Farside buildings

[ArMaP]

I can lead you to water Armap, but I can't make you think!

I think that the problem is that you cannot make me think like you would like me to think.

My thoughts are my own, like yours are yours alone, although all our thoughts may be influenced by someone else's thoughts.

[ArMaP]

Armaps course to the moon

Wrong again, I am not proposing any course, I was asking what was the course because I don't see it likely that they turned an orbit around the Earth into a straight line to Moon, then change it again to orbit the Moon, but apparently nobody here knows what was the real course.

[ArMaP]

P.S. CAD stands for Computer Aided Drafting not Drawing.

I always thought that it meant Computer Aided Design.

At least that's what I remember from first hearing about that type of software, some 20 years or more ago.

[deuem]

ArMap, some place on line there is a list of the coordinates for the entire Apollo 11 flight path. I saw it once but forgot to bookmark it. You might have to drive down to it. I can not get it on a search. You could help out here and locate it, then post the link. Maybe it was minute by minute or hour by hour, I forget.

When we finish with the gravity drawings either Sarge or I could lay in the path and make a movie or gif to watch. Most likely any satellite or craft to the moon would take a very similar flight path so I bet we could use it for them all. At least as a start, then correct as needed.

Deuem, going off to home, good night all.....

[ArMaP]

Can you think of any reason why they would not take a straight path once out of Earth's orbit except when approaching lunar orbit to fine tune the trajectory?  Any other path would be stupid surely?

What I am thinking is that they started by orbiting the Earth, right? From that, what would they do, a 90º turn to the Moon, or would they make more and more elliptical orbits to get to a point from where they would accelerate to move away from Earth orbit?

I am thinking about something like this:


Was the real route something like it?
If it was, was it enough to give different results, as the position may not have been exactly between the Earth and the Moon?
How do we get the neutral point for positions out of that straight line between the Earth and the Moon? It shouldn't be difficult to get, for those that know how to do it.

PS: yes, I know, my drawing abilities are not good at all. In things like this I work better with a compass and some rulers or set squares or whatever they are called than with a computer. D

Edit: now I am proposing a route.

[Linda Brown]

I can't get any of these videos but it seems to me from what little I understand that it only makes sense to sort of "use the inertia of the orbit to " sling" a capsule out toward the moon.... In fact wasn't that one of the main talking points for having satellites in the first place..... that it would provide that Jump out spot for efforts bound to the moon?

I will leave you guys to all of these videos. Too much input for me at once!!!  Linda

[Pimander]

Hey Sarge, unless one of the big guns steps in and adds the Sun, I think we can do this test with just the Earth/Moon plane.

Thee objects are always in the same plane relative to one another (i.e. a triangle), although the may move in a third dimension relative to that plane.  The moon does not rotate in the same plane as the Earth's orbit around the sun otherwise we would have an eclipse twice a month.

[Pimander]

What I am thinking is that they started by orbiting the Earth, right? From that, what would they do, a 90º turn to the Moon, or would they make more and more elliptical orbits to get to a point from where they would accelerate to move away from Earth orbit?

I am thinking about something like this:

That is pretty much what I would expect, with maybe minor adjustments when approaching orbit.

I am going to take a look at the relative gravitational effect of the Sun on Earth relative to the effect of the Moon on Earth.  I suspect the third body (Sun) will have a negligible effect on the calculation so isn't especially significant when calculating the Lunar gravity.  BRB

[Sgt.Rocknroll]

I always thought that it meant Computer Aided Design.

At least that's what I remember from first hearing about that type of software, some 20 years or more ago.

When I got into the business in 1978, it was referred to as Drafting because that's what we were using it for. Major companies like McDonnell Douglas Automation Company (McAuto) were using it more for design purposes, hence the CADesign....it just was called by what you were using it for....
The link below is a pretty good history of CAD/CADAM...(I started using the Autotrol-AD380 system.) But if you read the article you'll see that CAD was around since the 50's....

http://mbinfo.mbdesign.net/CAD-History.htm

[Pimander]

The sun and the earth
Earth: mass 5.97*10^24 kg
Radius 6,380,000 meters
Sun: mass 1.99*10^30
Radius 696,000,000
Distance Between Sun &Earth: 1.5*10^11 meters (149,668,992,000 meters)

(5.97*10^24)/1.5*10^11

=3.98*10^13


The moon and the earth
Moon: mass 7.35 *10^22 kg
Radius 1,740,000 meters
Earth: mass 5.97*10^24 kg
Radius: 6,380,000 meters
Distance between moon and earth: 384,000,000 meters

5.97*10^24/384,000,000
= 1.55*10^16


Effect Sun/Effect of Moon
(3.98*10^13)/(1.55*10^16)=0.002567742

Put another way the Moon has about 500 times the effect on Earth as the Sun does.  Yes the distances between the three bodies change in a relatively small way but that would not effect the result all that much.

I suppose I really need to give this more attention.  It does look as though using this method suggested by John that the Moon's gravity would be approximately 0.7 relative to Earth's.  Not exactly but nowhere near the NASA figure of 0.166667.  The figure is about 4.2 times that of the NASA one.

Food for thought....


