0708-1300/Class notes for Thursday, September 20 - Revision history

Drorbn: /* a non combinatorial, cut-and-paste topology solution */

2007-11-15T12:43:25Z

a non combinatorial, cut-and-paste topology solution

← Older revision		Revision as of 08:43, 15 November 2007
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	===a non combinatorial, cut-and-paste topology solution===		===a non combinatorial, cut-and-paste topology solution===
	which, conveniently, gives the same answer as my combinatorial solution above.		which, conveniently, gives the same answer as my combinatorial solution above.
			{\| align=center
⚫	[[Image:0708-1300roach1.jpg~~\|thumb\|center~~\|200px]]
			\|- align=center
⚫	[[Image:0708-1300roach2001.jpg~~\|thumb\|center~~\|200px]]
		⚫	\|[[Image:0708-1300roach1.jpg\|200px]]
		⚫	\|[[Image:0708-1300roach2001.jpg\|200px]]
			\|}
	please e-mail me if you have questions, comments or a burning desire for more rigour.		please e-mail me if you have questions, comments or a burning desire for more rigour.

Katiemann: /* a non combinatorial, cut-and-paste topology solution */

2007-11-05T15:43:25Z

a non combinatorial, cut-and-paste topology solution

← Older revision		Revision as of 11:43, 5 November 2007
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	===a non combinatorial, cut-and-paste topology solution===		===a non combinatorial, cut-and-paste topology solution===
	which, conveniently, gives the same answer as ~~the~~ solution above.		which, conveniently, gives the same answer as my combinatorial solution above.
	[[Image:0708-1300roach1.jpg\|thumb\|center\|200px]]		[[Image:0708-1300roach1.jpg\|thumb\|center\|200px]]
	[[Image:0708-1300roach2001.jpg\|thumb\|center\|200px]]		[[Image:0708-1300roach2001.jpg\|thumb\|center\|200px]]

Katiemann: /* a non combinatorial, cut-and-paste solution */

2007-11-05T15:42:05Z

a non combinatorial, cut-and-paste solution

← Older revision		Revision as of 11:42, 5 November 2007
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	k = <math>1 + 2^{n-3}(n-4)</math> -holed torus.		k = <math>1 + 2^{n-3}(n-4)</math> -holed torus.

	===a non combinatorial, cut-and-paste solution===		===a non combinatorial, cut-and-paste topology solution===
			which, conveniently, gives the same answer as the solution above.
	will be posted as soon as I can get access to a scanner. In the mean time, I reccommend Diestal's ''Graph Theory'' if you want to know what Euler genus is.
			[[Image:0708-1300roach1.jpg\|thumb\|center\|200px]]
⚫	--[[User:Katiemann\|Katiemann]] 09:39, 2 November 2007 (~~EDT~~)
			[[Image:0708-1300roach2001.jpg\|thumb\|center\|200px]]
			please e-mail me if you have questions, comments or a burning desire for more rigour.

			a webpage to go to if you want info about how to turn my recursive formula for Cn into my non-recursive one is:
			http://web.cs.wpi.edu/~cs504/s99/notes/recurrence/solve/solve.html

		⚫	--[[User:Katiemann\|Katiemann]] 10:42, 5 November 2007 (EST)

	==Enrichment: Configuration Spaces of Mechanical Systems==		==Enrichment: Configuration Spaces of Mechanical Systems==

Katiemann: /* Exercise */

2007-11-02T13:39:10Z

Exercise

← Older revision		Revision as of 09:39, 2 November 2007
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	--[[User:Drorbn\|Drorbn]] 19:46, 24 September 2007 (EDT)		--[[User:Drorbn\|Drorbn]] 19:46, 24 September 2007 (EDT)

			===A combinatorial solution to the roach problem===

			The answer:
			An n-legged roach has configuration space = a torus with <math>1+2^{n-3}(n-4)</math> holes!

			The reason:
			The configuration space of a roach can be represented as a graph as follows. Each face is the domain the body moves in when we fix which way each leg is bent. Since each of the n legs can be bent in 2 directions, we have <math>2^n</math> faces. Each edge is the boundary where one leg is straight (it borders the face where the leg is bent in the other direction). The vertices therefore correspond to positions where two adjacent legs are straight. This means that each vertex has degree four. (for those of you unfamiliar with graph theory lingo, degree 4 means that we have 4 edges from each vertex. In this case we have also four faces) We can now calculate the number of edges, vertices and faces (E, V, F) of an n-legged roach:

			F = <math>2^n</math>

			E = <math>nF/2</math> = <math>2^{n-1}n</math>

			V = <math>2E/degree</math> = <math>2^{n-2}n</math>

			(the above calculation of E in terms of F and V in terms of E is a fun counting argument, if you can’t figure it out ask me)
			Now we compute the Euler genus which tells us what kind of surface this graph is:
			Euler genus = <math>2 - V + E - F = 2 + 2^{n-2}(n-4)</math>
			Since a k-holed torus has genus 2k (the sphere has genus 0), the configuration space of an n-legged roach is a
			k = <math>1 + 2^{n-3}(n-4)</math> -holed torus.

