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mikerr
21-03-2011, 02:25 PM
Hi I'm new to the forum and getting back into astronomy after about a 15 year break.

I have an Orange tube Celestron 8 SCT which has just had the optics cleaned and collimated.

On checking the wedge angle with one of the new digital levels, I was surprised to see that the angle was set at about 56 degrees. My Lattitude is about 33.5 degrees S.

All of my books say to set the wedge angle to the same as your Lattitude.

This morning I found the following excellent article by Chris James on this board...

http://www.iceinspace.com.au/63-454-0-0-1-0.html

Chris says to set the angle to "complimentary angle for your lattitude" which using his formula of (90-your lattitude), would put me at an angle of about 56 degrees, which is where mine is set.

Why shouldn't I be setting it at 33 degrees??:shrug:

PS. I am a Math Free Zone:thanx:

Rgds

Michael

jenchris
21-03-2011, 02:40 PM
because the angle is measure from zenith and not the horizon

[1ponders]
21-03-2011, 04:08 PM
:thumbsup:

mikerr
21-03-2011, 04:30 PM
Thanks for the prompt replies.

I just always took it to be measured from the levelled top of the mount up. I even asked in one of the astronomy shops and was told that my understanding was correct!

Glad I saw the Chris James article and asked here.:thumbsup:

Many thanks.

Rgds

Michael.

astroron
21-03-2011, 05:46 PM
When your scope is pointed too Your Zenith it should read -33 DEGS which is taken from the Equator not the Zenith .
It should not read 90 deg.
IF you read the article he said the PROTRACTOR should read 58 deg.
Your scope gauge on the mount should read 33deg, well mine reads -26.38 anyway and I have had mine permanently mounted for nearly twenty years.
Maybe the newer ones start from the pole and work back:shrug:
The Pole is aproximatly 58deg from your location so you should read 33degs or so.
Cheers

mikerr
22-03-2011, 07:42 AM
Thanks for your reply Ron.

Just looking at E J Hartung's book and he says that his Polar support plate is machined "approxmately to the co-latitude of the observatory-in my case 52 deg 38 min". I guess the expression (co-latitude) is what we are talking about.

Thanks again for the help.

Rgds

Michael