--- Martin Farmer <[EMAIL PROTECTED]> wrote: > > I have a couple of questions that the group could possibly help me > with these involve the Eyepiece menu and adding your own > > I have the following eyepiece selection :- > > Focal Length > Field of View > > Vixen > 13mm > 65� > > Vixen > 40mm > 42� > > Vixen > 4mm > 45� > > TAL > 40mm > 45� > > TAL > 25mm > 52� > > TAL > 7.5mm > 52� > > I have a 1200mm focal length tube and I have put into the field of > vision colum the following values:- > > 13mm = 42 > > 40mm = 84 > > 4mm =9 > > 40mm =90 > > 25mm = 65 > > 7.5mm=19 > > > Are these the right figures to go into the Field of Vision box for > these eyepieces,
I checked the first two and they're correct, so assuming you used the same method for all, they should all be correct. >Also I am confused on the CCD Field of what to put > in there I have a Philips TouCam and the specification of the CCD is > as follows:- > > � Image size: Diagonal 4.5mm (Type 1/4) > > � Number of effective pixels: 659 (H) � 494 (V) approx. 330K pixels > > � Total number of pixels: 692 (H) � 504 (V) approx. 350K pixels > > � Chip size: 4.60mm (H) � 3.97mm (V) > > � Unit cell size: 5.6�m (H) � 5.6�m (V) > > > What do I put from this information into the CCD options please. You need to convert the chip size into arcminutes (at the focal plane). The starting point for making sense of scope optics is that two stars 1� apart in the sky will focus to points in the focal plane that are 1� apart, as measured from the primary (ie, as seen from a distance of 1200mm). A full circle is 360�, and a full circle of radius 1200mm would have a circumference of 1200*2*pi ~= 7540mm, so the two star images would be 7540/360 ~= 21mm apart. Thus the chip width of 4.6mm is 4.6/21 degrees at the focal plane, or about 13 arcminutes. The height is (3.97/21)*60 ~= 11'. This assumes you're imaging at prime focus (no barlow, etc). But you might want to check those numbers against images of actual star fields. Multiplying the pixel size of 5.6� square by the 659X494 pixel numbers gives a chip size of about 3.7mm X 2.8mm, smaller than the chip size listed above. So either there are parts of the chip surface around the perimeter not used for imaging (likely), or the pixels are larger than 5.6�, and 5.6 is only the size of the light sensitive part of each pixel (also possible). A little experimentation should show the true size of the field. Or ask on QCUIAG. BTW the chip is 659X494 pixels, but I believe the actual image output from a Toucam is only 640X480 pixels. This gives an aspect ratio of 4:3, so the actual used total light sensitive part of the chip surface (and the field of view) should have a 4:3 ratio (that's what tipped me off that something was wrong with the numbers). -John __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com ------------------------ Yahoo! Groups Sponsor --------------------~--> $9.95 domain names from Yahoo!. Register anything. http://us.click.yahoo.com/J8kdrA/y20IAA/yQLSAA/1.XolB/TM --------------------------------------------------------------------~-> To Post a message, send it to: [email protected] To Unsubscribe, send a blank message to: [EMAIL PROTECTED] Yahoo! Groups Links <*> To visit your group on the web, go to: http://groups.yahoo.com/group/skychart-discussion/ <*> To unsubscribe from this group, send an email to: [EMAIL PROTECTED] <*> Your use of Yahoo! Groups is subject to: http://docs.yahoo.com/info/terms/
