Step into my parlour said the spider to the fly😊.....i reckon most alleys are safe to walk down but there are those that lurk in the shadows....your comment is viable and needs no justification 🤝@Momonga-s7o
We have videos explaining that. First they figured out the size of the Earth, by walking a distance from north to south and seeing how much the shadow of a stick changed. (or the angle of the Sun relative to the Earth). Then they determined the size of the Moon (relative to the Earth) by timing a Lunar eclipse.
Even though it was not 100% during the lunar eclipse, I found the diameter of the moon to be 1/4 of the earth's diameter. Once I found this value, it was very easy for me to measure the distance of the moon.
How to calculate the real radius if the object is not really facing you? For example, Triangulum Galaxy M33 has an apparent size of 70.8 × 41.7 arc min. How do I relate the 2 apparent sizes to the real diameter? Distance to us 840±105 kpc
If you know through how much of angle the object is turned from being perpendicular to the line of sight you have to divide the apparent diameter by the cos(angle)
does anyone know how to calculate the distance to the moon using the pencil method when the moon is about 21 degrees above the horizon? I'm having trouble figuring out where to plug in the degrees into the formula.
Thanks for the great lectures, One question please: I understand that moon rotates around the earth with the balance between gravity and centrifugal force. However, it is unclear why the moon can follow the earth in its motion around the sun. this is different from the case of having a string in my hand with some weight attached to its end because if I move my hand, the tension in the string will increase and bring the weight to a new location. but gravity is inversely proportional to the distance; consequently, if the earth moves a bit away from the moon, an unbalance should happen and the moon should get free from the earth's gravity. So, Why the moon stays with earth, and the same question, why does the earth stay with the sun when the sun travels around the galaxy?
It is similar to a satellite that we place in orbit. The rocket takes off and the force created by the exhaust being expelled by the rocket engines push the rocket to orbit. Once in orbit, the engines are turned off, and the rocket or satellite begins to fall towards the Earth like a brick. But since it has ebough forward velocity (and due to the curvature of the Earth), it will not strike the Earth. The falling of the satellite is compensated by its velocity so even though the satellite is continuously falling towards the Earth, it will not strike the Earth (unless the satellite slows down). The same with the Moon. The Moon is fallign towards the Earth, but because of it forward motion, it will never strike the Earth. The same with the Earth and the Sun. The Earth is falling towards the Sun, but because of its forward motion (almost 30 km / sec).......
Why not use the same method you used to find the height of a mountain? Have 2 people say 4k miles apart take angle measurments and use simple trigonometry.
@@MichelvanBiezen You make a statement with out explaining why it won't work. .Are you implying that trigonometry only works at certain distance ? If so what is the cut off . You also assume the distance is great before the meassurment is done
How would a person get closer to make the measurement? And how would one person take the measurement along the same line of sight while being closer than the other person?
@@MichelvanBiezen very simple Pick two people at two cities at same longitude some distance apart .Each takes a angle reading at same time then work out the triangles just like the video you posted about sine laws. I give you another easier triangle with known measurements , the north star is exactly above the north pole ,London sits at 50.5N and is 2659.95 miles from the north pole. So you have London ,north pole and star forming an orthogonal triangle . why can't we use your sine law method to determine the height of the star?
We can use a coin at arms length that just covers the diamter of the Moon (at arms length) to determine that the ratio of the distance to the coin to the diameter of the coin is 108/1 which must be the same for the Moon.
@@smachohalla If the coin is 1cm in diameter and it matches the Moon at 108cm distance to the eye, then the ratio is 1/108. For example, 2cm diameter and 216cm distance will also be that ratio.
the diameter of the coin must be very small . Couse the lenght of the human arm normally don't pass a meter . they say the ratio of the distance to the object is 108 times the diameter or size of the object . if the diameter of the coin is 16,25mm wich is the diameter of a 1 cent ( european coin ) then the distanceto it turns into 1,75meters . This is just a note that i made to play a little . I guess the method continues to be nice
It does not seem the presenter is claiming they literally held the coin with their hands. For precise calculations, it is more logical to have the coin held in some kind of support on a table and move it backward and forwards until it matched the Moon in size from a fixed standpoint.
This video's content is based on Aristarchus and his calculations even though the speaker says he is "way off"! 😂 hopefully no one takes this seriously. thanks for the humorous content! 🕊️
Michel van Biezen Haha amazing, I live like 25 minutes away from Kapellen, your name sounded familiar to me. I live in Lier, also near Antwerpen. Can you still speak dutch (flemish)?
