man i was watching this video thinking about what id put in my comment explaining how a translation is a rotation around the point at infinity just for this to be the top comment
@@sid98geekwant more “facts”? A set of parallel lines that is invariant under the translation… converge at the point at infinity rotated 90 degrees away from the center of translation! This is the same point, in both left and right directions (this also implies there’s no positive and negative infinity) Thus all lines intersect in two points, including parallel lines?? Furthermore there is also a line going through every single point at infinity! In every direction! Not a circle, it’s a line. Hell, a parabola is just an ellipse with one focus at infinity! And a hyperbola? Beyond infinity? Google RP2 projective plane for more “facts”
This is wonderful, I would LOVE for this to win the SOME3 competition this year, the visual style and explanatory power of your video was on point! It really felt like I was engaged in proof making throughout the whole video and took something I had some intuition about and really went deep. Great work
Absolutely great! The illustrations in particular give this video so much personality! If you keep posting content with the same vibe, I can really see this unique artstyle and approach become your core identity, instantly recognizable from the rest.
Just wanted to share in case it's not obvious at 9:09: If B' *does* lie on the line A' C, then during the transformation it must have moved closer to C (inside the circle), further away from C (outside the circle), or fall on the same point as A'. In the first two cases, the distance between C and B isn't preserved, and in the third case the distance between A and B isn't preserved (they smoosh together).
I had similar conclusion from tangential reasoning - B is defined (by you, not the demon) as the midpoint around the circle between A and A'. ACB cannot lie on a straight line, as that would require A and A' to be the same point (either for B to be across the diameter from A, or for B=A, which are the only points on the circle on the line AC) which would be a contradiction, as A would be a second fixed point and the setup requires exactly one. Therefore the line AB is a chord of the circle which is not a diameter, so cannot pass through C (the centre). And since A'B'C is a congruent shape to ABC (otherwise you hit a contradiction due to some side not having preseved length), the same must be true for the line A'B'
The difference between a rotation and a translation is that a rotation is a reflection across two intersecting lines and a translation is a reflection across two parallel lines. Though parallel lines can be interpreted as intersecting at infinity. That confirms the other comment about a translation being a rotation about a point at infinity.
even though this was intuitively true to me at first glance (just like, the spatial reasoning that since you can choose any point of rotation I can visualize how it'd work out) I quickly realized I'd have no idea how to go about proving it short of "it makes sense to me" so it was really cool seeing the actual math that goes into proving this kind of thing! goes to show how understanding something on one level doesn't necessarily mean you understand all the optics of it
This video isn't actually useless at what it first seems, it unironically helped me as I was animating something in after effects. I was debugging my animation, as the center point was offset, and I was trying to fix it. I had both rotation and position animated, and it was causing issues with the other animations in the scene (reason why is very complicated, expressions and stuff). Suddenly I remembered this video, and I realized I could just simplify the movement to a single rotate keyframe.
You can do this with stretches and slides too! Edit, to clarify: for any homothety and translation that is equal to a homothety from a different point.
From a linear algebra point of view it is also easy to show using SE(2). Namely, it is easy to show that any matrix in SE(2) has an eigenvalue of one and the eigenvector associated with the eigenvalue of one is the rotation point. From this argument it is also easy to show that the same also holds for SE(3) and thus rotations and translations in 3D.
a quicker but uninspired proof using Projective Geometric Algebra. a rotation about the origin by angle a is represented by the rotor cos(a/2)+sin(a/2)e12, and a translation by d orthogonal to the point at infinity xe20+ye01 is represented by 1 + d/2(xe20+ye01). then applying a rotation and then a translation would be: (1+d/2(xe20+ye01))(cos(a/2)+sin(a/2)e12) and expanding that all out gives: cos(a/2)+sin(a/2)((d/2xcot(a/2)-d/2y)e20+(d/2x+d/2ycot(a/2))e01+e12) which exactly represents a rotation by a around the point (d/2xcot(a/2)-d/2y,d/2x+d/2ycot(a/2)) Projective Geometric Algebra gives a powerful tool for doing geometry like this, but it shouldn't be a substitute for more inspired proofs. the upside is that by proving this using PGA you actually get the point that its a rotation about, but it's just simplifying expressions so it doesn't give you much intuition behind why
In my Aerospace classes, we always kept rotations and translations separate since physical objects tend to rotate around their center of mass. However it is really interesting to see that an object can be seen as doing one fluid rotation about a fixed point if you look at just the start and ending. I wonder what this means about the physical properties.
