# Introduction to Robotics Mechanics & Control, John J Craig., 3rd Ed., Forward transformation problem Examples 2.2 and 2.4

I am reading the book "Introduction to Robotics Mechanics & Control", John J Craig., 3rd Ed., Forward transformation problem Examples 2.2 and 2.4.

Ex. 2.2 (Page 29): Frame {B} is rotated relative to frame {A} about X axis by 60 degrees clockwise, translated 20 units along Y axis and 15 units along z axis. Find P in frame {A} where P in frame {b} = [0 8 7]

The book's answer is [0.18.062 3.572]. But my answer is [0 30.062 11.572].

Ex. 2.4 (Page 33): Vector P1 has to be rotated by 60 degrees clockwise, about X axis and translated 20 units along Y axis, and 15 units along Z axis. If P1 is given by [0 8 7], find P2.

Essentially Ex.2.2 and 2.4 are the same problem. However, the Transformation matrix for Ex 2.4, has [0 8 7] as translation vector (The 4th column of T) instead of [0 20 15]. And, the given answer is [0.18.062 3.572].

I am not sure if it is just typo, or I am missing some genuine operation. Please let me know your opinion.

Thanks.

• It will be helpful if you show your work, step by step.
– Paul
May 31 '15 at 18:30
• I don't think all of those (of frame {A}) statements are necessary; it's implied in the statement "Frame {B} is rotated relative to frame {A}". Jun 2 '15 at 21:31
• @Paul , I have solved it the same way Chuck has done. Jun 4 '15 at 14:59
• Hey @Chuck, I would love to, but thought of posting the solution given in the book before that so that you can look at it. Jun 4 '15 at 22:43
• Sounds good. Just edit your question. Jun 5 '15 at 1:03

I think I would solve this as I believe you did, where I find a rotation matrix $E$, a translation vector $\overrightarrow {dr}$, and I locate point P in frame B by adding it to the origin of frame B; $\overrightarrow P_{B} = \overrightarrow P + F_B$. Recall that you are "unrotating" from frame B to frame A, so you need to transpose the rotation matrix. Assemble the $T$ matrix: $$E_{A\rightarrow B} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & cos(-60) & sin(-60) \\ 0 & -sin(-60) & cos(-60) \end{bmatrix} \\~\\~\\ \\ E_{B\rightarrow A} = E_{A\rightarrow B}^T\\~\\~\\ \overrightarrow {dr} = \begin{bmatrix} 0 \\ 20 \\ 15 \end{bmatrix} \\ T_{B\rightarrow A}= \begin{bmatrix} E_{B\rightarrow A} & \overrightarrow {dr} \\ 0 & 1 \end{bmatrix}$$ Now, multiply $T \overrightarrow {P_B}$ to get $\overrightarrow {P_A}$, which is the same answer you gave in your statement. I believe you did everything correct. I actually drew the problem on a piece of paper to verify the solution. You are also correct that the second problem (2.4) is the same problem, just reworded. Again, try drawing it on a sheet of paper.