Does shorter mean faster?

While browsing the user manual for my yet-to-be-built OF Woodworker, I read that OF actually suggests building up the spoilboard to 2.5 inches or more, to 'reduce the moment force' on the router, z axis bracket, etc. - in other words, shorten the "lever" formed by an advancing bit, the edge of the material being cut,  and the carriage.  I had not previously considered these force vectors...am I correct that one implication is that placing the router in the lowest bracket position and using as short a bit as possible results in a higher safe/effective/possible ipm than having the router mounted higher with a longer bit?  I'm not saying I'm old, but my last physics class was the semester after 'Zac got bonked on the head with that apple...

Sort of right… the ideal position would be the center (or upper, IMO) position to reduce the lever forces against the gantry. However, this thing is built so beefy, this is more a theoretical conversation because, honestly, the forces of the router cutting are a miniscule fraction of what it would take to flex the assembly. (Hope that makes sense, it does in my head :laughing:)

It’s always preferable to have the cutter as short as possible with the shank fully chucked into the collet. Whatever position you have the Z assembly placed, it should be with respect to a good operating reach for the bit & workpiece.

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Thanks for replies. Regarding flex and cutting too fast, that is actually the opposite of what led me to consider this issue…as I model various items I plan to cut once my machine arrives (using Vcarve Pro), I keep running into crazy fast IPMs in order to meet the recommended chip loads for typical end mills to avoid excessive heat build up.

Thanks, much appreciated.

Thanks for a reassuring post for somebody that is new to this and worried about burning up bits and obsessed about finding just the right router speed and cut rate.