One of my goals in high fantasy is to make my world as real as possible. High fantasy requires a lot of world-building. The author essentially becomes an architect of a new world, in addition to a story-teller.

My YA high fantasy series is set on the planet Tellia. I’ve always envisioned it being larger than earth, shrouded in a thick cloud-cover with no external light penetration, and teeming with bioluminescent life.

(I’ll admit when I saw the movie AVATAR back in 2008, I was not happy that James Cameron used all that bioluminescence on Pandora at night. I was afraid my book would sound like a Johnny-come-lately copy. But this aspect has been a part of my ideas for Tellia since 2003, so I’m sticking with it. Besides, on Tellia, there is NO light from their star, ever).

Most of the brighter white-light comes from a common medium-sized shrub called Ayastroh (light tree). Ayastroh are planted near the windows of buildings so the light can get in. They’re also potted inside of buildings. Every plant emits bioluminescence. Most fruit trees give off an orange-tinted light.

Another tree is used to tell time.

Tellia has its own calendar. Minutes and hours are equivalent to Earth minutes and hours. But Tellian days are 20 hours long. A week is 5 days (100 hours). A month is 25 days (500 hours). A segment is 8 months (4,000 hours). A year is 65 segments, with a bit of a quirk at the end, so it’s actually 259,980.387896312064 hours. Time on Tellia is counted by the Dimertroh (time tree), which has a natural 20-hour bioluminescent cycle. Its color changes every five hours. It glows green during the five morning hours, white for the five day hours, blue for the five evening hours, and dark-purple during the five night hours.

In the final minutes of each segment, the Dimertroh dims to black. The new segment is counted at the beginning of the green cycle that follows. The Dimertroh also signals the end of a Tellian year. The last day of the 65^{th} segment is shortened by nearly a full day (19 hours, ~37 minutes). The Dimertroh glows pink for the remaining ~23 minutes (0.387896312064 hours) of that last day. The New Year begins at the start of the green cycle that follows.

Ages on Tellia are counted by the segment rather than the year. A Tellian year is 29.657296 Earth years. An Earth year is 2.191538196 Tellian segments. So if you wanted to figure out what your age is in Tellian segments, you would multiply your age by 2.191538196. To make it easier, I’ve made a chart.

Something else I had to figure out was exactly how big Tellia is as a planet. I have people traveling from here to there by foot and by ship in specific amounts of time. One thing was certain: I had a lot of research to do, and it would require more math.

My first question was, “*How large is Tellia in the story itself?*” If my main character, Ashura, walks from her city to the capital city, and if the trip takes two full days of travel, how far is the capital from her home? If it takes her 3 months to sail from her home to the Isle of Shifqu, how many kilometers is the trip? I drew a map of Tellia with meridian and latitudinal lines. I needed to determine how far apart the meridian lines are at the equator. So I had to look at sailing speeds and walking speeds.

I wanted sailing to be fairly quick, but I wasn’t sure what decent sailing speeds are. I’ve always been enamored with clipper ships, so I did some research. *Flying Cloud* was a clipper that set a record by sailing from New York to San Francisco in 89 days, 8 hours in 1853 (full of people and cargo, mind you!). That record stood for 135 years until a few ships (built for speed and NOT carrying cargo!) beat *Flying Cloud*’s record. The anchor-to-anchor average speed for the 24,683 km journey was 11.51 kph.

On Tellia, I have ships being able to sail between meridian lines in 100 hours. I decided ships on Tellia would sail at an average of 12 kph. At that speed, a ship would cross 1,200 km of ocean in 100 hours. But I’ll make my meridian distances slightly smaller so it’s not such a neat number. I went with 1,198 km. On my map there are 42 meridian lines. So 42 * 1,198 km gives us Tellia’s equatorial circumference at 50,316 kilometers. (For reference, Earth’s is 40,075 km).

Now, the average human walking speed is 5 kph (3.10686 mph). If distance between meridians at the equator is 1,198 km, it will take 239.6 hours (11.83 days) to cover the distance while walking. However, I need to adjust for breaks and sleeping at night. No one can walk for 1,198 km without sleep. Travelers would walk for 10 hours/day max in Tellia’s 20 hour days (at least, that’s all the further I would want to walk). This will help me determine how far apart locations are on my map, and, therefore, how long it takes for my characters to walk there.

Now, because I know Tellia’s circumference is 50,316 km, I can calculate its diameter and radius. The formula for finding the circumference of a circle is C=πd. I need to divide the circumference by π to find the diameter.

