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Jun 16, 2008, 2:49:23 AM6/16/08

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Hi there,

I am looking for some technical help for a horary software that I am

creating.

In KP horary, we choose a random number and the value of ascendant is

derived on the basis of that number. We then need to calculate the

other houses with that ascendant in view. I need someone who can guide

me in calculating the placidus house systems with this fixed

ascendant, and using other data like latitude, etc.

If anyone of you great guys is willing to help, it'd be wonderful.

Even links to useful resources, and ideas will help.

Thank you in anticipation friends

Regards

Cyril Gupta

Jun 16, 2008, 8:33:52 PM6/16/08

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Cyril Gupta wrote:

> other houses with that ascendant in view. I need someone who can guide

> me in calculating the placidus house systems with this fixed

> ascendant, and using other data like latitude, etc.

Excerpts from Michael P. Munkasey's "An Astrological House Formulary"...

CALCULATION CONVENTIONS

The following standard abbreviations are used in the mathematics which

follow:

e represents the obliquity of the ecliptic

f represents the terrestrial latitude

ASC is the ascendant

MC is the MC

RAMC is the Right Ascension of the MC

F, G, J, K, and L are working terms, unimportant astronomically

+. -, x (or, *), ÷, = represent their normal arithmetic functions

SIN, COS, TAN, COT, etc. represent the trigonometric functions

For calculator purposes: COT = (1 ÷ TAN ) and vice-versa, etc.

ARCSIN, ARCCOS, ARCTAN, etc. represent the trig inverses

H11, etc. stands for the offset to compute the cusp of house eleven, etc.

C11, etc. stands for the value of the cusp of house eleven, etc.

Standard computer notation parenthesis nesting conventions are used

throughout the formulations. That is, three left parenthesis must be

balanced by three right parenthesis. Calculations are always performed

within the inner parenthesis first, and then outward to the outer

parenthesis. Persons attempting the mathematics herein should refer to

reasonable reference books if they are unfamiliar with trigonometric

procedures. Particularly, the process of adjusting house cusp

calculations for the correct trigonometric quadrant can be somewhat

tricky if not performed with care. House cusps which are over 360° or

under 0° should be converted to lie between 0° and 360° . That is, if

you compute a house cusp as being 372° then this should be changed to 12

Aries. Add 360° to any negative values or results. House cusps with

values between 0° and 29.99° lie in Aries; between 30° and 59.99° in

Taurus; between 60° and 89.99° degrees in Gemini, and so forth around

the zodiac and through the signs.

PRELIMINARY CALCULATIONS AND THE PERSONAL SENSITIVE POINTS

1. The RAMC (the right ascension of the midheaven) is computed from

Local Sidereal Time (LST) by converting time units to degree units. An

example of this calculation follows:

Given an LST of 12H 15M 00S, then first convert this to a decimal form

of time, or 12.25 hours. 12.25 x 15 = 183.75° which is the RAMC.

Given an LST of 6H 27M 14S, then convert this to a decimal form of time,

or 6.453889 hours. 6.453889 x 15 = 96.808333°.

2. MC = ARCTAN ( TAN (RAMC) ÷ COS e )

3. ASC = ARCCOT (- ( (TAN f x SIN e) + (SIN RAMC x COS e) ) ÷ COS RAMC)

4. EQA = ARCCOT ( - ( TAN RAMC x COS e) )

5. VTX = ARCCOT (- ( (COT f x SIN e) - (SIN RAMC x COS e) ) ÷ COS RAMC)

6. CAS = ARCCOT (- ( (COT f x SIN e) + (SIN RAMC x COS e) ) ÷ COS RAMC)

7. PAS = ARCCOT ( ( (TAN f x SIN e) - (SIN RAMC x COS e) ) ÷ COS RAMC)

8. ARI The Aries Point is always zero of Aries.

9. The declination of any point on the ecliptic can be calculated from:

declination = ARCSIN ( SIN (zodiacal longitude of point ) x SIN e)

10. The obliquity of the ecliptic, for any date in modern times, is

calculated by:

e = 23o 27' 08.26" - 46.845" x T - .0059" x T2 + .00181" x T3

where T is in fractions of a century starting from Jan 1, 1900

[snip]

THE PLACIDIAN HOUSE SYSTEM FORMULATION

1. Compute the RAMC, MC, and ASC in the normal manner. Use the MC as the

cusp of the tenth house and the ASC as the cusp of the first house. This

is a very fast converging algorithm adapted from a work by M.

