p ~ pi/4 *a *((b/a)^2 +2(b/a) +5) is much closer pc ~ pi/4 *((b/a)^2 +2(b/a) +5) yields the approximated analog to 2*pi for the ellipse in question. this is because the quadratic can be transformed into (1 +b/a)^2 + (1 +1)^2, so when b=a we simply get 4 twice. when this is divided by 4 we get 2, ultimately yielding 2pi when the ellipse is a circle. and for the same reason that Kepler's second law works, this yields a good approximation of the relationship of the ellipse perimeter relative to its area all the way out to an eccentricity of 1.
@@elyeshachichait would be nice if you do a another video about it try out to find a better formula. But I like it. I plot it Geogebra with the best formula from Remujan(sry if wrong) and it's close to be a parallel line to it
Of course, all finite formulas for an elliptic integral can only be approximations. At best, an infinite series is the best one can do to represent elliptic integrals.
While your formula meets the limit conditions, there is no proof the progression is correct. (b/a)^2 also meets the limit conditions, as do higher powers. Thanks for sharing your work.
it's quite bad, actually. almost every other conventional approximation is better except, possibly, at b=0 this simple quadratic is much, much closer: p ~ pi/4 *a *((b/a)^2 +2(b/a) +5)
b/a is linear but the power is not linear so as the others have said (b/a)^2 would also fit, but we just don't know what the curve is for how the power b/a increases.
You did a very good job a proving the boundary conditions, but there is a family of functions that will satisfy these. Also, I don't believe you can just call p = cos(θ). I like the xy plane projection approach though.
A few years back I figured out an easy calculation to use in a pinch that is about 98-99% accurate. Assume "a" is the longer axis, "b" is the shorter axis. Axis being the radius distance from the center. For ellipses with an a/b ratio up to around 5/1, use Perimeter = [(3.7a/b)+2.4]b. If you get into more extreme ellipses, for an a/b ratio up to 10/1, use P = [(3.84a/b)+2]b. For a/b ratio up to 15/1, use P = [(3.9a/b)+1.7]b. For up to 20/1, P = [(3.93a/b)+1.55]b. You can also use a polynomial function (for whatever crazy reason). For a/b ratios up to around 5/1 try, P = [(0.072a/b)^2 + 3.26a/b + 2.93]b. For a/b ratios up to 20/1, try P = [(0.007a/b)^2 + 3.78a/b + 2.1]b.
First of all, I have to say that your view point on this was very innovative in my opinion. Sadly, this much isn't enough to claim that you found the formula to calculating the circumference of an ellipse. You see, there is an infinite amount of functions whose share those characteristics: •f(0) = 4a •f(a) = 2πa •f(x) is continuous in [0,a] if x1,x2 are in [0,a] and x1
Thank you very much for your constructive feedback and interest. In fact I found the error margin and I'm still working on it as it became a passion solve this problem. I'm learning math on my own as I don't have time for regular courses due to other activities. However if you would like to contribute and help in this math problem, you are very welcome.
ISSUE with this is around 16m30 where you set p = cos(theta). That is just a "wild" guess or assumption which is not proven. Exactly that is the whole issue wrt ellips circumference. Yes the length of b is cos(theta). But the circumference of the projected circle into an ellips might follow some more complex rules. Question is also if the projection is an ellips or something that just looks like an ellips?! Cutting a cone at an angle gives an ellips so I am assuming that this projection does give an ellips. But does the cos(theta) directly go with the projection of the circumference? My guess is that a (small) correction factor might be needed!
That's the "small" factor in still searching... I'm suspecting a factor related to the "deformation" of the circle... I came up with other formulas bit arranging this formulas in all directions, I get closer but then I diverge too much from a certain ratio and those derived formulas unlike the one I published I have no explanation to them, simply by readjusting this one... I had for examples C=4*[(pi/2)^(2(b/a))]*a - 3b or C=4*[(pi/2)^(2(b/a))]*a - pi*b... I can't explain those...
If you instead say, C/4= etc... then your angled perspective trick is only applied to a quarter section of the total circumference. When b goes to zero, it's still just a single line segment. This was my approach when I ventured down this rabbit hole. Use the symmetry to your advantage. The existence of a and b axes justify and validate the symmetry of quartering the ellipse circumference.
You should learn about the special functions called "elliptic integrals" invented in the 18th century precisely for calculating exactly the length of elliptic arcs.
I know it but still it is not exact... The idea here was to propose the exact formula like for a circle C=2*pi*r I am aware mine is an approximation and has an error margin up to 4% from the integral.
@@elyeshachicha Sorry, but the definition of elliptical integrals K(k) and E(k) is mathematically exact, like pi, sin(x) or the product of two real numbers. The computer computations with real numbers are approximate. So 2*pi*r is in practice rarely exact like the computation of E(k). By the way the exact formula is C = 4*a*E(e), where e =sqrt(1-b^2/a^2) (see en.wikipedia.org/wiki/Elliptic_integral).
@@danpfwith elliptic integral you just substitute a sum to an integral. Anyway, we are in the 21 century and maybe it is time to acknowledge that there is an exact formula, it just includes an integral… we are past the arithmetics
@@dominiquecolin4716 Well the Elliptic Integrals are nowadays approximated with floating point numbers with dedicated and efficient algorithms, more efficient than general algorithms for definite integration. For example in Octave (Matlab clone) it takes 0.13 microsecond on a laptop to calculate both E and K, as the following code does: >> n = 1000000; m = 0:1/n:1; tic, [K,E] = elliptke(m); t=toc t = 0.1318
Excellent start. We need to see circumference computed by elliptic integral, which is exact, compared to your proposition. If your proposition is not precise, perhaps there is an alpha so that replacing (b/a) with (b/a)^alpha will be precise. Please share result of your investigation.
