’t right now.
                – kingledion
                Nov 29 at 18:39

              • With latitude: $-pi/2 le theta le pi/2$, longitude: $-pi le phi le pi$ I got i.stack.imgur.com/cTq0e.png (python pastebin.com/VDijE4WN) which is a pretty intriguing pattern! Is this right? Is there a source or derivation for your expression somewhere? Thanks!
                – uhoh
                Nov 30 at 2:49

              • 1

                @uhoh That looks right, the key finding is that for any point on the interior of the hollow sphere, the light intensity integrated over a full rotation is the same, 0.5! So for the derivation, I did a 2-d integration over the surface of the sphere with respect to $theta$ and $phi$. I did it myself, so not 100% that it is right. My answer is already too long, I’ll do it up in latex and share with you if you ping me in chat, if you want.
                – kingledion
                Nov 30 at 4:21

              • For equipment I wonder if you could use something like a pinhole camera. it would let you see which parts of the light source are occluded by the shade, though it wouldn’t really help with light levels except by the most skilled observers.
                – AndyD273
                Nov 30 at 18:00

              up vote
              13
              down vote

              What light looks like from the surface

              The sun in your world is half-bright, half-dark. Assume that it is a perfect sphere with any radius greater than zero. Therefore, for any point on the surface, the percentage of the sun that is facing that point at any given time is

              $$cos^2phicos^22theta + frac{1}{2}sin^2phi.$$

              Here are three plots of what that looks like at the equator:

              enter image description here

              At a point in the temperate zone at about 40 degrees latitude:

              enter image description here

              And at the poles:

              enter image description here

              How to measure latitude between two points

              Without sunrise over a horizon or a pole star, it is much harder to measure latitude. The best you can do is to measure the brightness of the sun at various points in the day, and to compare with the trigonometeric properties of its brightness from the last section.

              The the best of my knowledge, there is no way to scientifically measure brightness until you have a photographic plate. That will not be available to people of Renaissance technology. However, you can at the very least dead-eye reckon brightness, so we will assume some sort of brightness metric. One potential way to measure brightness is that brightness is proportional to the ‘bright’ portion of the sphere of the sun that is visible to you from your location. If you look at the sun through a dark lens, you may be able to measure this, depending on how large the radius of the sun is.

              In this case, you can calculate latitude relative to the equator easily. Maximum brightness at latitude $phi$ ($max(B_phi)$) is $1 – 1/2sin^2phi$ times that at the equator ($max(B_{eq})$), solve backwards for $phi$:

              $$phi = sin^{-1}sqrt{2left(1 – frac{max(B_phi)}{max(B_{eq})}right)}.$$

              It is possible that maximum brightness at the equator is a well known standard value in your scientific community, even for those who have never been there. You have to know, or at least be able to estimate, the maximum brightness to be able to calculate latitude. As an alternate measurement, not2 that equatorial max brightness is twice the constant brightness at the poles. You can re-work everything in terms of that value as
              $$max(B_{eq}) = 2B_{pole}.$$

              Now there are two ways to measure radius

              Assuming the hollow world is a perfect sphere, there are two ways to measure the radius. you can compare latitude of two points to the distance between them, or you can compare the time it takes the terminator to travel between two points.

              The terminator line

              The terminator is important here, because with no stars in the background, the only way to ensure that two points are at exactly the same longitude is to visually signal when the terminator passes by the points. This can be done using one lighthouse to signal another point within visual range of that lighthouse. Since your world is hollow and thus concave from the point of view of someone on the surface, the line of sight of a lighthouse is actually very long, limited only by atmospheric attenuation of light (due to water vapor, or whatever).

              The terminator can be exactly identified by looking at the sun through a telescope with a dark lens. As soon as there is no bright patch visible, the terminator line has passed.

              Comparing latitude method

              The latitude way is take two points that you know are at the same longitude, and calculate their latitude using the above method and distance ($d$) between them, using what ever method.

              If the latitude delta is $alpha$, then the polar radius of the hollow Earth is $$r_{pole} = d/alpha$$ in radians.

