Perception of the Heavens
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- Опубліковано 26 лип 2024
- In which I show you how to calculate the apparent size and brightness of your stars and planets.
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Inverse-square law: en.wikipedia.org/wiki/Inverse-...
Luminosity and distance: www.e-education.psu.edu/astro...
Albedo: en.wikipedia.org/wiki/Albedo
Angular Diameter: en.wikipedia.org/wiki/Angular_...
CORRECTIONS: At 4:56 - I say, and draw 0.05 degrees. That figure should be 0.005 degrees. - Наука та технологія
I remember in one of Arthur C Clark's books, the natives of Europa had interesting mythology. Jupiter had been turned into a mini-star, so that was very dominant in the sky and didn't move, whereas the actual sun was relatively dim and moved in a great arc, so the Europans believed that the suns were brothers and that one of them had to walk around the edge of the universe as punishment for some crime
Froy233 oh god that’s what I’m trying to do (my system is an s-type binary)
In my setting, I also have a habitable moon in orbit around a gas giant.
The fact that the planet would apparently vanish from the sky each time the sun rose, and would shine brightly at night, gave rise to the idea that the planet was a trickster who had angered the sun and was hiding amongst the stars to avoid being caught.
This contributes to an important trend in their culture, where light is an aggressive, hostile force that consumes peace, and darkness is viewed as a form of shelter. Very different from the respective cleansing beacon of virtue and harbor of sinister dangers that exist in our culture.
Nuada the silver hand that’s very interesting! I want to see how that would play out more
Aren't gas giants' satellites supposed to always have the same half facing the planet, the other half never seeing it?
@@LeDingueDeJeuxVideos, yes, but the gas giant still has phases, thus vanishing. And only the people on the hemisphere nearest the gas giant have that myth.
I applied this to my system and I found an interesting result. For my star, I calculated how big it would look from the planet surface and I got roughly 2.3 degrees, which makes it look bigger than our sun would look from Earth. But, that isn't the interesting part, I did the same calculation with a nearby Gas Giant in the system and I got 1.7 degrees. This meant, from my planet, you can just look up into the sky and see the gas giant in greater relative size than the moon from our own planet. I check the Gas Giant's brightness and got 1.9E-1 w/m^2, which meant that you can look up into the sky, during the day, and you can still clearly see the Gas Giant. This is amazing to think about for my story.
Cool beans! :)
Wowza, you sure you arrent gonna need to include this gas giant in your tidal considerations?
How do you get the light reflectivity of your gas giant so you can watch it in your planet? how many W/m2 do you get to obtain the value?
"Coincidence? I think so!"
Thumbs up, just for that!
OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO GOD
"Coincidence? I think so!" is probably one of the best lines I've ever heard
On the Uranus and Saturn brightness comparison, I do not think it is the 1.3 vs 1.2 that matters but the -7 vs -9 exponent.
An interesting perception of world building. Quite few of such worldbuilders rarely consider the possible impact culturally and religiously such stars and planets through the eyes of the fictional aliens that inhabit these systems and can potentially give birth to many mythological stories that add depth to a world that could also potentially reveal the mindset or even the paradigm of that culture. And considering that just as many cultures have different stories about the stars in the sky, the possibilities of "myth building" (or whatever that word J.R.R. Tolkien made up) are endless.
Speaking of stars, I can only assume that the constellations are either the same as our own (with some differences depending on local galactic positions, or entirely fictitious due to the location of said planet being somewhere on the far side of the galaxy (or universe if galaxy is made up as well). Personally I'd hate to see the system that would have to show what stars could be in a constellation and whatnots, especially since they're largely products of the imagination of the observer.
To my mind, worldbuilders should always construct constellations. The morphologies imagined by fictional aliens/peoples will have a serious impact on their culture. I don't think its necessary to mathematical construct them however that seems like overkill to me.
Interesting point though, how a galaxy is perceived can also be quiet important. I believe, that the aboriginal peoples of Australia, believed that the milky way was a giant emu and this emu was/is a central pillar of their mythology.
Now I could be wrong about the details of this. Brady Haran did a video a while back about it for deepskyvideos, its been awhile since I saw that video. My memory is a bit hazy here.
Take home point would be that EVERY celestial object can have massive cultural implications.