Actual exact formula of the circumference of an ellipse: C=4a×E(1-b²/a²) With E(x)=int_0^(π/2)(sqrt(1-x×sin²(θ))dθ) (Using Wolfram alpha's definition of E(x))
When the ellipse is flat and the circumference is 4 when b = 0 the equivalent radius of a circle with the same circumference has a radius that is irrational, where as when b = a making the power = 1 the radius of the circle is rational, as the power increases from 0 to 1 the radius of the equivalent circle with the same circumference goes from irrational to rational, only being rational when the power is equal to 1 so for all rational powers of your equation the corresponding equivalent radius would be irration except when the power equals 1, so 0
@@BigJim777 I thought about this too. First let me emphasize that power must be >=0. I used power=0 only for limit purpose but in other publications I explain why it can't be =0. Anyway, yes I agree with your approach and even tought about review the infinite serie to calculate pi itself where to power applies to the infinite serie that gives pi (something with the zeta function or something)
What is worth answering, or proving is this: Is it ALWAYS possible to describe ANY graph with a finite set of procedures (that can involve elementary functions) ? Basically for every graph, to be able to be presented with an F(x) so that F(x) = y. And this F(x) to have a finite structure that might consist of elementary functions.... For instance if you have 1*x + 2*x + 3*x + 4*x + 5*x + ...to infinity... Instead of writing all this you can simply write : x*n*(n+1)/2 where n is the number of numbers you are adding up... When we say "sin(x)" this is an elementary function but it's an infinite sum (search Taylor series)... how else could you even describe sin(3.89) ? These functions are the building blocks that we have already calculated... and we build from there. "Elementary functions". Sadly, I don't think those are possible to be described with a finite structure. (because irrationality plays a key role). Pi is irrational, so you can't write it in one form. You can write it with elementary functions... like.. arccos(-1) = pi.. BUT arccos(x) is an infinite sum... Now the perimeter of the ellipse is also an infinite sum but the function is not "elementary". It's the Eliptical function, search it up. Yet I believe, that it CAN AND SHOULD be presented with a finite set of elementary functions, by changing the rules of math. The rules of math don't have to make sense. The procedure doesn't have to make sense. As long as it gives us the right result, it should make sense. Did you know that for the formula of the cubic equation they used "i" ? and i^2 = -1, which was counter-intuitive at the time, but we said "let this be correct" and it worked. And we later realized it did make sense. Without this rule or procedure it couldn't be possible. So you might want to create your own rules in mathematics to reduce a weird infinite sum to a finite procedure mechanism that could involve elementary functions. This mechanism doesn't exist so far, that's why there isn't a formula with elementary functions. So you have to figure out this new rule somehow.
I have a counter example: x^2±x*y+y^2=1 a=sqrt(2) b=sqrt(6)/3 4*sqrt(2)*(pi/2)^(sqrt(3)/3) ≈ 7.3418389387375 Actual perimeter is 7.1343450992015408312.. Approximated by exhaustion to 3145728 sides 7.134345099200
Your numbers are a not a counter example, they simply describe the error margin for a ratio b/a = (sqrt(6)/3)/sqrt(2) (=sqrt(3)/3)). As I explained in previous comment, I noticed that my formula as an error margin. Here's more detail about my error margin for a ratio b/a starting from 0 to 1. It is 0% when b/a=0 It increases to a maximum of approx. 4.3962% when b/a is around 0.28, then it decrease to 0% until b/a=1. I'll share more of those detailed analysis on my website (on the link in description). Basically, with the number you gave, you can try it with any number b/a= sqrt(3)/3, you will always have an error of 2.9% of circumference. Although I am still looking for the factor or constant to bring to correct for that error margin. Thank you very much for your interest.
at the start of the video you claim that you're calculating the 'exact circumference of an ellipse with a formula that [you] have established'. so when you start talking about an error margin you're just lying. it can't be exact and have an error, dude. which means that anyone providing numbers that indicate an error, like thomasolson7447 did, is absolutely a counter example. you should apologize for lying and trying to gaslight thomasolson7447 with your nonsense.
I too am thinking along those lines but instead of tiltina circle I cut diagonally acros a cylinder. I am still working on my idea (as and when I have a moment). I think your need to prove that the exponent cos(theta) is the correct function required. It is an interesting approach but the result needs to be proven. I wish you success.
I guess it could be useful in certain circumstances - but I don't know what they are. In any case he has to claim to have found an APPROXIMATE formula.
In order to check your formulae, all I need to do is take the limit of a->0 but b remains fixed. When a is very small clearly the periments should be 4b. A finite number. Your formulae blows up in the limit. You get a number growing exponentially and a number converging to 0 linearly. The exponential growth will win.
if a becomes less than b then b is the semi great axis and a becomes the semi small axis. If you check my analysis on my website (in the link in the description) a cannot be lower than b, because by definition a is the semi great axis, so I limited a to b for 'circular' ellipse. I also explain that b can tend to 0, but cannot be equal to 0, I made it equal to 0 just for the demo here. So my formula is basically C=4*((pi/2)^(semi small axis/semi great axis)) *semi great axis Also b can't be 0, because if you search for another ellipse with the same C with a different and b is 0 , you can find, but finding the same C starting with b=0, you can't, so b tend to 0 but never reaches 0
@@elyeshachicha for 'a' or 'b' to not equal 0 and still tend to 0, the answer should be close enough. Thats not really an issue. So here is another limit. If you addup a bunch of these circumferences, you should get the area of the ellipse '\pi ab'. Since all the ellipses u should add up are scaled version of each other, it makes sense that 'b/a' is a constant. Thus after adding up, you should retain '(\pi)^(b/a)' in the formulae; however the area of the ellipse doesn't have that term. You should do the calculation and show that integral of differential circumferences gives you the desired area '\pi ab'. Also If I'm not mistaken the elliptic function were originally used to compute the circumference for an ellipse. Im guessing you have checked this. But just in case, you don't know this, "www.chrisrackauckas.com/assets/Papers/ChrisRackauckas-The_Circumference_of_an_Ellipse.pdf"
@@shohamsen8986 thank you for the link, yep I saw that and the fact he concluded that we need to compute an infinite serie that pushed to really work on this problem. Where I don't agree regarding the area, for the same circumference when B changes in proportion with A, there's no need for the area to remain the same, as an example, a rectangle with size 4 and 1 has same perimeter of size 3 and 2, although the area is 4m2 for the first and 6m2 for the second. If you wanna see how the ratio is changed in mine, I'm using the Lambert function to find the factor n for A when B changes by a factor m for the same circumference. I really invite you to read my analyses on this pdf from page 8 wixlabs-pdf-dev.appspot.com/assets/pdfjs/web/viewer.html?file=%2Fpdfproxy%3Finstance%3DwNO47M79mzeUZa4aUz1kJSrTMAHJ9goNx3p_omKSuhI.eyJpbnN0YW5jZUlkIjoiYzJkNjIzZjctYTE5Ni00ZDRjLTk1ZjMtM2RjM2VkMzkxZGIzIiwiYXBwRGVmSWQiOiIxM2VlMTBhMy1lY2I5LTdlZmYtNDI5OC1kMmY5ZjM0YWNmMGQiLCJtZXRhU2l0ZUlkIjoiYTI3ODA4OTYtMjQ3YS00YTI1LTljZDYtYWI3MjYwMjUxMjU5Iiwic2lnbkRhdGUiOiIyMDIzLTAyLTI3VDEzOjU3OjEyLjgzNFoiLCJkZW1vTW9kZSI6ZmFsc2UsImFpZCI6Ijc4NTJhYmRhLWMwOGItNDc2OS04MzBmLTZiYWE1NjcwMmQ1YSIsImJpVG9rZW4iOiI2MGFlMmI2MS04NWVjLTA3NjktMDkyNS05NmIxOGQxYzBmZWEiLCJzaXRlT3duZXJJZCI6Ijc4ZTc4YzdhLTQ0OWQtNGI1YS1iMjNmLWQzY2FlMDI4YjhmYyJ9%26compId%3Dcomp-lehoo02w%26url%3Dhttps%3A%2F%2Fdocs.wixstatic.com%2Fugd%2F78e78c_253a422562414d8f9c4d7f39253641a4.pdf#page=1&links=true&originalFileName=How_to_calculate_an_ellipse_Elyes_Hachicha%5B23FEB2023%5D&locale=fr&allowDownload=false&allowPrinting=false