              Timing the terminator method

              To time the terminator, you will need to get two points at the same latitude (confirmed using the methods above), and measure the distance ($d$) between them and the time it takes the terminator to pass between them ($t$). You also need to know the length of the day $t_{day}$.

              Then, the equatorial radius of the hollow Earth is $$r_{eq} = 2pi dfrac{ t_{day}}{t}.$$ Note this only works if the points are less than $pi$ radians apart on the surface, that is, in the same hemisphere.

              These numbers might not be the same!

              These two methods could give you different answers, polar and equatorial radius. If you have a perfect sphere, the two calculated radii should be the same, but if your planet is inscribed within an oblate spheroid with an equatorial bulge(like our Earth), or even a polar bulge, then the two numbers will not be the same.

              Conclusion

              It is actually pretty difficult to time astronomical pheomena with only one available astronomical object. But, you can use the unique configuration of the sun’s surface to do this.

              This does require some photometric skills more advanced than what was available in the Renaissance, which I’m not sure how to replicate with Renaissance technology, but I have faith that the Galileo’s and Newton’s of the world could figure it out. Also, there needs to be some sort of standard measurement of equatorial brightness to make these calculations, but given this number’s importance in observational ‘heliometry,’ I would expect this number to be a well established topic in a Hollow Earth’s scientific community.

              share|improve this answer

              • 1

                I have serious doubts about people being able to judge latitude based on brightness – the human eye is very bad at judging between subtle changes in brightness. Would it be possible to use something different, like using the dark lens to see how much light is visible at the top/bottom of the sun during night (or darkness during day)?
                – Rob Watts
                Nov 29 at 18:15

              • 1

                @RobWatts A good idea! I specifically left the brightness measurement assumed. As i mention in the conclusion, given the importance of equatorial max brightness in astronomy, I assume that a Galileo or Newton would find a good measure, even if I can’t right now.
                – kingledion
                Nov 29 at 18:39

              • With latitude: $-pi/2 le theta le pi/2$, longitude: $-pi le phi le pi$ I got i.stack.imgur.com/cTq0e.png (python pastebin.com/VDijE4WN) which is a pretty intriguing pattern! Is this right? Is there a source or derivation for your expression somewhere? Thanks!
                – uhoh
                Nov 30 at 2:49

              • 1

                @uhoh That looks right, the key finding is that for any point on the interior of the hollow sphere, the light intensity integrated over a full rotation is the same, 0.5! So for the derivation, I did a 2-d integration over the surface of the sphere with respect to $theta$ and $phi$. I did it myself, so not 100% that it is right. My answer is already too long, I’ll do it up in latex and share with you if you ping me in chat, if you want.
                – kingledion
                Nov 30 at 4:21

              • For equipment I wonder if you could use something like a pinhole camera. it would let you see which parts of the light source are occluded by the shade, though it wouldn’t really help with light levels except by the most skilled observers.
                – AndyD273
                Nov 30 at 18:00

              up vote
              13
              down vote

              up vote
              13
              down vote

              What light looks like from the surface

              The sun in your world is half-bright, half-dark. Assume that it is a perfect sphere with any radius greater than zero. Therefore, for any point on the surface, the percentage of the sun that is facing that point at any given time is

              $$cos^2phicos^22theta + frac{1}{2}sin^2phi.$$

              Here are three plots of what that looks like at the equator:

              enter image description here

              At a point in the temperate zone at about 40 degrees latitude:

              enter image description here

              And at the poles:

              enter image description here

              How to measure latitude between two points

              Without sunrise over a horizon or a pole star, it is much harder to measure latitude. The best you can do is to measure the brightness of the sun at various points in the day, and to compare with the trigonometeric properties of its brightness from the last section.

              The the best of my knowledge, there is no way to scientifically measure brightness until you have a photographic plate. That will not be available to people of Renaissance technology. However, you can at the very least dead-eye reckon brightness, so we will assume some sort of brightness metric. One potential way to measure brightness is that brightness is proportional to the ‘bright’ portion of the sphere of the sun that is visible to you from your location. If you look at the sun through a dark lens, you may be able to measure this, depending on how large the radius of the sun is.

