Theory: the moon, sun and stars are just reflections

Quote from: sceptimatic on July 24, 2014, 04:22:35 PM

Quote from: Shmeggley on July 24, 2014, 04:03:47 PM

But if you look at the Sun through a filter or make a projection you can see details, like sunspots, that look nothing like the Moon's surface, and furthermore these change over time while the Moon doesn't. I thought you were all about using your eyes and your brain Scepti; even a little kid can see the Moon and Sun are two different things!

Yea and if you look at the moon from the reflection of the sun you can actually seen the craters which is the inside of the sun glow which is (guessing) some kind of graphite core.

Show me where I can see this graphite core of the Sun which matches the features of the lunar surface.

Unless you have some unreal craft to take us tehre and go into it at the centre of Earth's circle, then I'm afraid you will have to view it from the reflection in the sky that you call a moon.

Quote from: rottingroom on July 24, 2014, 01:08:47 PM

Quote from: sceptimatic on July 24, 2014, 01:00:22 PM

This 9.8m/s/s makes no sense.

What sense is there to make about it? Objects literally fall 9.8 meters every second for every second. It's like saying that I have 5 fingers on my right hand. I say that cause my right hand quite literally has 5 fingers on it. I was able to measure exactly 5 fingers units. Similarly, one can quite literally measure that during, say a minute, an object will increase it speed by 9.8 meters during each second of that minute.

So what you mean is, an object falling will fall 588 metres in one minute or am I getting this wrong?

If it is free to continue accelerating then yes. It isn't once it reaches terminal velocity though

Quote from: sceptimatic on July 24, 2014, 04:14:03 PM

Quote from: rottingroom on July 24, 2014, 01:08:47 PM

Quote from: sceptimatic on July 24, 2014, 01:00:22 PM

This 9.8m/s/s makes no sense.

What sense is there to make about it? Objects literally fall 9.8 meters every second for every second. It's like saying that I have 5 fingers on my right hand. I say that cause my right hand quite literally has 5 fingers on it. I was able to measure exactly 5 fingers units. Similarly, one can quite literally measure that during, say a minute, an object will increase it speed by 9.8 meters during each second of that minute.

So what you mean is, an object falling will fall 588 metres in one minute or am I getting this wrong?

If it is free to continue accelerating then yes. It isn't once it reaches terminal velocity though

You're losing me. Explain this simply so I don't get confused.

The way I'm reading it is, 9.8m/s/s means that an object falling will fall at 9.8 metres every second it's falling. So if it's free to continue falling I could say, whoops there's 1 second gone and that's 9.8 metres covered...then I could say, there's 5 seconds gone and it's covered 5 x 9.8 metres which it's now covered 49 metres. Would this be right?

Quote from: rottingroom on July 24, 2014, 04:33:14 PM

Quote from: sceptimatic on July 24, 2014, 04:14:03 PM

Quote from: rottingroom on July 24, 2014, 01:08:47 PM

Quote from: sceptimatic on July 24, 2014, 01:00:22 PM

This 9.8m/s/s makes no sense.

What sense is there to make about it? Objects literally fall 9.8 meters every second for every second. It's like saying that I have 5 fingers on my right hand. I say that cause my right hand quite literally has 5 fingers on it. I was able to measure exactly 5 fingers units. Similarly, one can quite literally measure that during, say a minute, an object will increase it speed by 9.8 meters during each second of that minute.

So what you mean is, an object falling will fall 588 metres in one minute or am I getting this wrong?

If it is free to continue accelerating then yes. It isn't once it reaches terminal velocity though

You're losing me. Explain this simply so I don't get confused.

The way I'm reading it is, 9.8m/s/s means that an object falling will fall at 9.8 metres every second it's falling. So if it's free to continue falling I could say, whoops there's 1 second gone and that's 9.8 metres covered...then I could say, there's 5 seconds gone and it's covered 5 x 9.8 metres which it's now covered 49 metres. Would this be right?

No after 5 seconds its speed is now 49 m/s. Not the distance. It will have traveled much further than that since it was able to travel 49 m in just that 5th second.

I actually misread your first answer. I thought you said 588m/s but you didn't you said 588 m which is wrong.

Quote from: rottingroom on July 24, 2014, 04:33:14 PM

Quote from: sceptimatic on July 24, 2014, 04:14:03 PM

Quote from: rottingroom on July 24, 2014, 01:08:47 PM

Quote from: sceptimatic on July 24, 2014, 01:00:22 PM

This 9.8m/s/s makes no sense.

