# Angle setting on HH motor

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#### Blake Lane

##### Member
Hello,
I search the mathematical formula that determines the Angle setting on HH motor. This angle is different from the azimuth angle in the case of a fixed installation.

At Belfast for example, the satellite Astra 19,2 ° East
the Azimuth angle: 150.08 ° (29.92 East) and angle setting on HH motor : 27,43° East

You can see this type of calculation to the following address: _http://www.pteelien.net/c50b

Blake #### futumsch

##### Regular Member
Hello Blake

Azimuth angle can be calculated by:

180+ ArcTan(Tan (satellite longitude - site longitude)/sine site latitude)

a tip: keep your longitudes in degrees west, i.e. so Dubai might be 305 west for example. and BADR4 at 26E would be 334 degrees west.

Remember to convert your degrees to radians especially if you are using Excel.
This assumes that the satellite has 0 degrees inclination. And this also assumes tha the Earth is a perfect sphere, which it isn't.
This is going to give you azimuth values that you might expect from the majority of satellite pointing angle sites like _dishpointer dot com, which are good enough for small diameter dishes with 1-2 degree beamwidths.

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#### Blake Lane

##### Member
Hello futumsch,

Thank you for your answer but unfortunately that is not the formula that I want.

The calculation that I search for the position of the HH Motor arm. If you look at the position it where you oriented your antenna you will see that the scale is different from the calculation of the azimuth.

That's why I gave the example of a instalation in Belfast with an antenna pointed at Astra 19.2 ° E.
With the calculation : 180+ ArcTan(Tan (satellite longitude - site longitude)/sine site latitude). you get: 150.08 degrees East, while with the calculation HH STAB you get: 152.57 degrees East.

The difference is that the HH Stab is not flat but inclined of the value of the latitude.

Blake #### Huevos

##### Satellite Freak
There is a calculator at _http://www.satsig.net/sf.htm

One of the things it works out is polar mount rotation.

The code below is what it uses so you should be able to fish out the formula from that:
Code:
``````function rota(satlong,eslat,eslong) {
pi=3.1415926;
horb = 35786;
alfa=eslat*(pi/180);
omega=satlong*(pi/180);
mu=eslong*(pi/180);
beta=omega-mu;
os=rcentrum;
oc=rm*Math.cos(beta);
ac=rm*Math.sin(beta);
cs=rcentrum-oc;
distx=Math.atan(ac/cs);
landa=beta+distx;
if (Math.abs(landa*180/pi) >90) return "n/a";
esrota=landa*180/pi;
return esrota;
}`````` #### Robbo

##### Retired Mod
Thanks, for that, it works very well, I have just input the equation into my favourite programming language. Also as expected if you input a theoretical 90 degrees latitude, and 0 degrees longitude the result is the orbital position of the satellite. #### Huevos

##### Satellite Freak
Robbo said:
Thanks, for that, it works very well, I have just input the equation into my favourite programming language.
Is that PHP? If so I wouldn't mind having a copy.

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#### Blake Lane

##### Member
Huevos said:
There is a calculator at _http://www.satsig.net/sf.htm

One of the things it works out is polar mount rotation.

The code below is what it uses so you should be able to fish out the formula from that:
:-worshipThanks thanks thanks and.......... thanks Blake #### Robbo

##### Retired Mod
Huevos said:
Is that PHP? If so I wouldn't mind having a copy.
No, 'fraid it's not in PHP, Labview actually, I'm not a proper programmer. B

#### Blake Lane

##### Member
This algorithm is one used by motor equipped with the function USALS and Goto X.

Blake #### Huevos

##### Satellite Freak
Blake Lane said:
This algorithm is one used by motor equipped with the function USALS and Goto X.
Not the motor. The receiver works it out and sends the angular position to the motor. #### futumsch

##### Regular Member
Well, I like a challenge and with half an A4 notepad full of spherical trigonometry trying to work it out directly, I gave up. I could never work out inverse cosines anyway.

Then I opened my 'Practical Astronomy with your Calculator' by Peter Duffett-Smith. In there, is a chapter on horizon to equatorial coordinate conversion.

First find declination;

d = Arcsin ((sin a * sin L) +(cos a * cos L * cos A))

Then find Hour Angle (Polar angle),

H = ArcCos ((sin a - sin L * sin d) / (cos L * cos d))

Where
a=satellite elevation
A=satellite Azimuth
L=Observer Latitude

Then find Sine of the Azimuth. If sin A is postive, then the answer is 180-H. I can get the sums to work fine if the satellite is East of the observer, but if it is west, then you have to use 180-H.

Excuse my for a while while i detox from all that trigonometry...

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#### Snuffer

##### Regular Member
Do you have this program working and can i get a copy of it????