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View Full Version : Re: Calculation of 2D Angles ('CAST' problem)

Young-hoo Kwon, Ph.d.
01-08-2003, 05:13 AM
Dear Ton and all,

I'd like to point out a mistake in your summary.

> As in Kwon3D, we label these matrix elements as t11, t12, t13, etc.
> Using such an attitude matrix, obtained from measurements, the joint
> angles can be easily extracted as follows:
>
> ph = atan2(-t32,t33)
> th = atan2(t31,sqrt(t32*t32+t33*t33))
> ps = atan2(-t21,t11)
> (..sqrt is the square root function)
>

ph = atan2(-t32,t33)

is not right because

atan2(-t32,t33) = atan2(sin(ph)cos(th),cos(ph)cos(th))

is not equivalent to

atan2(sin(ph),cos(ph)).

It depends on the sign of cos(th). If this is negative, you will have

atan2(-sin(ph),-cos(ph))

instead. Likewise,

atan2(t31,sqrt(t32*t32+t33*t33)) = atan2(sin(th),sqrt(cos(th)^2))

is not equivalent to

atan2(sin(th),cos(th))

because sqrt(a2) can be either +a or -a, etc.

I agree that the use of ATAN2 is relatively more convenient. I have used FORTRAN, QuickBASIC, C, C++, and now Visual C++. The function has evolved accordingly and atan
2 was forgotten in the process somehow. I will update my function using atan2. (^_^)

Young-Hoo Kwon
------------------------------------------------------
- Young-Hoo Kwon, Ph.D.
- Biomechanics Lab, Texas Woman's University
- ykwon@twu.edu
- http://kwon3d.com
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