What sense is there to make about it? Objects literally fall 9.8 meters every second for every second. It's like saying that I have 5 fingers on my right hand. I say that cause my right hand quite literally has 5 fingers on it. I was able to measure exactly 5 fingers units. Similarly, one can quite literally measure that during, say a minute, an object will increase it speed by 9.8 meters during each second of that minute.

So what you mean is, an object falling will fall 588 metres in one minute or am I getting this wrong?

If it is free to continue accelerating then yes. It isn't once it reaches terminal velocity though

You're losing me. Explain this simply so I don't get confused.

The way I'm reading it is, 9.8m/s/s means that an object falling will fall at 9.8 metres every second it's falling. So if it's free to continue falling I could say, whoops there's 1 second gone and that's 9.8 metres covered...then I could say, there's 5 seconds gone and it's covered 5 x 9.8 metres which it's now covered 49 metres. Would this be right?

No, and this is basic high school physics. 9.8 m/s/s means that for every second the speed of an object falling will increase by 9.8 m/s. If it starts at zero, then it covers 4.9 m in the first second, 19.6 m in 2 seconds, 44.1 m in 3 seconds and so on. It will have a speed of 9.8 m/s after 1 second, 19.6 m/s after 2 seconds and so on (disregarding air friction). You're confusing speed and acceleration which are incredibly simple concepts and shows that you have no idea what you're talking about.

Quote from: sceptimatic on July 24, 2014, 05:22:34 PM

Quote from: rottingroom on July 24, 2014, 04:33:14 PM

Quote from: sceptimatic on July 24, 2014, 04:14:03 PM

Quote from: rottingroom on July 24, 2014, 01:08:47 PM

Quote from: sceptimatic on July 24, 2014, 01:00:22 PM

This 9.8m/s/s makes no sense.

What sense is there to make about it? Objects literally fall 9.8 meters every second for every second. It's like saying that I have 5 fingers on my right hand. I say that cause my right hand quite literally has 5 fingers on it. I was able to measure exactly 5 fingers units. Similarly, one can quite literally measure that during, say a minute, an object will increase it speed by 9.8 meters during each second of that minute.

So what you mean is, an object falling will fall 588 metres in one minute or am I getting this wrong?

If it is free to continue accelerating then yes. It isn't once it reaches terminal velocity though

You're losing me. Explain this simply so I don't get confused.

The way I'm reading it is, 9.8m/s/s means that an object falling will fall at 9.8 metres every second it's falling. So if it's free to continue falling I could say, whoops there's 1 second gone and that's 9.8 metres covered...then I could say, there's 5 seconds gone and it's covered 5 x 9.8 metres which it's now covered 49 metres. Would this be right?

No after 5 seconds its speed is now 49 m/s. Not the distance. It will have traveled much further than that since it was able to travel 49 m in just that 5th second.

I actually misread your first answer. I thought you said 588m/s but you didn't you said 588 m which is wrong.

Oh, ok, I see what you're saying. So it's a gain of 9.8 metres added for every second.

So for instance, a fall for 1 minute an object would be travelling at 588 metres per second? Am I getting this right?

Yes. That would be the speed. Not the distance which is more difficult to figure out. There is a formula that I could find for you if you want. It's an exponential function.

Keep in mind though that because there is an atmosphere there is a limit to how long the acceleration goes on for which is known as the terminal velocity at which point the acceleration is zero.

Yes. That would be the speed. Not the distance which is more difficult to figure out. There is a formula that I could find for you if you want. It's an exponential function.

Keep in mind though that because there is an atmosphere there is a limit to how long the acceleration goes on for which is known as the terminal velocity.

Ok fair enough. Keep me on the right lines. So the 588 m/s after 1 minute falling would equate to approximately 1315 mph, give or take the oddments. Would I be correct here?

Quote from: rottingroom on July 24, 2014, 05:43:12 PM

Yes. That would be the speed. Not the distance which is more difficult to figure out. There is a formula that I could find for you if you want. It's an exponential function.

Keep in mind though that because there is an atmosphere there is a limit to how long the acceleration goes on for which is known as the terminal velocity.

Ok fair enough. Keep me on the right lines. So the 588 m/s after 1 minute falling would equate to approximately 1315 mph, give or take the oddments. Would I be correct here?

Yes and I can already tell where you are going with this. I sense that you are preparing to completely ignore terminal velocity.

Quote from: sceptimatic on July 24, 2014, 05:47:32 PM

Quote from: rottingroom on July 24, 2014, 05:43:12 PM

Yes. That would be the speed. Not the distance which is more difficult to figure out. There is a formula that I could find for you if you want. It's an exponential function.

