It's embarrassing to ask such a basic question for me but.. I just spent a good amount of time wading through the 'net trying to find what I'm looking for but can't seem to locate it (here either!) - Does ANYONE know a simple formula for deriving the effective focal length (bellows draw) from magnification and nominal focal length ALONE?

So here's the deal - Let's say I'm trying to make a photograph of an object I know I need a magnification of 10x on in terms of what I want the aerial image height (fancy talk for image size) on the negative to be. One of the lenses I have at my disposal is a 105mm componon - but I think that's going to be way too long. Basic rule of thumb tells me that at 1:1, the focal length will be 210mm, and at 2:1 twice that, it will be 420mm, at 4:1, 840mm, 8:1 it will be 1680mm... and 10:1 would be SOMEWHERE in the region of 2400mm (7.5 feet) I'm guessing. So in that case I would try to find myself a 50mm lens to keep the bellows extension around 1.2m since it's more convenient...

So fine - I can work through this manually as above... but I have a dozens of such scenarios to work out so I'm wondering if there's a good simple formula that can help me automate my shot planning... (?)

Thanks if anyone can help...