Vijayaraghavulu.

2. Determine the following house cusp intervals:

H11 = RAMC + 30° H2 = RAMC + 120°

H12 = RAMC + 60° H3 = RAMC + 150°

3. Set the Semi-arc ratios:

F11 = 1 ÷ 3 F2 = 2 ÷ 3

F12 = 2 ÷ 3 F3 = 1 ÷ 3

4. Compute the cuspal declinations:

D11 = ARCSIN ( SIN e x SIN H11 ) D2 = ARCSIN ( SIN e x SIN H2 )

D12 = ARCSIN ( SIN e x SIN H12 ) D3 = ARCSIN ( SIN e x SIN H3 )

5. Compute the first intermediate values:

A11 = F11 x ( ARCSIN ( TAN f x TAN D11 ) )

A12 = F12 x ( ARCSIN ( TAN f x TAN D12 ) )

A2 = F2 x ( ARCSIN ( TAN f x TAN D2 ) )

A3 = F3 x ( ARCSIN ( TAN f x TAN D3) )

6. Compute the house cusp positions as follows:

M11 = ARCTAN ( SIN A11 ÷ ( COS H11 x TAN D11) )

M12 = ARCTAN ( SIN A12 ÷ ( COS H12 x TAN D12) )

M2 = ARCTAN ( SIN A2 ÷ ( COS H2 x TAN D2) )

M3 = ARCTAN ( SIN A3 ÷ ( COS H3 x TAN D3) )

7. Compute the intermediate house cusps:

R11 = ARCTAN ( ( TAN H11 x COS M11 ) ÷ COS ( M11 + e) )

R12 = ARCTAN ( ( TAN H12 x COS M12 ) ÷ COS ( M12 + e) )

R2 = ARCTAN ( ( TAN H2 x COS M2 ) ÷ COS ( M2 + e) )

R3 = ARCTAN ( ( TAN H3 x COS M3 ) ÷ COS ( M3 + e) )

8. Substitute: D11 = R11; D12 = R12; D2 = R2; and D3 = R3. Then repeat

steps 5 thru 8 again. Substitute the R's for the D's a third time and

repeat steps 5 thru 8. The answer for R on the third try is the cusp you

desire.

9. Compute the individual house cusps as follows:

C11 = R11 C5 = 180° + C11

C12 = R12 C6 = 180° + C12

C2 = R2 C8 = 180° + C2

C3 = R3 C9 = 180° + C3

End of excerpt...

Hope that helps. :)

Todd

--

Free Astrology Resources (FAR)

http://carnesoft.quotaless.com

Jun 17, 2008, 8:39:46 AM6/17/08

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I don't know Todd...

What I need to do is to relate the ascendant to this. Like for a given

Ascendant, I want to calculate teh position of the houses for a given

latitude.

Hmm.. I think I want to create something similar to one of those table

of houses books that are available in the market.

Is there anything that can help me go forward?

Jun 17, 2008, 8:19:16 PM6/17/08

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Cyril Gupta wrote:

> I don't know Todd...

>

> What I need to do is to relate the ascendant to this. Like for a given

> Ascendant, I want to calculate teh position of the houses for a given

> latitude.

> I don't know Todd...

>

> What I need to do is to relate the ascendant to this. Like for a given

> Ascendant, I want to calculate teh position of the houses for a given

> latitude.

The "real" ascendant is dependent on time & location. If you follow the

algorithm, you'll see that Cusp 10 (and thus Cusp 4 as well) are the

same regardless of latitude. The MC changes with longitude in the time

conversion step (because the time on our clocks is based on the

longitude) and the ascendant is "normally" based on the computed MC

(which means it's automatically corrected for longitude in that step as

well).

