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Thread: Trig help for accurate bevel angles

  1. #1
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    Trig help for accurate bevel angles

    When I am setting initial blade angles, I know:
    the height of the jig or the tool rest
    what bevel angle I want.
    I need to know how far onto the stone / sander / sandpaper the tip of the blade should be to get this angle.

    Trigonometry is the answer, but I had Algebra and Geometry, but no trig. Yesterday I found this easy calculator:

    http://www.carbidedepot.com/formulas-trigright.asp

    Enter the height of your jig or tool rest in "side a".
    Enter the angle you want in "angle A".
    Click on "Calculate. Read the "side b" box.

    In my case, the tool rest on my belt sander is 1.6 inches high.
    I want a 25 degree bevel angle.
    Click on Calculate - now I know that I can measure 3.4 inches from the tool rest (Actually 3.43121107 inches, but I can't measure that accurately) , make a mark, and keep the tip at that mark.

    It's the help I need until I learn trig.

    Brian
    Veni Vidi Vendi Vente! I came, I saw, I bought a large coffee!

  2. #2
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    It's not real trig unless you're dealing with radians. Miters at Pi/4, sharpen plane blades to Pi/6, dovetails at Pi/12
    There are 10 types of people in the world, those who understand binary and those who don't.

  3. #3
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    Wow! More fun stuff to learn. Thanks!
    Veni Vidi Vendi Vente! I came, I saw, I bought a large coffee!

  4. #4
    When I am faced with solving these kinds of problems, I don't figure out blade projection. Instead I solve for the angles in my "imaginary" triangle. I then use the angles to set the tool rest, and eyeball any finer adjustments.

    The sum of the angles of any triangle will be 180 degrees. A common right triangle form is 30/60/90. If we want a 30 degree bevel, the tool rest should be set at a 60 degree angle. For a 25 degree bevel, it should be set at 65 degrees. You could consider picking up a plastic 30/60/90 drafting triangle, and using that to guide you. A sliding bevel will work fine too.

    There is an article by Joel Moscowitz about grinding that may help you. Its more simplistic(and definitely better) than what I attempted to describe. Here-http://www.toolsforworkingwood.com/M...html&BlogID=48

    I hope this helps, and that I didn't screw anything up.
    Last edited by Chris Padilla; 04-03-2009 at 1:23 PM.

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    Dang, I must have missed something. Somehow even though doing trig is not that difficult, I have not found it necessary to sharpen a blade.

    Working by feel and using the eyes has worked well so far. A protractor or drafting triangle only occasionally find their way into the mix.

    Keeping it simple is a much better approach than having to get into scientific notation.

    jim

  6. #6
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    Good idea, Liam, but I am still using an angle measure which I can do on the 25 or 30° side. I'm going to mark the sides of the sander next to the belt, so that every time I want 25° I go to that line. Same with a 30° line.

    Jim, that's what I do too, but in establishing a new angle I can figure this out once and grind to the line. All my honing is freehand now.
    Veni Vidi Vendi Vente! I came, I saw, I bought a large coffee!

  7. #7
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    Well, it is a matter of understanding the sin, cos, and tan trig functions along with some other interesting properties of a right triangle, namely the Pythagorean Theorm.

    Using the picture of the right triangle posted on the link you provided, here are some formulas:

    (1) a² + b² = c²
    (2) sin A = a/c, sin B = b/c
    (3) cos A = b/c, cos B = a/c
    (4) tan A = a/b, tan B = b/a

    In general, the sin of any angle of a right triangle is: the opposite side over the hypontenuse.

    In general, the cos of any angle of a right triangle is: the adjacent side over the hypontenuse.

    In general, the tan of any angle of a right triangle is: the opposite side over the adjacent side.

    Some quickie stuff:

    sin 45° is 0.707 or 1/sqrt(2)
    cos 45° is 0.707 or 1/sqrt(2)
    sin 30° is 1/2
    cos 60° is 1/2
    tan 45° is 1
    Wood: a fickle medium....

