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DaveH (01-20-2014)
There was a question on another forum about how to accurately measure a chamfer and since I've been meaning to add something similar to the my website, I decided that now was as good a time as any to crank one out.
The following chart can be used to accurately calculate the size and location of an angled surface, using a pin and a bit of trig. It will work for a dovetail, a chamfer, or any angled surface. I'll likely be shooting a video on how to use it tomorrow.
Tom
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That's great Tom, my last experience cutting a dovetail was disappointing. The dovetail looked great but was way oversized. Some tips on measuring them will be fantastic. Looking forward to seeing the video.
Shawn
Shawn, proud to be a member of MetalworkingFun Forum since Nov 2013.
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DaveH (01-20-2014)
Thanks Shawn.
My best laid plans didn't happen this weekend, so it'll have to wait until next. I may just cut a dovetail and use the chart to measure it.
Tom
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No worries Tom, turns out I have to silver solder a bandsaw blade anyways. ? I ain't going anywhere.
Shawn, proud to be a member of MetalworkingFun Forum since Nov 2013.
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Thanks for a great posting Tom.
It might be worth pointing out that the math requires that the angle between CH and CL is a solid 90 degrees. (The trig functions used here assume that this angle is 90 deg).
Another clarification for those that don't work with a lot of trig:
Tan A/2 is to be taken as Tan (A/2) not (Tan A)/2. In other words, Tan of 1/2 the angle, not 1/2 the Tan of A. (gives two different results)
Not to worry, if you do it wrong it will be obvious because the result will be way out of line.
Terry S.
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(01-21-2014, 10:38 AM)Terry S Wrote: Thanks for a great posting Tom.
It might be worth pointing out that the math requires that the angle between CH and CL is a solid 90 degrees. (The trig functions used here assume that this angle is 90 deg).
Another clarification for those that don't work with a lot of trig:
Tan A/2 is to be taken as Tan (A/2) not (Tan A)/2. In other words, Tan of 1/2 the angle, not 1/2 the Tan of A. (gives two different results)
Not to worry, if you do it wrong it will be obvious because the result will be way out of line.
Terry S.
Terry,
Thanks for pointing out my lack of parentheses ( I knew I should have had someone double check my work. I'll add some when I get home tonight.
Not sure what you mean by the angle between CL and CH needing to be 90º. The constructed triangle br will always be a right triangle regardless of angle A.
Tom
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(01-21-2014, 12:17 PM)TomG Wrote: (01-21-2014, 10:38 AM)Terry S Wrote: Thanks for a great posting Tom.
It might be worth pointing out that the math requires that the angle between CH and CL is a solid 90 degrees. (The trig functions used here assume that this angle is 90 deg).
Another clarification for those that don't work with a lot of trig:
Tan A/2 is to be taken as Tan (A/2) not (Tan A)/2. In other words, Tan of 1/2 the angle, not 1/2 the Tan of A. (gives two different results)
Not to worry, if you do it wrong it will be obvious because the result will be way out of line.
Terry S.
Terry,
Thanks for pointing out my lack of parentheses ( I knew I should have had someone double check my work. I'll add some when I get home tonight.
Not sure what you mean by the angle between CL and CH needing to be 90º. The constructed triangle br will always be a right triangle regardless of angle A.
Tom
True that the constructed angle will always be 90 deg. However, when you solve for CL and solve for h, the math assumes a 90 degree angle between the CL direction and the CH direction.
For example, if I were trying to measure a upper LH chamfer (as shown in your sketch) and had the part resting on granite table (or machine table), I would need to place another vertical, flat surface to support the LH side of the rod. It is this surface that needs to generate an accurate 90 deg to the table. I think all of use would use a machined 90 degree setup piece to do this, but it is worth pointing out that if one were to use a piece of cold-rolled angle or something other than 90 deg, the accuracy would suffer.
Terry S.