The instant center of rotation (also known as instantaneous velocity center, instantaneous center, or pole of planar displacement) of a body undergoing planar movement is a point that has zero velocity at a particular instant of time. At this instant, the velocity vectors of the other points in the body generate a circular displacement field around this center of rotation which is identical to what is generated by a pure rotation. Planar movement of a body is often described using a plane figure moving in a two-dimensional plane. The instant center is the point in the moving plane around which all other points are rotating at a specific instant of time. The continuous movement of a plane has an instant center for every value of the time parameter. This generates a curve called the moving centrode. The points in the fixed plane corresponding to these instant centers form the fixed centrode. The generalization of this concept to 3-dimensional space is that of a twist around a screw. The screw has an axis which is a line in 3D space (not necessarily through the origin), the axis of rotation; the screw also has a finite pitch (a fixed translation along its axis corresponding to a rotation about the screw axis).
Pole of a planar displacement 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 ( P x , P y ) {\displaystyle \left(P_{x},P_{y}\right)} can be arranged as follows: The midpoint between A 1 {\displaystyle A^{1}} and A 2 {\displaystyle A^{2}}
has the Cartesian coordinates
A x m = 1 2 ( A x 1 + A x 2 ) ; A y m = 1 2 ( A y 1 + A y 2 ) , {\displaystyle A_{x}^{m}={\frac {1}{2}}\left(A_{x}^{1}+A_{x}^{2}\right);\quad A_{y}^{m}={\frac {1}{2}}\left(A_{y}^{1}+A_{y}^{2}\right),}
and the midpoint between B 1 {\displaystyle B^{1}} and B 2 {\displaystyle B^{2}} has the Cartesian coordinates
B x m = 1 2 ( B x 1 + B x 2 ) ; B y m = 1 2 ( B y 1 + B y 2 ) . {\displaystyle B_{x}^{m}={\frac {1}{2}}\left(B_{x}^{1}+B_{x}^{2}\right);\quad B_{y}^{m}={\frac {1}{2}}\left(B_{y}^{1}+B_{y}^{2}\right).}
The two angles from A 1 {\displaystyle A^{1}} to A 2 {\displaystyle A^{2}} and from B 1 {\displaystyle B^{1}} to B 2 {\displaystyle B^{2}} measured counter-clockwise relative to the horizontal are determined by
tan τ A = A y 2 − A y 1 A x 2 − A x 1 , tan τ B = B y 2 − B y 1 B x 2 − B x 1 {\displaystyle \tan \tau _{A}={\frac {A_{y}^{2}-A_{y}^{1}}{A_{x}^{2}-A_{x}^{1}}},\quad \tan \tau _{B}={\frac {B_{y}^{2}-B_{y}^{1}}{B_{x}^{2}-B_{x}^{1}}}}
Find the position of P ( P x , P y ) {\displaystyle P\left(P_{x},P_{y}\right)}
Method 1:
Taking the correct branches of the tangent. Let the center ( P x , P y ) {\displaystyle \left(P_{x},P_{y}\right)} of the rotation have distances d A {\displaystyle d_{A}} and d B {\displaystyle d_{B}} to the two midpoints. Assuming clockwise rotation (otherwise switch the sign of π / 2 {\displaystyle \pi /2} ):
P x = A x m + d A cos ( τ A − π 2 ) = B x m + d B cos ( τ B − π 2 ) ; P y = A y m + d A sin ( τ A − π 2 ) = B y m + d B sin ( τ B − π 2 ) . {\displaystyle {\begin{aligned}P_{x}&=A_{x}^{m}+d_{A}\cos \left(\tau _{A}-{\frac {\pi }{2}}\right)\\&=B_{x}^{m}+d_{B}\cos \left(\tau _{B}-{\frac {\pi }{2}}\right);\\P_{y}&=A_{y}^{m}+d_{A}\sin \left(\tau _{A}-{\frac {\pi }{2}}\right)\\&=B_{y}^{m}+d_{B}\sin \left(\tau _{B}-{\frac {\pi }{2}}\right).\end{aligned}}}
Rewrite this as a 4 × 4 inhomogeneous system of linear equations with 4 unknowns (the two distances d {\displaystyle d} and the two coordinates P {\displaystyle P} of the center):
( 1 0 − sin τ A 0 1 0 0 − sin τ B 0 1 cos τ A 0 0 1 0 cos τ B ) ( P x P y d A d B ) = ( A x m B x m A y m B y m ) . {\displaystyle {\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_{x}^{m}\\B_{x}^{m}\\A_{y}^{m}\\B_{y}^{m}\end{pmatrix}}.}
The coordinates of the center of the rotation are the first two components of the solution vector
