i've never used the expression "blown away" in my life, but there's no other way to describe how clear, consice, and informative this video was. thank you dan walsh.
This is a really clear explanation. Thanks. In my field, accelerator physics, including particle accelerator design, we depend heavily on Frenet-Serret theory.
immediately one of my favortie videos. I really struggled with space curves in school but now I have better visualization and I feel like I can go back and approach the material differently. Also, I will never confuse pitch yaw and roll again!
Thank you! One important piece of errata: an airplane's control yoke actually cannot control yaw. That's done with the rudder pedals. The yoke only controls pitch and roll. Just in case you ever find yourself flying a plane!
Robert Ghrist introduced me to F-S frames in his Calculus Blue series. Your presentation (along with a nice series by BillCookMath - YT channel) solidifies what I have learned even more, so thanks. By the way, say "Hi" to Santa Barbara, where I resided for the last decade and a half of the 20th Century.
Sure, so by the way it’s constructed, the normal N points perpendicular to the plane spanned by T and B. We can then intuit that the rate of change of N is controlled by how quickly we’re rotating that plane. T and B together define the orientation of the plane - in fact kappa tells us how fast T is rotating, and tau tells us how fast B is rotating. So up to sign, these two constants, multiplied by the frame vector in which the tip of N would move under pure curvature (along T) and pure torsion (along B), define the rate of change of N.
Those sounded like Euler angles near the end! Could the failure for curves that have 0 acceleration at a given point have something to do with gimbal lock? P.S. I don't think the frame is 100% geometrically significant, because the Right Hand Rule is not intrinsically geometrically significant. Regardless of which direction you pick, all it's going to change is a minus sign in a few places. Speaking of, I'm curious if this would also work in 4D. Since 3D requires a non-zero acceleration everywhere so that we have a second derivative, and I think it would also break down in 2D if there was anywhere with 0 velocity, it would make sense if 4D required the curve to have non-zero "jerk" everywhere.
Interesting comment - I don't know enough about Euler angles and gimbal lock to really say, but skimming articles about them, I have a feeling you're on to something there. If a space curve has an inflection point (0 acceleration), then the Frenet frame & fundamental theorem of space curves will apply to the pieces of the curve immediately "before" and "after" that point, but there's a degree of rotational freedom at the point itself. So those two pieces can be attached in all sorts of different ways at the inflection point, and the FS formulas won't be able to notice this. And yes, true point about the binormal vector being defined by a convention up to minus sign. Finally, yes indeed, the Frenet frame was generalized to n dimensions by Camille Jordan (of Jordan curve theorem fame). It uses the Gram-Schmidt algorithm to add in more orthogonal unit vectors, and your intuition is right on about it needing non-zero higher derivatives. Thanks for watching and leaving a thoughtful comment!
@@franciscusrebro1416 I may or may not be more familiar with an alternate method of generating an orthogonal basis than the Gram-Schmidt algorithm. One that uses a "rejection" formula, e₂' = (e₂ ∧ e₁)e₁, and then you can follow that with e₃' = (e₃ ∧ e₂ ∧ e₁)(e₂ ^ e₁) and just continue as long as you remember to normalize everything. A variation of this rejection formula can be derived for quaternions if you want to stick with the traditional cross product: vu = v×u - v·u. v - (v·u)u = v + (-v×u + vu)u = v - (v×u)u + vu² = (assuming |u| = 1) -(v×u)u + v - v = -(v×u)u which just has a negative compared to the old one, which doesn't actually make a difference. (You can actually follow this up to derive the two-sided rotation formula for quaternions.)
There are three parameters needed in four dimensions. Two of which remain the curvature and torsion, and the final one is denoted sigma, which dictates how rapidly the newly introduced unit vector (called D) rotates in the direction of B.
7:42 - I have a question. why exactly T.T = 1? I think you're forcing it to be this way but then you need T = y'/|y|. I think I understood what you did but I'm not sure if I'm correct and you just took shortcuts or if there's actually a reason for the derivative of that function to have always size 1.