It's actually impossible - the "fixed axis" might be very far from the transformed point(s) but it will be present unless you fail to preserve distances.
I think the best proof is probably going to be change of basis. It gets quite clear that any distance and orientation preserving transformation is a rotation.
Wait, what if you rotate by 0 or 360 degrees and then move eg 400 units away? That would make the circles infinite in size and you couldn't rotate them back.
That’s what I was thinking, the rotation =/= 0 or any multiple of 2pi radians unless you allow the point of rotation to exist at infinity, which kinda just makes rotation and translation synonymous and this 10 minute video much shorter
@@iamforsaken_1personally, I think that is kind of the point. This video breaks down the barrier between what a rotation and translation really means to us.
I gotta say, when I read the title + thumbnail, I thought this was going to be a ten minute genuine schizophrenic rant about nothing. Pleasantly surprised; this was a lot more interesting than I expected.
It's tempting to say that *translation alone is the same as rotation around infinity.* Sometimes* this is reasonable, but in general there is a major problem with this: translation in the plane is described by 2 numbers, while rotation around a point-once you've chosen that fixed point-is described by only 1 number (the angle). Formally, the group of translations is ℝ² and the group of rotations is ℝ/(2πℤ) ≅ S¹. \* For example, horizontal translation in the half-plane model of the hyperbolic plane ℋ² really is rotation around the single point ∞. Since this is horizontal translation only, it is indeed one-dimensional.
Key: "once you've chosen that fixed point". Prior to choosing that fixed point, you have 3 degrees of freedom, which is more than the 2 for a translation. If you fix the direction of the translation, then you only have 1 degree of freedom for the distance, just like the 1 degree of freedom for the angle of rotation. The center of rotation can be represented in polar form rather than Cartesian form as a direction away from the origin and a distance. Make the distance ∞ and the remaining two degrees of freedom uniquely define any translation.
Yes, any 3D spatial displacement may be reduced to a rotation about some axis fixed to some point and a translation in a direction parallel to that axis. Chasles' theorem
Rotation plus translation of a poi t x is Rx+t With abit of algebra we get Rx+t=R(x+R*t) That means it's just a rotation with a different origin point.
If I'm the demon, I win your game. Because my mystery transformation is a rotation and a reflection. Obviously the fact that I need a reflection to beat you proves the thesis of the video.
Well it's obvious that rotation by arbitrary point is just translation to that point and rotation around the center and translation back to the proper distance... Just like transposition, flipX, flipY and rotation are all interconnected...
All of the demon analogy, could be solved with the demon using the mirror transformation. ergo, B and B´ is in the same position, so the demon has won against you, cause you assumed B´ would be in a different position.
5:07 all this faffing around with preserving distances, rotations and such, while I could immediately defeat the demon saying "your fingers are nachos, therefore your argument is invalid" Jokes aside, great video, it blew my mind!
Ohhh I'm trying to figure out how to use this... I'm trying to draw Orbital segments, which are just an arc of an ellipse which is rotated and translated around the focal point. I keep messing it up though, especially since my start and end angles for the arc are referenced from the focal point of the ellipse not the center.
Could you show that the coordinate plane preserves just the center point as fixed. And then show that any movement of the coordinate plane will map one point back on to the og.
Shouldn't this require stating that it must be a non-0 (or equivalent) rotation? A 0 degree (360/720/etc) rotation followed by a translation does not leave a fixed point.
I feel like I'm missing something obvious... Can someone explain how this would work if you rotate 90, then slide straight up? I don't see how that could be undone with a single rotation. Where would the rotation point be?
Let's say your image is a fish. Duplicate it and move the duplicate as you like. Draw a line segment between the original's eye and the copy's eye. Draw a line segment between the original's tail tip and the copy's tail tip. Perpendicularly bisect each segment. The axis of rotation is where the two bisectors meet.