              In this case, you can calculate latitude relative to the equator easily. Maximum brightness at latitude $phi$ ($max(B_phi)$) is $1 – 1/2sin^2phi$ times that at the equator ($max(B_{eq})$), solve backwards for $phi$:

              $$phi = sin^{-1}sqrt{2left(1 – frac{max(B_phi)}{max(B_{eq})}right)}.$$

              It is possible that maximum brightness at the equator is a well known standard value in your scientific community, even for those who have never been there. You have to know, or at least be able to estimate, the maximum brightness to be able to calculate latitude. As an alternate measurement, not2 that equatorial max brightness is twice the constant brightness at the poles. You can re-work everything in terms of that value as
              $$max(B_{eq}) = 2B_{pole}.$$

              Now there are two ways to measure radius

              Assuming the hollow world is a perfect sphere, there are two ways to measure the radius. you can compare latitude of two points to the distance between them, or you can compare the time it takes the terminator to travel between two points.

              The terminator line

              The terminator is important here, because with no stars in the background, the only way to ensure that two points are at exactly the same longitude is to visually signal when the terminator passes by the points. This can be done using one lighthouse to signal another point within visual range of that lighthouse. Since your world is hollow and thus concave from the point of view of someone on the surface, the line of sight of a lighthouse is actually very long, limited only by atmospheric attenuation of light (due to water vapor, or whatever).

              The terminator can be exactly identified by looking at the sun through a telescope with a dark lens. As soon as there is no bright patch visible, the terminator line has passed.

              Comparing latitude method

              The latitude way is take two points that you know are at the same longitude, and calculate their latitude using the above method and distance ($d$) between them, using what ever method.

              If the latitude delta is $alpha$, then the polar radius of the hollow Earth is $$r_{pole} = d/alpha$$ in radians.

              Timing the terminator method

              To time the terminator, you will need to get two points at the same latitude (confirmed using the methods above), and measure the distance ($d$) between them and the time it takes the terminator to pass between them ($t$). You also need to know the length of the day $t_{day}$.

              Then, the equatorial radius of the hollow Earth is $$r_{eq} = 2pi dfrac{ t_{day}}{t}.$$ Note this only works if the points are less than $pi$ radians apart on the surface, that is, in the same hemisphere.

              These numbers might not be the same!

              These two methods could give you different answers, polar and equatorial radius. If you have a perfect sphere, the two calculated radii should be the same, but if your planet is inscribed within an oblate spheroid with an equatorial bulge(like our Earth), or even a polar bulge, then the two numbers will not be the same.

              Conclusion

              It is actually pretty difficult to time astronomical pheomena with only one available astronomical object. But, you can use the unique configuration of the sun’s surface to do this.

              This does require some photometric skills more advanced than what was available in the Renaissance, which I’m not sure how to replicate with Renaissance technology, but I have faith that the Galileo’s and Newton’s of the world could figure it out. Also, there needs to be some sort of standard measurement of equatorial brightness to make these calculations, but given this number’s importance in observational ‘heliometry,’ I would expect this number to be a well established topic in a Hollow Earth’s scientific community.

              share|improve this answer

              What light looks like from the surface

              The sun in your world is half-bright, half-dark. Assume that it is a perfect sphere with any radius greater than zero. Therefore, for any point on the surface, the percentage of the sun that is facing that point at any given time is

              $$cos^2phicos^22theta + frac{1}{2}sin^2phi.$$

              Here are three plots of what that looks like at the equator:

              enter image description here

              At a point in the temperate zone at about 40 degrees latitude:

              enter image description here

              And at the poles:

              enter image description here

              How to measure latitude between two points

              Without sunrise over a horizon or a pole star, it is much harder to measure latitude. The best you can do is to measure the brightness of the sun at various points in the day, and to compare with the trigonometeric properties of its brightness from the last section.

              The the best of my knowledge, there is no way to scientifically measure brightness until you have a photographic plate. That will not be available to people of Renaissance technology. However, you can at the very least dead-eye reckon brightness, so we will assume some sort of brightness metric. One potential way to measure brightness is that brightness is proportional to the ‘bright’ portion of the sphere of the sun that is visible to you from your location. If you look at the sun through a dark lens, you may be able to measure this, depending on how large the radius of the sun is.