Yeah, constructing a culture around a world is probably one of my favorite things to do. It can give your world a much more alien feel, as well as make it easier to construct characters for xenofiction (if that's what you're doing).
Sabersonic yeah! My system is an s-type, so my species on it worship the king, and his exiled brother, who will destroy the world at armageddon
the name is Bond, Albedo bond
Your name is Jeb, Jeb Kerman
Lol
Names......lol
Things I learned from this video:
Jeeze, venus is bright!
+LarlemMagic
There's a reason it's called the Morning Star
That’s why it’s the second brightest thing ever. Second to the MOON.
@@plantsempire * in the night
missed an opportunity to say jesus venus
Great! This video answers a question I asked on a previous one. Thanks!
I can now develop a trinary star system, with a habitable planet orbiting the third, distant star. The idea being that the planet, tidally locked, still has relatively bright light coming to its night side, from the primary and secondary stars, a few hundreds of AUs away. If I did my math right, light on the night side would be something like 10 times brighter than Earth's full moon.
Getting down to the nitty gritty! I've considered eclipse effects, but not the rest. Very cool to know!
God, this channel is great. God? Did you hear me? You should have done it like this guy here, and maybe you could have also added an explanation video like this one
Maybe it's like a college that we didn't get into or something.
For those who, like me, had problems comparing their work to other celestial objects, you can multiply the w/m^2 value by 683 to get the brightness in lux. a full moon on earth has a lux of 0.25.
Indeed some other useful values for comparison severe thick storm clouds can reduce illumination to as little as 5 lux, around 1,000-2,000 lux is typical overcast conditions, clear conditions are closer to 120,000 lux those values all for solar noon but gives a good idea and also helps make the logarithmic relation of percieved brightness more obvious. Course it also explains why the quoted 0.2x solar luminosity (24,000 lux at noon on a clear day) is far from "dark".
Shocked you don't have more views. Keep it up, man.
I cannot tell you how much the apparent brightness equation helped me out here. Thank you!
Short version: creating an earth-like planet with an orbital period of 10 Earth years, but receives just as much light and solar energy as Earth does from the Sun. I tried once before, creating this ungodly Frankenstein of an equation using solar radiance and similar equations found on Wikipedia, and ended up with the smallest number I've ever seen (10^-30, so I know messed up the math somewhere). With the apparent brightness equation, I assumed that the AB of my planet's star is equal to that of the Sun on Earth, essentially making AB = 1 relative to Sol (big assumption, so I would not be surprised if I got my science wrong here). From there it was a matter of substituting variables so I could calculate the star's mass from the orbital period (P = M^2.125, for those interested), and then everything else fell into place using the equations in the star-building video.
Ended up with a B-type blue star about three times the mass of Sol, which wouldn't last long enough in the main sequence to allow sapient life to evolve, but that's where applying the handwavium and following that to its logical conclusions is much easier.
4:15 filling in for DnD
This seems like a good place to ask a question - if I wanted to build a system where the rings of a Saturn-like planet were visible to the naked eye from an Earth-like planet, how far away would I need to put the gas giant? And, more to the point, how would I calculate that distance for myself?
+Werrf1 I'll try to answer the question ! ( Sorry if my english is bad :/ )
To distinguish the rings of the planet, it means you are able to see the shape of this planet, so its size in the sky ( in degrees ) must be superior to the resolution of the eye.
I found that the resolution of the eye is 55 arc seconds( the formula is : angle = 1.22 time the wavelength, here I took 560 nanometer, which is the middle of the visible spectrum. divided by the diameter of the pupil ), which is close to 1 arc minute = 1/60 of a degree. So to be sure we "distinguish" the rings, let's say the planet needs to be in the sky a 2 minute arc object, or 1/30 of degree ( 15 times smaller than the Sun )
Saturn + its rings makes an object around 250 000 km in diameter. You end up with the equation tan (1/30) = 250 000/ D (D being the distance in kilometers) D= 250 000/Tan (1/30)
I find 430 000 000 kilometers, just to distinguish the rings, without atmospheric blurr, or any eye problem.
The distance between the Earth and saturn is 3 times that distance. 1 200 000 000 km.
