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Martin,
I'm with you on this. No offense meant to Wojciechf, but what matters is content. Keep up the good work!
- Andy.
2007-12-18 17:58:49 UTC in mjbWorld
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Hi, Martin - I'm all in favor of the change. I reiterate my curse on spammers, who have poisoned the greatest communal well of the last century. Anyway, I agree it's nice to have a name on the posts. And thanx again for maintaining such a wonderfully informative web site.
Cheers - Andy.
2006-06-19 13:40:36 UTC in mjbWorld
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OK... here's the algebraic analysis. First, remember that quaternions are a general mechanism, and rotation is just one application. Not all quaternions are valid rotations, and the mapping of specific operations on quaternions to the effect on rotations may not be completely obvious.
So... first, let's accept that
-Q = -(s, V) = (-s, -V)
From this, it follows pretty trivially that.
2006-04-11 16:43:16 UTC in mjbWorld
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Paul, your equation -q = (-s,-V) does not represent rotation in the direction opposite of q. The conjugate of a quaternion (representing the opposite rotation) is -q = (s,-V). See Martin's page http://www.euclideanspace.com/maths/algebra/realNormedAlgebra/quaternions/functions/index.htm
for his description of the conjugate.
Did you get -q = (-s,-V) from one of Martin's pages? If so, where?.
2006-04-09 18:03:29 UTC in mjbWorld
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Let me add my support to Martin here...
I went researching this stuff a couple years ago and also was unable to find a clear definition of the "NASA coordinate system" on the web. I took my best guess based on what I could find and was also wrong. (I was close - I had the axes in the right orientation but used a left-handed system instead of a right-handed one. See...
2005-12-16 16:44:55 UTC in mjbWorld
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> Anyone know how to deal with this?
Yup: don't do it.
Once again you've run afoul of the fact that you can't readily do arithmetic with Eulers. Simple summing of the change in roll angle in the external frame of reference does not yield a meaningful result.
The first thing you need to do is to come up with a precise definition of what it is you're trying to compute and in what frame...
2005-10-31 18:44:41 UTC in mjbWorld
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The answer to the above two possibilities is "probably neither". The reason has to do with frame of reference.
Anthony, you haven't said so, but I'm going to make a guess that you want your "difference angle" (i.e., the correction quaternion) expressed in the frame of reference of the airplane. (After all, that's how control inputs are applied - ailerons apply a roll along...
2005-10-26 19:15:07 UTC in mjbWorld
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Anthony,
Reverse the order of multiplication and you'll get the right answer. You didn't give a formal expression of the quaternion product and I can't afford the time right now to reverse-compute it from your algebra. (Sorry - I know how tedious this stuff is. I've spent may hours cranking through multiplications.) However...
Going back to basic quaternion theory, the way you rotate a...
2005-10-24 13:58:15 UTC in mjbWorld
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Hi, Anthony...
As Martin says, once you're into quaternion form things are fairly simple. Given Q1 as your current attitude and Q2 as your desired attitude, the correction Qx you need (i.e., the difference between Q1 and Q2) is simply Q2 multiplied by the conjugate of Q1. To apply the correction gradually, you can divide the rotation angle represented by Qx by going back to the definition of...
2005-10-23 16:07:25 UTC in mjbWorld
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The basic definition of a rotation quaternion is as follows:
For a rotation r about some axis, the applicable quaternion (w,x,y,z) is (cos(r/2), sin(r/2)*(x,y,z)), where x,y,z is a unit vector pointing along the axis of rotation. When you make the axis of rotation one of the axes of your coordinate system, this simplifies to, for example for the X axis, (cos(r/2), sin(r/2), 0, 0).
For an...
2005-09-09 15:09:58 UTC in mjbWorld