              In this case, you can calculate latitude relative to the equator easily. Maximum brightness at latitude $phi$ ($max(B_phi)$) is $1 – 1/2sin^2phi$ times that at the equator ($max(B_{eq})$), solve backwards for $phi$:

              $$phi = sin^{-1}sqrt{2left(1 – frac{max(B_phi)}{max(B_{eq})}right)}.$$

              It is possible that maximum brightness at the equator is a well known standard value in your scientific community, even for those who have never been there. You have to know, or at least be able to estimate, the maximum brightness to be able to calculate latitude. As an alternate measurement, not2 that equatorial max brightness is twice the constant brightness at the poles. You can re-work everything in terms of that value as
              $$max(B_{eq}) = 2B_{pole}.$$

              Now there are two ways to measure radius

              Assuming the hollow world is a perfect sphere, there are two ways to measure the radius. you can compare latitude of two points to the distance between them, or you can compare the time it takes the terminator to travel between two points.

              The terminator line

              The terminator is important here, because with no stars in the background, the only way to ensure that two points are at exactly the same longitude is to visually signal when the terminator passes by the points. This can be done using one lighthouse to signal another point within visual range of that lighthouse. Since your world is hollow and thus concave from the point of view of someone on the surface, the line of sight of a lighthouse is actually very long, limited only by atmospheric attenuation of light (due to water vapor, or whatever).

              The terminator can be exactly identified by looking at the sun through a telescope with a dark lens. As soon as there is no bright patch visible, the terminator line has passed.

              Comparing latitude method

              The latitude way is take two points that you know are at the same longitude, and calculate their latitude using the above method and distance ($d$) between them, using what ever method.

              If the latitude delta is $alpha$, then the polar radius of the hollow Earth is $$r_{pole} = d/alpha$$ in radians.

              Timing the terminator method

              To time the terminator, you will need to get two points at the same latitude (confirmed using the methods above), and measure the distance ($d$) between them and the time it takes the terminator to pass between them ($t$). You also need to know the length of the day $t_{day}$.

              Then, the equatorial radius of the hollow Earth is $$r_{eq} = 2pi dfrac{ t_{day}}{t}.$$ Note this only works if the points are less than $pi$ radians apart on the surface, that is, in the same hemisphere.

              These numbers might not be the same!

              These two methods could give you different answers, polar and equatorial radius. If you have a perfect sphere, the two calculated radii should be the same, but if your planet is inscribed within an oblate spheroid with an equatorial bulge(like our Earth), or even a polar bulge, then the two numbers will not be the same.

              Conclusion

              It is actually pretty difficult to time astronomical pheomena with only one available astronomical object. But, you can use the unique configuration of the sun’s surface to do this.

              This does require some photometric skills more advanced than what was available in the Renaissance, which I’m not sure how to replicate with Renaissance technology, but I have faith that the Galileo’s and Newton’s of the world could figure it out. Also, there needs to be some sort of standard measurement of equatorial brightness to make these calculations, but given this number’s importance in observational ‘heliometry,’ I would expect this number to be a well established topic in a Hollow Earth’s scientific community.

              share|improve this answer

              share|improve this answer

              share|improve this answer

              edited Nov 29 at 18:53

              answered Nov 29 at 17:55

              kingledion

              72.5k26244428

              72.5k26244428

              • 1

                I have serious doubts about people being able to judge latitude based on brightness – the human eye is very bad at judging between subtle changes in brightness. Would it be possible to use something different, like using the dark lens to see how much light is visible at the top/bottom of the sun during night (or darkness during day)?
                – Rob Watts
                Nov 29 at 18:15

              • 1

                @RobWatts A good idea! I specifically left the brightness measurement assumed. As i mention in the conclusion, given the importance of equatorial max brightness in astronomy, I assume that a Galileo or Newton would find a good measure, even if I can’t right now.
                – kingledion
                Nov 29 at 18:39

