So The Lt 10 is really ripping along well and I just had a friend of mine drop off a few cord of cedar to play. I'm looking to get the most I can out of these trees, so I dusted off the old geometry book and looked up how to measure the area of a circumscribed square, and solved for the length of the side of the square. Now I take this number and subtract it from the diameter of the narrow end, divide by 2 and I now know what I'm peeling off to get my square cant. In other words:
L = 1.414R where L is length of side of square and R is radius of small end of log.
(D-l)/2 = total removed per side for a square cant.
I was playing with this today, it seemed to work, but I ended up doing 2 cuts per side so I'd be able to rip that extra 1x whatever out of there. Anybody else out there do it like this? Is there a better way? thanks in advance for any feedback.
The math if anybody wants to see what I did:
Let d = diameter of log, and Diagonal of square
Let L= Length of side of square
L^2+L^2=D^2= (2r)^2
(2L^2=4r^2)/2
(L^2 = 2r^2)sqrt
L = sqrt2R = 1.414R
(d-1.414r)/2
L = 1.414R where L is length of side of square and R is radius of small end of log.
(D-l)/2 = total removed per side for a square cant.
I was playing with this today, it seemed to work, but I ended up doing 2 cuts per side so I'd be able to rip that extra 1x whatever out of there. Anybody else out there do it like this? Is there a better way? thanks in advance for any feedback.
The math if anybody wants to see what I did:
Let d = diameter of log, and Diagonal of square
Let L= Length of side of square
L^2+L^2=D^2= (2r)^2
(2L^2=4r^2)/2
(L^2 = 2r^2)sqrt
L = sqrt2R = 1.414R
(d-1.414r)/2