T.T = 1 just holds by assumption because we're only looking at unit speed curves. T.T = 1 is a way of saying it's unit speed. It's possible to work out the same theory for non unit speed curves, it's just messier.
Let me try to explain the motivation logically step by step. Our objective is to find some canonical set of 3 orthogonal axis that can be drawn along a curve in 3D space. The first axis naturally is the tangential axis. This is the direction tangent to the curve. It is also the direction of the velocity vector T of a particle moving on this curve. Note that the curve shape itself only restricts the direction of the vector. If I move on this curve twice the speed, then I would have T be double the length. Hence I am free to set what speed i move on this curve. The curve shape doesn't change. The next axis we naturally choose is the radial direction. Lets segway abit into tangential and radial acceleration. Tangential acceleration would speed up a particle without changing direction. Think of pushing a car from straight behind. Radial acceleration would change direction but not speed. Think of circular motion like a planet orbit. Both these directions are perpendicular. We want to extract the radial acceleration direction. To do this we want to set up the speed of the particle in such a way such that the accelration is only radial and we have no tangential acceleration. Then we take the direction of acceleration as our 2nd axis. But no tangential acceleration just means speed is constant. Meaning T.T = constant and differentiating with time gives T.T' = 0 like in the video. Hence T' is the direction of axis 2. Axis 3 is trivial. We have 3D. Just take the remaining direction. We can find it by T' X T. TLDR, we set T.T = 1 so that we find T' is perp to T and can find our 2nd perp direction.
Could I use this equation to explain things in the physical sciences? Lets say a foot untwisting and retwisting as we walk caused by axial rotation above it?
trying to explore solutions to problems noone is looking to solve.. using math and physics to create a model to predict joint angles at other joints by identifying foot characteristics and vice versa... might no make sense, but I am a sport scientist and performance therapist for high level athletes.... people in research just want to re-do the same ol stuff@@danthewalsh
I got lost at 11:08 "Since c1 is the coefficient of N´ in the direction of the unit vector T," [I´m cool with that] "we can find it by computing N´*T" Why?
I have figured it follows from N´T+NT´=0, that c1=-k, since we can use N´=c1T+c2B and TT=1, BT=0. But the video suggests that there is a more intuitive way, bc it shows N´T=c1 earlier.
Because T and B are orthogonal (by construction of B as the cross product TxN). Whenever a vector is expressed as a linear combination of orthogonal, normalized vectors, we can extract any of its coefficients by computing its dot product with the vector multiplying that coefficient. In fact, this is the key idea behind Fourier analysis. To see this, try computing N’*T, or (c1 T + c2 B)*T. Distribute T into each term, giving c1 (T*T) + c2 (B*T). Now note that the second term vanishes due to the orthogonality of B and T, and T*T = 1 in the first term because T has unit length, showing that N’*T = c1, as claimed.
Hello Daniel. My son and I have been interested in a problem for a number of years and cannot seem to make any headway. Perhaps you can help? So, if you take apart a baseball, you see that it its insides are mostly composed of a really long string of rubber that appears randomly wound around a core. We wonder if there is any formally defined space curve that would accomplish this winding, in that it is not biased in any dimension, and yields constant 'density' of rubber as the diameter grows larger. Sort of the 3 dimensional equivalent of a simple (Archimedean) 2 dimensional spiral. Any ideas? Thanks.
Yes. I suggest you read about space-filling curves, pioneered by Giuseppe Peano. There are many such examples of these, especially in two and three dimensions. en.m.wikipedia.org/wiki/Space-filling_curve
I understand what is the radius of curvature, but what does the curvature measure k = 1/r mean by itself? Why do we need both of them when only the radius has an intuitive meaning (geometrically)?
There are a few ways to interpret k. Knowing about radius of curvature, k being the reciprocal of that means low values of k correspond with large osculating circles, and large k goes with small ones. That's one way of making the idea of tightly or widely turning precise. It can also be interpreted as how fast the "turning angle" of a point is changing as it moves along a plane curve. Or, how fast the point locally gains distance away (orthogonally) from staying along a straight line. Thanks for watching!
