The Maths of the Solar System - Aristarchus and Jantar Mantar
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- Опубліковано 7 вер 2024
- Oxford Mathematician Dr Tom Crawford visits Jantar Mantar in Jaipur, India, and explains the ancient calculations of Aristarchus of Samos in his seminal work "On the Sizes and Distances (of the Sun and Moon)".
The video begins with a history of Jantar Mantar in Jaipur, and a tour of the many instruments used for astronomical measurements. We then look at the history of heliocentrism and the begin an exploration of the work of Aristarchus.
The first calculation relates to the relative distances of the Sun and the Moon from the Earth. Using simple trigonometry at a 'Half-Moon' we obtain the same estimates for the distances as Aristarchus.
The second calculation is more involved and looks at the relative sizes of the Sun and Moon in comparison to the Earth. This requires the use of the properties of similar triangles during a Lunar Eclipse. We also use the fact that during a total eclipse, the sun and moon take up the same proportion of the sky, which means the ratios of their sizes relative to the Earth, is the same as that of their distances from Earth.
Finally, we end with a discussion of the translation of the work of Aristarchus and how his calculations may in fact have been much more accurate than first thought...
Useful links for more information / resources:
Zij-i-Muhammed Shahi: raymondm.co.uk...
Geocentric model animation: • Ptolemy's Geocentric C...
Translation of Aristarchus' work: archive.org/de...
Measuring the relative size of the Earth and the Moon at home: websites.umich...
Produced by Dr Tom Crawford at the University of Oxford. Tom is Public Engagement Lead at the Oxford University Department of Continuing Education: www.conted.ox....
For more maths content check out Tom's website tomrocksmaths....
You can also follow Tom on Facebook, Twitter and Instagram @tomrocksmaths.
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Get your Tom Rocks Maths merchandise here: www.beautifule...
With thanks to
Chuck Creager Jr.
Scritch314
Learn about the Maths of Ancient Egypt here: ua-cam.com/video/lX_f5oB83YI/v-deo.html
This is my break of studying maths… i love this
ikr!, I was solving a few log questions when I wanted to take break. And guess who recently uploaded :D
@@beastnighttv for me i was doing calcules! I love our life
I went there back in 2011. What an experience as an astronomy fan.
👍Please more of this Physics related Maths!
I’m terrible at math, always have been. But for some reason I enjoy watching your content, extremely high level math has always impressed me seeing as how I’m so terrible at it (even really low level math) keep up the fantastic work!
Error - 8mins. 1/18 > 1/20
Doesn’t affect the understanding though
Loved your videos on Math vs. Animation. And now this wonderful piece. Just a small correction. Trigonometry was well in use at the time, 7:39. Infact, Hiparachus if I am not wrong published table of values for common angles. So I think by 2nd or 3rd Century it would have been a thing.
d/Rm = 2 ... close enough for UA-cam! :-) It doesn't really matter HOW close he was, though closer is obviously better. What's amazing is that with ancient mathematical techniques and a few observations, they were able to come up with any value. Definitely an improvement over the sun is out there somewhere
Everyone always forgets about D/Re.
I don’t understand how he got 87 degrees. That means he measured the angle between sun and moon with an error of over 5 moon widths. But if you just had two straight sticks with a hinge, put your eye at the hinge, and lined up the 2 sticks with the left edges of sun and moon, then I would expect an error of a small fraction of a moon width. Not 5 whole moon widths.
And if you align the sun stick by ensuring its shadow is on the hinge, then you can just look down the moon stick to align it with the moon. You don’t even need to put your eye at the hinge. You should easily be able to get an error of less than a tenth of a moon width.
It doesn’t seem possible that he actually thought it was 87 degrees.
It's probably hard to determine when the Moon is exactly at half phase.
True, recognizing the exact half moon date is probably harder than recognizing the full or new moon dates.
But if in a given month you know the date of the full and new moons, then the half moon is the date halfway in between. And even if you thought there were 3 consecutive days of full moon, you could choose the middle of those days as the true full moon date.
