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Makercase uses a different finger size algorithm which gives nice looking joints that are convenient for boxes.
When looking at the joint between the front and bottom; fingers and notches have the same size, except the first and last finger which are at least twice the thickness.
Size formula
Given $len =$length of the joint, $fsize =$inner finger and notch size, $minfirstsize=minlastsize= $ min size of the first / last finger,
We want to find $n =$number of inner notches and fingers, $firstsize$the size of the first and last fingers
Makercase uses a different finger size algorithm which gives nice looking joints that are convenient for boxes.
When looking at the joint between the front and bottom; fingers and notches have the same size, except the first and last finger which are at least twice the thickness.
Size formula$len =$ length of the joint, $fsize =$ inner finger and notch size, $minfirstsize=minlastsize= $ min size of the first / last finger,$n =$ number of inner notches and fingers, $firstsize$ the size of the first and last fingers
Given
We want to find
with $ floorToClosestOdd(x) = floor(x) - {(({floor(x)+1}) mod 2}) $
finally $ firstsize = lastsize = {{len - n*fsize} \over 2}$
Would someone be interested in such an algorithm? Thank you!
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