Robin Roziers Natuurlijk, maar het is wel lang geleden. Veel jaren geleden heb ik van Berchem tot Lier en terug naar Berchem gelopen in 2 uur and 4 minuten. Als we hier zeggen: "That was a life time ago"
@@MichelvanBiezen the lunar eclipse is not caused by the shadow of the earth. I know that this is something astronomers want to believe so the world makes more sense to them but it isn't the case. In fact you can see the blue sky through the moon during an eclipse so the science community is completely in denial when it comes to our cosmology..
How can we be sure that Aristarchus was wrong and maybe the moon and the sun diameters are the same. So this means most likely that the sun isnt 96mil miles away but much closer..
@@lualalsa yes it is far...farther than the moon.....but the whole scale of the solar system is completey made up by NASA..the king of all scammers... I suppose you're going to tell me the hasseblad cameras were taking pictures on the moons surface at 500 degrees...try putting one in your oven with film....and let me know how that works out for you...
@@garlicdawg The scale matches with what we observe everyday. It is totally worthless to criticise the current accurate model without presenting a better one.
Lets start at the beginning. You cannot determine that the moon's diameter to be 3.5 times the earths diameter using the shadows by simple reason of the fact that you do not know the distance of the moon and its size. Distance makes big things look small. Same applies in a solar eclips. This can only be done if the moon was right next to the earth, giving relevance to the time it passes through the earths shadow. Secondly, there are still no values only comparisons. The moon covering the sun is nothing. How far is the sun and how big it is once again matters. The sun being 400 times further then the moon cannot be calculated due to a lack of values. These people are bluffing you. In order for the coin thing to work, you need to know the exact distance to the moon first to make that ratio standard. It cannot be a ratio if you dont know the distance. The sun might be super large and super far away, making it look the size of a truck tyre. The ratio only works and is standard for things of which we know actual size and distance, which are things on earth. I have yet to see the math on how we eventually figured out distance to the moon apart from these lowsy chance taking ancient astronomers. Guess what? There are no calculations. You cannot do maths or physics if you dont have the values. Search and see for yourself. They bluffing. You need to know the distance to the moon to get there. To determine distance, you need to compare it to the distance of other things you already know the distance of. We dont have this luxury with stellar bodies. Same thing using trigonometry. They assume the distance to the sun. Even phycists are tricked into believing this rubbish. The not critical about the obvious.
I think you have to look at the videos before this one , by Michel van Blazon. Also check other sources. I also do not know if your are asking questions or making statements about the achievements of the ancient Greeks. Of course Newton and Galileo improved on these but there is no denying the achievements of Ancient Greek astronomers.
Of course we can determine that Moon's diameter to be 3.5 times the Earth diameter, using the shadow, the fact that both the Sun and the Moon are of the same angular size of 0.5 degree, and knowledge of trigonometry. Distance of the Sun from the Earth can also be calculated, by some observations, and trigonometry. Coin "thing" works regardless of the knowledge of distance to the Moon, since we can observe things here on Earth, and when we cover some object, in the distance, with a coin, then measure the distance from the coin to that object and height of that object, we will find out that distance from coin to the object, divided by height of the object equals 108. This confirms us, that coin "thing" works, while your knowledge and capacity to learn sucks big time. If you was critical with yourself, as you are trying to be with scientists, you wouldn't even try to write anything, before you fill in all the gaps in your knowledge, and, boy are they huge...
I can tell you with 100% certainty, the result of these calculations are all completely wrong and all based upon a single false data point ASSUMPTION, the "R" value of Earth Radius. Earth Radius cannot be measured. The reason Eratosthenes was wrong is an assumption, the Earth is a Globe. The same result would occur with a Flat Earth Plane with diverging Sun rays, as would with a Globe with parallel Sun rays. I can show you how, using simple trigonometry, to measure the real distance to the Moon is 2,278 miles, not 240k miles. Knowing the distance from your location, to the Zenith spot location on Earth (lat/long coords, result in miles) underneath the Sun or Moon, and the inclination angle from your location up to the Sun or Moon, and a Right Triangle Calculator. At 12pm noon, the Sun is at Meridian at your location (same longitude) and measure the angle up to the Sun using a vertical yardstick and the shadow length relative to 36 inches tall stick.