              • With latitude: $-pi/2 le theta le pi/2$, longitude: $-pi le phi le pi$ I got i.stack.imgur.com/cTq0e.png (python pastebin.com/VDijE4WN) which is a pretty intriguing pattern! Is this right? Is there a source or derivation for your expression somewhere? Thanks!
                – uhoh
                Nov 30 at 2:49

              • 1

                @uhoh That looks right, the key finding is that for any point on the interior of the hollow sphere, the light intensity integrated over a full rotation is the same, 0.5! So for the derivation, I did a 2-d integration over the surface of the sphere with respect to $theta$ and $phi$. I did it myself, so not 100% that it is right. My answer is already too long, I’ll do it up in latex and share with you if you ping me in chat, if you want.
                – kingledion
                Nov 30 at 4:21

              • For equipment I wonder if you could use something like a pinhole camera. it would let you see which parts of the light source are occluded by the shade, though it wouldn’t really help with light levels except by the most skilled observers.
                – AndyD273
                Nov 30 at 18:00

              • 1

                I have serious doubts about people being able to judge latitude based on brightness – the human eye is very bad at judging between subtle changes in brightness. Would it be possible to use something different, like using the dark lens to see how much light is visible at the top/bottom of the sun during night (or darkness during day)?
                – Rob Watts
                Nov 29 at 18:15

              • 1

                @RobWatts A good idea! I specifically left the brightness measurement assumed. As i mention in the conclusion, given the importance of equatorial max brightness in astronomy, I assume that a Galileo or Newton would find a good measure, even if I can’t right now.
                – kingledion
                Nov 29 at 18:39

              • With latitude: $-pi/2 le theta le pi/2$, longitude: $-pi le phi le pi$ I got i.stack.imgur.com/cTq0e.png (python pastebin.com/VDijE4WN) which is a pretty intriguing pattern! Is this right? Is there a source or derivation for your expression somewhere? Thanks!
                – uhoh
                Nov 30 at 2:49

              • 1

                @uhoh That looks right, the key finding is that for any point on the interior of the hollow sphere, the light intensity integrated over a full rotation is the same, 0.5! So for the derivation, I did a 2-d integration over the surface of the sphere with respect to $theta$ and $phi$. I did it myself, so not 100% that it is right. My answer is already too long, I’ll do it up in latex and share with you if you ping me in chat, if you want.
                – kingledion
                Nov 30 at 4:21

              • For equipment I wonder if you could use something like a pinhole camera. it would let you see which parts of the light source are occluded by the shade, though it wouldn’t really help with light levels except by the most skilled observers.
                – AndyD273
                Nov 30 at 18:00

              1

              1

              I have serious doubts about people being able to judge latitude based on brightness – the human eye is very bad at judging between subtle changes in brightness. Would it be possible to use something different, like using the dark lens to see how much light is visible at the top/bottom of the sun during night (or darkness during day)?
              – Rob Watts
              Nov 29 at 18:15

              I have serious doubts about people being able to judge latitude based on brightness – the human eye is very bad at judging between subtle changes in brightness. Would it be possible to use something different, like using the dark lens to see how much light is visible at the top/bottom of the sun during night (or darkness during day)?
              – Rob Watts
              Nov 29 at 18:15

              1

              1

              @RobWatts A good idea! I specifically left the brightness measurement assumed. As i mention in the conclusion, given the importance of equatorial max brightness in astronomy, I assume that a Galileo or Newton would find a good measure, even if I can’t right now.
              – kingledion
              Nov 29 at 18:39

              @RobWatts A good idea! I specifically left the brightness measurement assumed. As i mention in the conclusion, given the importance of equatorial max brightness in astronomy, I assume that a Galileo or Newton would find a good measure, even if I can’t right now.
              – kingledion
              Nov 29 at 18:39

              With latitude: $-pi/2 le theta le pi/2$, longitude: $-pi le phi le pi$ I got i.stack.imgur.com/cTq0e.png (python pastebin.com/VDijE4WN) which is a pretty intriguing pattern! Is this right? Is there a source or derivation for your expression somewhere? Thanks!
              – uhoh
              Nov 30 at 2:49

