# Accelerated reference frames equation derivation question

1. Feb 26, 2013

### charliepebs

My question regards how the approximation becomes an equality.

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2. Feb 27, 2013

### Fredrik

Staff Emeritus
I think you will need to provide more details. What's Q' supposed to be, and what's going on in the very first step?

Last edited: Feb 27, 2013
3. Feb 27, 2013

### charliepebs

Explanation of Diagram:

Q' is a representation of a 3D vector, vector Q, after a rotation of δθ about an axis, that axis being represented by vector n hat in the diagram. The angle between vector Q and vector n hat is represented as α.

Explanation of the very first step:

During the rotation depicted in the diagram, the arc traced by the head of the vector is a partial circle.
The partial circle has a center on the path of vector n hat, and a radius perpendicular to the path of vector n hat. therefore, the arc length traced by the rotation can be represented by the arc length formula for a circle in radians, s = rθ.

The equation in the very first step is based on two assertions. The first assertion is that when δθ is very small, the arc length can be represented as a vector, vector s, whose magnitude = rθ (from the arc length formula for a circle), and whose direction is in the direction of n hat cross Q. That vector is represented by the second part of the RHS of the very first equation, (|Q|sinα)δθ(in direction of n hat cross Q), where (|Q|sinα)δθ is the magnitude and corresponds to rθ in that |Q|sinα = the radius of the partial circle (r), and δθ = the angle displacement from Q to Q' (θ). The second assertion is that vector Q' can be represented as the vector sum of vector Q and vector s. However, given the stipulation of δθ being very small, this representation of Q' is regarded as an approximation.

4. Feb 27, 2013

### charliepebs

Also, zooming out a little, the diagram and proceeding equations and derivations are a part of depicting the rate of change of the principal axis vectors, which share an origin with the body axis vectors, of the accelerated reference frame.

Also, to clarify, my question is regarding why we are able to transition from an approximation to an equality later in the derivation.

5. Feb 27, 2013

### Fredrik

Staff Emeritus
OK, I understand. When dt is close to 0, we have
$$\frac{Q(t+dt)-Q(t)}{dt}\approx \frac{\delta\theta}{dt}\times Q =\frac{\theta(t+dt)-\theta(t)}{dt}\times Q.$$ If we take the limit $dt\to 0$, the left-hand side becomes dQ/dt, and the right-hand side becomes $d\theta/dt\times Q$. But it's far from obvious that ≈ becomes =.

To prove it, what you would have to do is to keep track of the error caused by the approximation $Q'-Q\approx \delta\theta|Q|\sin\alpha\,\hat e$ where $\hat e$ is a unit vector in the direction of $\hat n\times Q$. I'm not going to do that, because it would take too much of my time, so I'll just say that if you want to do it yourself, you should try to obtain a result of the form
$$\frac{Q(t+dt)-Q(t)}{dt}=\frac{\theta(t+dt)-\theta(t)}{dt}\times Q+\frac{E(dt)}{dt},$$ and then try to prove that $|E(dt)/dt|\to 0$ as $dt\to 0$. Note that you may not have to find an exact formula for E(dt). It would suffice to show e.g. that it's a sum of terms that all contain at least two factors of dt.

6. Mar 1, 2013

### charliepebs

thank you! so wonderful to have resources like this now days!

7. Mar 1, 2013

### charliepebs

I hope people appreciate that questions that could once stump someone for years can be cleared up in days!

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