Keep in mind though that because there is an atmosphere there is a limit to how long the acceleration goes on for which is known as the terminal velocity.

Ok fair enough. Keep me on the right lines. So the 588 m/s after 1 minute falling would equate to approximately 1315 mph, give or take the oddments. Would I be correct here?

Yes and I can already tell where you are going with this. I sense that you are preparing to completely ignore terminal velocity.

No, I won't ignore anything. I'd like you to keep me right here. So what would terminal velocity actually be?

It is the velocity at which the sum of the drag force and buoyancy equals the downward force of gravity. At this moment the acceleration is zero and the speed becomes constant.

Quote from: Shmeggley on July 24, 2014, 04:24:56 PM

Quote from: sceptimatic on July 24, 2014, 04:22:35 PM

Quote from: Shmeggley on July 24, 2014, 04:03:47 PM

But if you look at the Sun through a filter or make a projection you can see details, like sunspots, that look nothing like the Moon's surface, and furthermore these change over time while the Moon doesn't. I thought you were all about using your eyes and your brain Scepti; even a little kid can see the Moon and Sun are two different things!

Yea and if you look at the moon from the reflection of the sun you can actually seen the craters which is the inside of the sun glow which is (guessing) some kind of graphite core.

Show me where I can see this graphite core of the Sun which matches the features of the lunar surface.

Unless you have some unreal craft to take us tehre and go into it at the centre of Earth's circle, then I'm afraid you will have to view it from the reflection in the sky that you call a moon.

Isn't the Sun also a reflection in your model? If so then why doesn't the Moon just look like a mirror image of the Sun? And if the dome is such a perfect reflector then why is the Moon so much dimmer than the Sun? As I said before, it's obvious to almost anyone that they are two different things. People have realized this for aeons before your alleged indoctrination ever started.

It is the velocity at which the sum of the drag force and buoyancy equals the downward force of gravity. At this moment the acceleration is zero and the speed becomes constant.

The downward force of gravity? I thought gravity was a pulling force from the centre of Earth and it gets stronger the closer you get?

For instance, let's assume near space as we are told. Ok, we are told it's a close vacuum and we are also told that gravity is weaker at that altitude, then gets stronger. So I'm a little bit puzzled by this bit.

Here's the problem I'm having, see if you can make sense of it.
We now know that 1 minute of falling at 9.8m/s/s equates to approximately 1315 mph.

What i'd like to know, is why doesn't this apply to Felix?



Now as you can see from this Baumgartner space jump, he is exactly 1 minute into his free fall in a near vacuum, falling with the gravity we are told about, in which he should be falling at approximately 1315 mph going by the 9.8m/s/s we are told about.
Something is clearly wrong with this as we can see that he is 586mph short of this speed and is only doing 729 mph.

Can you explain this?

Quote from: sceptimatic on July 24, 2014, 04:28:49 PM

Quote from: Shmeggley on July 24, 2014, 04:24:56 PM

Quote from: sceptimatic on July 24, 2014, 04:22:35 PM

Quote from: Shmeggley on July 24, 2014, 04:03:47 PM

But if you look at the Sun through a filter or make a projection you can see details, like sunspots, that look nothing like the Moon's surface, and furthermore these change over time while the Moon doesn't. I thought you were all about using your eyes and your brain Scepti; even a little kid can see the Moon and Sun are two different things!

Yea and if you look at the moon from the reflection of the sun you can actually seen the craters which is the inside of the sun glow which is (guessing) some kind of graphite core.

Show me where I can see this graphite core of the Sun which matches the features of the lunar surface.

Unless you have some unreal craft to take us tehre and go into it at the centre of Earth's circle, then I'm afraid you will have to view it from the reflection in the sky that you call a moon.

Isn't the Sun also a reflection in your model? If so then why doesn't the Moon just look like a mirror image of the Sun? And if the dome is such a perfect reflector then why is the Moon so much dimmer than the Sun? As I said before, it's obvious to almost anyone that they are two different things. People have realized this for aeons before your alleged indoctrination ever started.

You do realise how big the dome is, right?