+khenricx Wonderful, thank you so much! I'm afraid I have no head for the numbers, can't keep them straight, so that's a huge help :) So, Saturn would need to be roughly halfway between Mars and Jupiter for the rings to be visible to the naked eye? That should work for my needs, thank you :)
Werrf1 Well, I suggest you to verify my calculation, or to ask someone else, but it seems coherent to me. Also I took the 2 arc minute value, but I don't know if it's the perfect value, it could have been 1.5 or 2.5 arc minutes. And it also mean you have very, very good eyes, in very, very good conditions. But overall I think the value I gave you is a good order of magnitude.
Also keep in mind that with that distance, you could only see that the planet is wider in the axis of the equator than in the axis of the poles (as the rings would appears as bulges around the equator)
So you would not really "see" the ring, but rather... guess their presence.
"Concidence? I think so!" - 5:25 HAHA you got me there Edgar, suck on it Theists!!!
It doesn't directly, but it addresses the argument that is commonly made by theists: many of the constants in the universe, our solar system, or our planet, all line up too perfectly in order for life to exist, hence an intelligent being must have made it so. (eg. the value of Newton's constant, the type of star that our sun is, the distance between the sun and the earth, and many other factors)
That may be true, nevertheless, coincidence can and do exist, because of the sheer number of stars and planets and galaxies there are in the universe, there has been plenty of opportunities for the die-throw to be in our favour, no matter how improbable it may be.
That's definitely true, either way, we're unbelievably lucky. (:
Didn't the Greeks have 7 planets? Mercury, Venus, Mars, Jupiter, Saturn, The Sun, and The Moon?
well, the sun isn't a planet, and the moon is a moon. I don't know much about anything else.
Yes, but the Greeks thought they were planets.
But, as we don't, we say they knew of only 5.
They're big balls in the sky; in fact so far as we see without knowing any better, they're the biggest balls of them all. For their purposes, planets.
The ancient Greek definition of a "planet" was any naked-eye object which appeared to "wander" in the sky relative to the stars.
Algebra Game A+
Great video man!
Shoddycast actually touched on this in its Skyrim Lore series (specifically s1 ep10 "Mundus"), with the constellations'/stars'/moons'/planets' effect on culture and history.
Damn, thanks to you man I can combine My two favorite things on this world
imagination
and science.
AND THATS WONDERFULL
Lol, "Coincidence? I think so." Got me.
sweet, whats that cool writing at the beginning?
It's "Artifexian" written in an early medieval irish alphabet called Ogham.
Ogham doesn't have an "x" character per say so it's only and approximation and I've also highly stylised it.
www.ireland-information.com/articles/oghamlanguage.htm
Oh cool thanks :)
Adventurenauts No probs at all. Thanks for watching :)
There is an issue with the relative apparent brightness at 1:45. The conversion in luminosity is linear, but the conversion in distance is quadratic. Thus, the numbers calculated by the formula given will always be waaayyy too small. I was scratching my head wondering why my planet, orbiting in the innermost portion of the habitable zone of an F-class star, was coming out to receive much, _much_ less light than the Earth. Then I plugged in the Sun's luminosity and conversion of AU to meters and found that it actually receives slightly more light.
The whole moon being roughly the side of the sun in the sky is a bit less of a coincidences when you factor in that the size of the moon varies with it orbit. So it not always you get a prefect eclipse. Someone one can also use in ones world building to make it all seem a bit more plausible of course. As well as set up interesting cycles.
If your thumb is only 0.5° at arm's-length, you either have very small thumbs or very long arms...
I don't know how long ago this started, or from where, but when I tested it (by sighting over my hand at the full moon) I found the thumb somewhere around 3x the Moon... my little finger was closest.
Assuming an 8mm fingernail and a 900mm arm (5/16" & 36") the angle is ~0.5°
Hello ! Can you explain me how to calculate the luminosity in watts?
Love it
Me with irregularly wide thumbs:
"Coincidence? Not even true!"
4:14 your small-angle approximation is oversimplified to the point of wrong. Solving for the sun gives an angle of something like 0.009°. A better question would be: α, the observed size in arcseconds, equals D, the diameter of the observed object, multiplied by 206,265, the number of arcseconds in a radian, all over d, the distance to the observed object: α=(D*206,265)/d . Divide the answer by 3600 to convert to degrees. (Solving for the sun now gives 0.53275933523, which is a bit off, but way closer). (Adapted from this formula: astronomyonline.org/Science/SmallAngleFormula.asp )
Question: How do you know whether an object is visible during the day? The moon has an apparent magnitude of -12, and is easily visible. The next brightest object is Venus, with a magnitude of -5 (2000 times dimmer) and is only visible during sunsets. What is the cutoff point of an object being visible during the day or not?
is there a way to calculate magnitudes?