@@elyeshachicha I dont think u understood my suggested limiting process. Think of a circle, you can compute the area from the circumference of a circle. \int 2\pi r dr = \pi r^2. You are adding a bunch of circumferences (of similar circles) and getting the area of the circle. I don't think if you apply that to your formulae, you will get the correct area of an ellipse '\pi a b'. Note that since all the ellipses are similar, you have b/a a constant. Anyway, I computed the formulae of an ellipse numerically and using your formulae, they don't match. In case you are interested, here is the Mathematica code that achieves this. r[a_, b_, u_] := {a Cos[u], b Sin[u]} perimeter[a_, b_] := NIntegrate[ Sqrt[FullSimplify[D[r[a, b, u], u] . D[r[a, b, u], u]]], {u, 0, 2 Pi}] Manipulate[ ParametricPlot[r[a, b, t], {t, 0, 2 Pi}, PlotLabel -> Grid[{{"perimeter:", perimeter[a, b]}, {Style["ArcLength", Bold], ArcLength[r[a, b, u], {u, 0, 2 Pi}]}, {"Closed Form:", 4*(Pi/2)^(Min[a, b]/Max[a, b])*Max[a, b]}}], PlotRange -> Table[{-3, 3}, 2]], {a, 1, 3}, {b, 1, 3}]
@@shohamsen8986 I'm not entirely sure you are correct on that one. It would have been my firse guess as well, but by that logic you should be able to get the circumference of an elipse by differentiating the area of it If we deside that we only take a look at elipses with ratio c = b/a = constant Then we get A(a) = πab = cπ*a² (c = b/a ==> b = c*a) ?==> C(a) = A'(a) = 2cπ*a This is cleaely not the circumference of an elipse, so there is some kind of a wrong assumption behind the scenes. I would assume it has something to do with the fact that unlike in a circle, a isn't perpendicular to all points on the fircumferense, and therefore the calculastion of an infinitesimal contribution to the overall area from an elipse of a of a major axis 0
I see what you mean but I really insist of the fact that the formula is C(semi-major axis, semi-minor axis) So if b becomes greater than a, then b is the semi-major axis and a the semi-minor axis
c=4a EllipticE[(a^2-b^2)/a^2]; of course this is just a new name for the integrations that have to be done - but e.g. Mathematica can calculate this to almost any desired precision.
Wait until you see my proof on why the circumferences of a circle and an ellipse are actually 2x2 matrices, where circles have "2pi * r" as their circumference only because the identity matrix is also equal to 1, so it can still be seen as a scalar.
The description of the video should be updated. On Dec 26, 2024, the description claims that you calculate the exect circumference. Yet in comments below that you posted months ago, you admit that when you posted you believed it was exact in general, but you have since realized it isn't.
This method is effectively a guess and check of the function form (pi/2)^p with the assumption of p = cos(θ) but neither of these are well motivated by anything other than their outputs match how we know a true circumference formula should behave on the boundaries 0 and 90 degrees. The true circumference formula, if it exists, likely has a much different profile between the boundary points. Unfortunately this is the case because the formula breaks down quite quickly. I'm trying to salvage something from the projection idea in my head because it seems interesting but I fear it might not be as meaningful as it seems. Projecting that way gives b = a.cos(θ) which is basically the same as saying b = n.a for some real number n without the idea of projecting in a third dimension since it's obvious that b is some fraction of a in 2D. Projecting just gives a handy parametrisation of n but the effect on the equation of an ellipse is the same. When attempting a line integral, you will end up with the cos(θ) attached to one of the terms so the expression can't be simplified.
Sure, I can, if you are interested, on my website I shared more details on the limits and analysis of this equation. The graph you are asking for, I can make it as I see many people are having interest to my work. Thank you
If you have checked your formula for accuracy as you state you did with the suggested graph, you would know if your formula is an approximation or the exact solution. You did not mention here if your solution is an approximation or the exact solution. There are numerous approximations to this problem. How does your approximation with your solution compare to the best approximations already found. I think it is disingenuous not to state your solution is an approximation, especially as you have stated that you have tested for accuracy.
No. Whilst your formula works in the limit as b -> 0 and b -> a it is only approximately true between these. There are many such formulae. Nice try but.
Thank you for your comment. I would like to invite you to see how I manage to reduce the error margin in between and the rest of my work in this video ruclips.net/video/Z7sDHY1PnWs/видео.htmlsi=aTf4NYa8dvyrMizN
Your video description is false. You are NOT "calculating the exact circumference of an ellipse". Your formula is merely an approximation. You should have stated that clearly at the outset. Viewers don't take kindly to misrepresentation.
When I published it I really thought it was the exact one, on another side, it allowed me to have some useful comments that pushed me to review and understanding where I need to bring a correction. There was no intention to mislead people. Meantime I would like to invite to see another video where I use that method for different applications, just to see that this approach or method (by tilting a circle) is not wrong...
Dude, you didn't demonstrate anything, you just found an aproximation. Who told you that (π/2)^cos(θ) is the ONLY formula that fits the extreme values of π/2 and 1?
You need to show two things you have not showed. 1. Is the projection of a tilted circle actually an ellipse (this is not obvious); 2. you need to show the numerical answers agree with the calculated answers from known calculus-derived formulae (which involve elliptic integrals). Without either of those, this is not a convincing proof.
It's not a proof, but rather an 'Idea' as was stated a couple of times, albeit without enough emphasis. Anyways... You said in your comment that it isn't obvious... Could you give me a counter argument? It feels like it should, but why won't it?
@@alexbennie If you want to show it's an ellipse, you need to show that it is of the form (x/a)² + (y/b)² = 1 If you want to prove it algebraicly. Otherwise, you might want to use the geometric diffinitiin, which is to show that the distance from the focal points located at f1,2 = ( +-c , 0 ) Where c = sqrt(a²-b²) (The focal points move as you rotate the circle plane with respect to x axis) Lastly, you might be able to show that any point on the circumference is of the form (x,y) = (acos(t),bsin(t)) , where 0
The third approach is by far the easiest imo, if you know some linear algebra and calculus Represent the circle as all the points of the form (x,y,z) = (acos(t),asin(t),0) for θ=0 Then, if you know that the matrix representation of a rotation of θ degrees around the x axis is R = 1 0 0 0 cosθ -sinθ 0 sinθ cosθ Then you get R(acos(t),asin(t),0) = (acos(t), acos(θ)sin(t) , asin(θ)sin(t)) = (acos(t), bsin(t) , asin(θ)sin(t)) We are only interested in the x,y components, and indeed (x,y) = (acos(t),bsin(t)) As needed
@313sib for your first point, I made a video to prove that rotating a circle becomes an ellipse ruclips.net/video/JhxFvpk3RpE/видео.html Regarding the second point, I will publish soon another video. Thank you for your interest in my work. I will keep you posted after posting the other video.
