m = <Dm*cos(ELm)*cos(AZm), Dm*cos(ELm)*sin(AZm), Dm*sin(ELm)>

s = <Ds*cos(ELs)*cos(AZs), Ds*cos(ELs)*sin(AZs), Ds*sin(ELs)>

where:

Dm is distance from earth to moon = 238,900 miles

Ds is distance from earth to sun = 92,960,000 miles

For my example, ELm=20.89, AZm=136.35, ELs=-0.24, AZs=294.62

m = <-160890, 154062, 85186>

s = <38726600, -84508300, -389400>

s cross m = <-7138932301000, -3236313581600, -7630242937800>

p = m cross (s cross m) = <-899841878721166000, -1835766873255628000, 1620528680300286000>

h = m cross z = <154062, 160890, 0>

p dot h = -433987971757638265212000

|p| = 2608805976556893954

|h| = 222757

(p dot h) / (|p|*|h|) = -0.74680042

taking the abs as per the convention in the paper to get the angle we want...

alpha = arccos(0.74680042) = 41.686 degrees!

The fraction of a degree we get in variance is due to different amounts of precision used in the calculations (round-off error).

If you object to the part of the math where we showed the distances are not necessary in this calculation, just forget we ever mentioned it. There's the math done with the actual distances. Can you accept it now?

Why doesn't the angle of the moon's phase change with that math when we place the sun one mile away from the earth and 92,960,000 miles away from the earth?

Surely the angle of the phase must change between those huge ranges.

At this point, it should be painfully clear to all readers that you are deliberately avoiding the suggestion that we move to the "hold up a ball" method instead of this math. You have repeatedly ignored the suggestion, and it should be very obvious to everyone that you are ignoring it. This strikes me as suspiciously deceitful. I trust that our readers see it this way as well.

Upon your insistence, back to the math. You ask, "Why doesn't the angle of the moon's phase change with that math when we place the sun one mile away from the earth and 92,960,000 miles away from the earth?"

Before I go any further, let's be clear. We are talking about the angle of the moon's tilt. We are not discussing the moon's phase.

Let's start with the 3D visualization.

At 41:40, I showed exactly what happens when we change those distances. In the left-hand view, we are looking at the triangle made by the Earth, Moon, and Sun. Right at 41:44, I change the distance to the sun from 100 to 1000 units. Notice that the angles of that triangle in the left-pane DO change. This is exactly what you would expect. (In your sample above, you are using a right-triangle calculator. This triangle is not likely to be a right-triangle. In my example it clearly is not... perhaps we can come back to that.) However, none of the angles of this triangle are the angle that we are looking for. None of these angles is the angle of the moon's tilt.

If we look in the right-hand pane, we'll see the angle we ARE looking for. The angle we are interested in is the blue line in the right-hand diagram. Remember what that blue line in the right-hand pane is? That's the line connecting the sun and the moon. Yes that IS one of the 3 sides of the triangle we were looking at in the left-hand pane. The difference here is that we're looking at that line from a different angle. In the right-hand view, we're looking up at the moon from our observer's location. Scroll back to 41:40 and watch as the distances are changed. Notice that the blue line in the right-hand pane gets skinnier, but its angle in this view does not change.

At 41:55 I increase the distance by another factor of 10 (for a total of 100 times further than it started). Again we see the triangle in the left-hand pane changes, but the angle of the blue line in the right-hand pane does not change.

So we've seen in the simulation that the angle of the moon's tilt doesn't change with those distances, but that answer may be unintuitive. That's understandable. But think about this. Back at 15:48, we took the intersection of 2 planes, and I talked about how that intersection would give us the apparent angle of the light hitting the moon. That may not be obvious, so let me walk through it slowly...

1) The light coming from the sun to the moon must lie within in the Observer-Moon-Sun plane. That seems fairly obvious. That light should be along the Moon-Sun edge of the triangle we used to make the plane.

2) The Observer is ALSO within the Observer-Moon-Sun plane. That also seems obvious. We used the observer as the 3rd point to create that plane.

3) What does a plane look like to a viewer inside the plane? It looks like an edge - a line.

4) So any part of that plane which lies in front of the observer has all collapsed into a single line. Any part of that plane is all the same line from our point of view.