But, and excuse me for challenging your proof, what if I were to take a picture and rotate it 90° (lets say counterclockwise) and then translate it along a straight line (lets say directly to the right) in such a way that rotation around a single point couldnt possibly return it to its original state? No matter how big or small a circle is, and no matter how far away the picture, the only way for a picture to rotate 90° solely from rotating around a fixed point is by moving along the circle's circumfrence by 1/4 of the corcumfrence's length (or, differently put, by rotating around said fixed point by 90°). There doesnt seem like there is a single fixed point, even in an infinite plane, where a single rotation around it (and only a single rotation around it) could simultaneously return the picture to its original orientation and its original location at the same time. It further seems impossible the farther one slides the picture away from its origin. I'm happy to be wrong, but I just want to know what possible point could revert such a change with just a single rotation about it. Does anyone know?
Is there a noniterative algorithm to find the center of rotation that replicates the translation? I'd like to calculate it without performing any additional transformations if possible.
Any rotation about a point can be decomposed into two reflections across lines passing through that point. Any translation can be decomposed into two reflections across lines orthogonal to the direction of translation (i.e parallel lines). In the case of rotation, the angle of rotation is twice the angle between the two lines of reflection. In the case of translation, the distance of translation is twice the distance between the two lines of reflection. So, first find the line passing through the origin which is orthogonal to the direction of translation, we'll call that L. Now, rotate L by half the rotation angle in the direction opposite from the rotation, we'll call that new line R. Lastly, translate the original L by half the distance of translation in the direction translation, we'll call that line T. The center of rotation for the composition of the rotation and translation is then the intersection point of R and T. Proof: The direction of rotation and translation depends on the order you do the reflections. So if we represent the translation by the operation TL (meaning we will reflect across L first and then T) then the operation LT undoes the transformation, because any repeated reflection ends up back where it started we have TLLT = TT = no transformation. From the way we constructed the lines for rotation, we must do those reflections like LR (first the rotated line, then the original). So, composing the translation with the rotation we get TLLR, and since the double L's cancel each other out, we end up with TR (a reflection first by the rotated line, then by the translated one). Since R is not parallel to T (assuming the angle of rotation was not 0), the lines must intersect and therefore these two reflections constitute a rotation about their intersection.
I would argue that a translation by 0 means no translation happened. Feels like getting a test back marked "wrong answer" on "x/2 = 5; x = 10" because x could also be "10 + 0 * 1" or "1 * 10 * 1 + 0 + 0".
This is true in hyperbolic and elliptic space, but it's also limited to 2D. In 1D it doesn't make sense since rotations that aren't translations don't exist, and in 3D, 1) rotations preserve lines rather than points, and 2) you can compose a translation and a rotation in such a way that the result does not preserve the points on a given line (except for possibly the end points on the horizon), though it _would_ preserve the line itself. Such a composition can however always be modeled as a rotation around a line and a translation asking the same line, and is often called a screw. 4D rotations preserve planes, and like in 3D, any distance and handedness preserving transformation can be formed from 2 distinct rotations. In general, in N dimensions, you need at most ceiling(N/2) rotations to model any combination of any number of rotations/translations.
I found this video randomly on YT and found it very interesting. Now, I have a newbie question. is there a simple formula to get the rotation axis point coordinates based on the coordinates of the "original" and "moved" objects? which coordinates would be necessary? only the ones for the "box" containing before and after objects? (i.e. their 4 corners), or just the opposite corners for each would suffice? sorry if it's a dumb question or I didn't phrase it correctly.
Hi! I’m glad you liked it. It’s a very natural question to ask. It turns out that you can figure out the position of the center of rotation if you know the before and after coordinates of just two (random) points on the plane, here’s how: Let A, B be your chosen points, and A‘ B‘ where they land. Then, if you know you’re dealing with a rotation, both A and A‘ lie on a disc around the center of rotation, and so do B and B‘. Now, picture a line L perpendicular to the line from A to A‘, running through the midpoint of A and A‘. Note that L will touch the center of rotation at some point. Draw another line, L‘, perpendicular to B B‘ and running through the midpoint between B and B‘. L‘ also runs through the center of rotation. Now the point at which L and L‘ intersect must be the center of rotation. Now, it’s a bit difficult to convey the idea through a RU-vid comment, but it’s a fun exercise to try to work out these steps on a piece of paper using ruler and compass!