Yeah... sure... Measure angular size of the Sun 6:00AM 9:00AM 12:00AM 03:00PM and 6:00PM, and try to explain why it is the same, although your "small and close" Sun is a vastly different distances... ROTFL
I don't think so. If we went to the moon we should know exactly how far it is to the moon without having to calculate it through math ratios that have been created on assumption.
SpottedSharks Have someone make an object and not tell you the size of the object. Put it in a black background at an unknown distance visible to the eye. Then have all the scientists in the world try to come up with the precise distance or size of the object from your their stand point. NEVER GONNA HAPPEN.
Your math has to be wrong. Nasa says Tycho Crater is 53 miles in diameter. We can see the entirety of Tycho with low powered binoculars. This is Not possible if the Moon is 238 thousand miles away. Either the size of Tycho is wrong, the distance to the Moon is wrong, the size of the Moon is wrong or all 3.
The moon's diameter is 2160 miles, so we should be able to fit 41 of Tycho's craters across the moon's diameter if NASA's figures are correct. I pulled up a random photo of a full moon and measured the diameter on my screen as 110 mm. The diameter of Tycho's crater on the same photo was 3 mm. That means I could fit 36.6 Tycho craters across the diameter of the full moon photo on my screen. Looks like NASA is right.
SpottedSharks what if they have the size of the crater wrong. What if it is one mile across how big does that make the moon and how far away would that make it. We are assuming they measured the crater size accurately. We are assuming they went there and measured it.
You don't have to go to the moon to measure its size accurately (thank you, trigonometry). The ancient Greeks figured out the approximate size of the moon ages ago. See video #3 in this series to see how they did it. Once you know the moon is about 2,100 miles in diameter you can estimate the approximate size of the lunar features such as Tycho's crater.
So, the Romans had to hire Greeks to more complex maths I'm thinking, since it seems their only fraction was 1/12th?? Did maths go backwards? How could the Romans build aqueducts with exact slopes over miles and not have better math symbols than the ancient Greeks?
@@MichelvanBiezen The Sun and Moon are in Inner Space . How would you get heat from the Sun from millions of mikes way in the icy cold depths of space like you lot believe
The moon is only 76,000 or so miles high…and so WAS the sun till God noticed it burning up his dinosaur creatures and moved it back to where he is trying to pull a fast one on everyone with the “108” and “coin figures” and NASA makes a pretty good tv show.
Nope. You can even test it with a water level made out of a transparent plastic flexible tube conveniently filled with water (maybe mixed with a bit of soap).
@@ZackWolfMusic I see your screaming and raise it to running observations with a topograhic level and seeing locations at a higher altitude than where I was at being below the level's horizontal line, something that shouldn't happen on a flat Earth. Oh, and I checked the other way around too and yes, it still was below the level's horizontal line. 157.88.229.30/curvatura for pictures of the event, if you are interested. Sorry, Earth is *not flat* : I tested it with precision instruments, something I highly doubt you ever did.
@@MaGaO Not screaming! Great you see locations at higher altitude with no curvature excellent that proves Flat Earth and no curvature can be seen. Also what do you use to determine curvature drop of the horzion??
rfn944 You forgot to add /s for sarcasm. 3K mi is one of those ultra stupid flat earth rants. Just showing his stupid FE is so that no one will believe them.
This brings to mind Mark Twain's quote, "The fascinating thing about science is the wholesale returns of conjecture one gets from such a trifling investment of fact". For fact-based evidence see you tube videos "math challenges the moon landings", pt.1 and pt.2. The moon's distance is less than 4000 miles(6436km). Don't take my word for it. Check the evidence. Then check the "data" current science presents to prove it's theories.
