# How to calculate the moon's illuminated fraction tilt?

I use "Astronomical algorithms" book by Jean Meeus for programming algorithms for the moon data calculations. I've already created a lot of methods but stuck on the calculation of the moon's illuminated fraction tilt.

Here is the example of how the moon tilt looks like (when you switch date the moon disk turns and so does the illuminated fraction).

I've already calculated: illumination (48.1), phase angle (48.4), position angle of the illuminated bright limb (48.5), parallactic angle and etc. So I think I'm pretty close but still can't get the sufficient result.

I think that the tilt depends on the position angle of the illuminated bright limb but the tilt looks almost the same during the day and the position angle changes its value rather drastically.

I would appreciate any assistance.

UPD

I think I've found the right formula for the tilt calculation (page 347, and in the image attached):

ZenithAngle = MoonPositionAngle - ParallacticAngle.

But I'm not sure if I understand where ZenithAngle exactly is. I made a picture where I marked ZenithAngle (ZOC) and AlphaAngle which is actually the angle I need to calculate.

So if ZenithAngle is ZOC then AlphaAngle = ZenithAngle - 90.

Also I created a sample page with angles values, moon illumination and limb turned by the angle.

• I think the images on the website change rather artistically, not drastically! In other words, the rotation of the Moon from New to New is shown changing smoothly in the anti-clockwise direction. (The position angle of the axis increases.) Then just after New Moon, it magically starts back at the original position (so it jumps position) and repeats the anti-clockwise motion. At the same time, the position angle of the limb follows the same pattern. I highly doubt that your calculation of the limb position angle shows the same behavior. – JohnHoltz Oct 1 '20 at 17:23
• Between what reference points is your "tilt" angle measured? – Mike G Oct 1 '20 at 18:58
• @JohnHoltz updated the question and added my calculations. Can you please tell me where is the Zenith angle on the picture I attached? – Andrey Zagoruyko Oct 1 '20 at 22:19
• @MikeG added the picture with the angles. I think I've understood how to calculate the angle but still not sure about zenith angle position. – Andrey Zagoruyko Oct 1 '20 at 22:40

Then the angle of the bright limb relative to a horizontal line ($$\alpha$$) is PA-q-90.