Regarding the numeral analysis with integral, here's a video I recently shared where I give more details ruclips.net/video/Z7sDHY1PnWs/видео.htmlsi=gV5YII9lLhiIFFW6
Nice try! I think it is well known the "circumference" of ellipses lead to elliptical integrals if one wishes to calculate via the arclength formula, which has no closed form in that case. Look at Micheal Penn's video *_a formula for the circumference of an ellipse_* He derives an infinite sum formula for the task.
Your 'method' gives c*=10.0265 for a=2 and b=1, whereas the correct formula results in c=4a EllipticE[(a^2-b^2)/a^2) = 9.68845: Your Result is - in this example - around 3.5 % too large.
That's right, for a ration of b/a=1/2, my result is 3.489% higher I noticed my error and I'm still figuring out that factor I'm missing. Here's more detail about my error margin for a ratio b/a starting from 0 to 1. It is 0% when b/a=0 It increases to a maximum of approx. 4.3962% when b/a is around 0.28, then it decrease to 0% until b/a=1. I'll share more of those detailed analysis on my website (on the link in description). Thank you for your interest.
@@rainerzufall42 at the end of my demonstration on my website I added another application involving theta only for a given circle c with radius r. I called it the 'observable circumference'. The ‘observable circumference’ of a circle is the circumference of a circle observed from a different angle or point of view. If we draw a circle c with a radius r on the y plane, and rotate the y plane by an angle θ around the x axis, the observable circumference C of the circle c observed from the z axis passing through its center is: C = 2*((pi/2)^|cos θ|)*r
The reason behind why there is no exact formula for an ellipse refers to multitude of factors that play on the focal points there too much formulas for stretched circles I can give you one hundred if you want.
أخي إلياس، لك مني تحية مبجلة على سيرك على طريق العلماء و محاولة إيجاد قانون جديد لمحيط الشكل البيضوي و البرهنة عليه. ولكن هل يمكنك أن تحسب لنا محيطه بقانونك الجديد كمثال : لما تكون زاوية الرؤية 30 درجة و نصف القطر 10 سم مع الشرح و التوضيح. لك مني كل الإحترام و التقدير.
I will reply in English as it is more comfortable for me to type on this keyboard in English. From a point of view of 30°, the semi-minor axis (b) will be equal to the radius (a) times the cosines of the angle b = a x cos α = 10 x cos 30° ≈ 8.66 We can now use the formula: C = 4 x ((pi/2)^(b/a)) x a The value of 'b' is an approximation in this example, so the circumference becomes also an approximation C ≈ 4 x ((pi/2)^(8.66/10)) x 10 C ≈ 59.14 Alternatively, as 'b/a = cos α', we can use the following formulas C = 4 x ((pi/2)^(cos α)) x a C = 4 x ((pi/2)^(cos 30°)) x 10 C ≈ 59.14 However, I discovered after sometime that ym formula bring us to an approximation. I keep believing that my way of solving is the right one, only missing one parameter... I am thinking of about a factor that I missed in the deformation of the circle.
Good try! But there are many many more fonction P=F(theta), that satisfy conditions that's why they call them elliptic fonctions. This is just mine which is of course faulse: C = 2(πb+2✓a^2-b^2) Greetings Youcef AMMAR KHODJA I invite you to watch Pr PARKER video on the subject: ruclips.net/video/5nW3nJhBHL0/видео.htmlsi=Hiw2BLwFR1vzT5ae
π isn't really a number. It's a construct. It can be constructed in many various ways. One of the ways is to find the 'exact' perimeter of a circle. So it's not surprising that it's tricky to find the 'exact' perimeter of an ellipse.
Honestly I believe that the pi gets "distorted" in an ellipse, what I mean by exact value, is assuming an acceptable value of pi, like for the circumference of a circle, we write it as equal when using pi as a symbol and not a rounded up number.
Since we define π as the ratio of any circle between it's circumference and it's diameter, then we can also define Π(e) "pi of e" where e is eccentricity, to be the ratio between an ellipse circumference and the length of it's major axis. Since we have the (extremely complicated) formula π/2 C = 4a S sqrt(1-e²sin²θ)dθ 0 Then π/2 Π(e) = 2 S sqrt(1-e²sin²θ)dθ 0 π/2 Π(0) = 2 S sqrt(1)dθ = 0 π/2 π/2 = 2 S 1 dθ = 2*θ| = 2(π/2-0) = π 0 0 π/2 Π(1) = 2 S sqrt(1-sin²θ)dθ = 0 π/2 = 2 S sqrt(cos²θ) dθ = 0 π/2 π/2 = 2 S cosθ dθ = 2*sinθ| = 2(1-0) = 2 0 0 Since e = 0 represents a circle and when e approaches 1 it the shape approaces a line, both agree with your resault of C(circle) = 2πa , C(line) = 4a So in some sence, yeah, π dies get "distorcted", since the ratio changes as a function of eccentricity. What you are trying to do is essentially solve the above integral, since e = sqrt(1-(b/a)²) So C = 2a*Π(sqrt(1-b²/a²)) is exactly the formula you are looking for
p ~ pi/4 *a *((b/a)^2 +2(b/a) +5) is much closer
pc ~ pi/4 *((b/a)^2 +2(b/a) +5) yields the approximated analog to 2*pi for the ellipse in question. this is because the quadratic can be transformed into (1 +b/a)^2 + (1 +1)^2, so when b=a we simply get 4 twice. when this is divided by 4 we get 2, ultimately yielding 2pi when the ellipse is a circle. and for the same reason that Kepler's second law works, this yields a good approximation of the relationship of the ellipse perimeter relative to its area all the way out to an eccentricity of 1.
great idea, byt this is not an exact formula, it's an approximation..
I noticed it after publishing... I still believe I'm missing a parameter with this method... Thank you for your comment :)
@@elyeshachichait would be nice if you do a another video about it try out to find a better formula. But I like it. I plot it Geogebra with the best formula from Remujan(sry if wrong) and it's close to be a parallel line to it
😅😮
Of course, all finite formulas for an elliptic integral can only be approximations. At best, an infinite series is the best one can do to represent elliptic integrals.
Pi is always approximated tho
While your formula meets the limit conditions, there is no proof the progression is correct. (b/a)^2 also meets the limit conditions, as do higher powers. Thanks for sharing your work.
it's quite bad, actually. almost every other conventional approximation is better except, possibly, at b=0
this simple quadratic is much, much closer:
p ~ pi/4 *a *((b/a)^2 +2(b/a) +5)
even p ~ 2pi v((a^2 +b^2)/2), which is famously bad, is better than the solution presented in the video up to an eccentricity of about 0.922
b/a is linear but the power is not linear so as the others have said (b/a)^2 would also fit, but we just don't know what the curve is for how the power b/a increases.
You did a very good job a proving the boundary conditions, but there is a family of functions that will satisfy these. Also, I don't believe you can just call p = cos(θ). I like the xy plane projection approach though.