Quote from: Shmeggley on July 24, 2014, 06:30:20 PM

Quote from: sceptimatic on July 24, 2014, 04:28:49 PM

Quote from: Shmeggley on July 24, 2014, 04:24:56 PM

Quote from: sceptimatic on July 24, 2014, 04:22:35 PM

Quote from: Shmeggley on July 24, 2014, 04:03:47 PM

But if you look at the Sun through a filter or make a projection you can see details, like sunspots, that look nothing like the Moon's surface, and furthermore these change over time while the Moon doesn't. I thought you were all about using your eyes and your brain Scepti; even a little kid can see the Moon and Sun are two different things!

Yea and if you look at the moon from the reflection of the sun you can actually seen the craters which is the inside of the sun glow which is (guessing) some kind of graphite core.

Show me where I can see this graphite core of the Sun which matches the features of the lunar surface.

Unless you have some unreal craft to take us tehre and go into it at the centre of Earth's circle, then I'm afraid you will have to view it from the reflection in the sky that you call a moon.

Isn't the Sun also a reflection in your model? If so then why doesn't the Moon just look like a mirror image of the Sun? And if the dome is such a perfect reflector then why is the Moon so much dimmer than the Sun? As I said before, it's obvious to almost anyone that they are two different things. People have realized this for aeons before your alleged indoctrination ever started.

You do realise how big the dome is, right?

How could anyone know that? You yourself have said you don't know.

Scepti. The modern theory of gravity is not a push or pull. It is an effect caused by the curvature of spacetime.

In any case moving on to your Felix problem. It seems that right after you agreed not to ignore terminal velocity, you decided to do exactly that.

So why does he max out at 729? Well, because that is the terminal velocity in that scenario.

Quote from: rottingroom on July 24, 2014, 05:54:46 PM

It is the velocity at which the sum of the drag force and buoyancy equals the downward force of gravity. At this moment the acceleration is zero and the speed becomes constant.

The downward force of gravity? I thought gravity was a pulling force from the centre of Earth and it gets stronger the closer you get?

For instance, let's assume near space as we are told. Ok, we are told it's a close vacuum and we are also told that gravity is weaker at that altitude, then gets stronger. So I'm a little bit puzzled by this bit.

Here's the problem I'm having, see if you can make sense of it.
We now know that 1 minute of falling at 9.8m/s/s equates to approximately 1315 mph.

What i'd like to know, is why doesn't this apply to Felix?



Now as you can see from this Baumgartner space jump, he is exactly 1 minute into his free fall in a near vacuum, falling with the gravity we are told about, in which he should be falling at approximately 1315 mph going by the 9.8m/s/s we are told about.
Something is clearly wrong with this as we can see that he is 586mph short of this speed and is only doing 729 mph.

Can you explain this?

RR, you called it:

Quote from: sceptimatic on July 24, 2014, 05:47:32 PM

Quote from: rottingroom on July 24, 2014, 05:43:12 PM

Yes. That would be the speed. Not the distance which is more difficult to figure out. There is a formula that I could find for you if you want. It's an exponential function.

Keep in mind though that because there is an atmosphere there is a limit to how long the acceleration goes on for which is known as the terminal velocity.

Ok fair enough. Keep me on the right lines. So the 588 m/s after 1 minute falling would equate to approximately 1315 mph, give or take the oddments. Would I be correct here?

Yes and I can already tell where you are going with this. I sense that you are preparing to completely ignore terminal velocity.

Quote from: sceptimatic on July 24, 2014, 06:35:57 PM

Quote from: Shmeggley on July 24, 2014, 06:30:20 PM

Quote from: sceptimatic on July 24, 2014, 04:28:49 PM

Quote from: Shmeggley on July 24, 2014, 04:24:56 PM

Quote from: sceptimatic on July 24, 2014, 04:22:35 PM

Quote from: Shmeggley on July 24, 2014, 04:03:47 PM

But if you look at the Sun through a filter or make a projection you can see details, like sunspots, that look nothing like the Moon's surface, and furthermore these change over time while the Moon doesn't. I thought you were all about using your eyes and your brain Scepti; even a little kid can see the Moon and Sun are two different things!

Yea and if you look at the moon from the reflection of the sun you can actually seen the craters which is the inside of the sun glow which is (guessing) some kind of graphite core.

Show me where I can see this graphite core of the Sun which matches the features of the lunar surface.

Unless you have some unreal craft to take us tehre and go into it at the centre of Earth's circle, then I'm afraid you will have to view it from the reflection in the sky that you call a moon.

Isn't the Sun also a reflection in your model? If so then why doesn't the Moon just look like a mirror image of the Sun? And if the dome is such a perfect reflector then why is the Moon so much dimmer than the Sun? As I said before, it's obvious to almost anyone that they are two different things. People have realized this for aeons before your alleged indoctrination ever started.

