The instant center can be considered the limiting case of the pole of a planar displacement. The
planar displacement of a body from position 1 to position 2 is defined by the combination of a planar
rotation and planar
translation. For any planar displacement there is a point in the moving body that is in the same place before and after the displacement. The displacement can be viewed as a rotation around this pole.
Construction for the pole of a planar displacement First, select two points A and B in the moving body and locate the corresponding points in the two positions; see the illustration. Construct the perpendicular
bisectors to the two segments A1A2 and B1B2. The intersection P of these two bisectors is the pole of the planar displacement. Notice that A1 and A2 lie on a circle around P. This is true for the corresponding positions of every point in the body. If the two positions of a body are separated by an instant of time in a planar movement, then the pole of a displacement becomes the instant center. In this case, the segments constructed between the instantaneous positions of the points A and B become the velocity vectors VA and VB. The lines perpendicular to these velocity vectors intersect in the instant center. The algebraic construction of the Cartesian coordinates \left(P_x, P_y\right) can be arranged as follows: The midpoint between A^1 and A^2 has the Cartesian coordinates : A^m_x = \frac{1}{2}\left(A^1_x + A^2_x\right); \quad A^m_y = \frac{1}{2}\left(A^1_y + A^2_y\right), and the midpoint between B^1 and B^2 has the Cartesian coordinates : B^m_x = \frac{1}{2}\left(B^1_x + B^2_x\right); \quad B^m_y = \frac{1}{2}\left(B^1_y + B^2_y\right). The two angles from A^1 to A^2 and from B^1 to B^2 measured counter-clockwise relative to the horizontal are determined by : \tan \tau_A = \frac{A^2_y - A^1_y}{A^2_x - A^1_x},\quad \tan \tau_B = \frac{B^2_y - B^1_y}{B^2_x - B^1_x}
Find the position of P\left(P_x, P_y\right) Method 1: Taking the correct branches of the
tangent. Let the center \left(P_x, P_y\right) of the rotation have distances d_A and d_B to the two midpoints. Assuming clockwise rotation (otherwise switch the sign of \pi/2): : \begin{align} P_x &= A^m_x + d_A \cos\left(\tau_A - \frac{\pi}{2}\right) \\ &= B^m_x + d_B \cos\left(\tau_B - \frac{\pi}{2}\right) ;\\ P_y &= A^m_y + d_A \sin\left(\tau_A - \frac{\pi}{2}\right) \\ &= B^m_y + d_B \sin\left(\tau_B - \frac{\pi}{2}\right) . \end{align} Rewrite this as a 4 × 4 inhomogeneous
system of linear equations with 4 unknowns (the two distances d and the two coordinates P of the center): : \begin{pmatrix} 1 & 0 & -\sin\tau_A & 0 \\ 1 & 0 & 0 & -\sin\tau_B \\ 0 & 1 & \cos\tau_A & 0 \\ 0 & 1 & 0 & \cos\tau_B \end{pmatrix} \begin{pmatrix} P_x \\ P_y \\ d_A \\ d_B \end{pmatrix} = \begin{pmatrix} A^m_x \\ B^m_x \\ A^m_y \\ B^m_y \end{pmatrix}. The coordinates of the center of the rotation are the first two components of the solution vector : \begin{pmatrix} P_x \\ P_y \\ d_A \\ d_B \end{pmatrix} = \frac{1}{\sin\left(\tau_A - \tau_B\right)} \begin{pmatrix} \left(B^m_y - A^m_y\right)\sin\tau_A\sin\tau_B + B^m_x \sin\tau_A\cos\tau_B - A^m_x \cos\tau_A\sin\tau_B \\ \left(A^m_x - B^m_x\right)\cos\tau_A\cos\tau_B + A^m_y \sin\tau_A\cos\tau_B - B^m_y \cos\tau_A\sin\tau_B \\ \left(B^m_x - A^m_x\right)\cos\tau_B + \left(B^m_y - A^m_y\right)\sin\tau_B \\ \left(B^m_x - A^m_x\right)\cos\tau_A + \left(B^m_y - A^m_y\right)\sin\tau_A \\ \end{pmatrix}. Method 2: Find equations of the bisectors of two segments A1A2 and B1B2 as follows The equation of a straight line in point- slope form is: y - y_0 = m (x -x_0) where (x_0,y_0) is the point and m is the slope. The equation of the bisector of A1A2 is y - A^m_y = \tan \tau_A\left(x - A^m_x\right) The equation of the bisector of B1B2 is y - B^m_y = \tan \tau_B\left(x - B^m_x\right) These two bisectors intersect at P\left(P_x, P_y\right) so a system of
2 equations with
2 unknowns and
coefficients can be written \begin{cases} P_y - A^m_y = \tan \tau_A\left(P_x - A^m_x\right)\\ P_y - B^m_y = \tan \tau_B\left(P_x - B^m_x\right) \end{cases} The solution of this system is P_x = \frac{B^m_y - A^m_y +A^m_x\tan \tau_A - B^m_x\tan \tau_B}{\tan \tau_A-\tan \tau_B},\qquad P_y = \frac{B^m_y - A^m_y +\tan \tau_A \left(A^m_x - B^m_x\right)}{\tan \tau_A-\tan \tau_B} + B^m_y
Pure translation If the displacement between two positions is a pure translation, then the perpendicular bisectors of the segments A1B1 and A2B2 form parallel lines. These lines are considered to intersect at a point on the
line at infinity, thus the pole of this planar displacement is said to "lie at infinity" in the direction of the perpendicular bisectors. In the limit, pure translation becomes planar movement with point velocity vectors that are parallel. In this case, the instant center is said to lie at infinity in the direction perpendicular to the velocity vectors. == Instant center of a wheel rolling without slipping ==