Need to calculate Placidus houses from ascendant

[Cyril Gupta]

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

[Todd Carnes]

Post by Cyril Gupta
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

[Cyril Gupta]

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?

[Todd Carnes]

Post by Cyril Gupta
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.)

Post by Cyril Gupta
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. :)

Post by Cyril Gupta
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. :)

Todd

[Todd Carnes]

Post by Cyril Gupta
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?

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. :)

[Cyril Gupta]

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

[Todd Carnes]

Post by Cyril Gupta
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

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

Todd

[Todd Carnes]

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.

[Todd Carnes]

And one last thing...

No, I can't count.

LOL :)

[Kjell Pettersson]

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

Post by Todd Carnes
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.)
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.

[Todd Carnes]

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

[c***@gmail.com]

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

[unknown]

Post by c***@gmail.com
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

Speaking for myself, I'm not sure what you're asking. Do you mean how do
two signs get into one house?

[a***@gmail.com]

Post by c***@gmail.com
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

I hope this helps. What you are calling 'submerged', I think, is 'subtended'. And what you are calling 'two houses' are 'two signs'. In astrology, 'subtended' is called 'interception'. A sign is intercepted if no house has than sign on it's cusp. #1) The 'signs' do not have equal amounts of R.A. They are close to 2 hours, but the equinoctal signs are small and the solstices large. #2) 'Houses' are different in different systems, but the most common perception is time, not space. Say you using a system with basically 2 hours per house. Aries can go from totally nocturnal to diurnal in that time. It can 'fit' within a house completely. Cancer can easily not fit in one at all. And it gets worse. Due to the inclination of the ecliptic (apparent path of Sol) and the celestial equator (projection of the plane perpendicular to the axis of rotation of Terra), at latitudes further from the equator by more than 2e, some signs may not rise at all, depending upon the season of the year. They are 'anti-circumpolar'. And some never set either. A good place to start is Munkasey, but beware. His formulas are approximate and some erroneous. I use a different system altogether, geodetic, a modification of topocentric using calculations from Alova (q.v.). And if anyone can give me a rational, step by step, algorithm for Placidus, I'd appreciated it! No one who uses that can seem to even explain it. And yes, I'm a degreed programmer, and can do differential calculus, so make it math if you would? (because that's what it is!). Blow off that other 'post'. Snooty answers don't show intelligence, just a bad upbringing :)

[a***@gmail.com]

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)

[unknown]

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!
--
A. B.

Post by a***@gmail.com
<>

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.

Post by a***@gmail.com
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)

[b***@gmail.com]

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

Post by Cyril Gupta
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

[a***@gmail.com]

Post by b***@gmail.com
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 ...

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

[a***@gmail.com]

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.

[Todd Carnes]

Post by b***@gmail.com
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

Post by Cyril Gupta
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

I already addressed that here -> https://groups.google.com/d/msg/alt.astrology.moderated/XPBdb-jqu1I/kI_6PekxB90J

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

[a***@gmail.com]

Actual algorithm!

Post by Cyril Gupta
Hi there,
I am looking for some technical help for a horary software that I am
creating.
Regards
Cyril Gupta

The Placidus system uses the TIME it takes for the Asc to culminate and the IC to rise. ASCENSION IS A COMPLEX TRIG FUNCTION WITH NO INVERSE.

The rate of ascension is NOT A CONSTANT. And dependent on latitude and sidereal time as well.

If you really want to compute the algorithm, you need to advance the time of the chart by very small increments, until the Asc culminates (interpolation will give a finer gradient within the increment interval, arcseconds).

Now do it again, to 1/3 of that time span. The degree reached by the Asc is the 12th. Again, 2/3, the 11th.

Now nocturnal, with the time that it takes the IC to rise. 1/3 of that time span. The degree of the IC is the 3rd. Again, 2/3, the 2nd.

7 houses done. Remaining non-angular in opposition.

You will need a compiler, any will do.

FPC 2.6.4 is free and will do the job (Pascal). You can get formulas for accurate sidereal time from SidTimeEn.xls (downloadable), JPL, and USN.

Use the Lazarus IDE with the compiler. Far better than the one provided.

See Wikipedia as well for formulas by Lasker for obliquity.

And don't believe commercial software, they are inaccurate (but close: Janus, Astrolog, Solar Fire).

Good luck!