Thank you for your feedback. I realised my mistake after publishing... Still trying to figure out the enigma
A few years back I figured out an easy calculation to use in a pinch that is about 98-99% accurate. Assume "a" is the longer axis, "b" is the shorter axis. Axis being the radius distance from the center. For ellipses with an a/b ratio up to around 5/1, use Perimeter = [(3.7a/b)+2.4]b.
If you get into more extreme ellipses, for an a/b ratio up to 10/1, use P = [(3.84a/b)+2]b. For a/b ratio up to 15/1, use P = [(3.9a/b)+1.7]b. For up to 20/1, P = [(3.93a/b)+1.55]b.
You can also use a polynomial function (for whatever crazy reason). For a/b ratios up to around 5/1 try, P = [(0.072a/b)^2 + 3.26a/b + 2.93]b. For a/b ratios up to 20/1, try P = [(0.007a/b)^2 + 3.78a/b + 2.1]b.
First of all, I have to say that your view point on this was very innovative in my opinion. Sadly, this much isn't enough to claim that you found the formula to calculating the circumference of an ellipse.
You see, there is an infinite amount of functions whose share those characteristics:
•f(0) = 4a
•f(a) = 2πa
•f(x) is continuous in [0,a]
if x1,x2 are in [0,a] and x1
Thank you very much for your constructive feedback and interest. In fact I found the error margin and I'm still working on it as it became a passion solve this problem. I'm learning math on my own as I don't have time for regular courses due to other activities. However if you would like to contribute and help in this math problem, you are very welcome.
ISSUE with this is around 16m30 where you set p = cos(theta). That is just a "wild" guess or assumption which is not proven. Exactly that is the whole issue wrt ellips circumference. Yes the length of b is cos(theta). But the circumference of the projected circle into an ellips might follow some more complex rules. Question is also if the projection is an ellips or something that just looks like an ellips?! Cutting a cone at an angle gives an ellips so I am assuming that this projection does give an ellips. But does the cos(theta) directly go with the projection of the circumference? My guess is that a (small) correction factor might be needed!
That's the "small" factor in still searching... I'm suspecting a factor related to the "deformation" of the circle... I came up with other formulas bit arranging this formulas in all directions, I get closer but then I diverge too much from a certain ratio and those derived formulas unlike the one I published I have no explanation to them, simply by readjusting this one... I had for examples C=4*[(pi/2)^(2(b/a))]*a - 3b or C=4*[(pi/2)^(2(b/a))]*a - pi*b... I can't explain those...
There is no "ellips"!
@@azztekethat’s correct, it is a ‘sideways’ projection of a circle.
If you instead say, C/4= etc... then your angled perspective trick is only applied to a quarter section of the total circumference. When b goes to zero, it's still just a single line segment. This was my approach when I ventured down this rabbit hole. Use the symmetry to your advantage. The existence of a and b axes justify and validate the symmetry of quartering the ellipse circumference.
You should learn about the special functions called "elliptic integrals" invented in the 18th century precisely for calculating exactly the length of elliptic arcs.
I know it but still it is not exact... The idea here was to propose the exact formula like for a circle C=2*pi*r
I am aware mine is an approximation and has an error margin up to 4% from the integral.
@@elyeshachicha Sorry, but the definition of elliptical integrals K(k) and E(k) is mathematically exact, like pi, sin(x) or the product of two real numbers. The computer computations with real numbers are approximate. So 2*pi*r is in practice rarely exact like the computation of E(k). By the way the exact formula is C = 4*a*E(e), where e =sqrt(1-b^2/a^2) (see en.wikipedia.org/wiki/Elliptic_integral).
@@danpfwith elliptic integral you just substitute a sum to an integral. Anyway, we are in the 21 century and maybe it is time to acknowledge that there is an exact formula, it just includes an integral… we are past the arithmetics
@@dominiquecolin4716 Well the Elliptic Integrals are nowadays approximated with floating point numbers with dedicated and efficient algorithms, more efficient than general algorithms for definite integration. For example in Octave (Matlab clone) it takes 0.13 microsecond on a laptop to calculate both E and K, as the following code does:
>> n = 1000000; m = 0:1/n:1; tic, [K,E] = elliptke(m); t=toc
t = 0.1318
@@elyeshachicha you know that mathematicians proved there is no exact formula, right?
Excellent start. We need to see circumference computed by elliptic integral, which is exact, compared to your proposition. If your proposition is not precise, perhaps there is an alpha so that replacing (b/a) with (b/a)^alpha will be precise. Please share result of your investigation.
Actual exact formula of the circumference of an ellipse:
C=4a×E(1-b²/a²)
With E(x)=int_0^(π/2)(sqrt(1-x×sin²(θ))dθ)
(Using Wolfram alpha's definition of E(x))
When the ellipse is flat and the circumference is 4 when b = 0 the equivalent radius of a circle with the same circumference has a radius that is irrational, where as when b = a making the power = 1 the radius of the circle is rational, as the power increases from 0 to 1 the radius of the equivalent circle with the same circumference goes from irrational to rational, only being rational when the power is equal to 1 so for all rational powers of your equation the corresponding equivalent radius would be irration except when the power equals 1, so 0
@@BigJim777 I thought about this too. First let me emphasize that power must be >=0. I used power=0 only for limit purpose but in other publications I explain why it can't be =0.
Anyway, yes I agree with your approach and even tought about review the infinite serie to calculate pi itself where to power applies to the infinite serie that gives pi (something with the zeta function or something)
What is worth answering, or proving is this:
Is it ALWAYS possible to describe ANY graph with a finite set of procedures (that can involve elementary functions) ? Basically for every graph, to be able to be presented with an F(x) so that F(x) = y.
And this F(x) to have a finite structure that might consist of elementary functions....
For instance if you have 1*x + 2*x + 3*x + 4*x + 5*x + ...to infinity... Instead of writing all this you can simply write :
x*n*(n+1)/2 where n is the number of numbers you are adding up...
When we say "sin(x)" this is an elementary function but it's an infinite sum (search Taylor series)... how else could you even describe sin(3.89) ? These functions are the building blocks that we have already calculated... and we build from there. "Elementary functions". Sadly, I don't think those are possible to be described with a finite structure. (because irrationality plays a key role). Pi is irrational, so you can't write it in one form. You can write it with elementary functions... like.. arccos(-1) = pi.. BUT arccos(x) is an infinite sum...
Now the perimeter of the ellipse is also an infinite sum but the function is not "elementary". It's the Eliptical function, search it up. Yet I believe, that it CAN AND SHOULD be presented with a finite set of elementary functions, by changing the rules of math.
The rules of math don't have to make sense. The procedure doesn't have to make sense. As long as it gives us the right result, it should make sense. Did you know that for the formula of the cubic equation they used "i" ? and i^2 = -1, which was counter-intuitive at the time, but we said "let this be correct" and it worked. And we later realized it did make sense. Without this rule or procedure it couldn't be possible. So you might want to create your own rules in mathematics to reduce a weird infinite sum to a finite procedure mechanism that could involve elementary functions. This mechanism doesn't exist so far, that's why there isn't a formula with elementary functions. So you have to figure out this new rule somehow.