@@shahineali5858 On some statements on crystalcodex.com/2018/03/13/distance-to-the-moon/ "Since the apparent size does not alter greatly, we must assume it does not change distance significantly on each orbit." The apparent diametre of the Moon at perigee is about 1.12 times the one at apogee. This difference is non-negligible. " If it rises and sets in 9 hours on some days, but can be visible for over 14 hours on other days, then the speed is not uniform." A hypothetical Moon following a perfectly circular orbit around Earth at a non-zero angle to Earth's equator will be visible more or less time depending on the time of the year and the observer's latitude. Moon's orbital speed isn't uniform but it doesn't depend on your premise. "A shadow can never be smaller than the object blocking a single source of light." Actually, it can if the source of light is larger than the object blocking its light. At least, the umbra can (the penumbra won't). Also, on crystalcodex.com/2019/07/08/the-moons-distance-calculations-explained/ "But the moon cannot be observed simultaneously over 180˚" It can, if the observers are high enough above sea level: e.g., they would if they were at 1mi asl (actually .72mi is the result of my calculations). Even if they were at sea level, they would only be 1.99º from being at 180º from each other. And that's assuming the Moon is at its perigee of 224000mi. "For example, when it is visible near the Arctic Circle (75˚ N. Lat), it is not seen in the Antartic (75˚ S. Lat) and vice versa." Actually, this can only happen if the Moon is above +15º declination (actually +14º if atmospheric refraction isn't accounted for and the southern observer is at sea level) and it has nothing to do with the latitude of the northern observer. These are just a few statements that don't match reality. There may be more but I won't keep looking. Your calculations are wrong, sorry.
@@MaGaO You have not addressed the 8 hour visibility, where the moon rises and sets in the northern hemisphere while being visible for over 12 hours in the southern hemisphere. And this reverses during the same cycle. It occurs every month, so can be verified. If the moon were closer this could be explained by the fact that when the visibility is only 8 hours in the north the moon's sub- lunar point is in the southern hemisphere. This then reverses when the moon moves into the northern hemisphere. I think one reason for the confusion is visualizing the moon orbiting Earth as though the Earth were stationary. If it is seen as spiralling round the Earth path completing a cycle every 27.32 days we would see it clearly. I have the model made- just haven't uploaded the video yet.
I'm new to Astronomy, but it's so fascinating how ancient greeks used such simple yet innovative ideas to answer these problems. Great Stuff!
Yes indeed. I am always amazed each time I think about how clever they were determine these facts.
So fasinating, these methods are filled with unrealistic math assumptions.
This is why I love astronomy.
This is why you love fakery!
@@ZackWolfMusic Why is it fake?
Step into my parlour said the spider to the fly😊.....i reckon most alleys are safe to walk down but there are those that lurk in the shadows....your comment is viable and needs no justification 🤝@Momonga-s7o
Sir I know this is an old video but would love an explanation as to how we figured out the 1/3.5 ratio for earth and the moon.
We have videos explaining that. First they figured out the size of the Earth, by walking a distance from north to south and seeing how much the shadow of a stick changed. (or the angle of the Sun relative to the Earth). Then they determined the size of the Moon (relative to the Earth) by timing a Lunar eclipse.
Even though it was not 100% during the lunar eclipse, I found the diameter of the moon to be 1/4 of the earth's diameter. Once I found this value, it was very easy for me to measure the distance of the moon.
Nice experiment.
How to calculate the real radius if the object is not really facing you? For example, Triangulum Galaxy M33 has an apparent size of 70.8 × 41.7 arc min. How do I relate the 2 apparent sizes to the real diameter? Distance to us 840±105 kpc
If you know through how much of angle the object is turned from being perpendicular to the line of sight you have to divide the apparent diameter by the cos(angle)
does anyone know how to calculate the distance to the moon using the pencil method when the moon is about 21 degrees above the horizon? I'm having trouble figuring out where to plug in the degrees into the formula.
It is the method of similar triangles. The tangent of the angle is equal to the ration of the height of the pencil the the distance to the pencil.
Thanks for the great lectures, One question please: I understand that moon rotates around the earth with the balance between gravity and centrifugal force. However, it is unclear why the moon can follow the earth in its motion around the sun. this is different from the case of having a string in my hand with some weight attached to its end because if I move my hand, the tension in the string will increase and bring the weight to a new location. but gravity is inversely proportional to the distance; consequently, if the earth moves a bit away from the moon, an unbalance should happen and the moon should get free from the earth's gravity. So, Why the moon stays with earth, and the same question, why does the earth stay with the sun when the sun travels around the galaxy?
It is similar to a satellite that we place in orbit. The rocket takes off and the force created by the exhaust being expelled by the rocket engines push the rocket to orbit. Once in orbit, the engines are turned off, and the rocket or satellite begins to fall towards the Earth like a brick. But since it has ebough forward velocity (and due to the curvature of the Earth), it will not strike the Earth. The falling of the satellite is compensated by its velocity so even though the satellite is continuously falling towards the Earth, it will not strike the Earth (unless the satellite slows down). The same with the Moon. The Moon is fallign towards the Earth, but because of it forward motion, it will never strike the Earth. The same with the Earth and the Sun. The Earth is falling towards the Sun, but because of its forward motion (almost 30 km / sec).......