For finding how bright my planet's moon would appear I used the equation at 2:42 and got about 11.7 out of it. What exactly is this number?
I'm assuming making moons follows the same distribution laws as for planets around a star?
Okay, just a quick question: what should I expect once I actually go through the calculations? I tried plugging the apparent brightness calculation in to check the results for a planet of mine, but got a number that only really started after seven decimal places. How should one analyze this?
I need a very bright planet for what I'm building. When I make the star bigger than the sun the distance a habitable world must be to be habitable means the apparent brightness is too low. When I make the star smaller than the sun the luminosity is too low for the AB to be >1 and the world to still be habitable. Am I just picking the wrong numbers by chance, or do I need to just take earth's star and set the planet closer?
But if there is a star that's really bright, but it's angle of apparent size is really small, will we be able to see it? Shouldn't there be an equation that defines it?
Is there an easier way to calculate the brightness of a planet?
so i guess black holes and stars have 0 albedo right? but is there an object with 1 albedo?
the small angle approximation won't give an answer in radians, you would have to take the arc tangent of d/D get the angle, otherwise it will be the tangent of the angle in radians.
so, if I got it correctly, your Tatooine system have an averange day with 60% of the britghness of a day on Earth, with lows on 32% when Star 1 eclipses Star 2, and 22% when Star 2 eclipses Star 1;
Also, Star 1 would ocupy 0.32º of the sky and Star 2 0.22º... (sound strange, I might have made a mistake)
and how do I measure the perceived distance between objects? would the inhabitants of "Tatooine" be able to see that there are 2 "Suns" in the sky, or would they notice a single one, pehaps expanding and contracting, but never clearly separating?
would it be possible to directly look at the stars during eclipses, given their lack of britghness?
I have a question about brightness when it comes to moons. If your moon is too far away, can you not see on the planet at night? On Earth, we have some ambient light at night even if the moon isn’t visible. How does this work, and how do we apply it for our universes?
My Kepler solar system has 4 rocky planets, 0 gas planets, 0 dwarf planets, 8 moons, 4 for both earth-like planets, and 4 stars. These 4 stars are all sun-like stars and the 2 earth-like planets are both in the habitable zone. The planet I will focus on is Kepler Bb. Both planets cooled relatively fast compared to earth and are tectonically active. Humanoids have survived every mass extinction on Kepler B#(more massive than Kepler Bb but still with reasonable gravity for human survival). However, these same humanoids have only been on Kepler Bb for a few thousand years. So on Kepler Bb, there is stone age technology but on Kepler B# there is futuristic technology, big difference.
Anyway, with this close double binary(a binary of binaries where all 4 stars are relatively close), the apparent size and brightness will be much more complicated to calculate since having 4 stars, each of 1 solar mass is much different from having 1 star of 4 solar masses.
You would have to take into consideration the distance between the binaries at their maximum and minimum separation and if you want to go further, the individual stars within the binaries.
For moons, it is easier to calculate the apparent size because all you need for that is the distance from the planet to the moon and the radius of the moon. Easy.
But the double binary, not so easy.
Planets, easy
Stars light years away, easy, unless they are also star systems.
So how would I go about calculating the apparent size and brightness of the double binary star system compared to our sun? It would most likely be similar given that it is a close star system and that the planet is in the habitable zone but is there a way to adapt those formulas for a single star to be used for a star system?
How about a detailed video about a planet's sky color? How atmosphere density, composition and star output affects the color. What are the requirements to have a green, yellow purple or red sky?
Planci Acanthaster The evolution of the eye looking at it would be an important factor. Our sky is a bit more violet than our eyes can perceive.
The premise is that the viewer is a human. -.-
@@JoakimfromAnka This isn't an obvious premisse, I think many worldbuilders are creating alien societies on the basis of these videos
Whether the power of negative number is important is not clearly explained. In fact, by saying, "greater than 1.3" and "less than 1.2" it sounds like they are irrelevant(but still not clearly so) as they aren't mentioned in that moment. Also, I'm still having a hard time visualizing these numbers which to me, are abstract and stiff.