Only Cusps 1 & 7 actually have anything to do with the ascendant and

they don't change with the latitude.

The other cusps are corrected for latitude in step 5.

(I'm not happy with my wording above, I hope it doesn't make things more

confusing.)

>

> Hmm.. I think I want to create something similar to one of those table

> of houses books that are available in the market.

I'm curious. Have you done any programming before?

Once you've calculated the house cusps, creating a table should be easy. :)

>

> Is there anything that can help me go forward?

>

Well...

Since you're deriving your ascendant from a random number, you aren't

using the true ascendant when you calculate your houses. I think this

would through off all your houses anyway, so why don't you just

substitute the formula you are using to derive the ascendant in for the

formula that I gave you and just calculate everything else as given?

The majority of the calculations depend on the MC, so I "think" that as

long as you don't change the MC, you "should" be ok...but I'm not sure.

(Except, I'm guessing, the cardinal and succedant houses might get

squished or stretched, since you are altering the relationship between

the Ascendant and the MC). Why not give it a try and find out?

Since you're using a false ascendant, instead of the real one you're not

going to get "real" Placidus Houses anyway, so why not call them Faux

Placidus or maybe Gupta Houses or something. :)

Jun 17, 2008, 9:09:18 PM6/17/08

to

After re-thinking things a bit (and re-reading your original question)

I think what you want is as follows...

You have a given Asc and a known latitude/longitude.

1. Solve the Ascendant equation (Equation 3, in PRELIMINARY CALCULATIONS

AND THE PERSONAL SENSITIVE POINTS) for RAMC.

2. Plug the RAMC you just calculated into the equation for the MC and

find the corresponding MC.

3. Use the ASC, MC and RAMC that you now have to calculate all the other

cusps as per the algorithm I already gave you.

Todd

P.S. You're on your own with doing the trig needed to solve those first

two equations. :)

Jun 21, 2008, 9:12:28 AM6/21/08

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Hello Todd,

I solved the ascendant equation to derive the RAMC, and then I was

able to calculate the rest of the houses using the RAMC.

Yes, you are correct.

Thank you very much for your help.

Cheers!

Cyril Gupta

Jun 26, 2008, 6:14:11 PM6/26/08

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You're welcome. I'm glad I was able to help.

Todd

Feb 26, 2013, 4:33:33 PM2/26/13

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This thread is nearly 5 YEARS old, yet I still get repeated private emails because of it.

I would like to post a few new things to this thread...

1. I already know about the errors in the posted algorithm. (The most obvious is the arctan thing, but there are others.)

2. I know the corrected algorithm is freely available on the internet.

3. No, I won't help you find it. Use Google.

4. I will NOT help anyone else with this problem via private email anymore. (Everyone keeps expecting me to do all their programming work for them, but no one wants to pay me for it.)

3. No, I don't care if I sound like an @ss, I'm tired of being nice to people and getting taken advantage of.

4. If you absolutely, positively MUST have my help on this issue, I will help you, but only if you are prepared to pay me real money for my help -- IN ADVANCE.

I would like to post a few new things to this thread...

1. I already know about the errors in the posted algorithm. (The most obvious is the arctan thing, but there are others.)

2. I know the corrected algorithm is freely available on the internet.

3. No, I won't help you find it. Use Google.

4. I will NOT help anyone else with this problem via private email anymore. (Everyone keeps expecting me to do all their programming work for them, but no one wants to pay me for it.)

3. No, I don't care if I sound like an @ss, I'm tired of being nice to people and getting taken advantage of.

4. If you absolutely, positively MUST have my help on this issue, I will help you, but only if you are prepared to pay me real money for my help -- IN ADVANCE.

Feb 26, 2013, 8:43:30 PM2/26/13

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And one last thing...

No, I can't count.

LOL :)

No, I can't count.

