Reducers -Optical layout

[Merlin66]

I see there are many threads about how to position reducers on the scope and how to obtain the necessary spacing between the reducer and the CCD chip.

One point is often overlooked and that is, where the reducer has to be placed relative to the prime focus.

The attached diagram may help.

If, say we have a x0.5 reducer which has a design backfocus of 50mm (A), then to achieve the necessary reduction A/B=0.5 so B=100mm.

This means that the reducer must be placed 100mm inside the original focus. Think about it, by the time you add filter wheels, OAG etc etc you can understand why reducers sometimes don't work on Newtonian reflectors or some of the smaller refractors. (No issues generally with SCT's). Check the available space between the focuser housing and the prime focus before you consider adding the reducer. The camera/ CCD will be (B-A)mm closer to the focuser than when at Prime focus without the reducer.

Hope this helps.

Ken

[jnp]

Spot on Ken. There is also the problem of increased vignetting too.

[BlueAstra]

I was inspired by your observation to look at the optical configuration more closely. I think I've calculated a general case where you have a primary focal length (F1), and a focal reducer focal length (F2), and a required focal reduction ratio ® where F = R*F1. The diagram shows the relationship between these parameters. In a Newtonian F1 would be the primary mirror focal length. In a refractor, F1 would be the effective focal length from the second principal plane.

Working through the equations we find P = F2*(1-R)/R, where R= F/F1 (the focal reducer factor) and P is the distance from the original focal plane to the reducer. Interestingly, for R=0.5, P=F2, so only for the special case of R=0.5 is the reducer located at its focal length (F2) from the original focus (P=F2). For other values of R, the distance from the original focus is given by the factor (1-R)/R times the reducer focal length. For R<0.5 the factor is greater than F2, for R>0.5 the factor is less than F2.

To give an example, my ED80 has F1=480mm and my focal reducer has R=0.8. This gives a resultant F=384mm. The location of the focal reducer depends on its design focal length (F2), but this is not stated in the reducer specification. However, it does state that it should be used at 56mm, so P=56mm. This gives a focal length of about 224mm. (These are approximations since neither F1 or F2 are 'simple' lenses). You can also use the equation for P to work out R if it is not used at its design location. For example, if I used it with P=40mm, R=0.85.

So to accurately locate the focal reducer you need to know its design focal length (F2) and reduction ratio ®. Its location is then given by P=F2*(1-R)/R. You can estimate the reducer focal length by imaging a distant object onto a card with just the reducer, but this will not be very accurate unless the lens is simple or you know where the principal plane is. In most cases the manufacturer specifies the working distance

[angusb1]

The two most useful resources I have found for trying to work out focal reducer spacing and unvignetted imaging circle size are the product details pages at Agena Astro where they specify the design focal lengths and clear apertures of lots of the focal reducers they sell, and the focal reducer calculator at Timo's Astro page. (I measured the clear aperture of my Atik x0.5 reducer the other day though and it is 26mm not the 28mm that Timo's calculator has.)

These two sites are responsible for a lot of lost man hours recently :-)