5) If any part of that plane is all the same line, then the line from the sun to the moon looks the same as any other part of that plane when viewed edge-on like this.

6) If we were to move the sun-moon line to another spot within this same plane, that line would look exactly the same - the same as any part of the plane does.

And now with that clear, what happens to the observer-moon-sun plane when we change the distance to either the moon or the sun? If we keep the ANGLE between the observer and the moon steady, but change the distance to the moon. Also if we keep the ANGLE from the observer to the sun steady while changing that distance. Try it out on a table-top. Place 3 objects on a table-top and call the triangle they make the plane. Well the plane is the table-top of course. How does this plane change when you change the distance between your 3 objects? It's still the table-top of course. Changing the distances between the 3 objects does not change the plane that contains them.

7) So changing the distance from the earth to the sun does not affect the observer-moon-sun plane.

The plane viewed edge-on still looks exactly the same.

9) That plane viewed edge-on IS the apparent lighting direction from the moon to the sun when viewed by the observer.

10) And it is unaffected by the distances between the bodies - you cannot change the orientation of that plane without changing the angles between the observer and the other bodies.

Hopefully that helps. But we could approach this with pure math as well. Feel free to just stop reading at this point because math is pretty boring to most people, and I think I'm beating a dead horse by now...

Back to this equation:

cos(alpha) = (p dot h) / (|p| * |h|)

We got this far using the full vectors for m and s. We have included all the proper distances up to this point.

Now let's substitute in the values we have for m and s:

m = <Dm*cos(ELm)*cos(AZm), Dm*cos(ELm)*sin(AZm), Dm*sin(ELm)>

s = <Ds*cos(ELs)*cos(AZs), Ds*cos(ELs)*sin(AZs), Ds*sin(ELs)>

Let's pull out those distances as constant factors:

m = Dm*<cos(ELm)*cos(AZm), cos(ELm)*sin(AZm), sin(ELm)> = Dm*mhat

s = Ds*<cos(ELs)*cos(AZs), cos(ELs)*sin(AZs), sin(ELs)> = Ds*shat

Ok? Now let's do the rest of the derivation like that...

s cross m = Dm*Ds*(shat cross mhat)

p = m cross (s cross m) = Dm*mhat cross (Dm*Ds*(shat cross mhat)) = Dm*Dm*Ds*(mhat cross (shat cross mhat))

h = m cross z = Dm*(mhat cross z) (z already has a unit length)

p dot h = (Dm*Dm*Ds*(mhat cross (shat cross mhat))) dot (Dm*(mhat cross z)) = Dm*Dm*Dm*Ds*( (mhat cross (shat cross mhat)) dot (mhat cross z) )

|p| = |Dm*Dm*Ds*(mhat cross (shat cross mhat))| = Dm*Dm*Ds*|mhat cross (shat cross mhat)|

|h| = |Dm*(mhat cross z)| = Dm*|mhat cross z|

Now let's divide:

(p dot h) / (|p|*|h|) = Dm*Dm*Dm*Ds*( (mhat cross (shat cross mhat)) dot (mhat cross z) ) / (Dm*Dm*Ds*|mhat cross (shat cross mhat)| * Dm*|mhat cross z|)

Quick rearrange:

= (Dm*Dm*Dm*Ds / Dm*Dm*Dm*Ds) * ( (mhat cross (shat cross mhat)) dot (mhat cross z) ) / ( |mhat cross (shat cross mhat)| * |mhat cross z| )

Let's cancel the constants

= ( (mhat cross (shat cross mhat)) dot (mhat cross z) ) / ( |mhat cross (shat cross mhat)| * |mhat cross z| )

And BOOM! proved mathematically.

So let's get real here... that's some fairly advanced math. I don't expect everyone to follow it. What I do expect is for you to VERIFY it EXPERIMENTALLY. Measure the angles to the sun and the moon (or look them up). Crunch these numbers using the final equation from the paper. Then go outside and measure the tilt angle of the moon. Does it match? How close is it? You can do that much with no math background at all.

And please please remember... The answers do not always match up with your first guess. Sometimes nature can be surprising. Embrace it. Science is much cooler when you discover the world is trickier than you thought it was.