LOL :)

Mar 1, 2013, 6:56:27 AM3/1/13

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I guess people find the thread via Google, as they are the ones who host older Usenet archives. The thought occurred to me that it might be possible for you to delete the post that is likely to lead people to email you by logging in to Google and then connecting your email -- if you still have the one the original post was made with - to Google. Google allows people to delete their old posts if made via Google, and perhaps they are satisfied with the account being connected.

As "on the Internet* everyone has ADD", it may be that your later post will not reach those most inclined to click "Reply to author" -- they may do it as soon as they feel the scent of a solution! -- which is why I thought I'd post this idea.

I cannot swear that deletion can be done if the original post was not also posted via Google's interface, but it may be worth a shot.

/Kjell

As "on the Internet* everyone has ADD", it may be that your later post will not reach those most inclined to click "Reply to author" -- they may do it as soon as they feel the scent of a solution! -- which is why I thought I'd post this idea.

I cannot swear that deletion can be done if the original post was not also posted via Google's interface, but it may be worth a shot.

/Kjell

Mar 9, 2013, 8:33:16 AM3/9/13

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I used to try to help everyone who asked, but it has simply become impossible to keep up...

Also, over the years people have become less and less civil about it. Instead of requesting my help, they now demand it as though I owed it to them. They act as if I have nothing better to do than work for them gratis.

Todd

Also, over the years people have become less and less civil about it. Instead of requesting my help, they now demand it as though I owed it to them. They act as if I have nothing better to do than work for them gratis.

Todd

Jun 28, 2013, 8:23:15 AM6/28/13

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dear Todd, Don't be pessimestic and harsh. Please remember beggers visit only the door from where they expect cherities. Consider the people who are ill equiped with lesa knwledge and be charitable to them.

Please explain as to how two houses gets submerged in one house in placidus system or suggest some article or literature.

I shall be deeply obliged if u clear my thid doubt.

yours

chawlatjs

Please explain as to how two houses gets submerged in one house in placidus system or suggest some article or literature.

I shall be deeply obliged if u clear my thid doubt.

yours

chawlatjs

Jul 1, 2013, 12:23:43 PM7/1/13

to

<chawl...@gmail.com> wrote on 28th June:

Speaking for myself, I'm not sure what you're asking. Do you mean how do

two signs get into one house?

--

A. B.

><>

My e-mail address is zen177395 at zendotcodotuk, though I don't check that

account very often.

Post unto others as you would have them post unto you.

two signs get into one house?

--

A. B.

><>

My e-mail address is zen177395 at zendotcodotuk, though I don't check that

account very often.

Post unto others as you would have them post unto you.

May 28, 2014, 4:00:01 AM5/28/14

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Jun 20, 2014, 12:30:02 PM6/20/14

to

Placidus Simple ...

We find the rising time of the MC (earlier)

We find the rising time of the IC (later)

Time of chart - rising time of MC = diurnal semi-arc

Rising time of IC - time of chart = nocturnal semi-arc

Add 2/3 of the diurnal semi-arc to the time of chart

Calc the Asc. It's the 11th house.

Add 1/3 of the diurnal semi-arc to the time of chart

Calc the Asc. It's the 12th house.

Subtract 1/3 of the nocturnal semi-arc from the time of chart

Calc the Asc. It's the 2nd house.

Subtract 2/3 of the nocturnal semi-arc from the time of chart

Calc the Asc. It's the 3rd house.

Since asc = atan(cos(ramc) / -(tan(lat) * sin(e) + sin(ramc) + cos(e)))

(and if the denominator is -, then add PI to the answer ...)

THERE IS NO one-to-one inverse function to determine the ramc needed

to produce the time of an objects ascension.

And this is dependent upon the latitude as well.

The simplest way is just to generate a temporary look-up table for the

latitude and interpolate. The table should be circular. 1 degree = 4 minutes.

Generate 145 records. One every 10 minutes and 1 more to make it

circular. The table values themselves are accurate to 1 degree of arc and

linear interpolation gives the output (on such a small dt, the function

approaches linear).

Stop fooling around. That's the way Placidus did it. Tables.

For the value of e, use the polynomial regression from JPL/NASA available

on the Web computed by difference from Julian Date (2000.5). Look it up.