Recall that in the video we define N = T'/||T'||. In the case where the curve has straight segments, T' vanishes, so N fails to be defined. This is mentioned in the video, which is why we restrict our attention to curves strictly positive kappa when working with the Frenet frame. Since we need N to define the torsion, and curves with straight regions fail to have N defined everywhere, we can no longer even define tau for these curves, let alone determine isometry between such curves by comparing these geometric invariants. You might wonder if it's possible to ignore these regions altogether; after all, we can define kappa, so whenever kappa=0 we have identified the straight regions of the curve. Conceptually, the issue with this is that the orientations of the osculating plane just before and just after the straight region can be chosen arbitrarily without changing kappa and tau. Essentially, passing through straight regions allow the derivatives to "forget" about the previous orientation, which allows two curves with the same curvature and torsion (defined everywhere where they make sense) to actually be non-isometric.
I got lost around 4:50. I don't understand how the magnitude of the acceleration with units L/T^2 can represent the curvature with units 1/L. I know I must be missing something but I cannot work out what.
The example around 4:50 has to do with uniform circular motion with constant speed v. The magnitude of the tangent vector (i.e. the speed) is v. The derivative of the tangent vector is the acceleration, and it has magnitude v^2 / R. Then if the circular motion is actually unit speed, it means v = 1, so the curvature is 1/R. In other words, the "1/L" you mentioned is really (L/T)^2 / L = L / T^2 just as needed.
I don’t see why not, although I’m not sure that such a construction would be particularly useful mathematically. I also wouldn’t refer to these maps as “projections”, since in general they don’t necessarily satisfy the typical requirements of a projection operator.
Ah, sorry for that confusion. In most higher math classes, textbooks and research papers, log t exclusively means the natural logarithm. But being clear on that would have been an improvement. Thanks for watching and giving feedback!
@@franciscusrebro1416 No worries. It just took me like one and a half paper to do all the calculations haha. Also why is that the case anyway, I mean writing "ln" is shorter than "log" and they mean different things in fundamental logarithms.
@@agabe_8989 I'm not really sure why this is, but my guess is log is just a little more appealing to read and pronounce out loud than ln, despite being one letter longer.
Thanks for your feedback! An oscillating plane would be a plane whose angle varies over time (think of an oscillating fan). While the plane described in the video (osculating plane) can oscillate for some curves, this is not the word we are using; rather, our word is a mathematical term used to describe a circle with curvature that matches the curve at a point. In this word, the ‘c’ is, in fact, pronounced. The word derives from the Latin word “osculum”, meaning “little mouth, or kiss”, because the osculating plane perfectly nestles in with the curve as if it were kissing it. Thanks for watching!
Can't expressed how much I was impressed by this. The ideas and presentation are simply flawless.
Excellent. I’m glad to hear that you enjoyed it.
i've never used the expression "blown away" in my life, but there's no other way to describe how clear, consice, and informative this video was. thank you dan walsh.
Thank you! This was a joint effort with me and my friend Franciscus Rebro.
@@danthewalsh tell him a physics freshman from israel thanks him:)
You're so underrated. Amazing and more high quality than official documentaries
That’s good to hear. What aspects of it did you like? What made it great?
This is a really clear explanation. Thanks. In my field, accelerator physics, including particle accelerator design, we depend heavily on Frenet-Serret theory.
This is such a concise video. Thank you!
I think this is my favorite some2 entry so far. And I’ve seen like 20+
immediately one of my favortie videos. I really struggled with space curves in school but now I have better visualization and I feel like I can go back and approach the material differently. Also, I will never confuse pitch yaw and roll again!
Thank you! One important piece of errata: an airplane's control yoke actually cannot control yaw. That's done with the rudder pedals. The yoke only controls pitch and roll. Just in case you ever find yourself flying a plane!
Nice one! Please keep making more of this kind of visualization! Instant subscribed!
Thank you for this clear and understandable introduction to this topic.
I’m glad you found it to be helpful. Hopefully others can find it to be of similar value, even those not familiar with the topic.