You do realise how big the dome is, right?

How could anyone know that? You yourself have said you don't know.

Just look around you and as you travel. It's a big dome. It doesn't matter about measuring. It's simply big.

Scepti. The modern theory of gravity is not a push or pull. It is an effect caused by the curvature of spacetime.

In any case moving on to your Felix problem. It seems that right after you agreed not to ignore terminal velocity, you decided to do exactly that.

So why does he max out at 729? Well, because that is the terminal velocity in that scenario.

Ohhh ok. 

Quote from: Shmeggley on July 24, 2014, 06:38:42 PM

Quote from: sceptimatic on July 24, 2014, 06:35:57 PM

Quote from: Shmeggley on July 24, 2014, 06:30:20 PM

Isn't the Sun also a reflection in your model? If so then why doesn't the Moon just look like a mirror image of the Sun? And if the dome is such a perfect reflector then why is the Moon so much dimmer than the Sun? As I said before, it's obvious to almost anyone that they are two different things. People have realized this for aeons before your alleged indoctrination ever started.

You do realise how big the dome is, right?

How could anyone know that? You yourself have said you don't know.

Just look around you and as you travel. It's a big dome. It doesn't matter about measuring. It's simply big.

OK then I'll consider it to be outside the observable universe, because its THAT BIG.

Quote from: rottingroom on July 24, 2014, 06:39:19 PM

Scepti. The modern theory of gravity is not a push or pull. It is an effect caused by the curvature of spacetime.

In any case moving on to your Felix problem. It seems that right after you agreed not to ignore terminal velocity, you decided to do exactly that.

So why does he max out at 729? Well, because that is the terminal velocity in that scenario.

Ohhh ok. 

Feel free to add something intelligent or even intelligible to this Scepti. 

Quote from: Macpie on July 24, 2014, 01:10:49 PM

Quote from: sceptimatic on July 24, 2014, 01:02:10 PM

Quote from: Macpie on July 24, 2014, 12:57:11 PM

Quote from: sceptimatic on July 24, 2014, 12:54:00 PM

Quote from: Macpie on July 24, 2014, 12:34:32 PM

Scepti, one of the most famous - for its innovativeness and precision, given the time when it was conducted - experiments related to gravitation is what is known as the Cavendish experiment. It was conducted right at the end of 1700's, and was able to provide the value of gravitational constant with a high precision given it all happened before even year 1800. It involved measuring the force twisting a wire upon which a set of lead orbs of known dimensions and mass was hanged, when a set of much heavier orbs was placed in their vicinity.

Have a read a bit about it if you have time to do this. How would you explain your "denpressure" causing the described behavior?

And, to keep it all on topic: if the Moon and the Sun are both reflections of the same light source(where the Moon is like a "second" bounce, if I remember correctly what you have said about it a while ago), how comes the Moon can even exist? If the dome is a perfect mirror, as in it reflects the light without scattering in other directions, and at any point of time the Sun is visible on any given point of the "dome" from some place on Earth(meaning that for any elevation and heading of your choosing relative to the surface, there is a place on Earth from which the Sun will be seen exactly at these coordinates), where on the dome is the "last" source for the light which's next "bounce" ends up as what we see as the Moon?

If all of the Earth's Sun is some reflection of the same single original light source, and the reflection from any point of dome ends up shining somewhere on Earth, where on the dome does the Moon's "previous" light actually come from? Or do you claim that light can "perfectly" bounce to several directions at once from a single point and direction?

Oh, and the last one about this: how does this match with the fact that relative positions of the Sun and the Moon change all the time?

The sun is on one side, high up and the moon is on the other side down low and vice versa.
At varying times you will see more reflections, known as sun dogs and such like. That's dependent on pressures and atmospheric changes, stuff like that.

I'm not sure I follow you here... Do you want to say that the Sun and the Moon are two different sources, kind of on the opposite sides of something, or that it is the same source reflected in a way that the Sun and the Moon appear on the sky on the opposite sides(like the Sun at day, and the Moon at night)? Or do I get you wrong?

No, you basically got it right.

Actually it was kind of an "either or" question - a) these are two separate sources, or b) one source, but two separate reflections. So it seems that it is one single source and sequential reflections perceived as the Sun and the Moon. You say that they appear on the opposite sides of the sky. Then why do we see the Moon at day half of each month?... Even right now, the Moon should be right next to the Sun for the next few days - till 2nd August, more or less. This is by no means "opposite".