              With latitude: $-pi/2 le theta le pi/2$, longitude: $-pi le phi le pi$ I got i.stack.imgur.com/cTq0e.png (python pastebin.com/VDijE4WN) which is a pretty intriguing pattern! Is this right? Is there a source or derivation for your expression somewhere? Thanks!
              – uhoh
              Nov 30 at 2:49

              1

              1

              @uhoh That looks right, the key finding is that for any point on the interior of the hollow sphere, the light intensity integrated over a full rotation is the same, 0.5! So for the derivation, I did a 2-d integration over the surface of the sphere with respect to $theta$ and $phi$. I did it myself, so not 100% that it is right. My answer is already too long, I’ll do it up in latex and share with you if you ping me in chat, if you want.
              – kingledion
              Nov 30 at 4:21

              @uhoh That looks right, the key finding is that for any point on the interior of the hollow sphere, the light intensity integrated over a full rotation is the same, 0.5! So for the derivation, I did a 2-d integration over the surface of the sphere with respect to $theta$ and $phi$. I did it myself, so not 100% that it is right. My answer is already too long, I’ll do it up in latex and share with you if you ping me in chat, if you want.
              – kingledion
              Nov 30 at 4:21

              For equipment I wonder if you could use something like a pinhole camera. it would let you see which parts of the light source are occluded by the shade, though it wouldn’t really help with light levels except by the most skilled observers.
              – AndyD273
              Nov 30 at 18:00

              For equipment I wonder if you could use something like a pinhole camera. it would let you see which parts of the light source are occluded by the shade, though it wouldn’t really help with light levels except by the most skilled observers.
              – AndyD273
              Nov 30 at 18:00

              up vote
              7
              down vote

              Two points, a known distance apart, will have a straight line-of-sight between them that will not be parallel to the curving surface. Stand a pole at the centre between the two points. Where the sight line intersects the pole will give you the height (er – depth) of the arc. That will allow them to calculate the radius of their world. Even if they don’t have the trig to calculate it, they can always just make a scale diagram and measure it.

              enter image description here

              share|improve this answer

              • 1

                You’d need a tower, not a pole, because the local curvature of the Earth can be quite different from the large scale curvature, for example, on a hillside.
                – kingledion
                Nov 29 at 22:51

              • 1

                Don’t try to measure the planet while on a hillside. 😐
                – DaveC426913
                Nov 30 at 2:30

              • You don’t really need a pole/tower. Angles could be measured quite accurately since ancient times. Just measure the angles between the straight line and the sun and use the fact that internal angles of a triangle add up to half circle to determine the difference in longitude/lattitude.
                – Jan Hudec
                Nov 30 at 9:09

              • 1

                You’ll end up with a triangle that is n miles at the base and 4000 miles tall. You’ll be working with an angle that is 0.15 degrees from square. A lot of room for error there.
                – DaveC426913
                Nov 30 at 16:15

              up vote
              7
              down vote

              Two points, a known distance apart, will have a straight line-of-sight between them that will not be parallel to the curving surface. Stand a pole at the centre between the two points. Where the sight line intersects the pole will give you the height (er – depth) of the arc. That will allow them to calculate the radius of their world. Even if they don’t have the trig to calculate it, they can always just make a scale diagram and measure it.

              enter image description here

              share|improve this answer

              • 1

                You’d need a tower, not a pole, because the local curvature of the Earth can be quite different from the large scale curvature, for example, on a hillside.
                – kingledion
                Nov 29 at 22:51

              • 1

                Don’t try to measure the planet while on a hillside. 😐
                – DaveC426913
                Nov 30 at 2:30

              • You don’t really need a pole/tower. Angles could be measured quite accurately since ancient times. Just measure the angles between the straight line and the sun and use the fact that internal angles of a triangle add up to half circle to determine the difference in longitude/lattitude.
                – Jan Hudec
                Nov 30 at 9:09