Robert Ghrist introduced me to F-S frames in his Calculus Blue series. Your presentation (along with a nice series by BillCookMath - YT channel) solidifies what I have learned even more, so thanks. By the way, say "Hi" to Santa Barbara, where I resided for the last decade and a half of the 20th Century.
The animations really helped me to understand these formulas, well done!!
Congrats really a great video and explanation.
fantastic entry. most of the other some2 entries are not good. this one is very good.
Thank you!!! One of best explanations I've seen!
That’s wonderful. I’m glad you enjoyed it and found it useful!
Beautiful explanation
Nice presentation!
Great work, keep it up! Would love more diff geo
Amazing!!!
Cool, thanks!
first intro ive herd that was top notch in logos
What does this mean?
really nice video!
Thank you! Hope you found it informative.
Wonderful video. Helped a lot. BTW, how do you visualize the second equation, which is a combination of curvature and torsion?
Sure, so by the way it’s constructed, the normal N points perpendicular to the plane spanned by T and B. We can then intuit that the rate of change of N is controlled by how quickly we’re rotating that plane. T and B together define the orientation of the plane - in fact kappa tells us how fast T is rotating, and tau tells us how fast B is rotating. So up to sign, these two constants, multiplied by the frame vector in which the tip of N would move under pure curvature (along T) and pure torsion (along B), define the rate of change of N.
@@danthewalsh Thank you!
Thanks u. Can you help me with the manim documentation that you create for this video?
woooooooow what a video
Those sounded like Euler angles near the end! Could the failure for curves that have 0 acceleration at a given point have something to do with gimbal lock?
P.S. I don't think the frame is 100% geometrically significant, because the Right Hand Rule is not intrinsically geometrically significant. Regardless of which direction you pick, all it's going to change is a minus sign in a few places. Speaking of, I'm curious if this would also work in 4D. Since 3D requires a non-zero acceleration everywhere so that we have a second derivative, and I think it would also break down in 2D if there was anywhere with 0 velocity, it would make sense if 4D required the curve to have non-zero "jerk" everywhere.
Interesting comment - I don't know enough about Euler angles and gimbal lock to really say, but skimming articles about them, I have a feeling you're on to something there. If a space curve has an inflection point (0 acceleration), then the Frenet frame & fundamental theorem of space curves will apply to the pieces of the curve immediately "before" and "after" that point, but there's a degree of rotational freedom at the point itself. So those two pieces can be attached in all sorts of different ways at the inflection point, and the FS formulas won't be able to notice this.
And yes, true point about the binormal vector being defined by a convention up to minus sign. Finally, yes indeed, the Frenet frame was generalized to n dimensions by Camille Jordan (of Jordan curve theorem fame). It uses the Gram-Schmidt algorithm to add in more orthogonal unit vectors, and your intuition is right on about it needing non-zero higher derivatives.
Thanks for watching and leaving a thoughtful comment!
@@franciscusrebro1416 I may or may not be more familiar with an alternate method of generating an orthogonal basis than the Gram-Schmidt algorithm. One that uses a "rejection" formula, e₂' = (e₂ ∧ e₁)e₁, and then you can follow that with e₃' = (e₃ ∧ e₂ ∧ e₁)(e₂ ^ e₁) and just continue as long as you remember to normalize everything.
A variation of this rejection formula can be derived for quaternions if you want to stick with the traditional cross product: vu = v×u - v·u. v - (v·u)u = v + (-v×u + vu)u = v - (v×u)u + vu² = (assuming |u| = 1) -(v×u)u + v - v = -(v×u)u which just has a negative compared to the old one, which doesn't actually make a difference. (You can actually follow this up to derive the two-sided rotation formula for quaternions.)
What about curves in 4th dimension? How many parameters do we need? Like 3?
There are three parameters needed in four dimensions. Two of which remain the curvature and torsion, and the final one is denoted sigma, which dictates how rapidly the newly introduced unit vector (called D) rotates in the direction of B.
7:42 - I have a question. why exactly T.T = 1?