50,316 km / 3.1415926535897932384626433832795 = 16,016.08 km diameter. That gives a radius of 8,008.04 km.

Now that I know Tellia’s mean radius, I need to determine its mass so the acceleration of gravity (g) will be nearly equal to the acceleration of gravity on Earth (g = 9.8 m/s^{2}). I want to be able to walk on Tellia without sweating each time I try to lift my foot.

In order to figure this stuff out, I needed to do some physics. The acceleration of gravity formula is g=G*M/R^{2}. G is the universal gravitational constant (6.67384×10^{11} N-m^{2}/kg^{2}). M is the mass. R^{2} is the square of the mean radius.

So, in order to calculate the acceleration of gravity (g) for Earth I needed to know earth’s radius and Earth’s mass. R_{Earth} = 6,371 km (6,371,000 m), so R_{Earth}^{2} = 40,589,641 km (40,589,641,000 m). M_{Earth} = 5.97219×10^{24} kg.

g (m/s^{2}) = ([G] 6.67384×10^{11} N-m^{2}/kg^{2}) * ([M_{Earth}] 5.97219×10^{24} kg) / ([R_{Earth}^{2}] 40,589,641,000 m).

The solution is 98,196,090,252,682,944,399,532,875.88821, but somehow that is equated to 9.8 m/s^{2}. Don’t ask me how. I’m not a physicist. It might be different from what you’ll read on Wikipedia, but apparently G has been adjusted slightly over time. I picked one value for G and rolled with it.

I followed this by trying several calculations using different variables for Tellia’s possible mass. Tellia’s squared radius is R_{Tellia}^{2} = 64,128,706.509 km. When I calculate Tellia’s force of gravity with a mass of 9.4360602×10^{24} kg (which is 1.58 times the mass of Earth), I get this:

g (m/s^{2}) = (6.67384×10^{11} N-m^{2}/kg^{2}) * (9.4360602×10^{24} kg) / (64,128,706,509.264 m). g = 9.8 m/s^{2}.

The solution is actually 98,200,571,059,499,970,799,449,914, but I’m guessing it equates to 9.82 m/s^{2} just like Earth’s equation above. If Tellia’s mass were 1.6 times that of Earth, then g=9.94 m/s^{2}.

I also calculated Tellia’s volume [V=4*π*R*R*R / 3] and surface area [A=4*π*R^{2}].

Tellia’s mean diameter is** 16,016.08 km**. Earth’s is 12,742 km.

Tellia’s mean radius is** 8,008.04 km**. Earth’s is 6,371 km.

Tellia’s circumference is** 50,316 km. **Earth’s = 40,075 km.

Tellia’s solar rotation speed is** 2,515.8 km/hr**. Earth’s is 1,674.4 km/hr.

Tellia’s surface area is** 805,865,093,014.879 km ^{2}**. Earth’s is 510,072,000 km

^{2}.

Tellia’s volume is** 2,151,133,237,173.37** **km ^{3}**. Earth’s is 1.08321×10

^{12}km

^{3}.

Tellia’s mass is** 9.4360602×10 ^{24} kg**. Earth’s is 5.97219×10

^{24}kg.

Tellia’s acceleration of gravity at its surface is** 9 .82 m/s^{2}**. Earth’s is 9.8 m/s

^{2}.

Of course, all of this assumes that I’ve done my math correctly. But math has never been one of my stronger skills. So if anyone finds a flaw in my calculations, I’d love to know about it.

I’ve done some work on Tellia’s atmosphere, too. From the outside, I want Tellia to look blue-ish (think Neptune!), so I need methane to be somewhat significant because it absorbs visible light on the red end of the spectrum. I imagine the upper Troposphere being subject to high wind speeds and persistent thick cloud-cover. From the inside, the sky would be grey-black swirling clouds, only visible because of the bioluminescent light, originating on the surface, that gets reflected back down.

Here’s what I came up with for the dry atmospheric composition: Nitrogen 65%; Oxygen 25.35%; Argon 3.23%; Methane 2.85%; Hydrogen 1.92%; Carbon Dioxide 1.0%; Helium 0.5%; Other gasses 0.15%.

I don’t have atmospheric pressure or thickness figured out. I’m going to go with the idea that it’s similar pressure to earth, if not a little heavier. For now, that math is beyond me.

So that’s that. World-building. It’s a lot of fun. It’s a lot of work. It’s frustrating at times, especially when you have to learn to calculate all the stuff above. But, in the end, I’m certain the work I’ve put into making my world as real as possible will help the story. That’s why I do it.

Christopher