( P x P y d A d B ) = 1 sin ( τ A − τ B ) ( ( B y m − A y m ) sin τ A sin τ B + B x m sin τ A cos τ B − A x m cos τ A sin τ B ( A x m − B x m ) cos τ A cos τ B + A y m sin τ A cos τ B − B y m cos τ A sin τ B ( B x m − A x m ) cos τ B + ( B y m − A y m ) sin τ B ( B x m − A x m ) cos τ A + ( B y m − A y m ) sin τ A ) . {\displaystyle {\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_{y}^{m}-A_{y}^{m}\right)\sin \tau _{A}\sin \tau _{B}+B_{x}^{m}\sin \tau _{A}\cos \tau _{B}-A_{x}^{m}\cos \tau _{A}\sin \tau _{B}\\\left(A_{x}^{m}-B_{x}^{m}\right)\cos \tau _{A}\cos \tau _{B}+A_{y}^{m}\sin \tau _{A}\cos \tau _{B}-B_{y}^{m}\cos \tau _{A}\sin \tau _{B}\\\left(B_{x}^{m}-A_{x}^{m}\right)\cos \tau _{B}+\left(B_{y}^{m}-A_{y}^{m}\right)\sin \tau _{B}\\\left(B_{x}^{m}-A_{x}^{m}\right)\cos \tau _{A}+\left(B_{y}^{m}-A_{y}^{m}\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 ) {\displaystyle y-y_{0}=m(x-x_{0})} where ( x 0 , y 0 ) {\displaystyle (x_{0},y_{0})} is the point and m {\displaystyle m} is the slope. The equation of the bisector of A1A2 is y − A y m = tan τ A ( x − A x m ) {\displaystyle y-A_{y}^{m}=\tan \tau _{A}\left(x-A_{x}^{m}\right)} The equation of the bisector of B1B2 is y − B y m = tan τ B ( x − B x m ) {\displaystyle y-B_{y}^{m}=\tan \tau _{B}\left(x-B_{x}^{m}\right)}
These two bisectors intersect at P ( P x , P y ) {\displaystyle P\left(P_{x},P_{y}\right)} so a system of 2 equations with 2 unknowns and coefficients can be written
{ P y − A y m = tan τ A ( P x − A x m ) P y − B y m = tan τ B ( P x − B x m ) {\displaystyle {\begin{cases}P_{y}-A_{y}^{m}=\tan \tau _{A}\left(P_{x}-A_{x}^{m}\right)\\P_{y}-B_{y}^{m}=\tan \tau _{B}\left(P_{x}-B_{x}^{m}\right)\end{cases}}}
The solution of this system is
P x = B y m − A y m + A x m tan τ A − B x m tan τ B tan τ A − tan τ B , P y = B y m − A y m + tan τ A ( A x m − B x m ) tan τ A − tan τ B + B y m {\displaystyle P_{x}={\frac {B_{y}^{m}-A_{y}^{m}+A_{x}^{m}\tan \tau _{A}-B_{x}^{m}\tan \tau _{B}}{\tan \tau _{A}-\tan \tau _{B}}},\qquad P_{y}={\frac {B_{y}^{m}-A_{y}^{m}+\tan \tau _{A}\left(A_{x}^{m}-B_{x}^{m}\right)}{\tan \tau _{A}-\tan \tau _{B}}}+B_{y}^{m}} 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
Consider the planar movement of a circular wheel rolling without slipping on a linear road; see sketch 3. The wheel rotates around its axis M, which translates in a direction parallel to the road. The point of contact P of the wheel with road does not slip, which means the point P has zero velocity with respect to the road. Thus, at the instant the point P on the wheel comes in contact with the road it becomes an instant center. The set of points of the moving wheel that become instant centers is the circle itself, which defines the moving centrode. The points in the fixed plane that correspond to these instant centers is the line of the road, which defines the fixed centrode. The velocity vector of a point A in the wheel is perpendicular to the segment AP and is proportional to the length of this segment. In particular, the velocities of points in the wheel are determined by the angular velocity of the wheel in rotation around P. The velocity vectors of a number of points are illustrated in sketch 3 and can be calculated using the following equation:
v → = ω → × r → A {\displaystyle {\vec {v}}={\vec {\omega }}\times {\vec {r}}_{A}}
where v → {\displaystyle {\vec {v}}} is velocity of point A, ω → {\displaystyle {\vec {\omega }}} angular velocity of wheel and r → A {\displaystyle {\vec {r}}_{A}} vector from point P to A. The further a point in the wheel is from the instant center P, the proportionally larger its speed. Therefore, the point at the top of the wheel moves in the same direction as the center M of the wheel, but twice as fast, since it is twice the distance away from P. All points that are a distance equal to the radius of the wheel 'r' from point P move at the same speed as the point M but in different directions. This is shown for a point on the wheel that has the same speed as M but moves in the direction tangent to the circle around P.
Relative center of rotation for two contacting planar bodies
If two planar rigid bodies are in contact, and each body has its own distinct center of rotation, then the relative center of rotation between the bodies has to lie somewhere on the line connecting the two centers. This is due to the Aronhold-Kennedy Theorem. In addition, pure rolling can only exist when the instant center of rotation is coincident with the point of contact between the two bodies since this point of co