@@Game_Masters I see your point... Thank you very much for this very constructive comment and suggestion... I'll try.. Thanks!
It’s proven that the answer is no. Proven a thousand of years ago.
@@vinny5004 How ? (I don't see how you can prove or disprove such a thing). As I said, the math doesn't have to make sense.
I have a counter example:
x^2±x*y+y^2=1
a=sqrt(2)
b=sqrt(6)/3
4*sqrt(2)*(pi/2)^(sqrt(3)/3) ≈ 7.3418389387375
Actual perimeter is 7.1343450992015408312..
Approximated by exhaustion to 3145728 sides
7.134345099200
Your numbers are a not a counter example, they simply describe the error margin for a ratio b/a = (sqrt(6)/3)/sqrt(2) (=sqrt(3)/3)).
As I explained in previous comment, I noticed that my formula as an error margin. Here's more detail about my error margin for a ratio b/a starting from 0 to 1.
It is 0% when b/a=0
It increases to a maximum of approx. 4.3962% when b/a is around 0.28, then it decrease to 0% until b/a=1. I'll share more of those detailed analysis on my website (on the link in description).
Basically, with the number you gave, you can try it with any number b/a= sqrt(3)/3, you will always have an error of 2.9% of circumference. Although I am still looking for the factor or constant to bring to correct for that error margin. Thank you very much for your interest.
at the start of the video you claim that you're calculating the 'exact circumference of an ellipse with a formula that [you] have established'. so when you start talking about an error margin you're just lying. it can't be exact and have an error, dude.
which means that anyone providing numbers that indicate an error, like thomasolson7447 did, is absolutely a counter example.
you should apologize for lying and trying to gaslight thomasolson7447 with your nonsense.
I too am thinking along those lines but instead of tiltina circle I cut diagonally acros a cylinder. I am still working on my idea (as and when I have a moment).
I think your need to prove that the exponent cos(theta) is the correct function required. It is an interesting approach but the result needs to be proven. I wish you success.
@@maqninezawesh8749 I thought about it too, I put all the calculations together, I didn't go anywhere, and I gave up that approach...
Is cooking up rough approximations for the circumference of an ellipse an important thing to do?
I guess it could be useful in certain circumstances - but I don't know what they are. In any case he has to claim to have found an APPROXIMATE formula.
In order to check your formulae, all I need to do is take the limit of a->0 but b remains fixed. When a is very small clearly the periments should be 4b. A finite number. Your formulae blows up in the limit. You get a number growing exponentially and a number converging to 0 linearly. The exponential growth will win.
if a becomes less than b then b is the semi great axis and a becomes the semi small axis. If you check my analysis on my website (in the link in the description) a cannot be lower than b, because by definition a is the semi great axis, so I limited a to b for 'circular' ellipse. I also explain that b can tend to 0, but cannot be equal to 0, I made it equal to 0 just for the demo here. So my formula is basically
C=4*((pi/2)^(semi small axis/semi great axis)) *semi great axis
Also b can't be 0, because if you search for another ellipse with the same C with a different and b is 0 , you can find, but finding the same C starting with b=0, you can't, so b tend to 0 but never reaches 0
@@elyeshachicha for 'a' or 'b' to not equal 0 and still tend to 0, the answer should be close enough. Thats not really an issue. So here is another limit. If you addup a bunch of these circumferences, you should get the area of the ellipse '\pi ab'. Since all the ellipses u should add up are scaled version of each other, it makes sense that 'b/a' is a constant. Thus after adding up, you should retain '(\pi)^(b/a)' in the formulae; however the area of the ellipse doesn't have that term. You should do the calculation and show that integral of differential circumferences gives you the desired area '\pi ab'. Also If I'm not mistaken the elliptic function were originally used to compute the circumference for an ellipse. Im guessing you have checked this. But just in case, you don't know this, "www.chrisrackauckas.com/assets/Papers/ChrisRackauckas-The_Circumference_of_an_Ellipse.pdf"
@@shohamsen8986 thank you for the link, yep I saw that and the fact he concluded that we need to compute an infinite serie that pushed to really work on this problem.
Where I don't agree regarding the area, for the same circumference when B changes in proportion with A, there's no need for the area to remain the same, as an example, a rectangle with size 4 and 1 has same perimeter of size 3 and 2, although the area is 4m2 for the first and 6m2 for the second. If you wanna see how the ratio is changed in mine, I'm using the Lambert function to find the factor n for A when B changes by a factor m for the same circumference. I really invite you to read my analyses on this pdf from page 8
wixlabs-pdf-dev.appspot.com/assets/pdfjs/web/viewer.html?file=%2Fpdfproxy%3Finstance%3DwNO47M79mzeUZa4aUz1kJSrTMAHJ9goNx3p_omKSuhI.eyJpbnN0YW5jZUlkIjoiYzJkNjIzZjctYTE5Ni00ZDRjLTk1ZjMtM2RjM2VkMzkxZGIzIiwiYXBwRGVmSWQiOiIxM2VlMTBhMy1lY2I5LTdlZmYtNDI5OC1kMmY5ZjM0YWNmMGQiLCJtZXRhU2l0ZUlkIjoiYTI3ODA4OTYtMjQ3YS00YTI1LTljZDYtYWI3MjYwMjUxMjU5Iiwic2lnbkRhdGUiOiIyMDIzLTAyLTI3VDEzOjU3OjEyLjgzNFoiLCJkZW1vTW9kZSI6ZmFsc2UsImFpZCI6Ijc4NTJhYmRhLWMwOGItNDc2OS04MzBmLTZiYWE1NjcwMmQ1YSIsImJpVG9rZW4iOiI2MGFlMmI2MS04NWVjLTA3NjktMDkyNS05NmIxOGQxYzBmZWEiLCJzaXRlT3duZXJJZCI6Ijc4ZTc4YzdhLTQ0OWQtNGI1YS1iMjNmLWQzY2FlMDI4YjhmYyJ9%26compId%3Dcomp-lehoo02w%26url%3Dhttps%3A%2F%2Fdocs.wixstatic.com%2Fugd%2F78e78c_253a422562414d8f9c4d7f39253641a4.pdf#page=1&links=true&originalFileName=How_to_calculate_an_ellipse_Elyes_Hachicha%5B23FEB2023%5D&locale=fr&allowDownload=false&allowPrinting=false
@@elyeshachicha I dont think u understood my suggested limiting process. Think of a circle, you can compute the area from the circumference of a circle. \int 2\pi r dr = \pi r^2. You are adding a bunch of circumferences (of similar circles) and getting the area of the circle. I don't think if you apply that to your formulae, you will get the correct area of an ellipse '\pi a b'. Note that since all the ellipses are similar, you have b/a a constant. Anyway, I computed the formulae of an ellipse numerically and using your formulae, they don't match.
