The links are actually only random images from an image search with the terms “solar eclipse through tree leaves” and “iss in front of sun”.

I think you have in mind, that the rays are not parallel because they have to arrive at different positions. As you say, this is negligible, and it can even be avoided by the tilting of the mirrors. However, the rays start from different parts of the sun, and as the sun is huge, this angle is not so small.

I’ll try to explain it in more detail, sorry for the wall of text, it got longer than expected. In this case, we can use simplified ray optics and ignore the wave nature of light. This means, the light of the different mirrors or even pieces of mirrors just adds to one another. An important point, even if obvious, is that each point on the mirror surface can only have one orientation. Now we “select” the orientation of this point, we orient it in a way, that it reflects the rays from the center of the sun¹ directly on the sun.² But until now, we have ignored the rays which come from the rim of the sun. These rays start at a different position, namely the sun radius (695700 km). Due to different starting position, the rays have a different angle to arrive at the power plant, arcsin(sun radius / sun-earth-distance), which is 0.27°. Now we already oriented the mirrors parts in a way, that the rays from the center of the sun are reflected onto the satellite, but the rays from the rim of the sun come at an 0.27° differing angle. If the incidence of the ray on a mirror is changed by an angle, the outbound ray is also changed by the same angle. This leads reflected rays leaving in a direction 0.27° offset from the direction to the satellite. Assuming the satellite is at a height of 300 km and directly above, it is 300 km away, the smallest realistic distance. With this angle, it leads to a miss of plant-satellite-distance * sin(angle) leading to 1.4 km. This thought is valid for all points on the rim. Similarly, the rays between the rim and the center land between the satellite and 1.4 km off target. Hence the plant projects an image of the sun onto the satellite with a radius of 1.4 km.

¹ Well they actually do not come directly from the sun, they still come from very close to the surface, but they seem to come from the center of the sun and for rays it is not important how far they have already traveled. We can just assume the sun is a disc.
² If we assume the mirror is optimally shaped, we can reflect every ray, which seems to come from the center of the sun, perfectly on the satellite. Such a mirror would be part of an ellipsoid, with focal points at the center of the sun and the center of the satellite. In practice, it would be practically indistinguishable from a paraboloid with the satellite (deviance of 1.5 µm with a guessed plant size of 1 km). This is possible as the rays through the center of the sun falling onto the plant are, as you say, almost parallel.

Yes, you are right, considering the rays emerge from a point. And yes, each panel or all panels in unison can act like a magnifying glass. However, if they focus the light on a point at the height of the satellite, they work like a magnifying glass, or telescope with a focal length of the satellite – power plant distance, so at least 300 km. Considering the angular size of the sun, this telescope would lead to an image of the sun, the size of 3 km.

No sun rays are not parallel. If you looked at the sun (don’t, it will burn your eyes), would you see it as a point or a disc? As a disc. Why? Because even looking in slightly different directions, you see the sun. So the rays from the sun are not almost parallel, the rays from other stars are, they look point like.

Two interesting images for you: A solar eclipse viewed trough tree leaves: You can see the partial sun disc by using the small free points in the tree cover as pinhole cameras. Sure, the tree cover does not have lenses, but they only make the image sharper, not smaller. In this image the focal length is only the height of the trees and the image is already a few cm across. It also shows that the rays from the sun are not parallel. If they were, all rays going through the small free spots in the tree cover would end up at the same spot on the ground.

International Space Station, ISS, flying in front of the Sun: As the sun and the satellite are far away, we can assume that the angular size of the original sun and the virtual sun image are approximately the same when viewed from the power plant. Hence, this image shows how the mirrors would form an image of the sun, where only a small part of it hits the sun.
As the sun is much larger than the ISS, the angle of rays which come from the sun is much larger than the angle of rays which hit the ISS.

Yes i am, because it is unimportant if the light comes from the sun or the moon or a 3km large satellite (assuming they would have the same radiance). It would be important if the power plant were ten times larger, the satellite would be closer or larger. However in this case the limit to the power is is the etendue of the light at the satellite. The maximum power is the etendue at the satellite times radiant flux of the sun.
If you want a fun and interesting read which does explain a related “problem”, there is a relevant xkcd

I could explain it to you in at least five different ways in detail, three of them i have already done in short here in the comments. However, you never argued directly against my point. You don’t talk to me seriosly but laugh about it.
This is not what a serious disussion looks like like. If you want an explaination, i would be motiviated to take the time and explain it in detail.
Note that i listened to your point, considered it and argued why it plays no role. You have not considered my explainations.

