Shear

[M@35TR0]

CAn someone please explain shear to me?

 

[Abdullahkhan92]

in shear the object changes shape but area remains the same..... The movement is parallel to the invarient line which may be Y - AXIS or any line parallel to Y AXIS for eg X = 4, X-AXIS or any line parallel to X-AXIS for Y = -1....
Shear factor can be taken out by this formula K = AA'/Al = BB'/Bl = CC'/Cl
In simpler terms K = Distance of object to Image / Distance of Object to Invarient line........ remember to count the distance from the image !!!!!!!!!!! if the object is at right side or down the image the value of AA' will be in negative !!!!!! try it and you will know the difference...... When asked Describe Completely or Describe fully, give your answer like this : A shear, with X-AXIS/ Y- AXIS, Y= 1/X=2 (etc) invarient line and shear factor K..... And remember when object and image is given and u have to explain what happend, just luk to the shape and see where it is going parallel to X axis or Y axis....... Then figure out the the invarient line by trial and error method choosing lines X AXIS or Y AXIS or lines parallel to either AXIS and then applying the formula i mentioned earlier
CHEERS..........

 

[Sijan92]

in shear the object changes shape but area remains the same..... The movement is parallel to the invarient line which may be Y - AXIS or any line parallel to Y AXIS for eg X = 4, X-AXIS or any line parallel to X-AXIS for Y = -1....
Shear factor can be taken out by this formula K = AA'/Al = BB'/Bl = CC'/Cl
In simpler terms K = Distance of object to Image / Distance of Object to Invarient line........ remember to count the distance from the image !!!!!!!!!!! if the object is at right side or down the image the value of AA' will be in negative !!!!!! try it and you will know the difference...... When asked Describe Completely or Describe fully, give your answer like this : A shear, with X-AXIS/ Y- AXIS, Y= 1/X=2 (etc) invarient line and shear factor K..... And remember when object and image is given and u have to explain what happend, just luk to the shape and see where it is going parallel to X axis or Y axis....... Then figure out the the invarient line by trial and error method choosing lines X AXIS or Y AXIS or lines parallel to either AXIS and then applying the formula i mentioned earlier
CHEERS..........

YOu said the scale factor is AA'/Al ....wat is Al???? wat does l stands for??? As for as i know the formula is distance of the object TO the image/ the distance of the invariant line TO the object. And the factor will be negative if the object is at the right side of the image or in if the object is below the invariant line (WHY BELOW THE IMAGE :roll: )

 

[hassam]

easy way to find the invariant line.
ExTRAPOLATE(extend) tone of the lines in original fig and nd the coresponding line in the image figure E-G AB nd A'B' these lines will meet at at a point on the invariant line. GUARANTEE this works
caution: the two lines being extrapolate must not be parallel or else the can never meet.

 

[XPF MASTER]

@:.hassam.. 100% agreed!!

 

[OakMoon!]

Yeah trial and error is always a risk... always try finding a proper method...