			===a non combinatorial, cut-and-paste solution===
			will be posted as soon as I can get access to a scanner. In the mean time, I reccommend Diestal's ''Graph Theory'' if you want to know what Euler genus is.
			--[[User:Katiemann\|Katiemann]] 09:39, 2 November 2007 (EDT)

	==Enrichment: Configuration Spaces of Mechanical Systems==		==Enrichment: Configuration Spaces of Mechanical Systems==

Drorbn at 20:35, 31 October 2007

2007-10-31T20:35:17Z

← Older revision		Revision as of 16:35, 31 October 2007
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	==Exercise==		==Exercise==
	{{~~07008~~-1300/Solution Warning}}		{{0708-1300/Solution Warning}}

	Configurations of a Generalized Cockroach (not entirely rigourous)		Configurations of a Generalized Cockroach (not entirely rigourous)

Bpym: Some discussion of configurations spaces

2007-09-29T06:23:19Z

Some discussion of configurations spaces

← Older revision		Revision as of 02:23, 29 September 2007
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	--[[User:Drorbn\|Drorbn]] 19:46, 24 September 2007 (EDT)		--[[User:Drorbn\|Drorbn]] 19:46, 24 September 2007 (EDT)

			==Enrichment: Configuration Spaces of Mechanical Systems==
			As Prof. Bar-Natan mentioned in class, manifolds are quite useful for studying the physics of mechanical systems like robots and steam engines. There is actually a very general mathematical theory of rigid body mechanics that makes plenty of use of differential geometry. See, for example,
			{{ref\|BulloLewis_04}} for a discussion of one of these models and its application to nonlinear control theory for mechanical systems. The following is a brief introduction to the idea of a mechanical system's <i>configuration space</i>, and makes use of the notation of {{ref\|BulloLewis_04}}.

			Imagine that you have some rigid body <math>B\!</math> floating around in space. The first thing to do is to set up an orthonormal frame <math>(O_{spatial},(s_1,s_2,s_3))\!</math> with respect to which we will describe <math>B\!</math>'s position. Here <math>O_{spatial}\!</math> is some point in Euclidean (affine) space which serves as the origin of our coordinate system and <math>(s_1,s_2,s_3)\!</math> is an orthonormal basis for Euclidean space centred at <math>O_{spatial}\!</math>.

			Now suppose that we rigidly attach a frame <math>(O_{body},(b_1,b_2,b_3))\!</math> to <math>B\!</math> in the sense that the frame's origin is fixed to a specific point on the body (such as the centre of mass) and the basis <math>(b_1,b_2,b_3)\!</math> rotates around with <math>B\!</math>. We can now specify <math>B\!</math>'s configuration with the vector

			::<math>r = O_{body} - O_{spatial} \in \mathbb{R}^3\!</math>

			and the matrix <math> R \in SO(3;\mathbb{R})\!</math> which gives the orientation of the basis <math>(b_1,b_2,b_3)\!</math> relative to <math>(s_1,s_2,s_3)\!</math>. Here <math>SO(3;\mathbb{R})\!</math> is the group of <math>3\times 3\!</math> real matrices with unit determinant---the group of rotations of <math>\mathbb{R}^3\!</math>. Concisely, the configuration space of a rigid body is the space <math>Q_{free}=SO(3;\mathbb{R}) \times \mathbb{R}^3\!</math>.

			Lucky for us, <math>SO(3;\mathbb{R})\!</math> is actually a manifold. It's an example of something called a <i>Lie group</i>, which is a group that is also a manifold and whose group operations are smooth according to the differentiable structure of the manifold. Hence <math>Q_{free}\!</math> is also a manifold.

			Now imagine we have <math>k\!</math> rigid bodies. If the bodies are free to move around independently, and if we briefly suspend disbelief and allow them to pass right through each other instead of colliding, our configuration space is <math>Q_{free}^k= \left(SO(3;\mathbb{R}) \times \mathbb{R}^3\right)^k</math>. If, on the other hand, the bodies are interconnected---ie., their motions are not independent---or if the motion of the bodies is constrained in some other way, then the configuration space for the system will be some subset <math>Q \subset Q_{free}^k\!</math>. In fact, it is (usually) a <i>submanifold</i>.

			This situation is precisely what's happening with the roach's configuration. In this case the system consists of 12 rigid bodies (six legs composed of two segments each) that are constrained to move in a plane, connected together at their ends, and pinned down.

			Hopefully, it is starting to become clear how we can model some of the physics now: we can describe the motion of a mechanical system with a curve <math>\gamma : I \rightarrow Q\!</math> where <math>I \subset \mathbb{R}\!</math> is an interval. The velocities of all the bodies (both translational and rotational) at time <math>t\!</math>are encapsulated in the tangent vector <math>\gamma'(t) \in T_{\gamma(t)}Q\!</math>. Similarly, there are nice differential-geometric ways of describing the kinetic and potential energies, forces and so on.

