posted October 03, 2016 09:52 PM            

quote:

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Originally posted by soren:
I just thought that they were misinformed of the actual geography of the whole eclpitic ring. I know as we stand on the horizon- it appears that as the sun makes its motion across the entire sky- and sets on the other side, somewhat opposite to its rising- that it seemed as if the rising and setting spots would be opposite in points to the signs.

You'd think the signs' would be centered around us. Like wherever you go- the signs are centered around you.

Edit: oh the signs likely are centered around us. But those 2 points rising and setting are not opposite. As you can see from the youtube

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?

Simply put, wherever two planes intersect, they create a 180° line.

So they choose to make a big deal of that horizon/ecliptic intersection, calling it the ASC/DSC axis.

posted October 03, 2016 09:57 PM            

quote:

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Originally posted by DopGang:
It's an interesting point.
I'm not sure though.
Like you say, it's not been anything other than 180 degrees.

Maybe we'd have to have a lengthy study on it. Get people to try it out.

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Yes...

But I'm still not clear on the math, because it seems like too much trouble, for too little gain in knowledge.

posted October 03, 2016 10:06 PM            

If what you at presenting is what you feel to be the missing link or an ah-ha moment, then your other recent theory on planets falling below or above the ascendant axis would be false. Right?

I appreciate your enthusiasm in looking for new ideas and love how excited you are with what you are discovering. It's obviously making a lot of people think here.

If you are born in Alaska and change your birth place to Uruguay, the ascendant sign changes. So location is factored in, at least for the ascendant.

Your theory now is questioning the descendant placement. And it has been represented in the past as the point opposite the ascendant point. But you are basing it on how long the daylight time would be in an particular point on the earth. Right? And that's not how the traditional sense of the ascendant has been viewed.

posted October 03, 2016 10:07 PM            

Actually- the ascendant in all charts is calculated as the exact spot on the ecliptic where it crosses the horizon (EG the same area where the sun rises each day, on your horizon, if you looked at your horizon in reality).

posted October 03, 2016 10:21 PM            

quote:

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Originally posted by soren:
@Faith those diagrams are the exact reason why i always imagined the ecliptic being exactly 180 degrees to the north of the horizon, and 180 to the south of it. Yet if that were true- I was surprised how the MH was rarely ever exactly square the ASC/DSC. Anyway from the youtube video, we see the diagram is just made to reach a broad audience.

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You mean 90° north of the horizon?

I never pictured it that way.

posted October 03, 2016 10:28 PM            

quote:

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Originally posted by soren:
the point *square* to where they cross

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posted October 03, 2016 10:31 PM            

How about instead of confusing novices here with overreaching statements requiring math you haven't yet worked out, you talk to a programmer/astrologer like David Cochrane and put this all to him?
https://www.youtube.com/user/DavidCochrane100

posted October 03, 2016 10:33 PM            

quote:

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Originally posted by soren:
It's not 90 degrees north of the horizon (zenith) its 90 degrees square the ascendant/descendant.

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posted October 03, 2016 10:34 PM            

quote:

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Originally posted by Kannon McAfee:
soren,

How about instead of confusing novices here with overreaching statements requiring math you haven't yet worked out, you talk to a programmer/astrologer like David Cochrane and put this all to him?
https://www.youtube.com/user/DavidCochrane100

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Don't refer to us as "novices" like we're not here.

posted October 03, 2016 10:35 PM            

quote:

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Originally posted by Kannon McAfee:
soren,

How about instead of confusing novices here with overreaching statements requiring math you haven't yet worked out, you talk to a programmer/astrologer like David Cochrane and put this all to him?
https://www.youtube.com/user/DavidCochrane100

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posted October 03, 2016 10:51 PM            

The sun rises. it takes exactly 24 hours to rise at that same spot again. Assuming the motion of the sun around the earth is constant- then 3 hours of time above the horizon, shows how much distance the sun moved. if the sun moves 360 degrees in 24 hours- assuming steady (same) motion, then if it moved those 3 hours above the horizon at the same speed, then 3 hours to 24 hours is the exact same ratio as x amount of degrees above the horizon, with 360 degrees.

I might be wrong. We'll be waiting on cochraine

more clues are that we can compare the times that the sun was located at the MH (square the asc and TRUE dsc) and IC, and DSC. if the sun is steady motion for all of them.

This is for a random city I chose:
Sunrise: 7:24...Sun at MH: 1:01
Sunset:6:38 PM...IC:1:02 AM
Following Sunrise:7:22

posted October 03, 2016 10:52 PM            

quote:

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Originally posted by soren:
Ah. Really sorry about that. When i said 180 north and south of the horizon, i wasn't talking about the MH/IC axis, i was talking about how much of the ecliptic path was above or below the horizon. From the antarctic video about 20 degrees was above the horizon, and the rest, 340, would be below it. So when i said 180 above and below, i meant yeah. Sorry about that.

