Hi,i have another problem to solve in my solar system simulation.
I want to keep the distances and dimensions in the right scale.
8 billion km become 1 million units in Esenthel so moon diameter becomes 0.43 and the distance of the moon from the earth is about 45.
But the apparent size of the moon, as seen from the earth, is much smaller than it should be,so the moon seems farther (and consequently visually smaller).
I have played with camera parameters but nothing.
Is there a solution ? (keeping the real scales)
Thank you
This is a screenshot of the situation. The moon is the small sphere on the left,i think it should be bigger,even considering the distance.

Have you tried changing field of view (FOV)?
I am not sure that it's what you need, just saying.

Right!!! I didn't do the obvious thing . Thank you very much !

If is not a camera fov problem than based on your data, your moon is slightly smaller than the real thing but is also a tad near than the real thing, so it should look visually, similar to what it does here on Earth, the only reasons I can think that it does not, are the following.

One is a computer game and you are looking at it from a flat screen (I assume is not a VR game) and that changes the way our brain sees distant objects, two in the real world, looking at it from the ground (that is not the case here) the Earth atmosphere, acts slightly as a lens and enlarges the moon but is not a huge effect, third and more probable reason, for what we are seeing the so called "Moon Illusion" https://www.scientificamerican.com/artic...and-the-s/

Even thou, personally never saw the moon from space, so that could be correct...

IMO in this case the only thing you can do to get the effect you want, is to really fake the distances.

edit:: didn't saw your new comment but I will let mine stay, it has imo useful info, even thou in this case it seems, this was just a camera FOV problem.

Hi! Following the proportions i did (8 billion km = 1 million unit),distances and diameters are close to those I wrote before. For example the average distance of the moon is 384400 km, so this value becomes x = 384400*1000000/8000000000 = 48 ( ok not 45 ,but this doesn't change much)
The same calculations for moon diameter.
I find very interesting your observations on the atmosphere and the screen,I didn't think about these problems.
I have seen some images of the moon taken from the international space station and other sources. In some cases the moon is small even if not as in my image, in others it is bigger.
So i have decided to keep the right distances and scales, modifying the camera fov :-)
Thanks anyway :-)