              • 1

                You’ll end up with a triangle that is n miles at the base and 4000 miles tall. You’ll be working with an angle that is 0.15 degrees from square. A lot of room for error there.
                – DaveC426913
                Nov 30 at 16:15

              up vote
              7
              down vote

              up vote
              7
              down vote

              Two points, a known distance apart, will have a straight line-of-sight between them that will not be parallel to the curving surface. Stand a pole at the centre between the two points. Where the sight line intersects the pole will give you the height (er – depth) of the arc. That will allow them to calculate the radius of their world. Even if they don’t have the trig to calculate it, they can always just make a scale diagram and measure it.

              enter image description here

              share|improve this answer

              Two points, a known distance apart, will have a straight line-of-sight between them that will not be parallel to the curving surface. Stand a pole at the centre between the two points. Where the sight line intersects the pole will give you the height (er – depth) of the arc. That will allow them to calculate the radius of their world. Even if they don’t have the trig to calculate it, they can always just make a scale diagram and measure it.

              enter image description here

              share|improve this answer

              share|improve this answer

              share|improve this answer

              edited Nov 29 at 21:28

              answered Nov 29 at 21:22

              DaveC426913

              1813

              1813

              • 1

                You’d need a tower, not a pole, because the local curvature of the Earth can be quite different from the large scale curvature, for example, on a hillside.
                – kingledion
                Nov 29 at 22:51

              • 1

                Don’t try to measure the planet while on a hillside. 😐
                – DaveC426913
                Nov 30 at 2:30

              • You don’t really need a pole/tower. Angles could be measured quite accurately since ancient times. Just measure the angles between the straight line and the sun and use the fact that internal angles of a triangle add up to half circle to determine the difference in longitude/lattitude.
                – Jan Hudec
                Nov 30 at 9:09

              • 1

                You’ll end up with a triangle that is n miles at the base and 4000 miles tall. You’ll be working with an angle that is 0.15 degrees from square. A lot of room for error there.
                – DaveC426913
                Nov 30 at 16:15

              • 1

                You’d need a tower, not a pole, because the local curvature of the Earth can be quite different from the large scale curvature, for example, on a hillside.
                – kingledion
                Nov 29 at 22:51

              • 1

                Don’t try to measure the planet while on a hillside. 😐
                – DaveC426913
                Nov 30 at 2:30

              • You don’t really need a pole/tower. Angles could be measured quite accurately since ancient times. Just measure the angles between the straight line and the sun and use the fact that internal angles of a triangle add up to half circle to determine the difference in longitude/lattitude.
                – Jan Hudec
                Nov 30 at 9:09

              • 1

                You’ll end up with a triangle that is n miles at the base and 4000 miles tall. You’ll be working with an angle that is 0.15 degrees from square. A lot of room for error there.
                – DaveC426913
                Nov 30 at 16:15

              1

              1

              You’d need a tower, not a pole, because the local curvature of the Earth can be quite different from the large scale curvature, for example, on a hillside.
              – kingledion
              Nov 29 at 22:51

              You’d need a tower, not a pole, because the local curvature of the Earth can be quite different from the large scale curvature, for example, on a hillside.
              – kingledion
              Nov 29 at 22:51

              1

              1

              Don’t try to measure the planet while on a hillside. 😐
              – DaveC426913
              Nov 30 at 2:30

              Don’t try to measure the planet while on a hillside. 😐
              – DaveC426913
              Nov 30 at 2:30

              You don’t really need a pole/tower. Angles could be measured quite accurately since ancient times. Just measure the angles between the straight line and the sun and use the fact that internal angles of a triangle add up to half circle to determine the difference in longitude/lattitude.
              – Jan Hudec
              Nov 30 at 9:09

              You don’t really need a pole/tower. Angles could be measured quite accurately since ancient times. Just measure the angles between the straight line and the sun and use the fact that internal angles of a triangle add up to half circle to determine the difference in longitude/lattitude.
              – Jan Hudec
              Nov 30 at 9:09

              1

              1

              You’ll end up with a triangle that is n miles at the base and 4000 miles tall. You’ll be working with an angle that is 0.15 degrees from square. A lot of room for error there.
              – DaveC426913
              Nov 30 at 16:15