Why not use the same method you used to find the height of a mountain? Have 2 people say 4k miles apart take angle measurments and use simple trigonometry.
Astronomical distances are at a scale where that method would not work
@@MichelvanBiezen You make a statement with out explaining why it won't work. .Are you implying that trigonometry only works at certain distance ? If so what is the cut off . You also assume the distance is great before the meassurment is done
How would a person get closer to make the measurement? And how would one person take the measurement along the same line of sight while being closer than the other person?
@@MichelvanBiezen very simple Pick two people at two cities at same longitude some distance apart .Each takes a angle reading at same time then work out the triangles just like the video you posted about sine laws. I give you another easier triangle with known measurements , the north star is exactly above the north pole ,London sits at 50.5N and is 2659.95 miles from the north pole. So you have London ,north pole and star forming an orthogonal triangle . why can't we use your sine law method to determine the height of the star?
@@MichelvanBiezen no comment?
How do we figure out the 108 ratio?
what type of trigonometry?
also great video, thanks!
We can use a coin at arms length that just covers the diamter of the Moon (at arms length) to determine that the ratio of the distance to the coin to the diameter of the coin is 108/1 which must be the same for the Moon.
@@MichelvanBiezen Thank you!
@@Mysoi123 can you show us how the figure 108 was figured. Thats really the point
@@smachohalla do the same thing in this video but with arctan instead of tan th-cam.com/video/WXjjOKwyhKg/w-d-xo.html
@@smachohalla If the coin is 1cm in diameter and it matches the Moon at 108cm distance to the eye, then the ratio is 1/108. For example, 2cm diameter and 216cm distance will also be that ratio.
the diameter of the coin must be very small . Couse the lenght of the human arm normally don't pass a meter . they say the ratio of the distance to the object is 108 times the diameter or size of the object . if the diameter of the coin is 16,25mm wich is the diameter of a 1 cent ( european coin ) then the distanceto it turns into 1,75meters . This is just a note that i made to play a little . I guess the method continues to be nice
It does not seem the presenter is claiming they literally held the coin with their hands. For precise calculations, it is more logical to have the coin held in some kind of support on a table and move it backward and forwards until it matched the Moon in size from a fixed standpoint.
Can't wait to major in astronomy in a couple months
It is a very exciting field of study indeed! 🙂
Thank you!
You're welcome!
What months do we get counter clockwise field rotation with the moon?
Both the Earth and the Moon rotate counterclockwise. (all the time). The Moon never moves across the sky in retrograde motion.
4:06 Mayan or Aztec calendar can still happen 😐
The 108:1 is what to what ratio?
The distance to the Moon / diameter of the Moon
How far the coin was from Aristarchus’ eye divided by the size of the coin that matched the size of the moon.
This video's content is based on Aristarchus and his calculations even though the speaker says he is "way off"! 😂 hopefully no one takes this seriously. thanks for the humorous content! 🕊️
🙂
Hi Michel, where you from?
Robin Roziers
Robin,
Originally, I am from Belgium
Michel van Biezen May I ask where exactly in Belgium?
Robin Roziers
Sure. I grew up in the small town of Kapellen near Antwerpen
Michel van Biezen Haha amazing, I live like 25 minutes away from Kapellen, your name sounded familiar to me. I live in Lier, also near Antwerpen. Can you still speak dutch (flemish)?
Robin Roziers
Natuurlijk, maar het is wel lang geleden.
Veel jaren geleden heb ik van Berchem tot Lier en terug naar Berchem gelopen in 2 uur and 4 minuten.
Als we hier zeggen: "That was a life time ago"
What means 1/3,5 ??
They found that the diameter of Earth is 3.5 x diameter of the Moon
Thank you! And 8000 miles are the diameter of the moon?!
8000 miles is the diameter of the Earth
@@MichelvanBiezen Please: how they found that figure ? Or just they supposed it?
I like the statement the Thing up in the sky that keep everybody warrm ...
🙂
Beware heathen....I worship big yellow ball in sky...👁🌅👁
wait how did they get the diameter of the moon
By the amount of time the Moon spent in Earth's shadow during a lunar eclipse.
