If T=ai+bj
since N is orthogonal to T and both has length 1,
so I set
N=bj-ai or N= -bi+aj (which is correct)
How can I make sure that N(s) is in the direction as dT/ds
(dT/ds points in the direction in which T turns as the curve bends)
Oh i thought you were saying that the two were correct.
Well, for me it's N= -bi+aj which is correct.
Derivate T.How can I make sure that N(s) is in the direction as dT/ds
$\displaystyle \vec{T}=a \vec{i}+b \vec{j}$
As a and b are constant, we have :
$\displaystyle \frac{d \vec{T}}{ds}=a \frac{d \vec{i}}{ds} + b \frac{d \vec{j}}{ds}$
According to what i wrote above :
$\displaystyle \frac{d \vec{T}}{ds}=a \vec{j} - b \vec{i}$, which is the same direction as N.
Outch, my eyes >.<
Well, if i got the pitch, T is the unit vector of V, which you can get by dividing V by its norm.
N is the normal vector of T, that is to say the derivative of T in t.
But you can also remember that if you consider a vector of (a,b) coordinates, one of its normal vector will have (b,-a) coordinates (scalar product). Another of its normal vectors will have (-a,b) coordinates. Never mind, the direction remains the same, it's just the sens that will change.
This explains the formula N=bj-ai or N=-bj+ai
Then for this :
I need more time... but it's not the thing i prefer to do :sHow can I make sure that N(s) is in the direction as dT/ds