No i am talking about all the mirrors as one surface, no matter they are really one or consist of small pieces

For the 65 W/m^2 i already used the size of the whole system, so all 10000 mirrors.

The sun has a angular diameter of 32 arcmin. (see here) Hence, the rays hitting one point of the one mirror, have come from different angles, namly filling a circle with this angular diameter. By reflection, the directions of the rays changes. But rays hitting the same spot on the mirror which were misaligned before by 32 arcmin are also misaglined by 32 arcmin after the mirror, independent of its shape. Therefore, the rays emerging from the power plant diverege by at least 32 arcmin. This is not a problem for operation, as this leads to a size of 4.6 m at an estimated maximum distance of 500 m between tower and mirrors. When the mirrors point at a satellite however, a distance of 300 km leads to a beam diameter of 2.8 km calculation

Even an ideal mirror can only project a point source onto a point. It is impossible to focus the rays of an extended source onto one point. See https://en.wikipedia.org/wiki/Etendue if you want to know details. With conservation of etendue you can also calculate this in a similar way.

The problem is the size of the sun. If you could look at the sun (don’t, try the moon its approximately the same size in the sky), you see it has a relatively large angular size. Its not just a point in the sky.

So the problem, the rays from one point of the sun are almost parallel. But the rays from the different points of the sun are not. So they also aren’t parallel after your mirror. They spread in an angle similar to the size of the sun on the sky. And this is much larger than a satellite. So you cannot focus all energy on a satellite.

It holds if the light spreads wider than the target. So also for directed light sources at large enough distances. Even a perfect mirror must spread the light in the same angle as it is incoming. Hence the beam would at least 3 km wide at the satellite. Therefore the satellite can only recieve a Illumination of ~65W/m^2 which is a few percent of the normal sun brightness of 1300 W/m^2 in space.

Another way to look at it, the mirrors cant make the sun seem brighter only larger. From the tower you see a large solid angle around you the mirror, therefore, it can seem like you are at the surface of the sun. However, fro. the position of a satellite, the power plant only takes a small solid angle, so it seems like a “smaller” sun. Assuming 400 MW and 1 kW/m^2 (at surface) solar power, it has an area of 400000 m^2, so a solid angle of 4.5e-6 sr from 300km while the sun has 70e-6 sr. So ten times smaller, therefore weaker. Note however here i did not account for attenuation in the atmosphere

It holds if the light spreads wider than the target. So also for directed light sources at large enough distances. Even a perfect mirror must spread the light in the same angle as it is incomming. Hence the beam would at least 3 km wide at the satellite. Therefore the satellite can only recieve a Illumination of ~65W/m^2 which is a few percent of the normal sun brightness of 1300 W/m^2.

Another way to look at it, the mirrors cant make the sun seem brighter only larger. From the tower you see a large solid angle around you the mirror, therefore, it can seem like you are at the surface of the sun. However, fro. the position of a satellite, the power plant only takes a small solid angle, so it seems like a “smaller” sun. Assuming 400 MW and 1 kW/m^2 (at surface) solar power, it has an area of 400000 m^2, so a solid angle of 4.5e-6 sr from 300km while the sun has 70e-6 sr. So ten times smaller, therefore weaker. Note however here i did not account for attenuation in the atmosphere

I cannot recommend mindfulness enough, as already suggested by many others in this thread.

I think, you said you are already in therapy? In this case, I would definitely talk with your therapist about this and things you want to adopt beforehand. If you want a simple concrete tip, you could try the “mindfulness coach” by the US department of veteran affairs. I liked it a lot and the apps from there get good privacy recommendations from mozilla.

I am a bit suprised by the many people recommending to just stop giving fucks. Is this what you really want? Or do you just want avoid the emotions of taking control?

You do not need to be that hard to yourself when your feeling “wrong”. Yes it is probably better for yourself if you don’t overreact. However you cannot really cotntrol your feelings. So it is still better to accept your anger. First, as you said, it drives up the frustration, because now you are also worried about your feelings. And second your original emotion wants to be “noticed”. I read and experienced a few times myself, the “wrong” emotion disappears often quickly when you accept it. It is an essential concept of mindfulness, to accept your emotions.

Edit: As far as i understand it and experienced it, saying to yourself “no i shouldn’t be angry about this” won’t change your thinking

Don’t be so pessimistic about it, i thought as you do. However, if you meet a person fitting to you, it is of no importance that you have no experience. I had my first date recently, probably ten years later than most. While i was embarrassed, it was no problem for her and she was very considerate to me.