In case you are interested, here is the Mathematica code that achieves this.
r[a_, b_, u_] := {a Cos[u], b Sin[u]}
perimeter[a_, b_] :=
NIntegrate[
Sqrt[FullSimplify[D[r[a, b, u], u] . D[r[a, b, u], u]]], {u, 0, 2 Pi}]
Manipulate[ ParametricPlot[r[a, b, t], {t, 0, 2 Pi}, PlotLabel ->
Grid[{{"perimeter:", perimeter[a, b]}, {Style["ArcLength", Bold],
ArcLength[r[a, b, u], {u, 0, 2 Pi}]}, {"Closed Form:",
4*(Pi/2)^(Min[a, b]/Max[a, b])*Max[a, b]}}],
PlotRange -> Table[{-3, 3}, 2]], {a, 1, 3}, {b, 1, 3}]
@@shohamsen8986
I'm not entirely sure you are correct on that one. It would have been my firse guess as well, but by that logic you should be able to get the circumference of an elipse by differentiating the area of it
If we deside that we only take a look at elipses with ratio c = b/a = constant
Then we get
A(a) = πab = cπ*a²
(c = b/a ==> b = c*a)
?==> C(a) = A'(a) = 2cπ*a
This is cleaely not the circumference of an elipse, so there is some kind of a wrong assumption behind the scenes. I would assume it has something to do with the fact that unlike in a circle, a isn't perpendicular to all points on the fircumferense, and therefore the calculastion of an infinitesimal contribution to the overall area from an elipse of a of a major axis 0
This is a true theory i applaud you
You came up with a formula, we can see that. But you didn't prove that it gives the circumference except in the limiting cases.
One of weakness of your formula is symmetricless - C(a, b) should be equal to C(b, a).
I see what you mean but I really insist of the fact that the formula is
C(semi-major axis, semi-minor axis)
So if b becomes greater than a, then b is the semi-major axis and a the semi-minor axis
Great work. it works for circles since we get a=b, hence C=2pia=2pia. I WILL CHECK IT LATER. THANK YOU AND GOOD LUCK TO YOU.
❶ 4 (π/2)¹ a = 2πa
❷ 4 (π/2)⁰ a = 4a
Are just EXACT to two extremes.
Situations in between are just logical. Amen!
π*(a+b)*(1 + h/4 + h^2/64 + h^3/256 + 25h^4/16384 + 49h^5/65536 + 441h^6/1048576 + ....)
h = (a-b)^2/(a+b)^2
there aint no closed formula for that
c=4a EllipticE[(a^2-b^2)/a^2]; of course this is just a new name for the integrations that have to be done - but e.g. Mathematica can calculate this to almost any desired precision.
@@hansulrichkeller303that isn't a closed form formula...
Wait until you see my proof on why the circumferences of a circle and an ellipse are actually 2x2 matrices, where circles have "2pi * r" as their circumference only because the identity matrix is also equal to 1, so it can still be seen as a scalar.
@@eagle32349 I look forward
why when
theta=0 c=4a
The description of the video should be updated. On Dec 26, 2024, the description claims that you calculate the exect circumference. Yet in comments below that you posted months ago, you admit that when you posted you believed it was exact in general, but you have since realized it isn't.
You should have shown the error of your approximation compared to the exact value, to benchmark it against other known approximations.
I found much better approximation C=4a + (2pi-4)*a*(b/a)^1.4555, the error doesn't exceed 0.5%.
This method is effectively a guess and check of the function form (pi/2)^p with the assumption of p = cos(θ) but neither of these are well motivated by anything other than their outputs match how we know a true circumference formula should behave on the boundaries 0 and 90 degrees. The true circumference formula, if it exists, likely has a much different profile between the boundary points. Unfortunately this is the case because the formula breaks down quite quickly.
I'm trying to salvage something from the projection idea in my head because it seems interesting but I fear it might not be as meaningful as it seems. Projecting that way gives b = a.cos(θ) which is basically the same as saying b = n.a for some real number n without the idea of projecting in a third dimension since it's obvious that b is some fraction of a in 2D. Projecting just gives a handy parametrisation of n but the effect on the equation of an ellipse is the same. When attempting a line integral, you will end up with the cos(θ) attached to one of the terms so the expression can't be simplified.
Will you show us some graphs to illustrate how closely your formula gets to the perimeter calculated with elliptic functions?
Sure, I can, if you are interested, on my website I shared more details on the limits and analysis of this equation. The graph you are asking for, I can make it as I see many people are having interest to my work. Thank you
@@elyeshachicha I will look out for it. Thank you.
As promised here is the video
ruclips.net/video/Z7sDHY1PnWs/видео.htmlsi=gV5YII9lLhiIFFW6
Please feel free to share your feedback and questions.
If you have checked your formula for accuracy as you state you did with the suggested graph, you would know if your formula is an approximation or the exact solution. You did not mention here if your solution is an approximation or the exact solution. There are numerous approximations to this problem. How does your approximation with your solution compare to the best approximations already found. I think it is disingenuous not to state your solution is an approximation, especially as you have stated that you have tested for accuracy.
@@elyeshachichaThank you. I didn't see your response till now. Great work !! A remarkable formula !!
With your formula I get a max error of 4.4% at a/b = 3.5
No. Whilst your formula works in the limit as b -> 0 and b -> a it is only approximately true between these. There are many such formulae. Nice try but.
Thank you for your comment. I would like to invite you to see how I manage to reduce the error margin in between and the rest of my work in this video
ruclips.net/video/Z7sDHY1PnWs/видео.htmlsi=aTf4NYa8dvyrMizN
I like it, it seems to result in general slightly larger than the others I've tried
I noticed later it's not exact :( I'm missing something/ a factor that I'm working on to find
@@elyeshachicha the one method where you have sqrt((3a+b)*(3b+a)) is a really good approximation probably the most precise I’ve found
Your video description is false. You are NOT "calculating the exact circumference of an ellipse". Your formula is merely an approximation. You should have stated that clearly at the outset. Viewers don't take kindly to misrepresentation.
When I published it I really thought it was the exact one, on another side, it allowed me to have some useful comments that pushed me to review and understanding where I need to bring a correction. There was no intention to mislead people. Meantime I would like to invite to see another video where I use that method for different applications, just to see that this approach or method (by tilting a circle) is not wrong...
This is not a proof that it is the exact formula of perimeter.
An analysis of approximation depending theta would be required
you are very strong my guy
I really like Ramanujan's 2nd approximation.
My point is to establish a formula for EXACT result.
Tnx. Very nice
OK, the best exponent is not cos(theta). Then the question becomes, what is the best function for the exponent?
Dude, you didn't demonstrate anything, you just found an aproximation. Who told you that (π/2)^cos(θ) is the ONLY formula that fits the extreme values of π/2 and 1?
You need to show two things you have not showed. 1. Is the projection of a tilted circle actually an ellipse (this is not obvious); 2. you need to show the numerical answers agree with the calculated answers from known calculus-derived formulae (which involve elliptic integrals).
Without either of those, this is not a convincing proof.
It's not a proof, but rather an 'Idea' as was stated a couple of times, albeit without enough emphasis.
Anyways... You said in your comment that it isn't obvious... Could you give me a counter argument? It feels like it should, but why won't it?