@@MichelvanBiezen Any doc or video link? 🙂
@@MichelvanBiezen the lunar eclipse is not caused by the shadow of the earth. I know that this is something astronomers want to believe so the world makes more sense to them but it isn't the case. In fact you can see the blue sky through the moon during an eclipse so the science community is completely in denial when it comes to our cosmology..
@@eirikmurito lol
@@eirikmurito idiot
Miles il parayathe kilometres il para
Good stuff.
Very good 🙏🙏🙏🙏. Thanks
So nice of you
Wooo....what a classic way..
How can we be sure that Aristarchus was wrong and maybe the moon and the sun diameters are the same. So this means most likely that the sun isnt 96mil miles away but much closer..
Our current technology and knowledge of science has enables us to know the size and distance to the Sun and all the planets and moons accurately.
@@MichelvanBiezen its all theory...and scammery...the sun in fact is not 93 mil miles...its much closer and much smaller
@@garlicdawg Moon's phases, crepuscular and anticrepuscular rays prove that the Sun is far away.
@@lualalsa yes it is far...farther than the moon.....but the whole scale of the solar system is completey made up by NASA..the king of all scammers... I suppose you're going to tell me the hasseblad cameras were taking pictures on the moons surface at 500 degrees...try putting one in your oven with film....and let me know how that works out for you...
@@garlicdawg The scale matches with what we observe everyday. It is totally worthless to criticise the current accurate model without presenting a better one.
Moon faces 😂😂😂 how to hide more secrets
😄😄
Lets start at the beginning. You cannot determine that the moon's diameter to be 3.5 times the earths diameter using the shadows by simple reason of the fact that you do not know the distance of the moon and its size. Distance makes big things look small. Same applies in a solar eclips. This can only be done if the moon was right next to the earth, giving relevance to the time it passes through the earths shadow. Secondly, there are still no values only comparisons. The moon covering the sun is nothing. How far is the sun and how big it is once again matters. The sun being 400 times further then the moon cannot be calculated due to a lack of values. These people are bluffing you.
In order for the coin thing to work, you need to know the exact distance to the moon first to make that ratio standard. It cannot be a ratio if you dont know the distance. The sun might be super large and super far away, making it look the size of a truck tyre. The ratio only works and is standard for things of which we know actual size and distance, which are things on earth. I have yet to see the math on how we eventually figured out distance to the moon apart from these lowsy chance taking ancient astronomers. Guess what? There are no calculations. You cannot do maths or physics if you dont have the values. Search and see for yourself. They bluffing. You need to know the distance to the moon to get there. To determine distance, you need to compare it to the distance of other things you already know the distance of. We dont have this luxury with stellar bodies. Same thing using trigonometry. They assume the distance to the sun. Even phycists are tricked into believing this rubbish. The not critical about the obvious.
We are all entitled to our opinion.
I think you have to look at the videos before this one , by Michel van Blazon. Also check other sources. I also do not know if your are asking questions or making statements about the achievements of the ancient Greeks. Of course Newton and Galileo improved on these but there is no denying the achievements of Ancient Greek astronomers.
Of course we can determine that Moon's diameter to be 3.5 times the Earth diameter, using the shadow, the fact that both the Sun and the Moon are of the same angular size of 0.5 degree, and knowledge of trigonometry.
Distance of the Sun from the Earth can also be calculated, by some observations, and trigonometry.
Coin "thing" works regardless of the knowledge of distance to the Moon, since we can observe things here on Earth, and when we cover some object, in the distance, with a coin, then measure the distance from the coin to that object and height of that object, we will find out that distance from coin to the object, divided by height of the object equals 108. This confirms us, that coin "thing" works, while your knowledge and capacity to learn sucks big time.
If you was critical with yourself, as you are trying to be with scientists, you wouldn't even try to write anything, before you fill in all the gaps in your knowledge, and, boy are they huge...
I can tell you with 100% certainty, the result of these calculations are all completely wrong and all based upon a single false data point ASSUMPTION, the "R" value of Earth Radius.
Earth Radius cannot be measured. The reason Eratosthenes was wrong is an assumption, the Earth is a Globe. The same result would occur with a Flat Earth Plane with diverging Sun rays, as would with a Globe with parallel Sun rays.