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  8. #8
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    Quote Originally Posted by Chris Padilla View Post

    (2) sin A = a/c, sin B = b/c
    (3) cos A = b/c, cos B = a/c
    (4) tan A = a/b, tan B = b/a

    In general, the sin of any angle of a right triangle is: the opposite side over the hypontenuse.

    In general, the cos of any angle of a right triangle is: the adjacent side over the hypontenuse.

    In general, the tan of any angle of a right triangle is: the opposite side over the adjacent side.
    So I am looking for the Tangent of 25°, which according to one webpage is .47.
    That's the bevel I want on a blade.
    So the opposite side (the tool rest) I know to be 1.615 inches high.
    Solving for the adjacent side - the distance where I want to put the tip of the blade resting on the tool rest - the adjacent side is 3.436… or about 3.44 inches. I'll measure out that far, draw a line that says "25°" and stick the end of the blade next to that.

    Later on when I get really good I can hold up a blade that I ground freehand and say, "Yup! That looks like 25° to me!"

    Brian

    PS: Last night my youth group told me the secret of SOHCAHTOA!
    Veni Vidi Vendi Vente! I came, I saw, I bought a large coffee!

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    Quote Originally Posted by Brian Kent View Post
    PS: Last night my youth group told me the secret of SOHCAHTOA!
    Many years ago, an instructor who taught shop and math told us of the Indian Chief who dropped a large stone on his foot and exclaimed, "SOH CAH TOA!"

    He also taught us Ohms law with a pictogram of an eye seeing an eagle soaring over a rabbit, I=E/R.

    That was over 45 years ago and the original lessons are still clear in my mind.

    jim

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    Jim, I like that.

    Now can you offer a pictogram to remind me that I = Current and E = Voltage? I think I can remember that R = Resistance.
    Veni Vidi Vendi Vente! I came, I saw, I bought a large coffee!

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    Quote Originally Posted by Brian Kent View Post
    Jim, I like that.

    Now can you offer a pictogram to remind me that I = Current and E = Voltage? I think I can remember that R = Resistance.
    The I comes from the French for intensity. E is electricity in volts. The line drawing of the eagle was just two curved wings that sort of looked like a flying V.

    Remember, it isn't the voltage that gets you, it is the intenseness of the current.

    It gets a little more weird when you get into Watts, power factors and phase angles or shifts. Then you have to start thinking about ELI the ICE man. Voltage leads the current through and inductor (L) and Current (I) leads voltage through a capacitor (C).

    Some of it is still in the old noggen, but will not likely have to use it much working wood unless I drift to a much bigger and powered shop.

    jim

  12. #12
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    The mathematical proof is when you plane the wood and it goes "ssssst" and the wood is shiny.
    Veni Vidi Vendi Vente! I came, I saw, I bought a large coffee!

  13. #13
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    Quote Originally Posted by Jim Koepke View Post
    The I comes from the French for intensity. E is electricity in volts. The line drawing of the eagle was just two curved wings that sort of looked like a flying V.

    Remember, it isn't the voltage that gets you, it is the intenseness of the current.

    It gets a little more weird when you get into Watts, power factors and phase angles or shifts. Then you have to start thinking about ELI the ICE man. Voltage leads the current through and inductor (L) and Current (I) leads voltage through a capacitor (C).

    Some of it is still in the old noggen, but will not likely have to use it much working wood unless I drift to a much bigger and powered shop.

    jim
    E came from emf: electromotive force. However when I was in college (early 90s), V was in full use for Ohm's law and to represent voltage in general. E was used mostly for electric field.
    Wood: a fickle medium....

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  14. #14
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    Quote Originally Posted by Jim Koepke View Post
    Remember, it isn't the voltage that gets you, it is the intenseness of the current.
    Actually it's not the voltage nor the amperage but rather the kilowatt-hours!

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