What is world building
I did the diameter over the distance as in the video but I got -0.03 so I'm more than a tiny bit confused.
The diameter of my star, in the solar units, is 0.85 and the distance between it and the planet I was working this out for is 0.88 AU. Have I seriously naffed up some calculations somewhere?
Do I just disregard the negative or...?
I learn more on this than in physical science
How would binary planets work?
If 2 planets of similar size and mass were orbiting each other around a star.
If they were the same distance apart than earth from the moon then its small angle aproximation would be more than 3x that of the moon.
Could the planets be that close without tidal forces tearing them apart?
How would that afect the solar exposure of both planets would eclipses happen more often and affect the temperatures due to prolonged shadow casting into one another?
i got 625 wats/m^2. did i do something wrong?
1.6th like exactly at 4:40 on mon may 6
Is there a mythological system where the creator of humans gave them thumbs of just the right size to block the sun and moon?
Question: Are there times we don't see a planet on the time of the year because we're on the other side of the sun or something?
***** Thanks. I never see this topic being spoken and I myself am terrible at identifying them on the sky.
How about a star that is slightly reddish? How will that affect brightness and the perception of its light colour?
In the 1000 stars video he says no
How does he call "0s" ?
Anyone know how to calculate absolute and apparent magnitude? I want to convert these values to absolute and apparent magnitude of my stars and planets. If it's in the video, I might've not realized it and didn't understand it so could someone word it where it's simple and easy to understand?
Latching onto this comment because I also wanted to know - I went through both the link in Edgar's description and a bunch of other articles. It looks like the math gets ridiculous very quickly when trying to convert these values. So unfortunately, it doesn't look like there is an easy to understand way to run these calculations.
If someone who knows more than me reads this anytime in the future, feel free to comment, I'd love to figure this out as well!
3:00 I don't think that equation has ever returned the same result for me twice
On my conworld, their sun would appear about 0.7 times as big as the sun, but just as bright!
And one moon would look about 1.5 times bigger than earth's moon, and the other about 0.7 times as big, but a tiny bit bigger than their Star.
And given the video about sky and plant color, their sky will look like a brighter and less opaque version of earth's sky, with pastel orange clouds and turquoise vegetation.
That would look so f-ing cool.
What's the unit for Apparent Brightness of an observed planet?
Watts per square meter
do you know any programs that will draw and simulate you color system for you?
Well he made a atmospheremcolour video and the description i think has a spreadsheet on atmosphere colours
Plants might be the colour of their star or a colour complementary to their star
Btw, Betelgeuse is red, not blue =8)
+Francois Lacombe all stars are white to the human eye.
+Millitron Whenever I look at Betelgeuse it seems to have a reddish tint to it.
+Dirtcubed To be fair it's more of a deeper orange than outright red. However, atmospheric scattering can push its color to a more redder one.
+Francois Lacombe Also it is pronounced "bettel guise" not "beetle juice"
+“Kennetron” John Kenwright It's pronounced "battle guys"
Small angle approximation is trash, use 2* arctangent of semimajor axis/radius
Is the narrator saying "naught" instead of zero or is that a term that some sort of astronomy term?
It's a U.K. Thing
+Josh Adams Partly true. The narrator is from Ireland.
Venus albedo is, from what I found, 77%, not 90%. I also found your calculations to be very hard to follow, given that so far we worked only with the orbit radius in AU, not in km, let alone m, and we haven't established the size of our planets yet, either. I'd say this video just came too soon and used hard to follow units, or rather, units that need to be converted and worked out. It would be better to have reference units to the star/to Venus brightness, as that would be easier to imagine and categorize.
5:41 I just spent that past half hour with that equation...
1:15 But... I just made the whole calculation to get the complete formula with watts and meters... :'l
What's the apparent brightness of Venus in watts per meter squared? No really, I've tried googling it six different ways.
What unit is the ab in?