I can show you how, using simple trigonometry, to measure the real distance to the Moon is 2,278 miles, not 240k miles.
Knowing the distance from your location, to the Zenith spot location on Earth (lat/long coords, result in miles) underneath the Sun or Moon, and the inclination angle from your location up to the Sun or Moon, and a Right Triangle Calculator.
At 12pm noon, the Sun is at Meridian at your location (same longitude) and measure the angle up to the Sun using a vertical yardstick and the shadow length relative to 36 inches tall stick.
Yeah... sure...
Measure angular size of the Sun 6:00AM 9:00AM 12:00AM 03:00PM and 6:00PM, and try to explain why it is the same, although your "small and close" Sun is a vastly different distances...
ROTFL
Radius = circumference/2pi. So, earth's circumference is unknown? Geographers and cartographers just missed that one?
I like how you ran away because you can't defend your position
So the moon is only a big coin. :D
I don't think so. If we went to the moon we should know exactly how far it is to the moon without having to calculate it through math ratios that have been created on assumption.
Astronomers knew the distance to the moon long before the Apollo missions.
SpottedSharks
Clearly you have no clue about how math works.
Ok, please show us your calculation for the distance to the moon.
SpottedSharks
You can't!!! That's the whole point. No one can. There is not enough information or points of reference to make such calculations.
SpottedSharks
Have someone make an object and not tell you the size of the object. Put it in a black background at an unknown distance visible to the eye. Then have all the scientists in the world try to come up with the precise distance or size of the object from your their stand point. NEVER GONNA HAPPEN.
Your math has to be wrong. Nasa says Tycho Crater is 53 miles in diameter. We can see the entirety of Tycho with low powered binoculars. This is Not possible if the Moon is 238 thousand miles away. Either the size of Tycho is wrong, the distance to the Moon is wrong, the size of the Moon is wrong or all 3.
How is your statement related to the video? (also note that the power of the binocular is not relevant to the resolution of the binocular)
The moon's diameter is 2160 miles, so we should be able to fit 41 of Tycho's craters across the moon's diameter if NASA's figures are correct. I pulled up a random photo of a full moon and measured the diameter on my screen as 110 mm. The diameter of Tycho's crater on the same photo was 3 mm. That means I could fit 36.6 Tycho craters across the diameter of the full moon photo on my screen. Looks like NASA is right.
Agreed. To work out one or the other you need to know something definate. All mathematical speculation.
SpottedSharks what if they have the size of the crater wrong. What if it is one mile across how big does that make the moon and how far away would that make it. We are assuming they measured the crater size accurately. We are assuming they went there and measured it.
You don't have to go to the moon to measure its size accurately (thank you, trigonometry). The ancient Greeks figured out the approximate size of the moon ages ago. See video #3 in this series to see how they did it. Once you know the moon is about 2,100 miles in diameter you can estimate the approximate size of the lunar features such as Tycho's crater.
So, the Romans had to hire Greeks to more complex maths I'm thinking, since it seems their only fraction was 1/12th?? Did maths go backwards? How could the Romans build aqueducts with exact slopes over miles and not have better math symbols than the ancient Greeks?
Like many other ancient engineering developments, Romans found out that slope through experience.
@@MaGaO : I dont think so, when tunnels meet midway through from opposite directions.
@@rogerscottcathey
So? The slope is the same.
@@MaGaO : So they used Roman numerals, right angles, squares and simple plumb bobs to build everything. Ok.
@@rogerscottcathey
Yes, they did. And medieval cathedral builders did the same: arabic numerals didn't reach Europe until the 10th century.
Ancient Indians have calculated more precisely... They were more accurate.
When did they do it?
@@MaGaO ... 4th or 5th Century... in texts of " Surya Sidhanta " By Varahamihira Sage
@@viplavrakhewar9538
Interesting. Thanks for the information.
@@MaGaO ... Though the values are approximate to the Modern values,.. but they surely found out.
Talking CRAP
Why? What don't you agree with?
CRAP adequately describes the stuff between your ears.
@@MichelvanBiezen The Sun and Moon are in Inner Space . How would you get heat from the Sun from millions of mikes way in the icy cold depths of space like you lot believe
The moon is only 76,000 or so miles high…and so WAS the sun till God noticed it burning up his dinosaur creatures and moved it back to where he is trying to pull a fast one on everyone with the “108” and “coin figures” and NASA makes a pretty good tv show.
