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unknown F-ratio

Started by rhinson11, 07/30/2002 09:20PM
Posted 07/30/2002 09:20PM Opening Post
I picked up a newt/dob at a yard sale today, but don't know how to figure out the F-ratio and need some help. Here's what I know about the scope. It's a 6 inch mirror from orion, with a 2 inch rack and pinion focuser. the distance between the face of the mirror and center of the foucser is 21.5 inches. I'm thinking it's around F.5, but I'm not sure. Tube diameter is 8.5 inches. It's a home made setup, but a lot of care and work went into it. I felt bad about paying what the woman wanted for it ($50.00) Fit and finish are very nice, it has a home made wooden push/pull cell that looks like it's made from oak. Rubberized end rings. A celestron 6x30 LER finder in a 6 point holder, and what looks like a U.O. spider holding a diag around 1.5 or so. I gave it a quick collmation using a 35 mm film tube and took it out for a short while tonight. It had pretty good views. Still needs fine tuning, but very useable. So how do I figure out the F-ratio?.
Thanks
Roger
Posted 07/31/2002 12:55AM #1
Sounds like it is an F6. The easiest way to confirm it would be to aim it at a very distant object and project the image onto a piece of paper. If it is an F6 it should come to focus about 4 inches from the outside of the tube. From what you have described I would be very surprized if the true F ratio is outside the F5.5 to 6.5 range.
You can also use an eyepiece with a known field stop diameter (Televue publishes theirs, others will have to be measured very carefully with a caliper, taking care not to nick the field stop) to measure it at night using a star on the celestial equator. Time the star's travel time across the field of view of the eyepiece you measured. It must travel directly across the center of the field of view. If the scope is an 6" F6, the star will take exactly 15 seconds for every millimeter of field stop. So say, for example that you had a Televue 32mm Plossl (which has a 27mm field stop diameter), the star would take 6 minutes and 45 seconds to drift across. If there is any variation from this, the difference is directly proportional to the focal length of the scope.

Joplin
Posted 07/31/2002 05:40AM #2
Roger,

Easy to figure, depending on how exact you want it. If it has an 8.5 inch tube, it is approximately half that distance from the secondary to the side of the tube--4.25 inches. Probably another inch (maybe two, you will neeed to measure if you want it exact) from the side of the tube to where the eyepiece focuses. Add that total--5.25 inches--to the 21.5 from the secondary to the primary, and you have a total focal length of 26.75. (I am assuming you meant the 21.5 as the measure from the primary to the secondary--which is right under the focuser.) DIvide that number by the 6 inches of the primary's diameter and you get--little less than 4.5 or so.

The focal ratio (F number) is merely the total length of the light path divided by the diameter of the primary objective.

Are there any optics in the focuser? SOme scopes have a built in barlow or corrector. That could change the calculation.

Alex
Posted 07/31/2002 09:36AM #3
Another good clue as to the f5 rating is that Orion sells one and I don't think they sell an f6 in the 6" size. I have one of the 6" f5 and like it.

Charlie

Fort Lewis Observatory: (37.238, -108.052) ~2360m (7744 ft.) elevation.
Darkness - typically 6.5+
Scope - Meade 16" LX200; f6.3 focal reducer
Focus - JMI Smart focus
Camera - SBIG ST-10XE; (~.5"/pixel)
Guiding - AO-8 and/or Meade 5"/DSI/PHD
http://www.fortlewis.edu/observatory