If you mean the one for star light, its in sol
@@smartart6841 lol
I have a question, though I think I won't get an answer: I was making a solar system, and when I run the small angle aproximation equation for my Sun and Habitable planet, i've got a value of 1,325 (relative to the sun). I'm not sure what this value means. I transformed it into degrees, and the value turned to 75,9. If I'm doing the comparison correctly, 75º is a lot compared to the 0.5º of our real sun, but I'm not sure of that.
when you evaluate d/D, turn it into a percent, treat that as your decimal, then convert to radians. something like this:
d/D
2.197/1.88 = 1.16861702
1.16861702 ---> .0116861702
.0116861702 rad = .669 deg
Yes. When I ran the real equation: L/((4*pi*d)^2) i got the right result. Turned out that I was missing a parethesis (my equation was L/(4*pi*d)^2; which Excel didn't understood well).
My Sun's aparent brightness is 1386,055 w/m2 (close to our sun, on purpouse).
Sorry :P I was talking about the aparent brightness. My sun's aparent diameter as seen from my habitable planet turns out to be 0,635935911 º
If my calculations are right, you don't have to go too crazy to have planets visible in a fair amount of detail from another planet. I set a venus-like superterran next to the sun, with another planet between it and my inhabited one, and it looks like it's going to be more than half the size and 10% the brightness of the Earth's moon.
You know, this vidio is eksteremle helping
You mean more than 10^{-7} and less than 10^{-9}.
I like your funny words magic man
Isn't it important to mention that our eyes work logarithmicly?
How so?
3:17 *farther
So my moon is the same size as earths but 0.00264 AU away from the planet. How big would it appear in the sky? I'm shit at math. (If somebody answers like this comment so I'll get notified)
Are you an angel?
Cause uranus is shining at 1.2*10^-9 watts/meter/meter
No a coincidence... a survivability need? "The sun, it is... no longer in my eyes... Thanks thumb!"
3:23 you mean 9 planets right ;)
Pluto was not a planet all this time...
But now we know for sure.
Koray Acar I was talking about the recently "discovered" ninth planet .
+Foxintoxx It's fake.
+Scarlet Encarnacion everything is fake . Nothing is true . Everything is permitted .
+Scarlet Encarnacion
Actually, all current evidence points to the fact that there is indeed a ninth planet.
how about you try to build a culture?
you could have put "cultures are complicated before walling me in with text o.o
What does Radians mean? I'm not fully understanding what that means
A measurement of angles equaling 57.3°.
@@redforest9269 Ok that's good and all, but i still don't fully understand, is there an example I could use that'd be better to understand?
@@saturnianwolf0365 Get a protractor, or a picture of one.
Find 57°.
If the protractor doesn't support decimals, then just move your finger between 57 and 58.
That is 57.3, a radian.
If the angle of something you are measuring is bigger than the protractor, then mentally scale it up.
@@redforest9269 So if I do some calculations and get an answer let's say.....0.125, for Radians, how do I apply that? Would that be some sort of percentage or degrees then?
@@saturnianwolf0365 I'm not an expert, but I'd imagine you would multiply 57.3 by 0.125 in that case. (Yes, you're supposed to get a smaller answer, don't panic)
Sooo, simply to build the world, I had had to listen to my Physics teacher more often :V
No' point no' no' no' no' no' no' no' no' Lol I NEVER heard someone say "zero" this way XD
ok
d(km)
---------- = radians
D(km)
That math in the climax of the video is really complicated
Interesting I accidentally made a planet that will require bio-luminescence.
4π = α (alpha)
well I've now realised that I made the moon which is very important to my society far too small..... back to the drawing board I guess
you don't need your thumb, your pinkie is enough
I had a binary system, but turns out they can't even see one of the stars lol
Quote from video:"So, how bright is my planet? Well, xkbxufajdaixybchokhcifxogsifdphxufsifxotxofsl ihhdifchpcitspyxo."
Sorry, I just can't focus on math in English lol.
uranus isn't that bright.
not not not not not...
4:00 lol sin(θ)=θ
who is the privilegiated dude who is the owner of this so referenced thumb that takes appearently the exact same angle in the sky as the sun and the moon when held from his arm distance from his face? And I'm assuming it's a dude because women were not considered in those ages.
Will someone tell them that every human body has different proportions?
Not point not not not not not not... dang dont be so negative :P
Yikes
meters? why not miles?
Less than a twentieth of the world population uses fucking miles and 99percent of the rest uses meters, that's why.