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Homework 06

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F11Gratz

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Unread post Mon Oct 17, 2011 10:00 pm

Homework 06

In regards to problem 3.4.7, what is the "third constraint equation of Eq. 3.4.20?" I only see two constraint equations there.
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Dan Negrut

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Unread post Tue Oct 18, 2011 8:10 am

Re: Homework 06

F11Gratz wrote:In regards to problem 3.4.7, what is the "third constraint equation of Eq. 3.4.20?" I only see two constraint equations there.


Josh - you meant problem 3.4.8, right?
In Eq. 3.4.20 the set of constraints \Phi associated with this joint includes three equations. The first expression in 3.4.20 has two components since it says that two points should be identical. That means that the x values of the two points and the y values of the two points should be the same.
The third constraint equation is the dot product of two vectors, which should be zero. The dot product of two vectors is a scalar, that's why this condition leads to one equation.
This adds up to three equations.
Post a follow up question if this is not clear yet.
Dan
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F11Wendy

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Unread post Tue Oct 18, 2011 2:09 pm

Re: Homework 06

For problem 3.4.9: Deriving the Jacobians from equation 3.4.29:

I am stuck on taking the partial derivative with respect to phi 1. What part of the first constraint is dependent on phi 1? Any hints on what matrix relations or equations to use to get there? Thanks!
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Dan Negrut

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Unread post Tue Oct 18, 2011 2:49 pm

Re: Homework 06

F11Wendy wrote:For problem 3.4.9: Deriving the Jacobians from equation 3.4.29:

I am stuck on taking the partial derivative with respect to phi 1. What part of the first constraint is dependent on phi 1? Any hints on what matrix relations or equations to use to get there? Thanks!



1. There seems to be a mistake, in that the book should have s_i_prime_Q instead of s_i_prime_P.
2. Use the fact that B=AR=RA and R*R=-I and therefore B_i^T*B_j= A_i^T*A_j=A_{ij}.

I hope this helps.
Dan
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F11Wendy

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Unread post Wed Oct 19, 2011 12:29 pm

Re: Homework 06

Dan Negrut wrote:
F11Wendy wrote:For problem 3.4.9: Deriving the Jacobians from equation 3.4.29:

I am stuck on taking the partial derivative with respect to phi 1. What part of the first constraint is dependent on phi 1? Any hints on what matrix relations or equations to use to get there? Thanks!



1. There seems to be a mistake, in that the book should have s_i_prime_Q instead of s_i_prime_P.
2. Use the fact that B=AR=RA and R*R=-I and therefore B_i^T*B_j= A_i^T*A_j=A_{ij}.

I hope this helps.
Dan


I am still not seeing where the rho_i*u'_i term comes from.
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Dan Negrut

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Unread post Wed Oct 19, 2011 1:02 pm

Re: Homework 06

F11Wendy wrote:
Dan Negrut wrote:
F11Wendy wrote:For problem 3.4.9: Deriving the Jacobians from equation 3.4.29:

I am stuck on taking the partial derivative with respect to phi 1. What part of the first constraint is dependent on phi 1? Any hints on what matrix relations or equations to use to get there? Thanks!



1. There seems to be a mistake, in that the book should have s_i_prime_Q instead of s_i_prime_P.
2. Use the fact that B=AR=RA and R*R=-I and therefore B_i^T*B_j= A_i^T*A_j=A_{ij}.

I hope this helps.
Dan


I am still not seeing where the rho_i*u'_i term comes from.


gets you from Q to the contact point on the boundary.
stop by during office hours if this continues to be confusing.
Dan
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F11Gratz

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Unread post Wed Oct 19, 2011 8:49 pm

Re: Homework 06

I'm stuck on the next part of problem 3.4.9 (the partial derivative of the second constraint with respect to phi_1):

If B_i^T*B_j = B_{ij} = A_{ij},

then to get the right solution, the partial derivative of A_{ij} with respect to phi has to be A_{ij}, and I can't find how this could be true.
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Dan Negrut

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Unread post Wed Oct 19, 2011 9:26 pm

Re: Homework 06

F11Gratz wrote:I'm stuck on the next part of problem 3.4.9 (the partial derivative of the second constraint with respect to phi_1):

If B_i^T*B_j = B_{ij} = A_{ij},

then to get the right solution, the partial derivative of A_{ij} with respect to phi has to be A_{ij}, and I can't find how this could be true.


Read point 2 at my Tue Oct 18, 2011 1:49 pm post. that should do it.
Also, in general, B_i^T*B_j is not B_{ij}, we never said that. This was the case for A only. Let's talk about this after class tomorrow, i can explain why that's the case.

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