@@alexbennie If you want to show it's an ellipse, you need to show that it is of the form
(x/a)² + (y/b)² = 1
If you want to prove it algebraicly. Otherwise, you might want to use the geometric diffinitiin, which is to show that the distance from the focal points located at
f1,2 = ( +-c , 0 )
Where c = sqrt(a²-b²)
(The focal points move as you rotate the circle plane with respect to x axis)
Lastly, you might be able to show that any point on the circumference is of the form
(x,y) = (acos(t),bsin(t)) , where 0
The third approach is by far the easiest imo, if you know some linear algebra and calculus
Represent the circle as all the points of the form
(x,y,z) = (acos(t),asin(t),0) for θ=0
Then, if you know that the matrix representation of a rotation of θ degrees around the x axis is
R =
1 0 0
0 cosθ -sinθ
0 sinθ cosθ
Then you get
R(acos(t),asin(t),0) =
(acos(t), acos(θ)sin(t) , asin(θ)sin(t)) =
(acos(t), bsin(t) , asin(θ)sin(t))
We are only interested in the x,y components, and indeed
(x,y) = (acos(t),bsin(t))
As needed
@313sib for your first point, I made a video to prove that rotating a circle becomes an ellipse
ruclips.net/video/JhxFvpk3RpE/видео.html
Regarding the second point, I will publish soon another video.
Thank you for your interest in my work. I will keep you posted after posting the other video.
Regarding the numeral analysis with integral, here's a video I recently shared where I give more details
ruclips.net/video/Z7sDHY1PnWs/видео.htmlsi=gV5YII9lLhiIFFW6
New proof method just droppen: proof by "suggestion"
@@bhbr-xb6po please detail
Nice try! I think it is well known the "circumference" of ellipses lead to elliptical integrals if one wishes to calculate via the arclength formula, which has no closed form in that case.
Look at Micheal Penn's video *_a formula for the circumference of an ellipse_* He derives an infinite sum formula for the task.
Yes, that's an excellent video by Michael Penn.
Your 'method' gives c*=10.0265 for a=2 and b=1, whereas the correct formula results in c=4a EllipticE[(a^2-b^2)/a^2) = 9.68845: Your Result is - in this example - around 3.5 % too large.
That's right, for a ration of b/a=1/2, my result is 3.489% higher
I noticed my error and I'm still figuring out that factor I'm missing. Here's more detail about my error margin for a ratio b/a starting from 0 to 1.
It is 0% when b/a=0
It increases to a maximum of approx. 4.3962% when b/a is around 0.28, then it decrease to 0% until b/a=1. I'll share more of those detailed analysis on my website (on the link in description). Thank you for your interest.
@@elyeshachicha You are assuming linearity, where there is none. The formula is correct for the boundaries 0 and \pi / 2 for \theta...
@@rainerzufall42 at the end of my demonstration on my website I added another application involving theta only for a given circle c with radius r. I called it the 'observable circumference'.
The ‘observable circumference’ of a circle is the circumference of a circle observed from a different angle or point of view. If we draw a circle c with a radius r on the y plane, and rotate the y plane by an angle θ around the x axis, the observable circumference C of the circle c observed from the z axis passing through its center is:
C = 2*((pi/2)^|cos θ|)*r
Nano-thing - title isn't a question so shouldn't have a question mark on the screen.
Interesting.
Are you a man of god? (As in, are you religious?)
It’s not circumference, it’s a perimeter
The reason behind why there is no exact formula for an ellipse refers to multitude of factors that play on the focal points there too much formulas for stretched circles I can give you one hundred if you want.
أخي إلياس، لك مني تحية مبجلة على سيرك على طريق العلماء و محاولة إيجاد قانون جديد لمحيط الشكل البيضوي و البرهنة عليه.
ولكن هل يمكنك أن تحسب لنا محيطه بقانونك الجديد كمثال : لما تكون زاوية الرؤية 30 درجة و نصف القطر 10 سم مع الشرح و التوضيح.
لك مني كل الإحترام و التقدير.
I will reply in English as it is more comfortable for me to type on this keyboard in English.
From a point of view of 30°, the semi-minor axis (b) will be equal to the radius (a) times the cosines of the angle
b = a x cos α = 10 x cos 30° ≈ 8.66
We can now use the formula:
C = 4 x ((pi/2)^(b/a)) x a
The value of 'b' is an approximation in this example, so the circumference becomes also an approximation
C ≈ 4 x ((pi/2)^(8.66/10)) x 10
C ≈ 59.14
Alternatively, as 'b/a = cos α', we can use the following formulas
C = 4 x ((pi/2)^(cos α)) x a
C = 4 x ((pi/2)^(cos 30°)) x 10
C ≈ 59.14
However, I discovered after sometime that ym formula bring us to an approximation. I keep believing that my way of solving is the right one, only missing one parameter... I am thinking of about a factor that I missed in the deformation of the circle.
Good try! But there are many many more fonction P=F(theta), that satisfy conditions that's why they call them elliptic fonctions.
This is just mine which is of course faulse:
C = 2(πb+2✓a^2-b^2)
Greetings
Youcef AMMAR KHODJA
I invite you to watch Pr PARKER video on the subject:
ruclips.net/video/5nW3nJhBHL0/видео.htmlsi=Hiw2BLwFR1vzT5ae
Thank you Youcef
Fake news...
π isn't really a number. It's a construct. It can be constructed in many various ways. One of the ways is to find the 'exact' perimeter of a circle. So it's not surprising that it's tricky to find the 'exact' perimeter of an ellipse.
Honestly I believe that the pi gets "distorted" in an ellipse, what I mean by exact value, is assuming an acceptable value of pi, like for the circumference of a circle, we write it as equal when using pi as a symbol and not a rounded up number.
Since we define π as the ratio of any circle between it's circumference and it's diameter, then we can also define Π(e) "pi of e" where e is eccentricity, to be the ratio between an ellipse circumference and the length of it's major axis.
Since we have the (extremely complicated) formula
π/2
C = 4a S sqrt(1-e²sin²θ)dθ
0
Then
π/2
Π(e) = 2 S sqrt(1-e²sin²θ)dθ
0
π/2
Π(0) = 2 S sqrt(1)dθ =
0
π/2 π/2
= 2 S 1 dθ = 2*θ| = 2(π/2-0) = π
0 0
π/2
Π(1) = 2 S sqrt(1-sin²θ)dθ =
0
π/2
= 2 S sqrt(cos²θ) dθ =
0
π/2 π/2
= 2 S cosθ dθ = 2*sinθ| = 2(1-0) = 2
0 0
Since e = 0 represents a circle and when e approaches 1 it the shape approaces a line, both agree with your resault of
C(circle) = 2πa , C(line) = 4a
So in some sence, yeah, π dies get "distorcted", since the ratio changes as a function of eccentricity. What you are trying to do is essentially solve the above integral, since
e = sqrt(1-(b/a)²)
So
C = 2a*Π(sqrt(1-b²/a²)) is exactly the formula you are looking for