You cannot determine obliquity if the latitude is closer to a Pole than 2e.

Unless ... You can use a quadratic regression of that (see Alova - Topocentrics)

We find the rising time of the MC (earlier)

We find the rising time of the IC (later)

Time of chart - rising time of MC = diurnal semi-arc

Rising time of IC - time of chart = nocturnal semi-arc

Add 2/3 of the diurnal semi-arc to the time of chart

Calc the Asc. It's the 11th house.

Add 1/3 of the diurnal semi-arc to the time of chart

Calc the Asc. It's the 12th house.

Subtract 1/3 of the nocturnal semi-arc from the time of chart

Calc the Asc. It's the 2nd house.

Subtract 2/3 of the nocturnal semi-arc from the time of chart

Calc the Asc. It's the 3rd house.

Since asc = atan(cos(ramc) / -(tan(lat) * sin(e) + sin(ramc) + cos(e)))

(and if the denominator is -, then add PI to the answer ...)

THERE IS NO one-to-one inverse function to determine the ramc needed

to produce the time of an objects ascension.

And this is dependent upon the latitude as well.

The simplest way is just to generate a temporary look-up table for the

latitude and interpolate. The table should be circular. 1 degree = 4 minutes.

Generate 145 records. One every 10 minutes and 1 more to make it

circular. The table values themselves are accurate to 1 degree of arc and

linear interpolation gives the output (on such a small dt, the function

approaches linear).

Stop fooling around. That's the way Placidus did it. Tables.

For the value of e, use the polynomial regression from JPL/NASA available

on the Web computed by difference from Julian Date (2000.5). Look it up.

You cannot determine obliquity if the latitude is closer to a Pole than 2e.

Unless ... You can use a quadratic regression of that (see Alova - Topocentrics)

Jun 21, 2014, 11:10:08 AM6/21/14

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Just to say, I can't comment on this because I still don't understand house

calculations myself; but I know people come through this group who do, so

thanks for posting!

calculations myself; but I know people come through this group who do, so

thanks for posting!

--

A. B.

><>

My e-mail address is zen177395 at zendotcodotuk, though I don't check that

account very often.

Post unto others as you would have them post unto you.

<adb...@gmail.com> wrote on 20th June:
A. B.

><>

My e-mail address is zen177395 at zendotcodotuk, though I don't check that

account very often.

Post unto others as you would have them post unto you.

Oct 19, 2014, 6:40:03 PM10/19/14

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Hello Cyril,

I cannot get this to work, despite the info of Todd Carnes. I hope you can help me and hopefully can send me the file in Excel.

Thanks

Hans

Op maandag 16 juni 2008 08:49:23 UTC+2 schreef Cyril Gupta:

I cannot get this to work, despite the info of Todd Carnes. I hope you can help me and hopefully can send me the file in Excel.

Thanks

Hans

Op maandag 16 juni 2008 08:49:23 UTC+2 schreef Cyril Gupta:

Jun 20, 2015, 5:10:02 PM6/20/15

to

On Sunday, October 19, 2014 at 6:40:03 PM UTC-4, bier...@gmail.com wrote:

> Hello Cyril,

> I cannot get this to work, despite the info of Todd Carnes. I hope you can help me and hopefully can send me the file in Excel.

Your answer is here ...
> Hello Cyril,

> I cannot get this to work, despite the info of Todd Carnes. I hope you can help me and hopefully can send me the file in Excel.

http://vytautus.com/files/File/pamoka.pdf

Jun 20, 2015, 5:10:05 PM6/20/15

to

The correct algorithm is given here ...

http://vytautus.com/files/File/pamoka.pdf

And it works just fine.

Make sure for south that you compute the asc and mc normally,

but for the other houses, you add PI to the ramc, and then

add PI again to the output (11th, 12th, 2nd and 3rd), calculating

as if for north (use a positive latitude).

Do each house individually.

It will take about 10 loops to converge to a minima.

Jun 29, 2015, 11:50:02 PM6/29/15

to

I suggest using the link to the pamoka.pdf file that has been given.

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