F = G m M / r2

F = m a


I suppose I should add.... If the NASA figure is wrong, any Lunar missions would have shown that CLEARLY.

[zorgon]

Hmmm I dunno...  I don't think any supposed Apollo flight could have got there with either of those courses


India claims they did it THIS way



But still waiting for the good pictures

[deuem]

What I am thinking is that they started by orbiting the Earth, right? From that, what would they do, a 90º turn to the Moon, or would they make more and more elliptical orbits to get to a point from where they would accelerate to move away from Earth orbit?

I am thinking about something like this:


Was the real route something like it?
If it was, was it enough to give different results, as the position may not have been exactly between the Earth and the Moon?
How do we get the neutral point for positions out of that straight line between the Earth and the Moon? It shouldn't be difficult to get, for those that know how to do it.

PS: yes, I know, my drawing abilities are not good at all. In things like this I work better with a compass and some rulers or set squares or whatever they are called than with a computer. D

Edit: now I am proposing a route.

I think if you change 2 things, you will be much closer. #1 you entered the moon on the wrong side ( if drawn looking down on the north pole ) The moons right side was facing the sun and they entered on the dark side and landed on the bright side facing Earth. #2 On leaving Earth I would project this in an up and over the pole area and out. In you photo about 1/3 to the right of center to avoid going through the worst of the belts.  If you can get that data I was talking about, we can plot it like they wrote it. In your picture and every book photo it looks like they left at the equator. This was not what was printed. It is just easy to draw.
Deuem

[deuem]

Deuems Step 2

Some standards I used and actual screen shots from the CAD model.

The Sun, diameter used 1,392,000 km



The Earth, diameter used 12,756 km



Sun to Earth, distance used 149,668,992 km

At this point in time we need to start positioning the moons in location to a date.

Remember the New Moon was on the 14 of July, so on the Sun Earth line, on the Sun side we place the 14^th.New Moon.

The Full moon was on the 29^th of the month so on the back side away from the sun we place the 29^th Full Moon on that same line. Both on the straight line through the Earth to the Sun. Neither of these placements mater right now because we have to wait for the elliptical orbit to be fixed, then moved to correct placements.

The Moon, diameter used 3,480 km



If I take the amount of days between 14^th New Moon and 29^th Full Moon, one can calculate the distance between them in the 180 degree arc and draw in day lines and numbers. Then do the same for a few days before the 14^th.  When the 13^th has been located, we can add in Apogee. Draw from Earths center. The Sun is to the left. Looking at the north pole.



Apogee Radius, distance used 203,301 km



Locating the 28^th on the orbit will give us the location of Perigee. Drawn from Earths center.

Perigee Radius, distance used 178,963 km



Apogee to Perigee, also known as the "Apsides line", distance 382,263 km



Apsides center line off set to the center of Earth, distance 12,169 km



With the Apogee and perigee known, we can calculate the average and use that for the third height of the ellipse. The center of the orbit ellipse worked out at the center of the Apsides.

Then day 19.5 was calculated and the distance found. Located its position on the ellipse.



So after all of that I have a distance center to center on the 19^th of July at 193,674 km

Call this "cM2cE" ( center moon to center earth )

For the distance Surface to Surface ( Ms2Es ), subtract both radius cM2cE-(Mr+Er) = 185,554 km.  193,674-(6380+1740)=185,554 km     

So when they say they have reached the neutral point is it based on the center to center numbers or the surface to surface numbers?  I would like to stay at center to center if we can. That way the math is relevant to a distance and not a size and distance.

It looks to me that the Neutral point is going to follow the Elliptical orbit also. It has to change from minute to minute. Also the Apsides line rotates counter clock wise at a very slow rate. One revolution in 3,232.6 days or 8.85 years. About 15 km per hour. So this model is only good for this Apogee/Perigee. The next one would change the numbers.

With this math model I get an average Moon Orbit circumference of 1,200,914 km / 28days or 672 hours. The moon is moving around 1,787 km per hour. 30 km per min and 0.5 km per second. So if you are entering orbit you are hitting a moving target moving this fast. So that moon is coming at you at over 1,000 miles an hour. Speeding ticket for sure! When we add the closing rates for a craft or satellite it adds up fast. One wrong move and you pass the moon or you splatter all over it. Just thinking out loud for a moment, I would think the return trip would be much safer. You would be shooting at a stationary target.

Please feel free to add or pick the math apart, I make mistakes too. If this post is too large a mod can move it to my research thread and I can just post the answer.

Can one of you take the distance center to center on the 19^th of July at 193,674 km and work out the neutral point?

Deuem

[ArMaP]

It is just easy to draw.

And that's why I did the drawing that way, it was easier and faster, and I only had to use two circles, three circumferences and one straight line. 

[ArMaP]

For the distance Surface to Surface ( Ms2Es ), subtract both radius cM2cE-(Mr+Er) = 185,554 km.  193,674-(6380+1740)=185,554 km     

So when they say they have reached the neutral point is it based on the center to center numbers or the surface to surface numbers?  I would like to stay at center to center if we can. That way the math is relevant to a distance and not a size and distance.

In a case like this, should the barycentre be used instead of the centre of the Earth, as that is the real centre of gravity of the Earth/Moon system?