A couple more errata have come in since the last update. From Tim Lowery comes:

“On page 117, the figures look mislabeled, or the axis is incorrect. For example, Figure 3.5a shows x-axis rotation, but the object is rotating around the y-axis. Likewise, for Figure 3.5b.”

In this case, the captions are correct (the intent is to go in *x*–*y*–*z* order), but the figures are reversed.

From k. avery comes:

“At the end of p. 623, cot theta should be csc theta / sec theta, not sec theta / csc theta.”

Indeed it should be. And I had thought I caught that one once before. Evasive devils, those cotangents.

Hello!

I couldn’t find any e-mail contact on this webpage so I’ll post my remarks here.

I’ve came across some doubts when reading Essential Mathematics For Games . I’ve checked errata but with no luck.

Page 32. Figure 1.13 (Dot product as projection). Assuming that the longer line segment under ‘w’ means ‘length of the projection of v onto w’ the label should go ‘||v||cos(tetha)’.

Next page (33). There is a mistake in all equations regarding Gram-Schmidt Orthogonalization. Projection of v_i onto w_j should have ||wj||^2 in the denominator.

I hope you will find it useful. Feel free to delete this post after reading it.

Comment by Piotr Orzechowski — 8/5/2006 @ 3:14 pm

Thanks for the corrections — I’ll add them to the errata and any future printings.

I’ll also leave the comment up for others, unless you want me to remove it.

-Jim

Comment by Jim Van Verth — 8/7/2006 @ 9:52 am

Hi,

I may have discovered a mistake in the second edition. On page 456, where it is describing how to automatically generate the control points for bezier piecewise curves, the formula for the second control point is different from that shown in figure 10.14. It would be also be better if figure 10.14 was located above the heading for section 10.2.7

Thanks for a great book

Comment by John — 10/4/2009 @ 7:10 am

Sorry about the confusion, both are technically correct. The formula as shown in the figure can be thought of as P_{i+1} + 1/3(P_i – P_{i+2}). If you rearrange the terms, you get P_{i+1} – 1/3(P_{i+2} – P_i), which is what is in the text.

Comment by Jim — 10/26/2009 @ 10:47 pm