|
|
Maximum cant sizeQuestion
Forum Responses
Example
So, basically a 12 x 12.
Draw a circle. Now draw a square inside the circle. Then draw a diagonal across the square. Notice that this diagonal is also the diameter of your circle. Measure the length of this diameter, now measure the length of one of the sides of the square. (Or use math.) You will discover that the side of your cant is .707 of the log. Round this off to 7/10 or 70% of the log. For example, a ten-inch log gives a seven-inch cant. A twenty-inch log gives a 14-inch cant. The above is right on the money for producing a square edge cant. You must allow for crook by reducing diameter. I measure the small end and multiply by 7. (Example: 8" log, 8*7=56. I drop the last digit. I know I can saw a 5" cant.) Measure the small end of the log, subtract 4", unless you have a crooked log. I've never put it to the math test, but I've found that that this is close enough without doing all that ciphering. The small end's diameter is 1.5 times larger than the square. For a 6x6, look for a 9" log, 8x8-12" log , 12" square an 18" log. As mentioned before, sweep reduces yield. In the post above, 1.5 is really 1.414, the approximate value of the square root of two and the .707 is really 1 divided by the square root of two. It is the value involved with a square inscribed in a circle or a circle circumscribed around a square. Would you like to add information to this article? Interested in writing or submitting an article? Have a question about this article? Have you reviewed the related Knowledge Base areas below? 
|