              You’ll end up with a triangle that is n miles at the base and 4000 miles tall. You’ll be working with an angle that is 0.15 degrees from square. A lot of room for error there.
              – DaveC426913
              Nov 30 at 16:15

              up vote
              6
              down vote

              Providing long distance travel is practical, his could be done with pure geometry ( or more easily, assuming the natives had advanced to usable trigonometry). One could measure the distance between two points on the inner surface that are visible from the same location, then travel to one point and directly measure the distance to the other point — or, lacking trigonometry, one could select two points that are sixty degrees apart (therefore forming an equilateral triangle between the two and your observation point), and know that the distance to either one is exactly one third of the circumference of the world.

              Angle bisection could be used, along with long distance signaling (with flashing mirrors, perhaps), to shorten the distance requiring measurement, at the cost of some loss of precision.

              If there’s not access to the antipodes, one could measure a distance across water (perhaps by surveying around the shore of a large lake or inland sea), and directly measure how much the surface dips below the sight line; this would allow deriving the radius of the sphere directly via similar triangles (same method as is used to calculate horizon distance on the outside of the Earth).

              share|improve this answer

                up vote
                6
                down vote

                Providing long distance travel is practical, his could be done with pure geometry ( or more easily, assuming the natives had advanced to usable trigonometry). One could measure the distance between two points on the inner surface that are visible from the same location, then travel to one point and directly measure the distance to the other point — or, lacking trigonometry, one could select two points that are sixty degrees apart (therefore forming an equilateral triangle between the two and your observation point), and know that the distance to either one is exactly one third of the circumference of the world.

                Angle bisection could be used, along with long distance signaling (with flashing mirrors, perhaps), to shorten the distance requiring measurement, at the cost of some loss of precision.

                If there’s not access to the antipodes, one could measure a distance across water (perhaps by surveying around the shore of a large lake or inland sea), and directly measure how much the surface dips below the sight line; this would allow deriving the radius of the sphere directly via similar triangles (same method as is used to calculate horizon distance on the outside of the Earth).

                share|improve this answer

                  up vote
                  6
                  down vote

                  up vote
                  6
                  down vote

                  Providing long distance travel is practical, his could be done with pure geometry ( or more easily, assuming the natives had advanced to usable trigonometry). One could measure the distance between two points on the inner surface that are visible from the same location, then travel to one point and directly measure the distance to the other point — or, lacking trigonometry, one could select two points that are sixty degrees apart (therefore forming an equilateral triangle between the two and your observation point), and know that the distance to either one is exactly one third of the circumference of the world.

                  Angle bisection could be used, along with long distance signaling (with flashing mirrors, perhaps), to shorten the distance requiring measurement, at the cost of some loss of precision.

                  If there’s not access to the antipodes, one could measure a distance across water (perhaps by surveying around the shore of a large lake or inland sea), and directly measure how much the surface dips below the sight line; this would allow deriving the radius of the sphere directly via similar triangles (same method as is used to calculate horizon distance on the outside of the Earth).

                  share|improve this answer

                  Providing long distance travel is practical, his could be done with pure geometry ( or more easily, assuming the natives had advanced to usable trigonometry). One could measure the distance between two points on the inner surface that are visible from the same location, then travel to one point and directly measure the distance to the other point — or, lacking trigonometry, one could select two points that are sixty degrees apart (therefore forming an equilateral triangle between the two and your observation point), and know that the distance to either one is exactly one third of the circumference of the world.

                  Angle bisection could be used, along with long distance signaling (with flashing mirrors, perhaps), to shorten the distance requiring measurement, at the cost of some loss of precision.

                  If there’s not access to the antipodes, one could measure a distance across water (perhaps by surveying around the shore of a large lake or inland sea), and directly measure how much the surface dips below the sight line; this would allow deriving the radius of the sphere directly via similar triangles (same method as is used to calculate horizon distance on the outside of the Earth).

                  share|improve this answer

                  share|improve this answer

                  share|improve this answer

                  answered Nov 29 at 16:45

                  Zeiss Ikon

                  65417

                  65417

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