So maybe that can reduce the error. But maybe the error is still too large. I’m not sure.
@@Anonymous-ob1kt You need more accuracy than just the date. The terminator angle will change quite a lot over the coarse of a few hours.
@@billthomas7644 that’s a good point. It explains how they could have had that large of an error.
The one thing which still intrigues me is how exactly would you try to measure phi from the earth's surface back then? I suspose by measuring the relative location of the sun and moon in the ecliptic at the point half moon happens?
more of math on tour!
For some reason speed x1.25 seems more like real life speed
You should definitely try the Scholarship Calculus exam for New Zealand students -- It's intended for the brightest students each year and only 3% of each cohort is expected to pass. Basically it's our high school syllabus, taken to the extreme in terms of thinking :)
Well, he is already a professor at Oxford. But incase you would like to give it a test, try some math questions from JEE advanced calculus section. Pretty standard high school stuff taken to extreme.
maybe you can explain this to me, how can you measure this any applicable time of the day?
1st fact is first quarter, Moon rises at Noon
either the moon or the sun, the one of their centers must be on the horizon, because the moon will orbit away about 3 degrees in 6 hours. this should amount to a BIG error, if the modern value turns out to be 89.98 decimal degrees
Shouldn't the right angle at 14:44 (and again at 17:57) be at the top, where the tangent meets the circle? The large diagram seems to imply that the radii are perpendicular to the centre line, but they're not, if you want to connect them to the tangent.
Except, that’s not a tangent. Looks like one the way it’s drawn, but it shouldn’t be if we begin with wanting the right angle to be at the bottom. And it doesn’t need to be a tangent either for any of the calculations.
@@ragad3 At 11:05 (when making the diagram) he says "it touches both of them". Doesn't that indicate it is indeed a common tangent?
I don't think there's any need for the right angle to be at the bottom for the maths to work.
@@tomspeltincx9567 Ah, thank you, I see the root of the confusion now! He describes the figure he has drawn accurately at 11:05 as one in which the two lines are tangents to both spheres. The figure that’s needed to proceed, however, is a slightly different one where the two lines are not tangents, but pass through the “north poles” and “south poles” of the two spheres. That’s the only way that the next figure that he draws (and all the calculations that rely on that subsequent figure) can make sense, IMO.
@@ragad3 I'm still not convinced. First, in all other diagrams to do with lunar eclipses that I've seen, those lines are drawn as tangents and the right angles are at the top. Second, and more importantly, the cone is supposed to represent the shadow cone of the Earth, as is mentioned or hinted at a few times. That only makes sense if they are tangents, I think. As for the maths, you still end up with two similar triangles and rather than adding the additional small triangle from the next few diagrams at the top, you'd have to add it at the bottom.
Maybe my namesake reads this at some point and he can shed light on this.
@@tomspeltincx9567 I think we both agree his diagram and what he says about the lines being tangent are not consistent with the rest of the steps that follow. Maybe we are just disagreeing on how to fix the mistake. My suggestion is a valid way to fix it and so is yours. The simplest analogy I can think of is the following: the equation 5 = 7 is incorrect; one way to fix it is to write 5 = 5 (this is what you are suggesting, that is, change everything that comes after the top diagram to match it - the placement of the right angles would be different but a similar approach will work, as you point out) and another way to fix the equation is to write 7 = 7 (this is what I was suggesting, that is, just change the top figure to match everything that comes after it so that the two lines won’t be tangents but lines connecting the “poles”). Both methods are valid!
Welch Labs, just recently published a beautiful documentary about Kipler’s amazing work.
why was 6 suspicious of seven ? because 7 gave 5 for free to 1
Were you filming a remake of "U Can't Touch This" while you were in India? 🛑!🔨⏲. Tom Rocks Pants! :) edit: Really enjoyed the video. It's very interesting learning about the way ancient people worked out complicated things like this without any modern mechanical or mathematical tools.