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Gibberish

Your words were "why i always imagined the ecliptic being exactly 180 degrees to the north of the horizon, and 180 to the south of it."

Referring to angular degrees, not zodiac degrees.

posted October 03, 2016 11:03 PM            

quote:

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Originally posted by soren:
Ah. Really sorry about that. When i said 180 north and south of the horizon, i wasn't talking about the MH/IC axis, i was talking about how much of the ecliptic path was above or below the horizon. From the antarctic video about 20 degrees was above the horizon, and the rest, 340, would be below it. So when i said 180 above and below, i meant yeah. Sorry about that.

@Kannon, i'll send him a messaage

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posted October 03, 2016 11:03 PM            

I entered in Dikson , a very northernly town, into astrotheme. the results are interesting.

Rise:9:25...MH:2:25
Set:7:21..IC:2:25
Edit:Try the town Longyearbyen instead, doesnt glitch like the other one formerly mentioned
As you could see from the antarctic viedeo if you saw it, the ASC and DESC were not far from one another, say 30 degrees- and the MH was right in between. It would make sense that it would take the sun at MH 24 hours to traverse to the IC, as the MH and IC are the only truly constant opposing forces. Besides the vertex.

posted October 03, 2016 11:24 PM            

Longyearbyen:
(Ignoring asc/dsc)
MH:1:04 PM.......Vertex:5:07 PM
IC: 1:00AM......Anti-Vertex:8:56 AM (sun opposite vertex)

Hmm... the results are astonishing.

posted October 03, 2016 11:46 PM            

There's still a chance that somehow this isn't true, that the sun does change speed. I dont know. But so far, from the calculations below, its looking like my original hypothesis is likely correct. So i have jupiter conjunct my descendant by 2 degrees. That's why i'm so weird

rise: 7 AM
MH: 12:44 pm
set: 6:28 pm
IC: 12:44 am
next rise (for anyone whos wants various info): 7 am

Rise to MH: 5.73
MH to Set: 5.73
Rise to set: 11.47 (exactly double)
Set to IC (i know this is redundant but idc): 6.26
IC to rise: 6.26
set to rise: 12.53

All fits

Ok so for anyone who wants more understanding.

Watch the sun. Watch it rise at 5 am. Now watch it move aaalll the way across the sky. See the path it makes? The energy it engrains into the atmosphere/ethers tailing behind it? Thats the ecliptic. Notice how the sun leaves the horizon at 4 PM in this example. This means that the entire eclpitic is only above the horizon for the equivelant distance that the sun travels in 11 hours. (5am - 4PM)

posted October 04, 2016 12:13 AM            

The mundane Asc is indeed the meeting of the eastern horizon and the ecliptic (sun's path) -- a point in space which always has an opposite/contra-parallel point as defined in that exact moment. The time arc you've referred to has been a consideration in how to trisect the hemispheric semiarcs for the purpose of house definnitions (if using Koch, Placidus).

Put your math to someone who works intimately with it for purposes of programming software. David Cochrane is a great guy, on fb and youtube.

posted October 04, 2016 12:25 AM            

"The Northern Hemisphere is comprised of Houses 1 through 6—the lower half of the chart (below the horizon)."
-cafeastrology

Based on the descendant point you speak of-180 d opposite the ascendant- it has nothing to do with the horizon.

Anything above it does not precisely mark where the horizon is. It is an imaginary point. Who knows if it has any importance.

People are reading lies when they look above that point to find their "above horizon" planets. You need to find my point, sunset point, to find if planets lie above or below the horizon. It's the only way.

posted October 04, 2016 12:41 AM            

Edit: actually i have it all figured out now. $_$

ok. imagine the earth. and the sun. envision them both in front of your face, 2 marbles. Now imagine they are both fixed- non moving. The earth starts to spin. As it makes one full circle- at a steady constant speed- the sun will appear from its viewpoint to circle around it.

The reason why the sun rises at 7:04 AM one day, 7:08 the next day, and so on (goes back and forth eventually) is due to the EARTH orbiting the sun.

So its a 100% proven fact. My OP says the true way to find the true second point of the ecliptic that is crossing the horizon at birth.

This information is vital for: finding if you have any conjunctions
Attracting energy of that degree
Finding if planets are above or below the horizon (based if they are above or below this point)
Chart configurations (squares, conjunctions, etc)

Based on having 2 still dots, having one of them circle around- the second dot's path would stay steady- the same, and at a regular speed. GG BOYS

Edit edit: if you really want to see it for yoruself. take 2 small circular objects- like 2 cherry tomatoes. make a marking on one of them with a small piece of black tape or something similar. so you can view this point as it rotates

now keep both of the points fairly fixed, non wavering. now rotate the marked cherry tomato. You can watch it spin. Exact speed the whole way through. Just some parts of it may get more sun exposure.