			==References==
			{{note\|BulloLewis_04}}Francesco Bullo and Andrew D. Lewis, <i>Geometric Control of Mechanical Systems: Modeling, Analysis, and Design for Simple Mechanical Control Systems</i>, Springer-Verlag, New York-Heidelberg-Berlin, 2004. Number 49 in <i>Texts in Applied Mathematics.</i>

Drorbn: /* another solution for n<5 */

2007-09-28T21:50:33Z

another solution for n<5

← Older revision		Revision as of 17:50, 28 September 2007
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	--[[User:Drorbn\|Drorbn]] 19:46, 24 September 2007 (EDT)		--[[User:Drorbn\|Drorbn]] 19:46, 24 September 2007 (EDT)


	===another solution for n<5===
	Here's a not-too-sketchy solution for n=3 and a sketch of how the same approach works for n=4. The strategy also works for a 2-legged roach but I really really can't do 5 or more. Perhaps someone else can help!
	Note that for n=4 we have a torus, not a sphere!
	p.s. Sorry if the image is upside-down or hard to get to, I am just learning how to deal with technology
	[[Image:roach.pdf\|\|center\|\|]]

Drorbn at 22:31, 25 September 2007

2007-09-25T22:31:19Z

← Older revision		Revision as of 18:31, 25 September 2007
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	In particular, what is <math>C_6</math>?		In particular, what is <math>C_6</math>?

	p<math>n=2</math>: Consider <math>C_2</math>.		<math>n=2</math>: Consider <math>C_2</math>.
	[[Image:0708-1300-cockroach-C2.jpg\|\|left\|\|200px]]		[[Image:0708-1300-cockroach-C2.jpg\|\|left\|\|200px]]
	As in the picture, label the angles of the joints <math>\theta_1</math> and <math>\theta_2</math> and the distances from the body to the feet <math>d_1</math> and <math>d_2</math> respectively.		As in the picture, label the angles of the joints <math>\theta_1</math> and <math>\theta_2</math> and the distances from the body to the feet <math>d_1</math> and <math>d_2</math> respectively.
Line 28:		Line 28:

	That is, the body must lie on the intersection of the two circles of radius <math>d_i</math> centred at the <math>i</math>th foot:		That is, the body must lie on the intersection of the two circles of radius <math>d_i</math> centred at the <math>i</math>th foot:
	[[Image:0708-1300-cockroach-twocircles.jg\|\|center\|\|]]		[[Image:0708-1300-cockroach-twocircles.jpg\|\|center\|\|]]

Katiemann: /* Exercise */

2007-09-25T18:00:35Z

Exercise

← Older revision		Revision as of 14:00, 25 September 2007
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	In particular, what is <math>C_6</math>?		In particular, what is <math>C_6</math>?

	<math>n=2</math>: Consider <math>C_2</math>.		p<math>n=2</math>: Consider <math>C_2</math>.
	[[Image:0708-1300-cockroach-C2.jpg\|\|left\|\|200px]]		[[Image:0708-1300-cockroach-C2.jpg\|\|left\|\|200px]]
	As in the picture, label the angles of the joints <math>\theta_1</math> and <math>\theta_2</math> and the distances from the body to the feet <math>d_1</math> and <math>d_2</math> respectively.		As in the picture, label the angles of the joints <math>\theta_1</math> and <math>\theta_2</math> and the distances from the body to the feet <math>d_1</math> and <math>d_2</math> respectively.
Line 28:		Line 28:

	That is, the body must lie on the intersection of the two circles of radius <math>d_i</math> centred at the <math>i</math>th foot:		That is, the body must lie on the intersection of the two circles of radius <math>d_i</math> centred at the <math>i</math>th foot:
	[[Image:0708-1300-cockroach-twocircles.~~jpg~~\|\|center\|\|]]		[[Image:0708-1300-cockroach-twocircles.jg\|\|center\|\|]]


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	--[[User:Drorbn\|Drorbn]] 19:46, 24 September 2007 (EDT)		--[[User:Drorbn\|Drorbn]] 19:46, 24 September 2007 (EDT)


			===another solution for n<5===
			Here's a not-too-sketchy solution for n=3 and a sketch of how the same approach works for n=4. The strategy also works for a 2-legged roach but I really really can't do 5 or more. Perhaps someone else can help!
			Note that for n=4 we have a torus, not a sphere!
			p.s. Sorry if the image is upside-down or hard to get to, I am just learning how to deal with technology
			[[Image:roach.pdf\|\|center\|\|]]

Drorbn: /* Exercise */

2007-09-24T23:57:15Z

Exercise

← Older revision		Revision as of 19:57, 24 September 2007
Line 10:		Line 10:

	==Exercise==		==Exercise==
			{{07008-1300/Solution Warning}}

	Configurations of a Generalized Cockroach (not entirely rigourous)		Configurations of a Generalized Cockroach (not entirely rigourous)