I think you're forcing it to be this way but then you need T = y'/|y|. I think I understood what you did but I'm not sure if I'm correct and you just took shortcuts or if there's actually a reason for the derivative of that function to have always size 1.
T.T = 1 just holds by assumption because we're only looking at unit speed curves. T.T = 1 is a way of saying it's unit speed. It's possible to work out the same theory for non unit speed curves, it's just messier.
Let me try to explain the motivation logically step by step. Our objective is to find some canonical set of 3 orthogonal axis that can be drawn along a curve in 3D space.
The first axis naturally is the tangential axis. This is the direction tangent to the curve. It is also the direction of the velocity vector T of a particle moving on this curve. Note that the curve shape itself only restricts the direction of the vector. If I move on this curve twice the speed, then I would have T be double the length. Hence I am free to set what speed i move on this curve. The curve shape doesn't change.
The next axis we naturally choose is the radial direction.
Lets segway abit into tangential and radial acceleration. Tangential acceleration would speed up a particle without changing direction. Think of pushing a car from straight behind. Radial acceleration would change direction but not speed. Think of circular motion like a planet orbit. Both these directions are perpendicular. We want to extract the radial acceleration direction. To do this we want to set up the speed of the particle in such a way such that the accelration is only radial and we have no tangential acceleration. Then we take the direction of acceleration as our 2nd axis. But no tangential acceleration just means speed is constant. Meaning T.T = constant and differentiating with time gives T.T' = 0 like in the video. Hence T' is the direction of axis 2.
Axis 3 is trivial. We have 3D. Just take the remaining direction. We can find it by T' X T.
TLDR, we set T.T = 1 so that we find T' is perp to T and can find our 2nd perp direction.
6:30 I spent 20 minutes trying to do it and I did not get that. :(
Could I use this equation to explain things in the physical sciences? Lets say a foot untwisting and retwisting as we walk caused by axial rotation above it?
With mathematics and physics, the sky is the limit. What did you have in mind to study with feet?
trying to explore solutions to problems noone is looking to solve.. using math and physics to create a model to predict joint angles at other joints by identifying foot characteristics and vice versa... might no make sense, but I am a sport scientist and performance therapist for high level athletes.... people in research just want to re-do the same ol stuff@@danthewalsh
I got lost at 11:08
"Since c1 is the coefficient of N´ in the direction of the unit vector T," [I´m cool with that]
"we can find it by computing N´*T" Why?
I have figured it follows from N´T+NT´=0, that c1=-k, since we can use N´=c1T+c2B and TT=1, BT=0. But the video suggests that there is a more intuitive way, bc it shows N´T=c1 earlier.
Because T and B are orthogonal (by construction of B as the cross product TxN). Whenever a vector is expressed as a linear combination of orthogonal, normalized vectors, we can extract any of its coefficients by computing its dot product with the vector multiplying that coefficient. In fact, this is the key idea behind Fourier analysis. To see this, try computing N’*T, or (c1 T + c2 B)*T. Distribute T into each term, giving c1 (T*T) + c2 (B*T). Now note that the second term vanishes due to the orthogonality of B and T, and T*T = 1 in the first term because T has unit length, showing that N’*T = c1, as claimed.
Hello Daniel. My son and I have been interested in a problem for a number of years and cannot seem to make any headway. Perhaps you can help? So, if you take apart a baseball, you see that it its insides are mostly composed of a really long string of rubber that appears randomly wound around a core. We wonder if there is any formally defined space curve that would accomplish this winding, in that it is not biased in any dimension, and yields constant 'density' of rubber as the diameter grows larger. Sort of the 3 dimensional equivalent of a simple (Archimedean) 2 dimensional spiral. Any ideas? Thanks.
Yes. I suggest you read about space-filling curves, pioneered by Giuseppe Peano. There are many such examples of these, especially in two and three dimensions. en.m.wikipedia.org/wiki/Space-filling_curve
@@danthewalsh Thank you!
I understand what is the radius of curvature, but what does the curvature measure k = 1/r mean by itself?
