Solving coefficients for cubic polynormial

**SterlingM** · May 1st 2013, 08:54 AM

I'm not sure if Advanced Algebra is the correct forum for this, but if not maybe it can be moved elsewhere?

Anyways, I am taking this class where we want to solve coefficients for a cubic polynomial. This handout reads:

Suppose we want to find the motion trajectory for a path with only two knot points - the start and goal, which satisfier the time constraint that the tie duration of motion is T, and in addition, satisfier the velocity constraint that the omotion starts from zero velocity and ends with zero velocity. Then, for each variable q_i, its trajectory as a function of time t should satisfy the following 4 constraints:

q_i(0) = s_i
q_i'(0) = 0
q_i(T) = g_i
q_i(T) = 0

Now the question is if we try to interpolate the trajectory of q_i, we should use a cubic polynomial of t

q_i(t) = a₀ + a₁t + a₂t² + a₃t³

because we can determine the 4 coefficients a_i, i=0,1,2,3 from the above 4 constraints. The following values for the coefficients can be easily derived:

a₀ = s_i
a₁ = 0
a₂ = (3/T²) * (g_i - s_i)
a₃ = (2/T³) * (g_i - s_i)

I don't understand how the a₀,a₁,a₂,a₃ are being derived. The things I have considered are:

For a₀: Are we setting t=0? So then we have: s_i = a₀ + a₁(0) + a₂(0) + a₃(0).
For a₁: We have s_i = s_i + a₁t + a₂(0) + a₃(0) ? I don't think that is right, but it's the only way I see to set it up so that a₁=0.
For a₂: I'm pretty lost here. I see that q_i - s_i is the displacement...but how are we getting 3/T²??
For a₃: Same problem as with a₂

Can anyone help me out here? I tried to summarize the handout text some, but I can try to copy more of it if the information is insufficient. Any help is appreciated.

**mathguy25** · May 1st 2013, 10:29 AM

To find $a_0$ , simply use the condition $q_i(0) = s_i$ . Then $s_i = q_i(0) = a_0 + a_1*0 + a_2*0^2 + a_3*0^3 = a_0$ . Thus, $a_0 = s_i$

To find $a_1$ , we have to calculate the derivative of $q_i(t)$ . Then $q_i'(t) = a_1 + 2a_2t + 3a_3t^2$ . Then use the condition $q_i'(0) = 0$ . Thus, $0 = q_i'(0) = a_1 + 2a_2*0 + 3a_3*0^2 = a_1$ . Thus, $a_1 = 0$ .

Now consider $q_i(T)$ and $q_i'(T)$ . We see that

$g_i = q_i(T) = a_0 + a_1*T + a_2*T^2 + a_3*T^3$ and $0 = q_i'(T) = a_1 + 2a_2*T + 3a_3*T^2$

This represents a system of two equations. We already know $a_0$ and $a_1$ so let's sub these values into our system. Then

$g_i = q_i(T) = s_i + a_2*T^2 + a_3*T^3$ and $0 = q_i'(T) = 2a_2*T + 3a_3*T^2$

Now, we just solve this system of two equations in two unknowns ( $a_ 2$ and $a_3$ ). Note that

$0 = 2a_2*T + 3a_3*T^2$
$0 = 2a_2*T + 3a_3*T^2$
$2a_2*T = -3a_3*T^2$
$a_2 =\frac{-3}{2}a_3*T$ . Note that $T \ne 0$

Subs this into the other equation

$g_i = s_i + a_2*T^2 + a_3*T^3$
$g_i = s_i + (\frac{-3}{2} a_3 * T *T^2) + a_3*T^3$
$g_i = s_i + \frac{-3}{2} a_3 * T^3 + a_3*T^3$
$g_i - s_i = \frac{-1}{2} a_3*T^3$
$\frac{-2}{T^3} (g_i - s_i) = a_3$

Now use this to find $a_2$ .

$0 = 2a_2*T + 3a_3*T^2$
$0 = 2a_2*T + 3(\frac{-2}{T^3} (g_i - s_i)*T^2$
$0 = 2a_2*T + \frac{-6}{T} (g_i - s_i)$
$2a_2*T = \frac{6}{T} (g_i - s_i)$
$a_2 = \frac{3}{T^2} (g_i - s_i)$

**SterlingM** · May 2nd 2013, 09:35 AM

Thanks! That's makes a lot of sense. The only thing I don't see clearly is when solving for a₁, why do we we have to calculate the derivative of q_i(t)? Is that approach solely determined so that we can use q'_i(0)=0? Is there no other way to do it?

**SterlingM** · May 2nd 2013, 10:25 AM

Hmm, well I also don't understand the next part. The problem we are trying to solve is finding motion trajectories of an m-degrees of freedom robot (m joints). I'm not sure how relevant that is, but I thought I would include it in case it helps. The next part says:

We want to pind a path of n(>2) knot points which satisfies the following constraints:

at t=t_j, j=1, ..., n, the joint q_i, i=1, ..., m, reaches position u_i,j with velocity q'_i(t_j) = u'_i,j, such that C_j = (u_1,j, ..., u_m,j)

where, normally, t₁=0, and q'_i(t₁)=q'_i(t_n)=0. Then, for each path segment (between two adjacent knot points), we can use cubic polynomials to interpolate joint trajectories. Without losing generality, for the path segment between C_j and C_j+1, let the joint variable q_i satisfy the cubic polynomial

q_i(t) = a₀ + a₁t + a₂t² + a₃t³,

and let u_i,j = s_i, u'_i,j = s'_i, u_{i, j+1}=g_i, u'_i,j+1=g'_i, and T=t_j+1 - t_j. Then the coefficients of the polynomial can be determined as the following:

a₀ = s_i
a₁ = s'_i
a₂ = (3/T²)*(g_i - s_i) - (2/T)*s'_i - (1/T)*g'_i
a₃ = -(2/T³)*(g_i - s_i) + (1/T²)*(s'_i + g'_i)

Okay, I understood most of the algebra from before (except that one part I replied about). I think what I am missing is how do these constraints help us setup the equations? I can't seem to see the connection between those constraints I listed and how our coefficients changed from the original post. I get that a₀ = s_i because that is where we start...but other than that easy case I don't really see how we get from our information to the coefficients. Thanks again for the help thus far and I'd be really grateful if you could help me here as well.

P.S. is there a way to make dots above letters? I did s' and g', but is that syntactically correct? I think ' is only for functions and the dots are for variables.

**mathguy25** · May 2nd 2013, 08:57 PM

Yes, you do need to find q'(t) since you have the intial condition q'(0) = 0.

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