I screwed it up with the triangles. I'm very ashamed. This just tells me I should move

on to administration because my mathematical skills are degrading. We're going to fix it

now. I was just missing one core argument in the big picture. If you have this simple

argument, then it's not that hard. That's what I'm going to explain to you now again.

Let me just go to the picture again. Listen, you will be surprised. Everything was going

great until Stefan screwed it up with some questions where I was totally confused. We

have parallel beam and we have fan beam geometry. Why is the light not on here? Let's draw the

situation again for fan beam. We have here our detector. We call it detector. Our detector

is characterized by, this is our coordinate system, by the rotation angle theta. That's

the rotation angle theta. Given the rotation angle theta, we index the pixels by s. That

means rotate the detector by theta. Then for a given theta, walk on this one straight line

to the point s. That's a 1D coordinate system. We have here the parallel projection lines.

For Ingvar, who is here with us the first time, these are the lines, the X-ray particles

are propagated along. Is there anybody in the audience except Ingvar who is not following

up on this? Nobody. Then we have fan beam. We have the following. We have our coordinate

system. We can move the coordinate system to a place that, again, the detector goes

through the origin. We call it, again, detector. Then we have an X-ray focus. We have an X-ray

focus. That means in the X-ray focus, all the X-ray particles are generated at a single

point. Then they are propagated through the object. It's a bundle of projection rays.

They intersect in the X-ray focus. If we consider the 2D situation, it's like a fan. That's

why it's called fan beam geometry. We have an X-ray focus. Then we have here our projection

rays. We measure the X-ray energy here. Quite often in CT systems, the detector is curved

that we have an equiangular sampling. If you have a flat detector, that means in the center

you have denser samples. The more you are away from the center, the more the projection

rays diverge from each other. The smaller or the sparser the sampling will be. We call

this detector I, like Ingvar. We call this detector here S, like Stefan, who screwed

it up. S, like screw it up. That was the situation we stopped last week.

Let's talk about the characterization of the fan beam geometry. We said that the angle

between the Y-axis and the central ray that intersects the detector in a rectangular angle,

so orthogonally, this is called beta. If I select a ray here, I call this angle here

gamma. If I give you a beta and a gamma, you know exactly what to do. You rotate your detector

such that the central ray that intersects the detector orthogonally has an angle beta

with the Y-axis. Then once the beta is selected, I give you a gamma. The gamma tells you how

much you deviate from the central projection ray. We call this intensity value here P of

S theta or theta S. This here we call gamma beta. Just for the ordering, I looked it up.

If I want to have this element on the detector, I have to find the rotation of the detector

such that I have this angle beta. Then I go for the ray gamma and I observe, was it gamma

beta or beta gamma? Gamma beta. Whatever I measure on the detector for parallel beam

and for fan beam, I can give you the theta and the S and you can look it up in all the

projections. I can give you the gamma and the beta and you can look it up for all the

projections. We have two bivariate functions, two functions in two variables that characterize

the situation. So far so good? Then let me just simplify the figure a little bit. I do

now the following. What do I do? I just clean it up a little bit that I have just this triangle

here. If you would still pay 500 euros, I would draw a new figure, but now you have

to deal with what you get here. This is the triangle I look at. Here is the orthogonal

relationship between the detector and the central ray. Max, do you agree so far? This

is the situation we have seen last week. Then I mixed the two figures without telling you

that I am mixing the two figures, but I was tremendously arrogant and was telling it is

so simple that my daughter can solve it and then I screwed it up and could not solve it.

You remember that was the situation last time. Think about the implication. That means if