Now you notice this definition doesn't specify s is given an m or

n frame components.

It just says it's a vector.

The t1 vector is equal to this vector.

So we can actually define these frames two different ways.

Here's a quick 3D visualization if this helps, I'm not sure.

But t1, magnetic was this one, yellow was there.

So the sun was there, this crossed this.

This one here would be be my t2 then.

So that's the t1, t2 and then t3 is orthogonal.

Rotating maybe helps but it's just 3D.

Yes?

>> And what do we do if s and m are colinear in this case?

>> You are a trouble maker aren't you?

No, absolutely.

So, that's an important thing.

If somebody tells me I'm in 3D tumbling in a gyroscope in this room, and

Warda is to my right and you're to my right,

both of you guys are exactly lined up with me, that doesn't help me.

because there's still the infinity of orientations that I have.

So, that's where these methods breakdown.

There is no estimation technique that will take and colinear observations,

I'll give you a full 3D attitude.

You will only always get a 2D measure of the attitude.

You will not know the rotation about the.

Yeah, so that's a fundamental thing we have to make sure.

So it's also an estimation.

That's where rate gyros come in nice, because if you're doing magnetic field,

it's possible the magnetic field at some point might line up with the sun at that

instant and then it changes again.

So you have to account for that in your code, yep.

But for now, we are assuming then to non colinear observations.

Yes, so that was something usually think people afterwards, but that's good.

You're thinking ahead.

So this matrix math, or this vector math we did earlier,

I can do in matrix components by assigning coordinate frames.

We measure s in the d frame, and we know s in the n frame.

I know where I am, I know right now in the inertial frame, the sun was in that way.

That's it, so I can go through this same matrix math in inertial components and

in matrix components.

And in defining the t1, 2, 3s in two different ways.

And now if you go back and look at the basic DCM definitions,

when we had the BN matrix, remember?

The BN matrix, or the C matrix as we called it at the time,

the first row was nothing but b1 transpose, right?

It was something n1, something n2, something n3.

That is b1 in n frame components.

Or vice versa the first column was n1, n2, n3.

So here we have bt, so it's like bn then I had n1, n2, n3.

You did your homework on that on Homework 1.

Here we need it but instead of n we have t, change letters that's it.

So this is going to be t1 in b frame components, the second column is going to

be t2 and the third column is t3, all in b frame components.

For the nt matrix it's the same stuff.

Now the t1s in the n frame components are going to be the first,

second and third of this matrix.

So, that's kind of the beauty of the triad method.

You can now, from the measurement how we define this, we get the coordinate frame

express in n, the t axis expressed at n and b frame components.

So we could the estimate b relative to t and the inertial t.

And the final step is attitude addition.

Now I have three frames, right?

I know how to go from n to b and b to f, I need n to f directly.

That is nothing but matrix multiplication.

So I'm looking for the attitude of n to my estimated body, right,

that's what I'm trying to go after.

And it's going to be nt transpose, because that flips the order and then tb.

And that's it.