@@martintrifonov thanks! that's pretty much how i thought of it, but I was getting overly complicated by thinking of points like (x,y) coordinates instead of just say A and B, and thinking of trigonometric formulas. But is much more easy to just visualize it as you explain it!
@@martintrifonov There is a navigation system called LORAN which uses something like this to find locations relative to pairs of transmitter stations. This is the particular case where the location is equidistant from the two members of each pair. (Differential distance is measured by speed of light and timing of pulses that are transmitted at the same time. The difference gives you a hyperbola, or in the case of zero difference, a straight line.)
I was expecting this would involve 3D rotation matrix, but I'm glad I'm wrong! One question I still have is, how do I define this mathematically? It has been a while since the last time I fiddle with math I need some help 😅
Rotation+Rotation is basically equivalent to Rotation+Translation. When you apply two rotations, the final image you have is basically rotated by some angle from its centre, and is moved from its position by some amount (translation). So then it can be represented by one rotation
A rotation about some point is literally a translation and rotation about the center, so to say a translation + rotation = rotation is not something to be proven with that definition
I love both the topic of this video and the manner which it is delivered. But I've watched it around ten times now and I just can't make any sense of the whole proof part; everything after 1.44 ! I have no maths background and all these concepts are quite abstract. I would have enjoyed a dummies version of the proof.
"Any distance-preserving transformation with a unique fixed point is a rotation" -- shouldn't you also specify linear? (I think that's the right condition.) Maybe you did and I missed it. For example, f(r, theta) -> (r, theta^3) is distance-preserving and has unique fixed point (0,0) but is not a rotation.
That function doesn't have a unique fixed point (the rays from the origin at theta = 0 or 1 are fixed) nor does it preserve distances (consider a point on a fixed ray and a point at a slightly different theta, the two either grow closer for theta=0 or farther for theta=1).
@@angeldude101 You can write a rotation around origin in the form of, e.g. a matrix multiplication: X' = aX+bY Y' = cX+dY To do a rotation around any other point you need to incorporate translation in some form, which requires at the very least adding a constant term.
@@API-Beast Or by offsetting your entire 2D plane 1 unit into a third dimension. Now, 2D translations become linear 3D shears, which can be handled entirely with 3x3 matrices. Said matrices aren't very efficient though, which is why planar-quaternions exist (a terrible name, but it's the standard outside of geometric algebra, where they're called 2D motors).
Poor wording. Distance preserving transformation is wrong, because it includes mirroring and the proof falls apart. Rigid transformation is the correct one.
Imagine an arrow pointing up. I rotate it along its center such that it points right. I then translate this right facing arrow straight down, such that it's just below where it used to be. I'm struggling to conceptualise what sort of rotation would bring this arrow back to its original location. Anyone able to help me out?
3:00 go back to this section. When you rotate your arrow 90° to the right, think of a bunch of vectors where the points were shifted by that rotation... It looks like a vortex of ever increasing vector lengths all spinning to the right. Next, remember your downward translation. What part of the vector field is the exact opposite as that translation vector? It's a point some distance to the left of the arrow's position. That's your fixed point, the center of your new rotation. Now think about the fact that the base of the arrow doesn't move left or right during the translation. Draw a vertical line over a circle, and observe where the two intersection points are... Equidistant from the horizontal center of the circle. Your arrow must have it's base be on those points before and after the transportation. Now note that the arrow originally did a 90° rotation, so it must do another 90° rotation the other way. The only place both those constraints are true is at ±45° from horizontal around the circle. So putting it all together, you want to rotate the arrow around a circle to match the 90° clockwise rotation and the downward translation. The circle must intersect the base of the arrow both before and after the transformation. The center of the circle must be exactly halfway vertically between the original and final positions of the base of the arrow. The arrow lies 45° tangent to the circle. Hope that helps!
Rotation has one problem that you’ve successfully ignored. Any rotation is in fact a smooth translation, because you can’t describe it without subdividing the distance by a certain amount, in other words you have to determine and approximate a unit of degree. We use 360 degrees as a base, but it could be any number essentially, that can be equally divided as approximation. So, more carefully is to say that there is no difference between the rotation and translation, them being a single way of positioning an object in certain determined space.
