algorithm for Bezier curve with eleven control points I would like to know the algorithm/ polynomial equation for a Bezier curve with eleven control points. Thanks in advance.
 A: The easiest way I've found to get the equation for a bezier curve of order / degree $N$ is to use the binomial theorem and pascal's triangle.
Basically, the generic formula for Bezier curves looks like $(As+Bt)^N$ where $N$ is the degree of the curve which has $N+1$ control points.
You could expand this by hand.  If you wanted to create a quadratic curve (degree 2) you would get this as a result:
$(As+Bt)^2 = A^2s^2 + 2ABst + B^2t^2$
You then replace all unique combinations of $A$ and $B$ with control points.  In this case we have $A^2, AB, B^2$ that we replace with $A,B,C$ to get:
$P = As^2 + 2Bst + Ct^2$
That is the quadratic bezier function!  Note that $s = (1-t)$, so you could also write it this way:
$P(t) = A(1-t)^2 + 2B(1-t)t + Ct^2$
This method gets a little more difficult with higher values of $N$ but that's where the binomial theorem comes in.  The binomial theorem gives us a shortcut for evaluating $(A+B)^N$.
If you look at the version with both $s$ and $t$, you may notice that the powers of $s$ on the terms goes 2,1,0.  $A^2s^2$ has $s$ as a power of 2, $2ABst$ has $s$ as a power of 1, and $B^2t^2$ has s in a power of 0 (remember anything raised to zero is 1, so $s^0 = 1$).
You might also notice that the powers of t go in reverse and go 0,1,2.
This is part of the binomial theorem!  In an 11th order equation, like the one you are looking for, the first term will have $s^{11}$, the next term will have $s^{10}$etc, until the last term has $s^0$ (ie it's not there).  Likewise, the first term will have $t^0$, the second term will have $t^1$, and so on, til the last term has $t^{11}$.
You might have noticed though, that the middle term of the quadratic beizer curve is multiplied by 2.  Where does that 2 come from?  well, if you look at the 3rd row pascal's triangle (row $N+1$), you'll see that it is 1,2,1.  Those values are the constants you multiply each term by, thus giving the middle term a multiplication by 2.
Now that you know how I'm going to do my trick here is your answer.  Here is the equation for an 11th order bezier triangle, with 12 control points $A$ through $L$, remembering that $s$ is just $1-t$.
$P = As^{11}+11Bs^{10}t+55Cs^9t^2+165Ds^8t^3+330Es^7t^4+462Fs^6t^5+462Gs^5t^6+330Hs^4t^7+165Is^3t^8+55Js^2t^9+11Kst^{10}+Lt^{11}$ 
Check this link out for more info!
http://en.wikipedia.org/wiki/Binomial_theorem
Edit: Oops, you asked for 11 control points not 12, here's that equation:
$P = As^{10}+10Bs^9t+45Cs^8t^2+120Ds^7t^3+210Es^6t^4+252Fs^5t^5+210Gs^4t^6+120Hs^3t^7+45Is^2t^8+10Jst^9+Kt^{10}$
A: The curve of degree $m$ with control points $P_0, \ldots, P_m$ has the equation:
$$
C(t) = \sum_{i=0}^m \varphi_i^m(t)P_i \quad\quad (0 \le t \le 1)
$$
where
$$
\varphi_i^m(t) = \displaystyle \binom{m}{i}(1 - t)^{m-i} t^{i}  \quad\quad (0 \le t \le 1)
$$
is the $i$-th Bernstein polynomial of degree $m$.
For $m=11$, the Bernstein polynomials are:
$$
\varphi_0(t) = (1 - t)^{11}  \\
\varphi_1(t) = 11t(1 - t)^{10} \\
\varphi_2(t) = 55t^2(1 - t)^{9} \\
\vdots \\
\varphi_{11}(t) = t^{11}  
$$
You can calculate points on the curve using the well-known deCasteljau algorithm.