Like I said, it depends on atmospheric conditions at the time. It's why you see it sometimes in day time. It is a dome after all, it's going  to produce varying reflections at times, just like when you see 3 suns and stuff. There's loads of weird stuff that can happen.

Wow. Are you saying that it is some pretty random atmospheric parameters, like maybe humidity or temperature, what causes:
1. An effect which is exactly the same regardless of your position on the Earth, altitude or local weather
2. Is cyclic, as perfectly cyclic as possible, allowing us to calculate the exact relative positions days, months or centuries ahead, and not once has it happened that the Moon was somewhere else than predicted.
What would this magical weather man be, who fine tunes the parameters to such a precise schedule?

Like I said, it depends on atmospheric conditions at the time. It's why you see it sometimes in day time. It is a dome after all, it's going  to produce varying reflections at times, just like when you see 3 suns and stuff. There's loads of weird stuff that can happen.

3 suns? WTF? You mean like a solar halo?

It's not weird at all and perfectly predictable.

Here's the moon phases:

http://www.moonconnection.com/moon_phases_calendar.phtml

Here's some upcoming comets in perihelion.

Comet Finlay
(Dec 27, 2014)

Comet Siding Spring
(Oct 25, 2014) Might even collide with mars but the odds are unlikely.

Comet Pons-Winnecke
(Jan 30, 2015)

Comet d'Arrest
(Mar 2, 2015)

Comet Churyumov-Grasimenko
(Aug 13, 2015)

If these were truly reflections why does these predictions work using Newtonian Physics regarding orbital paths of objects in the sky IOW Principia Mathimatica or Calculus?

Can you supply predictions such as these using denpressure and what not?

Name me one engineering project that uses gravity and I'll counteract it with denpressure.

How can you?  Engineers use maths to do their work, denpressure is just a bunch of words.

Anyway, here is some copy pasta showing the (very simplified) calculations for calculating the landing point of an artillery shell.  It is simplified, as it doesn't account for drag or the, eh, curvature of the earth.  Which actual gunners will need to do.  Anyway, please redo the equations using denpressure instead of gravity:

the cannon will shoot at a certain angle, a. It has a muzzle velocity of V, shown as the arrow at an angle.

Now, the velocity is a vector, meaning it has a speed and a direction. The velocity can be broken down into components, Vh and Vv, which are the vertical and horizontal arrows. The vertical and horizontal components are related to the velocity.

The components are orthogonal (at right angles), which means that a force acting in the direction of one component will not affect the other. Gravity is only vertical and so it does not affect the horizontal component.

So, let's start with the vertical component. We will focus only on the effect of gravity, since that is the main force acting on the shell. Gravity provides a constant acceleration, g, (the mass of the shell does not affect the acceleration). The relation between a constant acceleration and change in velocity (which we will call dv) are given as:
dv=g*t   or   t=dv/g

The bullet will reach the top of it's arc when the vertical velocity reaches zero, which is a change in velocity of Vv, and the time it takes is t=Vv/g.

Since the bullet is at the point only halfway to the target, the time to reach the ground is twice that, or t=2*Vv/g.

Looking now at the horizontal velocity, it doesn't have any forces on it in our simple example (this assumes no air resistance, etc...). So the acceleration is zero and the velocity doesn't change. For a constant velocity, the relation between position and time is:
d=v*t

Now, importantly, the time is the same time, so we can say:
d=Vh*t=Vh*2*Vv/g=2*Vv*Vh/g

So, if we know the vertical and horizontal components, we can calculate the distance. This is where trigonometry come in. Looking at the triangle, we can see that the components are related to the initial muzzle velocity and elevation angle:
Vh=V*cos(a)
Vv=V*sin(a)

This leaves us with the equation:
d=2*V*cos(a)*V*sin(a)/g

Combining terms, we get:
d=2*V^2/a*cos(a)*sin(a)

Using this equation, we can calculate the expected landing point based on the muzzle velocity and elevation angle"

...or something much simpler like water pressure.
pgh

Like I said, it depends on atmospheric conditions at the time.  It's why you see it [the moon] sometimes in day time.

Can you please explain what those "differences" are in the atmospheric conditions that cause the moon to disappear, or be apparent during the day?  Have you ever looked up at the sky from within a vertical mine shaft?

...just like when you see 3 suns and stuff. There's loads of weird stuff that can happen.

I assume you're referring to "sundogs"?



Are you not aware of what causes this phenomenon?  It's simply the sun's light interacting with atmospheric ice crystals.  Nothing "weird" about that at all to anyone vaguely familiar with optical theory.