If the Sun was only 76,000 miles away, there wouldn't be any life (or oceans) left on Earth. 🙂
flat earth?
Yup!
Nope. You can even test it with a water level made out of a transparent plastic flexible tube conveniently filled with water (maybe mixed with a bit of soap).
@@MaGaO EARTH IS FLAT DEAL WITH IT.
@@ZackWolfMusic
I see your screaming and raise it to running observations with a topograhic level and seeing locations at a higher altitude than where I was at being below the level's horizontal line, something that shouldn't happen on a flat Earth. Oh, and I checked the other way around too and yes, it still was below the level's horizontal line.
157.88.229.30/curvatura for pictures of the event, if you are interested.
Sorry, Earth is *not flat* : I tested it with precision instruments, something I highly doubt you ever did.
@@MaGaO Not screaming! Great you see locations at higher altitude with no curvature excellent that proves Flat Earth and no curvature can be seen. Also what do you use to determine curvature drop of the horzion??
Your off by 244,000 miles. The moon is 3000 miles from earth.
Can you tell me how did you measure that? lol
it must be much closer to be able to see the actual craters w/naked eye....NASA is scammers and every word they utter is a lie....
rfn944
You forgot to add /s for sarcasm.
3K mi is one of those ultra stupid flat earth rants. Just showing his stupid FE is so that no one will believe them.
This brings to mind Mark Twain's quote, "The fascinating thing about science is the wholesale returns of conjecture one gets from such a trifling investment of fact". For fact-based evidence see you tube videos "math challenges the moon landings", pt.1 and pt.2. The moon's distance is less than 4000 miles(6436km). Don't take my word for it. Check the evidence. Then check the "data" current science presents to prove it's theories.
Nope. Moon is about 240000 miles away. Verified the calculations myself. Sorry, but your TH-cam University degree is worthless.
How do you find the Moon is so close to the Earth? I'm interested in your proposed method and its maths.
@@MaGaO See the vidoes. Or see the math at www.crystalcodex.com/ science/ moon.
@@shahineali5858
On some statements on crystalcodex.com/2018/03/13/distance-to-the-moon/
"Since the apparent size does not alter greatly, we must assume it does not change distance significantly on each orbit."
The apparent diametre of the Moon at perigee is about 1.12 times the one at apogee. This difference is non-negligible.
" If it rises and sets in 9 hours on some days, but can be visible for over 14 hours on other days, then the speed is not uniform."
A hypothetical Moon following a perfectly circular orbit around Earth at a non-zero angle to Earth's equator will be visible more or less time depending on the time of the year and the observer's latitude. Moon's orbital speed isn't uniform but it doesn't depend on your premise.
"A shadow can never be smaller than the object blocking a single source of light."
Actually, it can if the source of light is larger than the object blocking its light. At least, the umbra can (the penumbra won't).
Also, on crystalcodex.com/2019/07/08/the-moons-distance-calculations-explained/
"But the moon cannot be observed simultaneously over 180˚"
It can, if the observers are high enough above sea level: e.g., they would if they were at 1mi asl (actually .72mi is the result of my calculations). Even if they were at sea level, they would only be 1.99º from being at 180º from each other. And that's assuming the Moon is at its perigee of 224000mi.
"For example, when it is visible near the Arctic Circle (75˚ N. Lat), it is not seen in the Antartic (75˚ S. Lat) and vice versa."
Actually, this can only happen if the Moon is above +15º declination (actually +14º if atmospheric refraction isn't accounted for and the southern observer is at sea level) and it has nothing to do with the latitude of the northern observer.
These are just a few statements that don't match reality. There may be more but I won't keep looking.
Your calculations are wrong, sorry.
@@MaGaO You have not addressed the 8 hour visibility, where the moon rises and sets in the northern hemisphere while being visible for over 12 hours in the southern hemisphere. And this reverses during the same cycle. It occurs every month, so can be verified. If the moon were closer this could be explained by the fact that when the visibility is only 8 hours in the north the moon's sub- lunar point is in the southern hemisphere. This then reverses when the moon moves into the northern hemisphere. I think one reason for the confusion is visualizing the moon orbiting Earth as though the Earth were stationary. If it is seen as spiralling round the Earth path completing a cycle every 27.32 days we would see it clearly. I have the model made- just haven't uploaded the video yet.
All calculations were wrong