Why do we need both of them when only the radius has an intuitive meaning (geometrically)?
There are a few ways to interpret k. Knowing about radius of curvature, k being the reciprocal of that means low values of k correspond with large osculating circles, and large k goes with small ones. That's one way of making the idea of tightly or widely turning precise. It can also be interpreted as how fast the "turning angle" of a point is changing as it moves along a plane curve. Or, how fast the point locally gains distance away (orthogonally) from staying along a straight line. Thanks for watching!
Does this method of checking isometrism of two curves fall apart, if curves have straight segments?
Recall that in the video we define N = T'/||T'||. In the case where the curve has straight segments, T' vanishes, so N fails to be defined. This is mentioned in the video, which is why we restrict our attention to curves strictly positive kappa when working with the Frenet frame. Since we need N to define the torsion, and curves with straight regions fail to have N defined everywhere, we can no longer even define tau for these curves, let alone determine isometry between such curves by comparing these geometric invariants. You might wonder if it's possible to ignore these regions altogether; after all, we can define kappa, so whenever kappa=0 we have identified the straight regions of the curve. Conceptually, the issue with this is that the orientations of the osculating plane just before and just after the straight region can be chosen arbitrarily without changing kappa and tau. Essentially, passing through straight regions allow the derivatives to "forget" about the previous orientation, which allows two curves with the same curvature and torsion (defined everywhere where they make sense) to actually be non-isometric.
@@danthewalsh thank you for clarification!
genius
I got lost around 4:50. I don't understand how the magnitude of the acceleration with units L/T^2 can represent the curvature with units 1/L. I know I must be missing something but I cannot work out what.
The example around 4:50 has to do with uniform circular motion with constant speed v. The magnitude of the tangent vector (i.e. the speed) is v. The derivative of the tangent vector is the acceleration, and it has magnitude v^2 / R. Then if the circular motion is actually unit speed, it means v = 1, so the curvature is 1/R.
In other words, the "1/L" you mentioned is really (L/T)^2 / L = L / T^2 just as needed.
Can we project a space curve into two planar curves so that the curvature of one is kappa and the curvature of the other is tau up to sign?
I don’t see why not, although I’m not sure that such a construction would be particularly useful mathematically. I also wouldn’t refer to these maps as “projections”, since in general they don’t necessarily satisfy the typical requirements of a projection operator.
6:38 I did that using log t as log_10(t), not the natural logarithm. Why didn't you just use ln(t) it just makes things confusing.
Ah, sorry for that confusion. In most higher math classes, textbooks and research papers, log t exclusively means the natural logarithm. But being clear on that would have been an improvement. Thanks for watching and giving feedback!
@@franciscusrebro1416 No worries. It just took me like one and a half paper to do all the calculations haha. Also why is that the case anyway, I mean writing "ln" is shorter than "log" and they mean different things in fundamental logarithms.
@@agabe_8989 I'm not really sure why this is, but my guess is log is just a little more appealing to read and pronounce out loud than ln, despite being one letter longer.
2:00 How come ( cos(t²), sin(t²) ) is slower than ( cos(t), sin(t) )?
This is true for 0 < t < 1 because in that interval, the square of a number is smaller than the number itself
@@franciscusrebro1416 Yes, true. but what about the values after 1?
@@777-b1f it would be faster
Great video! The C is silent in oscillating though.
Thanks for your feedback! An oscillating plane would be a plane whose angle varies over time (think of an oscillating fan). While the plane described in the video (osculating plane) can oscillate for some curves, this is not the word we are using; rather, our word is a mathematical term used to describe a circle with curvature that matches the curve at a point. In this word, the ‘c’ is, in fact, pronounced. The word derives from the Latin word “osculum”, meaning “little mouth, or kiss”, because the osculating plane perfectly nestles in with the curve as if it were kissing it. Thanks for watching!
Osculating and oscillating are different words, you oaf
🫡👏
There where a lot of heroic people working for NASA. Most of them men. I hope that you acknowledge that in one of your videos too
why don't you enable your dislikes
I didn't realize I had it turned off. Can you see them now?